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Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2013, Article ID 691740, 23 pages http://dx.doi.org/10.1155/2013/691740 Research Article Electrical Network Functions of Common-Ground Uniform Passive RLC Ladders and Their Elmore’s Delay and Rise Times D. B. KandiT, 1 B. D. Reljin, 2 and I. S. Reljin 3 1 Department of Physics & Electrical Engineering, School of Mechanical Engineering, University of Belgrade, Kraljice Marije 16, 11120 Belgrade, Serbia 2 Department of General Electrical Engineering, School of Electrical Engineering, University of Belgrade, Bulevar Kralja Aleksandra 73, 11000 Belgrade, Serbia 3 Department of Telecommunications, School of Electrical Engineering, University of Belgrade, Bulevar Kralja Aleksandra 73, 11000 Belgrade, Serbia Correspondence should be addressed to D. B. Kandi´ c; [email protected] Received 18 February 2013; Accepted 19 May 2013 Academic Editor: Xinkai Chen Copyright © 2013 D. B. Kandi´ c et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In the paper are presented the expressions for all network functions of common-ground, uniform passive ladders having, in general, complex terminations at both their ends. e Elmore’s delay and rise times calculated for selected types of RLC ladders have indicated their slight deviation from delay and rise times obtained according to their classical definitions. For common-ground, integrating type RC ladder with voltage-step input, the Elmore’s delay- and rise-times are produced in closed-form, both for ladder nodes and points. Furthermore, it is proposed a particular common-ground, uniform RLC ladder being amenable to application as delay line for pulsed and analog input signals. For this ladder, the Elmore’s delay and rise times relating to their node voltages are produced in a closed-form, enabling thus with the realization of artificial (a) pulse delay line with arbitrarily and independently specified overall Elmore’s delay and rise times and (b) true delay line with arbitrarily specified delay time for frequency bounded analog and/or pulsed input signals. In cases (a) and (b), precise procedures are formulated for calculation of ladder length and of all its RLC parameters. e obtained results are illustrated with several practical examples and are, also, verified through pspice simulation. 1. Introduction Modeling of digital MOS circuits by RC networks has become a well-accepted practice for estimating delays [16]. In digital integrated circuits, signal propagation delay through con- ducting paths with distributed resistance and capacitance is frequently a significant part of the total delay. ese con- ducting paths, or “interconnections,” can be modeled quite accurately by nonuniform, branched RC ladder networks, also known as “RC trees” [3]. Computationally simple bounds for signal delay in linear RC tree networks were found in [3] and have been used in several practical MOS timing analyzers reported in [6], but certain circuits used in MOS logic cannot be modeled as RC trees since they contain one or more closed loops of resistors, and these general RC networks are being referred to as “RC meshes”. In these networks, the time delay defined according to Elmore [7] is proved to be the valid estimate, and this fact has been used in [5] to advantage in an approach to MOS timing analysis of general RC networks containing RC meshes. Simple closed- form bounds for signal propagation delay in linear RC tree models for MOS interconnections derived in [3] are, also, valid for the more general class of linear networks known as RC meshes, which are useful as models for portions of MOS logic circuits that cannot be represented as RC trees [6]. Elmore’s delay is an extremely popular timing- perfomance metric which is used at all levels of electronic circuit design automation, particularly for RC tree analysis. e widespread usage of this metric is mainly attributable to its property of being a simple analytical function of
Transcript
Page 1: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 691740 23 pageshttpdxdoiorg1011552013691740

Research ArticleElectrical Network Functions of Common-Ground UniformPassive RLC Ladders and Their Elmorersquos Delay and Rise Times

D B KandiT1 B D Reljin2 and I S Reljin3

1 Department of Physics amp Electrical Engineering School of Mechanical Engineering University of Belgrade Kraljice Marije 1611120 Belgrade Serbia

2Department of General Electrical Engineering School of Electrical Engineering University of Belgrade Bulevar Kralja Aleksandra 7311000 Belgrade Serbia

3 Department of Telecommunications School of Electrical Engineering University of Belgrade Bulevar Kralja Aleksandra 7311000 Belgrade Serbia

Correspondence should be addressed to D B Kandic dbkandicgmailcom

Received 18 February 2013 Accepted 19 May 2013

Academic Editor Xinkai Chen

Copyright copy 2013 D B Kandic et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

In the paper are presented the expressions for all network functions of common-ground uniform passive ladders having in generalcomplex terminations at both their ends The Elmorersquos delay and rise times calculated for selected types of RLC ladders haveindicated their slight deviation from delay and rise times obtained according to their classical definitions For common-groundintegrating type RC ladder with voltage-step input the Elmorersquos delay- and rise-times are produced in closed-form both for laddernodes and points Furthermore it is proposed a particular common-ground uniform RLC ladder being amenable to applicationas delay line for pulsed and analog input signals For this ladder the Elmorersquos delay and rise times relating to their node voltagesare produced in a closed-form enabling thus with the realization of artificial (a) pulse delay line with arbitrarily and independentlyspecified overall Elmorersquos delay and rise times and (b) true delay line with arbitrarily specified delay time for frequency boundedanalog andor pulsed input signals In cases (a) and (b) precise procedures are formulated for calculation of ladder length and ofall its RLC parameters The obtained results are illustrated with several practical examples and are also verified through pspicesimulation

1 Introduction

Modeling of digitalMOS circuits byRC networks has becomea well-accepted practice for estimating delays [1ndash6] In digitalintegrated circuits signal propagation delay through con-ducting paths with distributed resistance and capacitance isfrequently a significant part of the total delay These con-ducting paths or ldquointerconnectionsrdquo can be modeled quiteaccurately by nonuniform branched RC ladder networksalso known as ldquoRC treesrdquo [3] Computationally simple boundsfor signal delay in linear RC tree networks were found in[3] and have been used in several practical MOS timinganalyzers reported in [6] but certain circuits used in MOSlogic cannot bemodeled asRC trees since they contain one ormore closed loops of resistors and these general RC networks

are being referred to as ldquoRC meshesrdquo In these networksthe time delay defined according to Elmore [7] is provedto be the valid estimate and this fact has been used in[5] to advantage in an approach to MOS timing analysis ofgeneral RC networks containing RC meshes Simple closed-form bounds for signal propagation delay in linear RC treemodels for MOS interconnections derived in [3] are alsovalid for the more general class of linear networks known asRC meshes which are useful as models for portions of MOSlogic circuits that cannot be represented as RC trees [6]

Elmorersquos delay is an extremely popular timing-perfomance metric which is used at all levels of electroniccircuit design automation particularly for RC tree analysisThe widespread usage of this metric is mainly attributableto its property of being a simple analytical function of

2 Mathematical Problems in Engineering

circuit parameters and its drawbacks are the uncertaintyof accuracy and restriction to being the estimate only forthe step-response delay An extension of Elmorersquos delaydefinition has been proposed in [8] to accommodate theeffect of nonunit-step (slow) excitations and to handlemultiple sources of excitation in order to show that delayestimation for slow excitations is no harder than for theunit-step input In [9] it has been shown that this extensionof Elmorersquos delay time offers a provision to deal with slowvarying excitations in timing analysis of MOS pass transistornetworks In addition in [10] it has been reported thatElmorersquos delay is an absolute upper bound on the actual50 delay of an RC tree response Also in [10] it has beenproved that this bound holds for input signals other thansteps and that actual delay asymptotically approaches toElmorersquos delay as the input signal rise-time increases It hasbeen emphasized in [11] that RC tree step responses alwaysare monotonic and this is why Elmorersquos definitions of bothdelay and rise time [7] are applicable on complex RC treenetworks

In this paperwewill firstly derive the general closed-formexpressions for input and transfer functions of common-ground uniform and passive ladders with complex doubleterminations making a distinction between the signal trans-fer to ladder nodes and to its points Then simplifications ofthe obtained results are produced for ladders with seven spe-cific pairs of complex double terminations being interestingfrom the practical point of view Thereafter for a uniformcommon-ground RLC ladder with (a) a relation between itsparameters resembling to analogous relation between per-unit-length parameters of distortionless transmission lineand (b) symmetric resistive double termination resemblingto characteristic impedance of distortionless line [12] it willbe shown that its Elmorersquos delay and rise times for pointvoltage transmittances can be efficiently calculated (but not inthe closed form) by using of the numerical scheme proposedherein In this case we will see that the obtained numericalvalues for Elmorersquos times differ slightly from the delay andrise times obtained according to their classical definitions andby using pspice simulation when ladder is excited by a stepvoltage

For the integrating type of distributed RC impedancecommon-ground ladder as a model of two-wire line it hasbeen suggested that it might be used as a true pulse delay line[13] provided that Schmitt triggers are used for reshapingthe delayed and edge-distorted transmitted signals This typeof delay line with step-input excitation has already beenthoroughly investigated in [14]mdashwhere Elmorersquos delay time isgiven in closed form only for ladder nodes and for Elmorersquosrise time is offered a conjecture relating to its lower boundfor overall network Elmorersquos rise times for all nodes ofintegrating RC open-circuited ladder are given in closedform in [15] and the obtaining of Elmorersquos delay and risetimes both for nodes and points is discussed to some extent in[16]mdashfor other types of open-circuited ladders In this paperwe are going to formulate the explicit closed-formexpressionsfor Elmorersquos delay and rise times both for the node and pointvoltages of integrating open circuited common-ground RCladder and will conclude that this type of network is not

recommendable for pulse delay line in its own right sinceElmorersquos rise time of each point voltage is not less than thetwice of its delay-time

And finally we will propose a type of uniform common-ground RLC ladder amenable for application as delay lineboth for pulsed andor analog input signals Elmorersquos delayand rise times of this ladder relating to the node voltagetransmittances are produced in closed form opening thuswith the following possibilities in ladder realization

(i) for pulsed inputs the overall Elmore delay and risetimes may be specified arbitrarily

(ii) for pulsed inputs theminimum ladder length (ie theminimumnecessary number of sections) is calculatedstraightforwardly by using only the overall Elmoredelay and rise times

(iii) the ladder RL parameters are calculated uniquelyfrom the assumed nonminimal ladder length overallElmore delay time and the assumed capacitancevalues

(iv) for realization of true delay for pulsed andor ana-log input signals with arbitrary variation in timeminimum ladder length is calculated from the spec-ified true delay time (which asymptotically tends toElmorersquos delay time when the number of sectionstends to infinity) and maximum frequency in thespectrum of the signal being transmitted along theladder purporting to represent delay line And againas in (iii) the ladder RL parameters are determinedfrom the ladder length overall ladder true delay time(being approximately equal to Elmorersquos delay time)and the assumed capacitance values The previousapproach and obtained results are illustrated andverified with realization examples of large true delaytime (=5 [ms]) for pulsed sine distorted sine CAMFM and chirp frequency and sweep amplitude inputsignals

Recall that in network synthesis the ladder topology is apreferable one since it has very low sensitivity to variationsof RLC parameters [16 17]

2 Network Functions of Common-GroundUniform and Passive RLC Ladders

Consider a uniform grounded and passive RLC ladder inFigure 1 with N identical sections whose impedances 119885

1=

1198851(119904) and 119885

2= 1198852(119904) are positive real rational functions

in complex frequency 119904 = 120590 + 119895120596 (119895 = radicminus1) There onare denoted the Laplace transforms 119864 = 119864(119904) of the voltageexcitation 119890 = 119890(119905) 119880

119896= 119880119896(s) and 119868

119896= 119868119896(s) of the point

voltages 119906119896= 119906119896(t) and currents 119894

119896= 119894119896(t) (119896 = 0119873) and

U1015840119898=U1015840119898(s) and 1198681015840

119898= 1198681015840119898(s) of node voltages 1199061015840

119898= 1199061015840119898(t) and

currents 1198941015840119898= 1198941015840119898(t) (119898 = 0119873 minus 1) The internal impedances

of the voltage excitation (voltage generator) and the load are119885119892= 119885119892(s) and 119885

119871= 119885119871(119904) respectively All the initial

conditions associated to the network reactive (LC) elementsare assumed to be zero

Mathematical Problems in Engineering 3

O

Zg

EU0

I0

Section 1

Section n

Section N

Z1 Z1Mnminus1

Z2

In

Un UNminus1Unminus1Inminus1

middot middot middot

middot middot middotmiddot middot middot

middot middot middot

UNZL

IN

Inminus1998400

Unminus1998400

Figure 1 The general common-ground uniform ladder with N sections

For ladder in Figure 1 the mesh transform equations forits 119899th section (119899 = 1119873) read

(1198851+ 1198852) sdot 119868119899minus1

minus 1198852sdot 119868119899= 119880119899minus1

1198852sdot 119868119899minus1

minus (1198851+ 1198852) sdot 119868119899= 119880119899

or in the recurrence form

[119880119899

119868119899

] =[[[

[

1 +1198851

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

sdot [119880119899minus1

119868119899minus1

]

(1)

The boundary values of voltages and currents in the set ofmatrix difference equations (1) are (U

0 I0) and (119880

119873 119868119873) For

119899 = 0119873 from (1) it immediately follows taht

[119880119899

119868119899

] =[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

119899

sdot [1198800

1198680

]

where 1198800= 119864 minus 119885

119892sdot 1198680and 119880119873= 119885119871sdot 119868119873

(2)

Since the roots (say 1205821and 120582

2) of the characteristic equation

corresponding to 2 times 2matrix appearing in (1)

det

[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

minus 120582 sdot U2

= 1205822minus 2 sdot (

Z1

Z2

+ 1) sdot 120582 + 1 = 0

(U2is 2 times 2 identity matrix)

(3)

satisfy the relations 1205821+ 1205822= 2 sdot (119885

11198852+ 1) and 120582

1sdot 1205822=

1 then by taking for convenience 1205821= exp(120591) and 120582

2=

exp(minus120591) we obtain cosh(120591) = 1 + 11988511198852 By using Cayley-

Hamiltonrsquos theorem we may put

[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

119899

= 1198600sdot U2+ 1198601sdot[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

(4)

where the ldquoconstantsrdquo1198600and119860

1are determined from system

of equations [exp(120591)]119899 = 1198600+ 1198601sdot exp(120591) and [exp(minus120591)]119899 =

1198600+ 1198601sdot exp(minus120591) whose solution is 119860

0= minus sinh[(119899 minus 1) sdot

120591]sinh(120591) and 1198601= sinh(119899 sdot 120591)sinh(120591)

By substituting1198600and119860

1in (4) and bearing inmind that

1 +1198851

1198852

= cosh (120591)

minus(11988521

1198852

+ 2 sdot 1198851) = 119885

2minus 1198852sdot (

1198851+ 1198852

1198852

)

2

= 1198852sdot [1 minus cosh2 (120591)] = minus119885

2sdot sinh2 (120591)

(5)

we finally obtain from (2) and (4) for 119899 = 0119873 that

[119880119899

119868119899

] =[[[

[

1 +1198851

1198852

minus(11988521

1198852

+ 2 sdot 1198851)

minus1

1198852

1 +1198851

1198852

]]]

]

n

sdot [1198800

1198680

]

=[[[

[

cosh (119899 sdot 120591) minus1198852sdot sinh (120591) sdot sinh (119899 sdot 120591)

minussinh (119899 sdot 120591)1198852sdot sinh (120591)

cosh (119899 sdot 120591)]]]

]

sdot [1198800

1198680

]

(6)

4 Mathematical Problems in Engineering

Since uniform ladder sections are electrically reciprocalthen their characteristic impedance 119885

119888and the quantity 119885

119888sdot

sinh(120591) can easily be produced in the form

119885119888= 1198852sdot sinh (120591) = radic119885

1sdot (1198851+ 2 sdot 119885

2)

119885119888sdot sinh (120591) = 119885

2sdot sinh2 (120591)

= 1198852sdot [cosh2 (120591) minus 1] = 119885

1sdot (

1198851

1198852

+ 2)

(7)

For the ladder depicted in Figure 1 the complete set ofthe network immittance and voltagecurrent transmittancefunctions can be produced by using the relations (2) (6) and(7)

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119871

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119868119899

1198680

= (119885119888sdot cosh [(119873 minus 119899) sdot 120591]

+119885119871sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119888sdot cosh (119873 sdot 120591) + 119885

119871sdot sinh (119873 sdot 120591))

minus1

119880119899

119864= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119880119899

1198800

= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119871sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591))

minus1

119880119899

119868119899

= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119888sdot cosh [(119873 minus 119899) sdot 120591]

+119885119871sdot sinh [(119873 minus 119899) sdot 120591])

minus1

sdot 119885119888 119899 = 0119873

(8)

Since 1198851= 119885119888sdot (sinh(120591)[1 + cosh(120591)]) = 119885

119888sdot tanh(1205912)

then the voltages and currents of impedances 1198852connecting

the middle nodes of ladder sections (119872119898) to common-node

119874 (Figure 1) are given as

1198801015840

119898=119880119898+ 119880119898+1

1 + cosh (120591)

1198681015840

119898=1198801015840119898

1198852

=119880119898+ 119880119898+1

1198852sdot [1 + cosh (120591)]

=119880119898+ 119880119898+1

1198851+ 2 sdot 119885

2

1198801015840119898

119864= ( (119885

119871minus 1198851) sdot cosh [(119873 minus 119898) sdot 120591]

+ [119885119888minus 119885119871sdot tanh(120591

2)]

sdot sinh [(119873 minus 119898) sdot 120591] )

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119898 = 0119873 minus 1

1198801015840119898

1198800

= ( (119885119871minus 1198851) sdot cosh [(119873 minus 119898) sdot 120591]

+ [119885119888minus 119885119871sdot tanh(120591

2)]

sdot sinh [(119873 minus 119898) sdot 120591] )

times (119885119871sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591))

minus1

(9)

Consider now the particular selection of 119885119892119885119871 to

investigate the generation of (i) finite length frequency-selecting ladders with various input and transfer immittancesand voltage and current transmittances and (ii) specificladder which can take the role of finite pulse delay linewithout pulse attenuation and with independently controlledpulse delay and rise times in Elmorersquos sense [7] calculable inclosed form We will assume that impedances 119885

119892and 119885

119871are

rational positive real functions in s so that they are realizableby passive transformerless RLC networks [17] All networkfunctions in (8) and (9) are real rational functions in s except

Mathematical Problems in Engineering 5

119880119899119868119899 which must be rational positive real function in s since

it is the input immittance of RLC network

Case A (119885119892-arbitrary and 119885

119871rarr infin [Ω]) (open-circuited

ladder) From (8) and (9) it follows

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

119885119888sdot cosh (119873 sdot 120591) + 119885

119892sdot sinh (119873 sdot 120591)

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(10)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)]

1198801015840119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(11)

Since we have cosh(120591) = 1 + 11988511198852 then by Property 2

(Appendix A) 119880119899U0(119899 = 1119873) (10) can be easily converted

into real rational function of 11988511198852or of s Similarly by

Property 3 (Appendix A) 1198681198981198680(119898 = 1119873 minus 1) (10) can be

also easily converted into real rational function of 11988511198852

or of s When 1198851and 119885

2are one-element-kind impedances

the zeros and poles of both 119880119899U0and 119868119898I0can be easily

determined in the closed form Since by Property 3 sinh[(119873minus

119898) sdot 120591] and sinh[(119873 minus 119898 minus 1) sdot 120591] contain the same factorsinh(120591) (119898 = 0119873 minus 1) then 1198801015840

119898U0(119898 = 0119873 minus 1) (11) can

be also easily converted into real rational function either of11988511198852or of the complex frequency s

Case B (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+1198852and119885119871rarr infin [Ω]) In this

case (10) and (11) simplify to

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

1198852sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=cosh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

sinh [(119873 minus 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(12)

1198801015840119898

119864

=2 sdot cosh [(119873 minus 119898 minus (12)) sdot 120591] sdot sinh (1205912)

sinh [(119873 + 1) sdot 120591]

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898 minus (12)) sdot 120591]

cosh (1205912) sdot cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(13)

Case C (119885119892-arbitrary and119885

119871= 0 [Ω] (short-circuited ladder))

From (8) and (9) it is obtained

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

119885119892sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591)

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=

sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(14)

1198801015840119898

119864

=119885119888sdot sinh [(119873 minus 119898) sdot 120591] minus 1198851 sdot cosh [(119873 minus 119898) sdot 120591]

119885119888sdot sinh (119873 sdot 120591) + 119885

119892sdot cosh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)]

6 Mathematical Problems in Engineering

1198801015840119898

1198800

=sinh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot cosh [(119873 minus 119898) sdot 120591]

sinh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(15)

According to Properties 2 and 3 respectively 119868119899I0(119899 = 1119873)

and 119880119898U0(119898 = 1119873 minus 1) (14) can be easily converted into

real rational functions either of11988511198852or of the complex fre-

quency s If1198851and119885

2are one-element-kind impedances the

zeros and poles of both 119868119899I0and 119880

119898U0can be determined

straightforwardly in closed formNow since119885119888sdotsinh(120591)119885

1=

2 + 11988511198852andor (1 + 119885

11198852)2minus sinh2(120591) = 1 it can be

seen from (15) that1198801015840119898U0(119898 = 0119873 minus 1) is also convertible

into real rational function either of 11988511198852or of the complex

frequency 119904 by using of both Properties 2 and 3

Case D (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+ 1198852and 119885119871= 0 [Ω]) From

(14) and (15) it readily follows that

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

1198852sdot cosh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=sinh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

cosh [(119873 + 1) sdot 120591]

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(16)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

cosh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(17)

By using of Properties 2 and 3 and sinh2(120591) = (1 + 11988511198852)2minus

1 it can be seen from (16) and (17) that the network functions119868119896I0(119896 = 1119873) 119880

119898E (119898 = 0119873 minus 1) 119880

119899U0(119899 =

1119873 minus 1) 1198801015840119901E (119901 = 0119873 minus 1) and 1198801015840

119902U0(119902 = 0119873 minus 1)

can be easily converted into real rational functions either of11988511198852or of the complex frequency 119904

Case E (119885119892= 119885119871= 1198851) From (8) and (9) after using the

relations119885119888= 1198852sdotsinh(120591) 119885

1= 119885119888sdot tanh(1205912) and cosh(120591) =

1 + 11988511198852 it can be easily obtained that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot 1198851sdot [ cosh (119873 sdot 120591)

+1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot 1198851sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591] + (1198851119885119888) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851119885119888) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119864

=cosh[(119873minus119899)sdot120591]+(1198851198881198851) sdot sinh [(119873minus119899) sdot 120591]

2sdot[cosh(119873sdot120591) + ((1198851+1198852) 119885119888) sdot sinh (119873sdot120591)]

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851198881198851) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] minus cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] minus cosh (119873 sdot 120591)

119880119899

119868119899

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888 119899 = 0119873

(18)

Mathematical Problems in Engineering 7

1198801015840119898

119864= (sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (1 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119898) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

= (2 sdot sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (2 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=2 sdot sinh [(119873 minus 119898) sdot 120591]

sinh[(119873+1) sdot 120591]+sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(19)

Case F (119885119892= 119885119871= 1198851+ 2 sdot 119885

2) From (8) and (9) after using

relations 119885119888= 1198852sdot sinh(120591) 119885

1= 119885119888 sdot tanh(1205912) cosh(120591) =

1 + 11988511198852 and 119885

119888sdot sinh(120591)(119885

1+ 21198852) = 119885

11198852 it readily

follows that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot (1198851+ 2 sdot 119885

2)

sdot [ cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot (1198851+ 2 sdot 119885

2) sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +1198851+ 2 sdot 119885

2

119885119888

sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 + 1) sdot 120591] + sinh (119873 sdot 120591)

119880119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times(2 sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +119885119888

(1198851+ 2 sdot 119885

2)sdot sinh (119873 sdot 120591))

minus1

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119868119899

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times ( sinh [(119873 minus 119899) sdot 120591 ]

+119885119888

(1198851+ 2 sdot 119885

2)sdot cosh [(119873 minus 119899) sdot 120591 ])

minus1

sdot 119885119888

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]sdot 119885119888

119899 = 0119873

(20)1198801015840119898

119864= (cosh [(119873 minus 119898) sdot 120591])

times ((1198851

1198852

+ 2)

sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh (120591) sdot cosh [(119873 minus 119898) sdot 120591]

[1 + cosh (120591)] sdot sinh [(119873 + 1) sdot 120591]

8 Mathematical Problems in Engineering

1198801015840119898

1198800

=2

(11988511198852) + 2

sdotcosh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)+(119885119888 (1198851+ 2 sdot 119885

2)) sdot sinh (119873sdot120591)

=2 sdot cosh [(119873 minus 119898) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(21)

Case G (119885119892= 119885119871= 119885119888= 119896 sdot 119885

1(119896 gt 1)) From (7)

it is easily obtained that 1198852= [(119896

2minus 1)2] sdot 119885

1 sinh(120591) =

2 sdot119896(1198962minus1) cosh(120591) = (1198962+1)(1198962minus1) exp(120591) = (119896+1)(119896minus1) and tanh(1205912) = 1119896 And from (8) and (9) it follows that

119868119899

119864=

1

2 sdot 119896 sdot 1198851

sdot (119896 minus 1

119896 + 1)

119899

119868119899

1198680

=119880119899

1198800

= (119896 minus 1

119896 + 1)

119899

119880119899

119864=1

2sdot (

119896 minus 1

119896 + 1)

119899

119880119899

119868119899

= 119885119888= 119896 sdot 119885

1 119899 = 0119873

(22)

1198801015840119898

119864=119896 minus 1

2 sdot 119896sdot (

119896 minus 1

119896 + 1)

119898

1198801015840119898

1198800

=119896 minus 1

119896sdot (

119896 minus 1

119896 + 1)

119898

119898 = 0119873 minus 1

(23)

From (22) and (23) it can be seen that even whenemf e [with Laplace transform 119864 = 119864(119904)] is pulsed thenode and point voltages with respect to common-ground119874 (Figure 1) are reproduced faithfully and without delaybut with geometrically progressive amplitude attenuationregardless of the impedance 119885

1

It is clear that relations (7)ndash(9) offer many other pos-sibilities for the selection of 119885

119892 119885119871 1198851 and 119885

2that lead

to generation of versatile ladders realizing different types offrequency selective networks An interesting one seems tobe the common-ground RLC ladder realizing (a) pulse delaywith independently selected Elmorersquos delay and rise timesandor (b) true delay for either pulsed or analog frequencylimited input signals These topics will be considered in thefollowing section

3 Elmorersquos Delay and Rise Times for SelectedTypes of Common-Ground Uniform RLCLadders

Consider the ladder in Figure 1 having 1198851= 119877 + 119871 sdot 119904 119884

2=

11198852= 119866 + 119862 sdot 119904 and 119885

119892= 119885119871= (119871119862)

12 Suppose that

it holds the relation 120572 = 119877119871 = 119866119862 similar to the oneassociated with per-unit-length parameters of distortionlesstransmission line [18 19] Let us define the auxiliary parameter120573 = 1(119871 sdot 119862)

12 and let us recast the network voltagetransmittances describing the transfer of emf E to the ladderpoints with voltages 119880

0 1198801 119880

119873(Figure 1)mdashin the form

suitable for calculating of Elmorersquos delay and rise times [7](Appendix B) Firstly we should observe that the followingholds

1198851

1198852

= (119904 + 120572

120573)

2

1198851

119885119871

=119885119871

119885119888

sdot sinh (120591) = 119904 + 120572

120573

119885119888

119885119871

sdot sinh (120591) = 119904 + 120572

120573sdot [(

119904 + 120572

120573)

2

+ 2]

1

119885119892+ 119885119871

sdot (119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (120591)

=119904 + 120572

2 sdot 120573sdot [(

119904 + 120572

120573)

2

+ 3]

(24)

and then by using (8) and the Properties 2 and 3 let us expressthe point voltage transmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873 minus 1)

and 119879119873(119904) = 119880

119873119864 in the following form

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [

2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587) ])

minus1

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 2: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

2 Mathematical Problems in Engineering

circuit parameters and its drawbacks are the uncertaintyof accuracy and restriction to being the estimate only forthe step-response delay An extension of Elmorersquos delaydefinition has been proposed in [8] to accommodate theeffect of nonunit-step (slow) excitations and to handlemultiple sources of excitation in order to show that delayestimation for slow excitations is no harder than for theunit-step input In [9] it has been shown that this extensionof Elmorersquos delay time offers a provision to deal with slowvarying excitations in timing analysis of MOS pass transistornetworks In addition in [10] it has been reported thatElmorersquos delay is an absolute upper bound on the actual50 delay of an RC tree response Also in [10] it has beenproved that this bound holds for input signals other thansteps and that actual delay asymptotically approaches toElmorersquos delay as the input signal rise-time increases It hasbeen emphasized in [11] that RC tree step responses alwaysare monotonic and this is why Elmorersquos definitions of bothdelay and rise time [7] are applicable on complex RC treenetworks

In this paperwewill firstly derive the general closed-formexpressions for input and transfer functions of common-ground uniform and passive ladders with complex doubleterminations making a distinction between the signal trans-fer to ladder nodes and to its points Then simplifications ofthe obtained results are produced for ladders with seven spe-cific pairs of complex double terminations being interestingfrom the practical point of view Thereafter for a uniformcommon-ground RLC ladder with (a) a relation between itsparameters resembling to analogous relation between per-unit-length parameters of distortionless transmission lineand (b) symmetric resistive double termination resemblingto characteristic impedance of distortionless line [12] it willbe shown that its Elmorersquos delay and rise times for pointvoltage transmittances can be efficiently calculated (but not inthe closed form) by using of the numerical scheme proposedherein In this case we will see that the obtained numericalvalues for Elmorersquos times differ slightly from the delay andrise times obtained according to their classical definitions andby using pspice simulation when ladder is excited by a stepvoltage

For the integrating type of distributed RC impedancecommon-ground ladder as a model of two-wire line it hasbeen suggested that it might be used as a true pulse delay line[13] provided that Schmitt triggers are used for reshapingthe delayed and edge-distorted transmitted signals This typeof delay line with step-input excitation has already beenthoroughly investigated in [14]mdashwhere Elmorersquos delay time isgiven in closed form only for ladder nodes and for Elmorersquosrise time is offered a conjecture relating to its lower boundfor overall network Elmorersquos rise times for all nodes ofintegrating RC open-circuited ladder are given in closedform in [15] and the obtaining of Elmorersquos delay and risetimes both for nodes and points is discussed to some extent in[16]mdashfor other types of open-circuited ladders In this paperwe are going to formulate the explicit closed-formexpressionsfor Elmorersquos delay and rise times both for the node and pointvoltages of integrating open circuited common-ground RCladder and will conclude that this type of network is not

recommendable for pulse delay line in its own right sinceElmorersquos rise time of each point voltage is not less than thetwice of its delay-time

And finally we will propose a type of uniform common-ground RLC ladder amenable for application as delay lineboth for pulsed andor analog input signals Elmorersquos delayand rise times of this ladder relating to the node voltagetransmittances are produced in closed form opening thuswith the following possibilities in ladder realization

(i) for pulsed inputs the overall Elmore delay and risetimes may be specified arbitrarily

(ii) for pulsed inputs theminimum ladder length (ie theminimumnecessary number of sections) is calculatedstraightforwardly by using only the overall Elmoredelay and rise times

(iii) the ladder RL parameters are calculated uniquelyfrom the assumed nonminimal ladder length overallElmore delay time and the assumed capacitancevalues

(iv) for realization of true delay for pulsed andor ana-log input signals with arbitrary variation in timeminimum ladder length is calculated from the spec-ified true delay time (which asymptotically tends toElmorersquos delay time when the number of sectionstends to infinity) and maximum frequency in thespectrum of the signal being transmitted along theladder purporting to represent delay line And againas in (iii) the ladder RL parameters are determinedfrom the ladder length overall ladder true delay time(being approximately equal to Elmorersquos delay time)and the assumed capacitance values The previousapproach and obtained results are illustrated andverified with realization examples of large true delaytime (=5 [ms]) for pulsed sine distorted sine CAMFM and chirp frequency and sweep amplitude inputsignals

Recall that in network synthesis the ladder topology is apreferable one since it has very low sensitivity to variationsof RLC parameters [16 17]

2 Network Functions of Common-GroundUniform and Passive RLC Ladders

Consider a uniform grounded and passive RLC ladder inFigure 1 with N identical sections whose impedances 119885

1=

1198851(119904) and 119885

2= 1198852(119904) are positive real rational functions

in complex frequency 119904 = 120590 + 119895120596 (119895 = radicminus1) There onare denoted the Laplace transforms 119864 = 119864(119904) of the voltageexcitation 119890 = 119890(119905) 119880

119896= 119880119896(s) and 119868

119896= 119868119896(s) of the point

voltages 119906119896= 119906119896(t) and currents 119894

119896= 119894119896(t) (119896 = 0119873) and

U1015840119898=U1015840119898(s) and 1198681015840

119898= 1198681015840119898(s) of node voltages 1199061015840

119898= 1199061015840119898(t) and

currents 1198941015840119898= 1198941015840119898(t) (119898 = 0119873 minus 1) The internal impedances

of the voltage excitation (voltage generator) and the load are119885119892= 119885119892(s) and 119885

119871= 119885119871(119904) respectively All the initial

conditions associated to the network reactive (LC) elementsare assumed to be zero

Mathematical Problems in Engineering 3

O

Zg

EU0

I0

Section 1

Section n

Section N

Z1 Z1Mnminus1

Z2

In

Un UNminus1Unminus1Inminus1

middot middot middot

middot middot middotmiddot middot middot

middot middot middot

UNZL

IN

Inminus1998400

Unminus1998400

Figure 1 The general common-ground uniform ladder with N sections

For ladder in Figure 1 the mesh transform equations forits 119899th section (119899 = 1119873) read

(1198851+ 1198852) sdot 119868119899minus1

minus 1198852sdot 119868119899= 119880119899minus1

1198852sdot 119868119899minus1

minus (1198851+ 1198852) sdot 119868119899= 119880119899

or in the recurrence form

[119880119899

119868119899

] =[[[

[

1 +1198851

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

sdot [119880119899minus1

119868119899minus1

]

(1)

The boundary values of voltages and currents in the set ofmatrix difference equations (1) are (U

0 I0) and (119880

119873 119868119873) For

119899 = 0119873 from (1) it immediately follows taht

[119880119899

119868119899

] =[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

119899

sdot [1198800

1198680

]

where 1198800= 119864 minus 119885

119892sdot 1198680and 119880119873= 119885119871sdot 119868119873

(2)

Since the roots (say 1205821and 120582

2) of the characteristic equation

corresponding to 2 times 2matrix appearing in (1)

det

[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

minus 120582 sdot U2

= 1205822minus 2 sdot (

Z1

Z2

+ 1) sdot 120582 + 1 = 0

(U2is 2 times 2 identity matrix)

(3)

satisfy the relations 1205821+ 1205822= 2 sdot (119885

11198852+ 1) and 120582

1sdot 1205822=

1 then by taking for convenience 1205821= exp(120591) and 120582

2=

exp(minus120591) we obtain cosh(120591) = 1 + 11988511198852 By using Cayley-

Hamiltonrsquos theorem we may put

[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

119899

= 1198600sdot U2+ 1198601sdot[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

(4)

where the ldquoconstantsrdquo1198600and119860

1are determined from system

of equations [exp(120591)]119899 = 1198600+ 1198601sdot exp(120591) and [exp(minus120591)]119899 =

1198600+ 1198601sdot exp(minus120591) whose solution is 119860

0= minus sinh[(119899 minus 1) sdot

120591]sinh(120591) and 1198601= sinh(119899 sdot 120591)sinh(120591)

By substituting1198600and119860

1in (4) and bearing inmind that

1 +1198851

1198852

= cosh (120591)

minus(11988521

1198852

+ 2 sdot 1198851) = 119885

2minus 1198852sdot (

1198851+ 1198852

1198852

)

2

= 1198852sdot [1 minus cosh2 (120591)] = minus119885

2sdot sinh2 (120591)

(5)

we finally obtain from (2) and (4) for 119899 = 0119873 that

[119880119899

119868119899

] =[[[

[

1 +1198851

1198852

minus(11988521

1198852

+ 2 sdot 1198851)

minus1

1198852

1 +1198851

1198852

]]]

]

n

sdot [1198800

1198680

]

=[[[

[

cosh (119899 sdot 120591) minus1198852sdot sinh (120591) sdot sinh (119899 sdot 120591)

minussinh (119899 sdot 120591)1198852sdot sinh (120591)

cosh (119899 sdot 120591)]]]

]

sdot [1198800

1198680

]

(6)

4 Mathematical Problems in Engineering

Since uniform ladder sections are electrically reciprocalthen their characteristic impedance 119885

119888and the quantity 119885

119888sdot

sinh(120591) can easily be produced in the form

119885119888= 1198852sdot sinh (120591) = radic119885

1sdot (1198851+ 2 sdot 119885

2)

119885119888sdot sinh (120591) = 119885

2sdot sinh2 (120591)

= 1198852sdot [cosh2 (120591) minus 1] = 119885

1sdot (

1198851

1198852

+ 2)

(7)

For the ladder depicted in Figure 1 the complete set ofthe network immittance and voltagecurrent transmittancefunctions can be produced by using the relations (2) (6) and(7)

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119871

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119868119899

1198680

= (119885119888sdot cosh [(119873 minus 119899) sdot 120591]

+119885119871sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119888sdot cosh (119873 sdot 120591) + 119885

119871sdot sinh (119873 sdot 120591))

minus1

119880119899

119864= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119880119899

1198800

= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119871sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591))

minus1

119880119899

119868119899

= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119888sdot cosh [(119873 minus 119899) sdot 120591]

+119885119871sdot sinh [(119873 minus 119899) sdot 120591])

minus1

sdot 119885119888 119899 = 0119873

(8)

Since 1198851= 119885119888sdot (sinh(120591)[1 + cosh(120591)]) = 119885

119888sdot tanh(1205912)

then the voltages and currents of impedances 1198852connecting

the middle nodes of ladder sections (119872119898) to common-node

119874 (Figure 1) are given as

1198801015840

119898=119880119898+ 119880119898+1

1 + cosh (120591)

1198681015840

119898=1198801015840119898

1198852

=119880119898+ 119880119898+1

1198852sdot [1 + cosh (120591)]

=119880119898+ 119880119898+1

1198851+ 2 sdot 119885

2

1198801015840119898

119864= ( (119885

119871minus 1198851) sdot cosh [(119873 minus 119898) sdot 120591]

+ [119885119888minus 119885119871sdot tanh(120591

2)]

sdot sinh [(119873 minus 119898) sdot 120591] )

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119898 = 0119873 minus 1

1198801015840119898

1198800

= ( (119885119871minus 1198851) sdot cosh [(119873 minus 119898) sdot 120591]

+ [119885119888minus 119885119871sdot tanh(120591

2)]

sdot sinh [(119873 minus 119898) sdot 120591] )

times (119885119871sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591))

minus1

(9)

Consider now the particular selection of 119885119892119885119871 to

investigate the generation of (i) finite length frequency-selecting ladders with various input and transfer immittancesand voltage and current transmittances and (ii) specificladder which can take the role of finite pulse delay linewithout pulse attenuation and with independently controlledpulse delay and rise times in Elmorersquos sense [7] calculable inclosed form We will assume that impedances 119885

119892and 119885

119871are

rational positive real functions in s so that they are realizableby passive transformerless RLC networks [17] All networkfunctions in (8) and (9) are real rational functions in s except

Mathematical Problems in Engineering 5

119880119899119868119899 which must be rational positive real function in s since

it is the input immittance of RLC network

Case A (119885119892-arbitrary and 119885

119871rarr infin [Ω]) (open-circuited

ladder) From (8) and (9) it follows

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

119885119888sdot cosh (119873 sdot 120591) + 119885

119892sdot sinh (119873 sdot 120591)

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(10)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)]

1198801015840119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(11)

Since we have cosh(120591) = 1 + 11988511198852 then by Property 2

(Appendix A) 119880119899U0(119899 = 1119873) (10) can be easily converted

into real rational function of 11988511198852or of s Similarly by

Property 3 (Appendix A) 1198681198981198680(119898 = 1119873 minus 1) (10) can be

also easily converted into real rational function of 11988511198852

or of s When 1198851and 119885

2are one-element-kind impedances

the zeros and poles of both 119880119899U0and 119868119898I0can be easily

determined in the closed form Since by Property 3 sinh[(119873minus

119898) sdot 120591] and sinh[(119873 minus 119898 minus 1) sdot 120591] contain the same factorsinh(120591) (119898 = 0119873 minus 1) then 1198801015840

119898U0(119898 = 0119873 minus 1) (11) can

be also easily converted into real rational function either of11988511198852or of the complex frequency s

Case B (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+1198852and119885119871rarr infin [Ω]) In this

case (10) and (11) simplify to

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

1198852sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=cosh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

sinh [(119873 minus 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(12)

1198801015840119898

119864

=2 sdot cosh [(119873 minus 119898 minus (12)) sdot 120591] sdot sinh (1205912)

sinh [(119873 + 1) sdot 120591]

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898 minus (12)) sdot 120591]

cosh (1205912) sdot cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(13)

Case C (119885119892-arbitrary and119885

119871= 0 [Ω] (short-circuited ladder))

From (8) and (9) it is obtained

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

119885119892sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591)

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=

sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(14)

1198801015840119898

119864

=119885119888sdot sinh [(119873 minus 119898) sdot 120591] minus 1198851 sdot cosh [(119873 minus 119898) sdot 120591]

119885119888sdot sinh (119873 sdot 120591) + 119885

119892sdot cosh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)]

6 Mathematical Problems in Engineering

1198801015840119898

1198800

=sinh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot cosh [(119873 minus 119898) sdot 120591]

sinh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(15)

According to Properties 2 and 3 respectively 119868119899I0(119899 = 1119873)

and 119880119898U0(119898 = 1119873 minus 1) (14) can be easily converted into

real rational functions either of11988511198852or of the complex fre-

quency s If1198851and119885

2are one-element-kind impedances the

zeros and poles of both 119868119899I0and 119880

119898U0can be determined

straightforwardly in closed formNow since119885119888sdotsinh(120591)119885

1=

2 + 11988511198852andor (1 + 119885

11198852)2minus sinh2(120591) = 1 it can be

seen from (15) that1198801015840119898U0(119898 = 0119873 minus 1) is also convertible

into real rational function either of 11988511198852or of the complex

frequency 119904 by using of both Properties 2 and 3

Case D (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+ 1198852and 119885119871= 0 [Ω]) From

(14) and (15) it readily follows that

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

1198852sdot cosh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=sinh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

cosh [(119873 + 1) sdot 120591]

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(16)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

cosh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(17)

By using of Properties 2 and 3 and sinh2(120591) = (1 + 11988511198852)2minus

1 it can be seen from (16) and (17) that the network functions119868119896I0(119896 = 1119873) 119880

119898E (119898 = 0119873 minus 1) 119880

119899U0(119899 =

1119873 minus 1) 1198801015840119901E (119901 = 0119873 minus 1) and 1198801015840

119902U0(119902 = 0119873 minus 1)

can be easily converted into real rational functions either of11988511198852or of the complex frequency 119904

Case E (119885119892= 119885119871= 1198851) From (8) and (9) after using the

relations119885119888= 1198852sdotsinh(120591) 119885

1= 119885119888sdot tanh(1205912) and cosh(120591) =

1 + 11988511198852 it can be easily obtained that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot 1198851sdot [ cosh (119873 sdot 120591)

+1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot 1198851sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591] + (1198851119885119888) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851119885119888) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119864

=cosh[(119873minus119899)sdot120591]+(1198851198881198851) sdot sinh [(119873minus119899) sdot 120591]

2sdot[cosh(119873sdot120591) + ((1198851+1198852) 119885119888) sdot sinh (119873sdot120591)]

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851198881198851) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] minus cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] minus cosh (119873 sdot 120591)

119880119899

119868119899

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888 119899 = 0119873

(18)

Mathematical Problems in Engineering 7

1198801015840119898

119864= (sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (1 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119898) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

= (2 sdot sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (2 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=2 sdot sinh [(119873 minus 119898) sdot 120591]

sinh[(119873+1) sdot 120591]+sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(19)

Case F (119885119892= 119885119871= 1198851+ 2 sdot 119885

2) From (8) and (9) after using

relations 119885119888= 1198852sdot sinh(120591) 119885

1= 119885119888 sdot tanh(1205912) cosh(120591) =

1 + 11988511198852 and 119885

119888sdot sinh(120591)(119885

1+ 21198852) = 119885

11198852 it readily

follows that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot (1198851+ 2 sdot 119885

2)

sdot [ cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot (1198851+ 2 sdot 119885

2) sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +1198851+ 2 sdot 119885

2

119885119888

sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 + 1) sdot 120591] + sinh (119873 sdot 120591)

119880119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times(2 sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +119885119888

(1198851+ 2 sdot 119885

2)sdot sinh (119873 sdot 120591))

minus1

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119868119899

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times ( sinh [(119873 minus 119899) sdot 120591 ]

+119885119888

(1198851+ 2 sdot 119885

2)sdot cosh [(119873 minus 119899) sdot 120591 ])

minus1

sdot 119885119888

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]sdot 119885119888

119899 = 0119873

(20)1198801015840119898

119864= (cosh [(119873 minus 119898) sdot 120591])

times ((1198851

1198852

+ 2)

sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh (120591) sdot cosh [(119873 minus 119898) sdot 120591]

[1 + cosh (120591)] sdot sinh [(119873 + 1) sdot 120591]

8 Mathematical Problems in Engineering

1198801015840119898

1198800

=2

(11988511198852) + 2

sdotcosh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)+(119885119888 (1198851+ 2 sdot 119885

2)) sdot sinh (119873sdot120591)

=2 sdot cosh [(119873 minus 119898) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(21)

Case G (119885119892= 119885119871= 119885119888= 119896 sdot 119885

1(119896 gt 1)) From (7)

it is easily obtained that 1198852= [(119896

2minus 1)2] sdot 119885

1 sinh(120591) =

2 sdot119896(1198962minus1) cosh(120591) = (1198962+1)(1198962minus1) exp(120591) = (119896+1)(119896minus1) and tanh(1205912) = 1119896 And from (8) and (9) it follows that

119868119899

119864=

1

2 sdot 119896 sdot 1198851

sdot (119896 minus 1

119896 + 1)

119899

119868119899

1198680

=119880119899

1198800

= (119896 minus 1

119896 + 1)

119899

119880119899

119864=1

2sdot (

119896 minus 1

119896 + 1)

119899

119880119899

119868119899

= 119885119888= 119896 sdot 119885

1 119899 = 0119873

(22)

1198801015840119898

119864=119896 minus 1

2 sdot 119896sdot (

119896 minus 1

119896 + 1)

119898

1198801015840119898

1198800

=119896 minus 1

119896sdot (

119896 minus 1

119896 + 1)

119898

119898 = 0119873 minus 1

(23)

From (22) and (23) it can be seen that even whenemf e [with Laplace transform 119864 = 119864(119904)] is pulsed thenode and point voltages with respect to common-ground119874 (Figure 1) are reproduced faithfully and without delaybut with geometrically progressive amplitude attenuationregardless of the impedance 119885

1

It is clear that relations (7)ndash(9) offer many other pos-sibilities for the selection of 119885

119892 119885119871 1198851 and 119885

2that lead

to generation of versatile ladders realizing different types offrequency selective networks An interesting one seems tobe the common-ground RLC ladder realizing (a) pulse delaywith independently selected Elmorersquos delay and rise timesandor (b) true delay for either pulsed or analog frequencylimited input signals These topics will be considered in thefollowing section

3 Elmorersquos Delay and Rise Times for SelectedTypes of Common-Ground Uniform RLCLadders

Consider the ladder in Figure 1 having 1198851= 119877 + 119871 sdot 119904 119884

2=

11198852= 119866 + 119862 sdot 119904 and 119885

119892= 119885119871= (119871119862)

12 Suppose that

it holds the relation 120572 = 119877119871 = 119866119862 similar to the oneassociated with per-unit-length parameters of distortionlesstransmission line [18 19] Let us define the auxiliary parameter120573 = 1(119871 sdot 119862)

12 and let us recast the network voltagetransmittances describing the transfer of emf E to the ladderpoints with voltages 119880

0 1198801 119880

119873(Figure 1)mdashin the form

suitable for calculating of Elmorersquos delay and rise times [7](Appendix B) Firstly we should observe that the followingholds

1198851

1198852

= (119904 + 120572

120573)

2

1198851

119885119871

=119885119871

119885119888

sdot sinh (120591) = 119904 + 120572

120573

119885119888

119885119871

sdot sinh (120591) = 119904 + 120572

120573sdot [(

119904 + 120572

120573)

2

+ 2]

1

119885119892+ 119885119871

sdot (119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (120591)

=119904 + 120572

2 sdot 120573sdot [(

119904 + 120572

120573)

2

+ 3]

(24)

and then by using (8) and the Properties 2 and 3 let us expressthe point voltage transmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873 minus 1)

and 119879119873(119904) = 119880

119873119864 in the following form

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [

2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587) ])

minus1

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 3

O

Zg

EU0

I0

Section 1

Section n

Section N

Z1 Z1Mnminus1

Z2

In

Un UNminus1Unminus1Inminus1

middot middot middot

middot middot middotmiddot middot middot

middot middot middot

UNZL

IN

Inminus1998400

Unminus1998400

Figure 1 The general common-ground uniform ladder with N sections

For ladder in Figure 1 the mesh transform equations forits 119899th section (119899 = 1119873) read

(1198851+ 1198852) sdot 119868119899minus1

minus 1198852sdot 119868119899= 119880119899minus1

1198852sdot 119868119899minus1

minus (1198851+ 1198852) sdot 119868119899= 119880119899

or in the recurrence form

[119880119899

119868119899

] =[[[

[

1 +1198851

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

sdot [119880119899minus1

119868119899minus1

]

(1)

The boundary values of voltages and currents in the set ofmatrix difference equations (1) are (U

0 I0) and (119880

119873 119868119873) For

119899 = 0119873 from (1) it immediately follows taht

[119880119899

119868119899

] =[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

119899

sdot [1198800

1198680

]

where 1198800= 119864 minus 119885

119892sdot 1198680and 119880119873= 119885119871sdot 119868119873

(2)

Since the roots (say 1205821and 120582

2) of the characteristic equation

corresponding to 2 times 2matrix appearing in (1)

det

[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

minus 120582 sdot U2

= 1205822minus 2 sdot (

Z1

Z2

+ 1) sdot 120582 + 1 = 0

(U2is 2 times 2 identity matrix)

(3)

satisfy the relations 1205821+ 1205822= 2 sdot (119885

11198852+ 1) and 120582

1sdot 1205822=

1 then by taking for convenience 1205821= exp(120591) and 120582

2=

exp(minus120591) we obtain cosh(120591) = 1 + 11988511198852 By using Cayley-

Hamiltonrsquos theorem we may put

[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

119899

= 1198600sdot U2+ 1198601sdot[[[

[

1 +Z1

Z2

minus(Z21

Z2

+ 2 sdot Z1)

minus1

Z2

1 +Z1

Z2

]]]

]

(4)

where the ldquoconstantsrdquo1198600and119860

1are determined from system

of equations [exp(120591)]119899 = 1198600+ 1198601sdot exp(120591) and [exp(minus120591)]119899 =

1198600+ 1198601sdot exp(minus120591) whose solution is 119860

0= minus sinh[(119899 minus 1) sdot

120591]sinh(120591) and 1198601= sinh(119899 sdot 120591)sinh(120591)

By substituting1198600and119860

1in (4) and bearing inmind that

1 +1198851

1198852

= cosh (120591)

minus(11988521

1198852

+ 2 sdot 1198851) = 119885

2minus 1198852sdot (

1198851+ 1198852

1198852

)

2

= 1198852sdot [1 minus cosh2 (120591)] = minus119885

2sdot sinh2 (120591)

(5)

we finally obtain from (2) and (4) for 119899 = 0119873 that

[119880119899

119868119899

] =[[[

[

1 +1198851

1198852

minus(11988521

1198852

+ 2 sdot 1198851)

minus1

1198852

1 +1198851

1198852

]]]

]

n

sdot [1198800

1198680

]

=[[[

[

cosh (119899 sdot 120591) minus1198852sdot sinh (120591) sdot sinh (119899 sdot 120591)

minussinh (119899 sdot 120591)1198852sdot sinh (120591)

cosh (119899 sdot 120591)]]]

]

sdot [1198800

1198680

]

(6)

4 Mathematical Problems in Engineering

Since uniform ladder sections are electrically reciprocalthen their characteristic impedance 119885

119888and the quantity 119885

119888sdot

sinh(120591) can easily be produced in the form

119885119888= 1198852sdot sinh (120591) = radic119885

1sdot (1198851+ 2 sdot 119885

2)

119885119888sdot sinh (120591) = 119885

2sdot sinh2 (120591)

= 1198852sdot [cosh2 (120591) minus 1] = 119885

1sdot (

1198851

1198852

+ 2)

(7)

For the ladder depicted in Figure 1 the complete set ofthe network immittance and voltagecurrent transmittancefunctions can be produced by using the relations (2) (6) and(7)

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119871

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119868119899

1198680

= (119885119888sdot cosh [(119873 minus 119899) sdot 120591]

+119885119871sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119888sdot cosh (119873 sdot 120591) + 119885

119871sdot sinh (119873 sdot 120591))

minus1

119880119899

119864= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119880119899

1198800

= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119871sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591))

minus1

119880119899

119868119899

= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119888sdot cosh [(119873 minus 119899) sdot 120591]

+119885119871sdot sinh [(119873 minus 119899) sdot 120591])

minus1

sdot 119885119888 119899 = 0119873

(8)

Since 1198851= 119885119888sdot (sinh(120591)[1 + cosh(120591)]) = 119885

119888sdot tanh(1205912)

then the voltages and currents of impedances 1198852connecting

the middle nodes of ladder sections (119872119898) to common-node

119874 (Figure 1) are given as

1198801015840

119898=119880119898+ 119880119898+1

1 + cosh (120591)

1198681015840

119898=1198801015840119898

1198852

=119880119898+ 119880119898+1

1198852sdot [1 + cosh (120591)]

=119880119898+ 119880119898+1

1198851+ 2 sdot 119885

2

1198801015840119898

119864= ( (119885

119871minus 1198851) sdot cosh [(119873 minus 119898) sdot 120591]

+ [119885119888minus 119885119871sdot tanh(120591

2)]

sdot sinh [(119873 minus 119898) sdot 120591] )

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119898 = 0119873 minus 1

1198801015840119898

1198800

= ( (119885119871minus 1198851) sdot cosh [(119873 minus 119898) sdot 120591]

+ [119885119888minus 119885119871sdot tanh(120591

2)]

sdot sinh [(119873 minus 119898) sdot 120591] )

times (119885119871sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591))

minus1

(9)

Consider now the particular selection of 119885119892119885119871 to

investigate the generation of (i) finite length frequency-selecting ladders with various input and transfer immittancesand voltage and current transmittances and (ii) specificladder which can take the role of finite pulse delay linewithout pulse attenuation and with independently controlledpulse delay and rise times in Elmorersquos sense [7] calculable inclosed form We will assume that impedances 119885

119892and 119885

119871are

rational positive real functions in s so that they are realizableby passive transformerless RLC networks [17] All networkfunctions in (8) and (9) are real rational functions in s except

Mathematical Problems in Engineering 5

119880119899119868119899 which must be rational positive real function in s since

it is the input immittance of RLC network

Case A (119885119892-arbitrary and 119885

119871rarr infin [Ω]) (open-circuited

ladder) From (8) and (9) it follows

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

119885119888sdot cosh (119873 sdot 120591) + 119885

119892sdot sinh (119873 sdot 120591)

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(10)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)]

1198801015840119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(11)

Since we have cosh(120591) = 1 + 11988511198852 then by Property 2

(Appendix A) 119880119899U0(119899 = 1119873) (10) can be easily converted

into real rational function of 11988511198852or of s Similarly by

Property 3 (Appendix A) 1198681198981198680(119898 = 1119873 minus 1) (10) can be

also easily converted into real rational function of 11988511198852

or of s When 1198851and 119885

2are one-element-kind impedances

the zeros and poles of both 119880119899U0and 119868119898I0can be easily

determined in the closed form Since by Property 3 sinh[(119873minus

119898) sdot 120591] and sinh[(119873 minus 119898 minus 1) sdot 120591] contain the same factorsinh(120591) (119898 = 0119873 minus 1) then 1198801015840

119898U0(119898 = 0119873 minus 1) (11) can

be also easily converted into real rational function either of11988511198852or of the complex frequency s

Case B (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+1198852and119885119871rarr infin [Ω]) In this

case (10) and (11) simplify to

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

1198852sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=cosh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

sinh [(119873 minus 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(12)

1198801015840119898

119864

=2 sdot cosh [(119873 minus 119898 minus (12)) sdot 120591] sdot sinh (1205912)

sinh [(119873 + 1) sdot 120591]

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898 minus (12)) sdot 120591]

cosh (1205912) sdot cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(13)

Case C (119885119892-arbitrary and119885

119871= 0 [Ω] (short-circuited ladder))

From (8) and (9) it is obtained

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

119885119892sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591)

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=

sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(14)

1198801015840119898

119864

=119885119888sdot sinh [(119873 minus 119898) sdot 120591] minus 1198851 sdot cosh [(119873 minus 119898) sdot 120591]

119885119888sdot sinh (119873 sdot 120591) + 119885

119892sdot cosh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)]

6 Mathematical Problems in Engineering

1198801015840119898

1198800

=sinh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot cosh [(119873 minus 119898) sdot 120591]

sinh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(15)

According to Properties 2 and 3 respectively 119868119899I0(119899 = 1119873)

and 119880119898U0(119898 = 1119873 minus 1) (14) can be easily converted into

real rational functions either of11988511198852or of the complex fre-

quency s If1198851and119885

2are one-element-kind impedances the

zeros and poles of both 119868119899I0and 119880

119898U0can be determined

straightforwardly in closed formNow since119885119888sdotsinh(120591)119885

1=

2 + 11988511198852andor (1 + 119885

11198852)2minus sinh2(120591) = 1 it can be

seen from (15) that1198801015840119898U0(119898 = 0119873 minus 1) is also convertible

into real rational function either of 11988511198852or of the complex

frequency 119904 by using of both Properties 2 and 3

Case D (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+ 1198852and 119885119871= 0 [Ω]) From

(14) and (15) it readily follows that

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

1198852sdot cosh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=sinh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

cosh [(119873 + 1) sdot 120591]

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(16)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

cosh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(17)

By using of Properties 2 and 3 and sinh2(120591) = (1 + 11988511198852)2minus

1 it can be seen from (16) and (17) that the network functions119868119896I0(119896 = 1119873) 119880

119898E (119898 = 0119873 minus 1) 119880

119899U0(119899 =

1119873 minus 1) 1198801015840119901E (119901 = 0119873 minus 1) and 1198801015840

119902U0(119902 = 0119873 minus 1)

can be easily converted into real rational functions either of11988511198852or of the complex frequency 119904

Case E (119885119892= 119885119871= 1198851) From (8) and (9) after using the

relations119885119888= 1198852sdotsinh(120591) 119885

1= 119885119888sdot tanh(1205912) and cosh(120591) =

1 + 11988511198852 it can be easily obtained that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot 1198851sdot [ cosh (119873 sdot 120591)

+1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot 1198851sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591] + (1198851119885119888) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851119885119888) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119864

=cosh[(119873minus119899)sdot120591]+(1198851198881198851) sdot sinh [(119873minus119899) sdot 120591]

2sdot[cosh(119873sdot120591) + ((1198851+1198852) 119885119888) sdot sinh (119873sdot120591)]

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851198881198851) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] minus cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] minus cosh (119873 sdot 120591)

119880119899

119868119899

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888 119899 = 0119873

(18)

Mathematical Problems in Engineering 7

1198801015840119898

119864= (sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (1 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119898) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

= (2 sdot sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (2 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=2 sdot sinh [(119873 minus 119898) sdot 120591]

sinh[(119873+1) sdot 120591]+sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(19)

Case F (119885119892= 119885119871= 1198851+ 2 sdot 119885

2) From (8) and (9) after using

relations 119885119888= 1198852sdot sinh(120591) 119885

1= 119885119888 sdot tanh(1205912) cosh(120591) =

1 + 11988511198852 and 119885

119888sdot sinh(120591)(119885

1+ 21198852) = 119885

11198852 it readily

follows that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot (1198851+ 2 sdot 119885

2)

sdot [ cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot (1198851+ 2 sdot 119885

2) sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +1198851+ 2 sdot 119885

2

119885119888

sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 + 1) sdot 120591] + sinh (119873 sdot 120591)

119880119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times(2 sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +119885119888

(1198851+ 2 sdot 119885

2)sdot sinh (119873 sdot 120591))

minus1

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119868119899

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times ( sinh [(119873 minus 119899) sdot 120591 ]

+119885119888

(1198851+ 2 sdot 119885

2)sdot cosh [(119873 minus 119899) sdot 120591 ])

minus1

sdot 119885119888

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]sdot 119885119888

119899 = 0119873

(20)1198801015840119898

119864= (cosh [(119873 minus 119898) sdot 120591])

times ((1198851

1198852

+ 2)

sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh (120591) sdot cosh [(119873 minus 119898) sdot 120591]

[1 + cosh (120591)] sdot sinh [(119873 + 1) sdot 120591]

8 Mathematical Problems in Engineering

1198801015840119898

1198800

=2

(11988511198852) + 2

sdotcosh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)+(119885119888 (1198851+ 2 sdot 119885

2)) sdot sinh (119873sdot120591)

=2 sdot cosh [(119873 minus 119898) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(21)

Case G (119885119892= 119885119871= 119885119888= 119896 sdot 119885

1(119896 gt 1)) From (7)

it is easily obtained that 1198852= [(119896

2minus 1)2] sdot 119885

1 sinh(120591) =

2 sdot119896(1198962minus1) cosh(120591) = (1198962+1)(1198962minus1) exp(120591) = (119896+1)(119896minus1) and tanh(1205912) = 1119896 And from (8) and (9) it follows that

119868119899

119864=

1

2 sdot 119896 sdot 1198851

sdot (119896 minus 1

119896 + 1)

119899

119868119899

1198680

=119880119899

1198800

= (119896 minus 1

119896 + 1)

119899

119880119899

119864=1

2sdot (

119896 minus 1

119896 + 1)

119899

119880119899

119868119899

= 119885119888= 119896 sdot 119885

1 119899 = 0119873

(22)

1198801015840119898

119864=119896 minus 1

2 sdot 119896sdot (

119896 minus 1

119896 + 1)

119898

1198801015840119898

1198800

=119896 minus 1

119896sdot (

119896 minus 1

119896 + 1)

119898

119898 = 0119873 minus 1

(23)

From (22) and (23) it can be seen that even whenemf e [with Laplace transform 119864 = 119864(119904)] is pulsed thenode and point voltages with respect to common-ground119874 (Figure 1) are reproduced faithfully and without delaybut with geometrically progressive amplitude attenuationregardless of the impedance 119885

1

It is clear that relations (7)ndash(9) offer many other pos-sibilities for the selection of 119885

119892 119885119871 1198851 and 119885

2that lead

to generation of versatile ladders realizing different types offrequency selective networks An interesting one seems tobe the common-ground RLC ladder realizing (a) pulse delaywith independently selected Elmorersquos delay and rise timesandor (b) true delay for either pulsed or analog frequencylimited input signals These topics will be considered in thefollowing section

3 Elmorersquos Delay and Rise Times for SelectedTypes of Common-Ground Uniform RLCLadders

Consider the ladder in Figure 1 having 1198851= 119877 + 119871 sdot 119904 119884

2=

11198852= 119866 + 119862 sdot 119904 and 119885

119892= 119885119871= (119871119862)

12 Suppose that

it holds the relation 120572 = 119877119871 = 119866119862 similar to the oneassociated with per-unit-length parameters of distortionlesstransmission line [18 19] Let us define the auxiliary parameter120573 = 1(119871 sdot 119862)

12 and let us recast the network voltagetransmittances describing the transfer of emf E to the ladderpoints with voltages 119880

0 1198801 119880

119873(Figure 1)mdashin the form

suitable for calculating of Elmorersquos delay and rise times [7](Appendix B) Firstly we should observe that the followingholds

1198851

1198852

= (119904 + 120572

120573)

2

1198851

119885119871

=119885119871

119885119888

sdot sinh (120591) = 119904 + 120572

120573

119885119888

119885119871

sdot sinh (120591) = 119904 + 120572

120573sdot [(

119904 + 120572

120573)

2

+ 2]

1

119885119892+ 119885119871

sdot (119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (120591)

=119904 + 120572

2 sdot 120573sdot [(

119904 + 120572

120573)

2

+ 3]

(24)

and then by using (8) and the Properties 2 and 3 let us expressthe point voltage transmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873 minus 1)

and 119879119873(119904) = 119880

119873119864 in the following form

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [

2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587) ])

minus1

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 4: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

4 Mathematical Problems in Engineering

Since uniform ladder sections are electrically reciprocalthen their characteristic impedance 119885

119888and the quantity 119885

119888sdot

sinh(120591) can easily be produced in the form

119885119888= 1198852sdot sinh (120591) = radic119885

1sdot (1198851+ 2 sdot 119885

2)

119885119888sdot sinh (120591) = 119885

2sdot sinh2 (120591)

= 1198852sdot [cosh2 (120591) minus 1] = 119885

1sdot (

1198851

1198852

+ 2)

(7)

For the ladder depicted in Figure 1 the complete set ofthe network immittance and voltagecurrent transmittancefunctions can be produced by using the relations (2) (6) and(7)

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119871

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119868119899

1198680

= (119885119888sdot cosh [(119873 minus 119899) sdot 120591]

+119885119871sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119888sdot cosh (119873 sdot 120591) + 119885

119871sdot sinh (119873 sdot 120591))

minus1

119880119899

119864= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119880119899

1198800

= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119871sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591))

minus1

119880119899

119868119899

= (119885119871sdot cosh [(119873 minus 119899) sdot 120591]

+119885119888sdot sinh [(119873 minus 119899) sdot 120591])

times (119885119888sdot cosh [(119873 minus 119899) sdot 120591]

+119885119871sdot sinh [(119873 minus 119899) sdot 120591])

minus1

sdot 119885119888 119899 = 0119873

(8)

Since 1198851= 119885119888sdot (sinh(120591)[1 + cosh(120591)]) = 119885

119888sdot tanh(1205912)

then the voltages and currents of impedances 1198852connecting

the middle nodes of ladder sections (119872119898) to common-node

119874 (Figure 1) are given as

1198801015840

119898=119880119898+ 119880119898+1

1 + cosh (120591)

1198681015840

119898=1198801015840119898

1198852

=119880119898+ 119880119898+1

1198852sdot [1 + cosh (120591)]

=119880119898+ 119880119898+1

1198851+ 2 sdot 119885

2

1198801015840119898

119864= ( (119885

119871minus 1198851) sdot cosh [(119873 minus 119898) sdot 120591]

+ [119885119888minus 119885119871sdot tanh(120591

2)]

sdot sinh [(119873 minus 119898) sdot 120591] )

times ( (119885119892+ 119885119871) sdot cosh (119873 sdot 120591)

+(119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (119873 sdot 120591))

minus1

119898 = 0119873 minus 1

1198801015840119898

1198800

= ( (119885119871minus 1198851) sdot cosh [(119873 minus 119898) sdot 120591]

+ [119885119888minus 119885119871sdot tanh(120591

2)]

sdot sinh [(119873 minus 119898) sdot 120591] )

times (119885119871sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591))

minus1

(9)

Consider now the particular selection of 119885119892119885119871 to

investigate the generation of (i) finite length frequency-selecting ladders with various input and transfer immittancesand voltage and current transmittances and (ii) specificladder which can take the role of finite pulse delay linewithout pulse attenuation and with independently controlledpulse delay and rise times in Elmorersquos sense [7] calculable inclosed form We will assume that impedances 119885

119892and 119885

119871are

rational positive real functions in s so that they are realizableby passive transformerless RLC networks [17] All networkfunctions in (8) and (9) are real rational functions in s except

Mathematical Problems in Engineering 5

119880119899119868119899 which must be rational positive real function in s since

it is the input immittance of RLC network

Case A (119885119892-arbitrary and 119885

119871rarr infin [Ω]) (open-circuited

ladder) From (8) and (9) it follows

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

119885119888sdot cosh (119873 sdot 120591) + 119885

119892sdot sinh (119873 sdot 120591)

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(10)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)]

1198801015840119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(11)

Since we have cosh(120591) = 1 + 11988511198852 then by Property 2

(Appendix A) 119880119899U0(119899 = 1119873) (10) can be easily converted

into real rational function of 11988511198852or of s Similarly by

Property 3 (Appendix A) 1198681198981198680(119898 = 1119873 minus 1) (10) can be

also easily converted into real rational function of 11988511198852

or of s When 1198851and 119885

2are one-element-kind impedances

the zeros and poles of both 119880119899U0and 119868119898I0can be easily

determined in the closed form Since by Property 3 sinh[(119873minus

119898) sdot 120591] and sinh[(119873 minus 119898 minus 1) sdot 120591] contain the same factorsinh(120591) (119898 = 0119873 minus 1) then 1198801015840

119898U0(119898 = 0119873 minus 1) (11) can

be also easily converted into real rational function either of11988511198852or of the complex frequency s

Case B (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+1198852and119885119871rarr infin [Ω]) In this

case (10) and (11) simplify to

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

1198852sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=cosh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

sinh [(119873 minus 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(12)

1198801015840119898

119864

=2 sdot cosh [(119873 minus 119898 minus (12)) sdot 120591] sdot sinh (1205912)

sinh [(119873 + 1) sdot 120591]

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898 minus (12)) sdot 120591]

cosh (1205912) sdot cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(13)

Case C (119885119892-arbitrary and119885

119871= 0 [Ω] (short-circuited ladder))

From (8) and (9) it is obtained

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

119885119892sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591)

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=

sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(14)

1198801015840119898

119864

=119885119888sdot sinh [(119873 minus 119898) sdot 120591] minus 1198851 sdot cosh [(119873 minus 119898) sdot 120591]

119885119888sdot sinh (119873 sdot 120591) + 119885

119892sdot cosh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)]

6 Mathematical Problems in Engineering

1198801015840119898

1198800

=sinh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot cosh [(119873 minus 119898) sdot 120591]

sinh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(15)

According to Properties 2 and 3 respectively 119868119899I0(119899 = 1119873)

and 119880119898U0(119898 = 1119873 minus 1) (14) can be easily converted into

real rational functions either of11988511198852or of the complex fre-

quency s If1198851and119885

2are one-element-kind impedances the

zeros and poles of both 119868119899I0and 119880

119898U0can be determined

straightforwardly in closed formNow since119885119888sdotsinh(120591)119885

1=

2 + 11988511198852andor (1 + 119885

11198852)2minus sinh2(120591) = 1 it can be

seen from (15) that1198801015840119898U0(119898 = 0119873 minus 1) is also convertible

into real rational function either of 11988511198852or of the complex

frequency 119904 by using of both Properties 2 and 3

Case D (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+ 1198852and 119885119871= 0 [Ω]) From

(14) and (15) it readily follows that

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

1198852sdot cosh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=sinh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

cosh [(119873 + 1) sdot 120591]

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(16)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

cosh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(17)

By using of Properties 2 and 3 and sinh2(120591) = (1 + 11988511198852)2minus

1 it can be seen from (16) and (17) that the network functions119868119896I0(119896 = 1119873) 119880

119898E (119898 = 0119873 minus 1) 119880

119899U0(119899 =

1119873 minus 1) 1198801015840119901E (119901 = 0119873 minus 1) and 1198801015840

119902U0(119902 = 0119873 minus 1)

can be easily converted into real rational functions either of11988511198852or of the complex frequency 119904

Case E (119885119892= 119885119871= 1198851) From (8) and (9) after using the

relations119885119888= 1198852sdotsinh(120591) 119885

1= 119885119888sdot tanh(1205912) and cosh(120591) =

1 + 11988511198852 it can be easily obtained that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot 1198851sdot [ cosh (119873 sdot 120591)

+1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot 1198851sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591] + (1198851119885119888) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851119885119888) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119864

=cosh[(119873minus119899)sdot120591]+(1198851198881198851) sdot sinh [(119873minus119899) sdot 120591]

2sdot[cosh(119873sdot120591) + ((1198851+1198852) 119885119888) sdot sinh (119873sdot120591)]

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851198881198851) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] minus cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] minus cosh (119873 sdot 120591)

119880119899

119868119899

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888 119899 = 0119873

(18)

Mathematical Problems in Engineering 7

1198801015840119898

119864= (sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (1 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119898) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

= (2 sdot sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (2 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=2 sdot sinh [(119873 minus 119898) sdot 120591]

sinh[(119873+1) sdot 120591]+sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(19)

Case F (119885119892= 119885119871= 1198851+ 2 sdot 119885

2) From (8) and (9) after using

relations 119885119888= 1198852sdot sinh(120591) 119885

1= 119885119888 sdot tanh(1205912) cosh(120591) =

1 + 11988511198852 and 119885

119888sdot sinh(120591)(119885

1+ 21198852) = 119885

11198852 it readily

follows that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot (1198851+ 2 sdot 119885

2)

sdot [ cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot (1198851+ 2 sdot 119885

2) sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +1198851+ 2 sdot 119885

2

119885119888

sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 + 1) sdot 120591] + sinh (119873 sdot 120591)

119880119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times(2 sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +119885119888

(1198851+ 2 sdot 119885

2)sdot sinh (119873 sdot 120591))

minus1

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119868119899

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times ( sinh [(119873 minus 119899) sdot 120591 ]

+119885119888

(1198851+ 2 sdot 119885

2)sdot cosh [(119873 minus 119899) sdot 120591 ])

minus1

sdot 119885119888

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]sdot 119885119888

119899 = 0119873

(20)1198801015840119898

119864= (cosh [(119873 minus 119898) sdot 120591])

times ((1198851

1198852

+ 2)

sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh (120591) sdot cosh [(119873 minus 119898) sdot 120591]

[1 + cosh (120591)] sdot sinh [(119873 + 1) sdot 120591]

8 Mathematical Problems in Engineering

1198801015840119898

1198800

=2

(11988511198852) + 2

sdotcosh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)+(119885119888 (1198851+ 2 sdot 119885

2)) sdot sinh (119873sdot120591)

=2 sdot cosh [(119873 minus 119898) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(21)

Case G (119885119892= 119885119871= 119885119888= 119896 sdot 119885

1(119896 gt 1)) From (7)

it is easily obtained that 1198852= [(119896

2minus 1)2] sdot 119885

1 sinh(120591) =

2 sdot119896(1198962minus1) cosh(120591) = (1198962+1)(1198962minus1) exp(120591) = (119896+1)(119896minus1) and tanh(1205912) = 1119896 And from (8) and (9) it follows that

119868119899

119864=

1

2 sdot 119896 sdot 1198851

sdot (119896 minus 1

119896 + 1)

119899

119868119899

1198680

=119880119899

1198800

= (119896 minus 1

119896 + 1)

119899

119880119899

119864=1

2sdot (

119896 minus 1

119896 + 1)

119899

119880119899

119868119899

= 119885119888= 119896 sdot 119885

1 119899 = 0119873

(22)

1198801015840119898

119864=119896 minus 1

2 sdot 119896sdot (

119896 minus 1

119896 + 1)

119898

1198801015840119898

1198800

=119896 minus 1

119896sdot (

119896 minus 1

119896 + 1)

119898

119898 = 0119873 minus 1

(23)

From (22) and (23) it can be seen that even whenemf e [with Laplace transform 119864 = 119864(119904)] is pulsed thenode and point voltages with respect to common-ground119874 (Figure 1) are reproduced faithfully and without delaybut with geometrically progressive amplitude attenuationregardless of the impedance 119885

1

It is clear that relations (7)ndash(9) offer many other pos-sibilities for the selection of 119885

119892 119885119871 1198851 and 119885

2that lead

to generation of versatile ladders realizing different types offrequency selective networks An interesting one seems tobe the common-ground RLC ladder realizing (a) pulse delaywith independently selected Elmorersquos delay and rise timesandor (b) true delay for either pulsed or analog frequencylimited input signals These topics will be considered in thefollowing section

3 Elmorersquos Delay and Rise Times for SelectedTypes of Common-Ground Uniform RLCLadders

Consider the ladder in Figure 1 having 1198851= 119877 + 119871 sdot 119904 119884

2=

11198852= 119866 + 119862 sdot 119904 and 119885

119892= 119885119871= (119871119862)

12 Suppose that

it holds the relation 120572 = 119877119871 = 119866119862 similar to the oneassociated with per-unit-length parameters of distortionlesstransmission line [18 19] Let us define the auxiliary parameter120573 = 1(119871 sdot 119862)

12 and let us recast the network voltagetransmittances describing the transfer of emf E to the ladderpoints with voltages 119880

0 1198801 119880

119873(Figure 1)mdashin the form

suitable for calculating of Elmorersquos delay and rise times [7](Appendix B) Firstly we should observe that the followingholds

1198851

1198852

= (119904 + 120572

120573)

2

1198851

119885119871

=119885119871

119885119888

sdot sinh (120591) = 119904 + 120572

120573

119885119888

119885119871

sdot sinh (120591) = 119904 + 120572

120573sdot [(

119904 + 120572

120573)

2

+ 2]

1

119885119892+ 119885119871

sdot (119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (120591)

=119904 + 120572

2 sdot 120573sdot [(

119904 + 120572

120573)

2

+ 3]

(24)

and then by using (8) and the Properties 2 and 3 let us expressthe point voltage transmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873 minus 1)

and 119879119873(119904) = 119880

119873119864 in the following form

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [

2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587) ])

minus1

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Stochastic AnalysisInternational Journal of

Page 5: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 5

119880119899119868119899 which must be rational positive real function in s since

it is the input immittance of RLC network

Case A (119885119892-arbitrary and 119885

119871rarr infin [Ω]) (open-circuited

ladder) From (8) and (9) it follows

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

119885119888sdot cosh (119873 sdot 120591) + 119885

119892sdot sinh (119873 sdot 120591)

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(10)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [cosh (119873 sdot 120591) + (119885119892119885119888) sdot sinh (119873 sdot 120591)]

1198801015840119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot sinh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(11)

Since we have cosh(120591) = 1 + 11988511198852 then by Property 2

(Appendix A) 119880119899U0(119899 = 1119873) (10) can be easily converted

into real rational function of 11988511198852or of s Similarly by

Property 3 (Appendix A) 1198681198981198680(119898 = 1119873 minus 1) (10) can be

also easily converted into real rational function of 11988511198852

or of s When 1198851and 119885

2are one-element-kind impedances

the zeros and poles of both 119880119899U0and 119868119898I0can be easily

determined in the closed form Since by Property 3 sinh[(119873minus

119898) sdot 120591] and sinh[(119873 minus 119898 minus 1) sdot 120591] contain the same factorsinh(120591) (119898 = 0119873 minus 1) then 1198801015840

119898U0(119898 = 0119873 minus 1) (11) can

be also easily converted into real rational function either of11988511198852or of the complex frequency s

Case B (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+1198852and119885119871rarr infin [Ω]) In this

case (10) and (11) simplify to

119868119899

119864=

sinh [(119873 minus 119899) sdot 120591]

1198852sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119864=cosh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

sinh [(119873 minus 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119868119899

= cotanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(12)

1198801015840119898

119864

=2 sdot cosh [(119873 minus 119898 minus (12)) sdot 120591] sdot sinh (1205912)

sinh [(119873 + 1) sdot 120591]

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898 minus (12)) sdot 120591]

cosh (1205912) sdot cosh (119873 sdot 120591)

=sinh [(119873 minus 119898) sdot 120591] minus sinh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(13)

Case C (119885119892-arbitrary and119885

119871= 0 [Ω] (short-circuited ladder))

From (8) and (9) it is obtained

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

119885119892sdot cosh (119873 sdot 120591) + 119885

119888sdot sinh (119873 sdot 120591)

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=

sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(14)

1198801015840119898

119864

=119885119888sdot sinh [(119873 minus 119898) sdot 120591] minus 1198851 sdot cosh [(119873 minus 119898) sdot 120591]

119885119888sdot sinh (119873 sdot 120591) + 119885

119892sdot cosh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot [sinh (119873 sdot 120591) + (119885119892119885119888) sdot cosh (119873 sdot 120591)]

6 Mathematical Problems in Engineering

1198801015840119898

1198800

=sinh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot cosh [(119873 minus 119898) sdot 120591]

sinh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(15)

According to Properties 2 and 3 respectively 119868119899I0(119899 = 1119873)

and 119880119898U0(119898 = 1119873 minus 1) (14) can be easily converted into

real rational functions either of11988511198852or of the complex fre-

quency s If1198851and119885

2are one-element-kind impedances the

zeros and poles of both 119868119899I0and 119880

119898U0can be determined

straightforwardly in closed formNow since119885119888sdotsinh(120591)119885

1=

2 + 11988511198852andor (1 + 119885

11198852)2minus sinh2(120591) = 1 it can be

seen from (15) that1198801015840119898U0(119898 = 0119873 minus 1) is also convertible

into real rational function either of 11988511198852or of the complex

frequency 119904 by using of both Properties 2 and 3

Case D (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+ 1198852and 119885119871= 0 [Ω]) From

(14) and (15) it readily follows that

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

1198852sdot cosh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=sinh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

cosh [(119873 + 1) sdot 120591]

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(16)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

cosh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(17)

By using of Properties 2 and 3 and sinh2(120591) = (1 + 11988511198852)2minus

1 it can be seen from (16) and (17) that the network functions119868119896I0(119896 = 1119873) 119880

119898E (119898 = 0119873 minus 1) 119880

119899U0(119899 =

1119873 minus 1) 1198801015840119901E (119901 = 0119873 minus 1) and 1198801015840

119902U0(119902 = 0119873 minus 1)

can be easily converted into real rational functions either of11988511198852or of the complex frequency 119904

Case E (119885119892= 119885119871= 1198851) From (8) and (9) after using the

relations119885119888= 1198852sdotsinh(120591) 119885

1= 119885119888sdot tanh(1205912) and cosh(120591) =

1 + 11988511198852 it can be easily obtained that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot 1198851sdot [ cosh (119873 sdot 120591)

+1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot 1198851sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591] + (1198851119885119888) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851119885119888) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119864

=cosh[(119873minus119899)sdot120591]+(1198851198881198851) sdot sinh [(119873minus119899) sdot 120591]

2sdot[cosh(119873sdot120591) + ((1198851+1198852) 119885119888) sdot sinh (119873sdot120591)]

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851198881198851) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] minus cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] minus cosh (119873 sdot 120591)

119880119899

119868119899

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888 119899 = 0119873

(18)

Mathematical Problems in Engineering 7

1198801015840119898

119864= (sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (1 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119898) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

= (2 sdot sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (2 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=2 sdot sinh [(119873 minus 119898) sdot 120591]

sinh[(119873+1) sdot 120591]+sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(19)

Case F (119885119892= 119885119871= 1198851+ 2 sdot 119885

2) From (8) and (9) after using

relations 119885119888= 1198852sdot sinh(120591) 119885

1= 119885119888 sdot tanh(1205912) cosh(120591) =

1 + 11988511198852 and 119885

119888sdot sinh(120591)(119885

1+ 21198852) = 119885

11198852 it readily

follows that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot (1198851+ 2 sdot 119885

2)

sdot [ cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot (1198851+ 2 sdot 119885

2) sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +1198851+ 2 sdot 119885

2

119885119888

sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 + 1) sdot 120591] + sinh (119873 sdot 120591)

119880119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times(2 sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +119885119888

(1198851+ 2 sdot 119885

2)sdot sinh (119873 sdot 120591))

minus1

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119868119899

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times ( sinh [(119873 minus 119899) sdot 120591 ]

+119885119888

(1198851+ 2 sdot 119885

2)sdot cosh [(119873 minus 119899) sdot 120591 ])

minus1

sdot 119885119888

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]sdot 119885119888

119899 = 0119873

(20)1198801015840119898

119864= (cosh [(119873 minus 119898) sdot 120591])

times ((1198851

1198852

+ 2)

sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh (120591) sdot cosh [(119873 minus 119898) sdot 120591]

[1 + cosh (120591)] sdot sinh [(119873 + 1) sdot 120591]

8 Mathematical Problems in Engineering

1198801015840119898

1198800

=2

(11988511198852) + 2

sdotcosh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)+(119885119888 (1198851+ 2 sdot 119885

2)) sdot sinh (119873sdot120591)

=2 sdot cosh [(119873 minus 119898) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(21)

Case G (119885119892= 119885119871= 119885119888= 119896 sdot 119885

1(119896 gt 1)) From (7)

it is easily obtained that 1198852= [(119896

2minus 1)2] sdot 119885

1 sinh(120591) =

2 sdot119896(1198962minus1) cosh(120591) = (1198962+1)(1198962minus1) exp(120591) = (119896+1)(119896minus1) and tanh(1205912) = 1119896 And from (8) and (9) it follows that

119868119899

119864=

1

2 sdot 119896 sdot 1198851

sdot (119896 minus 1

119896 + 1)

119899

119868119899

1198680

=119880119899

1198800

= (119896 minus 1

119896 + 1)

119899

119880119899

119864=1

2sdot (

119896 minus 1

119896 + 1)

119899

119880119899

119868119899

= 119885119888= 119896 sdot 119885

1 119899 = 0119873

(22)

1198801015840119898

119864=119896 minus 1

2 sdot 119896sdot (

119896 minus 1

119896 + 1)

119898

1198801015840119898

1198800

=119896 minus 1

119896sdot (

119896 minus 1

119896 + 1)

119898

119898 = 0119873 minus 1

(23)

From (22) and (23) it can be seen that even whenemf e [with Laplace transform 119864 = 119864(119904)] is pulsed thenode and point voltages with respect to common-ground119874 (Figure 1) are reproduced faithfully and without delaybut with geometrically progressive amplitude attenuationregardless of the impedance 119885

1

It is clear that relations (7)ndash(9) offer many other pos-sibilities for the selection of 119885

119892 119885119871 1198851 and 119885

2that lead

to generation of versatile ladders realizing different types offrequency selective networks An interesting one seems tobe the common-ground RLC ladder realizing (a) pulse delaywith independently selected Elmorersquos delay and rise timesandor (b) true delay for either pulsed or analog frequencylimited input signals These topics will be considered in thefollowing section

3 Elmorersquos Delay and Rise Times for SelectedTypes of Common-Ground Uniform RLCLadders

Consider the ladder in Figure 1 having 1198851= 119877 + 119871 sdot 119904 119884

2=

11198852= 119866 + 119862 sdot 119904 and 119885

119892= 119885119871= (119871119862)

12 Suppose that

it holds the relation 120572 = 119877119871 = 119866119862 similar to the oneassociated with per-unit-length parameters of distortionlesstransmission line [18 19] Let us define the auxiliary parameter120573 = 1(119871 sdot 119862)

12 and let us recast the network voltagetransmittances describing the transfer of emf E to the ladderpoints with voltages 119880

0 1198801 119880

119873(Figure 1)mdashin the form

suitable for calculating of Elmorersquos delay and rise times [7](Appendix B) Firstly we should observe that the followingholds

1198851

1198852

= (119904 + 120572

120573)

2

1198851

119885119871

=119885119871

119885119888

sdot sinh (120591) = 119904 + 120572

120573

119885119888

119885119871

sdot sinh (120591) = 119904 + 120572

120573sdot [(

119904 + 120572

120573)

2

+ 2]

1

119885119892+ 119885119871

sdot (119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (120591)

=119904 + 120572

2 sdot 120573sdot [(

119904 + 120572

120573)

2

+ 3]

(24)

and then by using (8) and the Properties 2 and 3 let us expressthe point voltage transmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873 minus 1)

and 119879119873(119904) = 119880

119873119864 in the following form

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [

2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587) ])

minus1

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 6: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

6 Mathematical Problems in Engineering

1198801015840119898

1198800

=sinh [(119873 minus 119898) sdot 120591] minus (1198851119885119888) sdot cosh [(119873 minus 119898) sdot 120591]

sinh (119873 sdot 120591)

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(15)

According to Properties 2 and 3 respectively 119868119899I0(119899 = 1119873)

and 119880119898U0(119898 = 1119873 minus 1) (14) can be easily converted into

real rational functions either of11988511198852or of the complex fre-

quency s If1198851and119885

2are one-element-kind impedances the

zeros and poles of both 119868119899I0and 119880

119898U0can be determined

straightforwardly in closed formNow since119885119888sdotsinh(120591)119885

1=

2 + 11988511198852andor (1 + 119885

11198852)2minus sinh2(120591) = 1 it can be

seen from (15) that1198801015840119898U0(119898 = 0119873 minus 1) is also convertible

into real rational function either of 11988511198852or of the complex

frequency 119904 by using of both Properties 2 and 3

Case D (119885119892= 1198852sdot 119888119900119904ℎ(120591) = 119885

1+ 1198852and 119885119871= 0 [Ω]) From

(14) and (15) it readily follows that

119868119899

119864=

cosh [(119873 minus 119899) sdot 120591]

1198852sdot cosh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)

119880119899

119864=sinh [(119873 minus 119899) sdot 120591] sdot sinh (120591)

cosh [(119873 + 1) sdot 120591]

119880119899

1198800

=sinh [(119873 minus 119899) sdot 120591]

sinh (119873 sdot 120591)

119880119899

119868119899

= tanh [(119873 minus 119899) sdot 120591] sdot 119885119888 119899 = 0119873

(16)

1198801015840119898

119864

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

cosh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

=cosh [(119873 minus 119898) sdot 120591] minus cosh [(119873 minus 119898 minus 1) sdot 120591]

sinh (120591) sdot sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(17)

By using of Properties 2 and 3 and sinh2(120591) = (1 + 11988511198852)2minus

1 it can be seen from (16) and (17) that the network functions119868119896I0(119896 = 1119873) 119880

119898E (119898 = 0119873 minus 1) 119880

119899U0(119899 =

1119873 minus 1) 1198801015840119901E (119901 = 0119873 minus 1) and 1198801015840

119902U0(119902 = 0119873 minus 1)

can be easily converted into real rational functions either of11988511198852or of the complex frequency 119904

Case E (119885119892= 119885119871= 1198851) From (8) and (9) after using the

relations119885119888= 1198852sdotsinh(120591) 119885

1= 119885119888sdot tanh(1205912) and cosh(120591) =

1 + 11988511198852 it can be easily obtained that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot 1198851sdot [ cosh (119873 sdot 120591)

+1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot 1198851sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

=cosh [(119873 minus 119899) sdot 120591] + (1198851119885119888) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851119885119888) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119864

=cosh[(119873minus119899)sdot120591]+(1198851198881198851) sdot sinh [(119873minus119899) sdot 120591]

2sdot[cosh(119873sdot120591) + ((1198851+1198852) 119885119888) sdot sinh (119873sdot120591)]

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591) + (1198851198881198851) sdot sinh (119873 sdot 120591)

=cosh [(119873 minus 119899 + 1) sdot 120591] minus cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] minus cosh (119873 sdot 120591)

119880119899

119868119899

=cosh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899) sdot 120591] + (1198851198881198851) sdot cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sdot 119885119888 119899 = 0119873

(18)

Mathematical Problems in Engineering 7

1198801015840119898

119864= (sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (1 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119898) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

= (2 sdot sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (2 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=2 sdot sinh [(119873 minus 119898) sdot 120591]

sinh[(119873+1) sdot 120591]+sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(19)

Case F (119885119892= 119885119871= 1198851+ 2 sdot 119885

2) From (8) and (9) after using

relations 119885119888= 1198852sdot sinh(120591) 119885

1= 119885119888 sdot tanh(1205912) cosh(120591) =

1 + 11988511198852 and 119885

119888sdot sinh(120591)(119885

1+ 21198852) = 119885

11198852 it readily

follows that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot (1198851+ 2 sdot 119885

2)

sdot [ cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot (1198851+ 2 sdot 119885

2) sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +1198851+ 2 sdot 119885

2

119885119888

sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 + 1) sdot 120591] + sinh (119873 sdot 120591)

119880119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times(2 sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +119885119888

(1198851+ 2 sdot 119885

2)sdot sinh (119873 sdot 120591))

minus1

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119868119899

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times ( sinh [(119873 minus 119899) sdot 120591 ]

+119885119888

(1198851+ 2 sdot 119885

2)sdot cosh [(119873 minus 119899) sdot 120591 ])

minus1

sdot 119885119888

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]sdot 119885119888

119899 = 0119873

(20)1198801015840119898

119864= (cosh [(119873 minus 119898) sdot 120591])

times ((1198851

1198852

+ 2)

sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh (120591) sdot cosh [(119873 minus 119898) sdot 120591]

[1 + cosh (120591)] sdot sinh [(119873 + 1) sdot 120591]

8 Mathematical Problems in Engineering

1198801015840119898

1198800

=2

(11988511198852) + 2

sdotcosh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)+(119885119888 (1198851+ 2 sdot 119885

2)) sdot sinh (119873sdot120591)

=2 sdot cosh [(119873 minus 119898) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(21)

Case G (119885119892= 119885119871= 119885119888= 119896 sdot 119885

1(119896 gt 1)) From (7)

it is easily obtained that 1198852= [(119896

2minus 1)2] sdot 119885

1 sinh(120591) =

2 sdot119896(1198962minus1) cosh(120591) = (1198962+1)(1198962minus1) exp(120591) = (119896+1)(119896minus1) and tanh(1205912) = 1119896 And from (8) and (9) it follows that

119868119899

119864=

1

2 sdot 119896 sdot 1198851

sdot (119896 minus 1

119896 + 1)

119899

119868119899

1198680

=119880119899

1198800

= (119896 minus 1

119896 + 1)

119899

119880119899

119864=1

2sdot (

119896 minus 1

119896 + 1)

119899

119880119899

119868119899

= 119885119888= 119896 sdot 119885

1 119899 = 0119873

(22)

1198801015840119898

119864=119896 minus 1

2 sdot 119896sdot (

119896 minus 1

119896 + 1)

119898

1198801015840119898

1198800

=119896 minus 1

119896sdot (

119896 minus 1

119896 + 1)

119898

119898 = 0119873 minus 1

(23)

From (22) and (23) it can be seen that even whenemf e [with Laplace transform 119864 = 119864(119904)] is pulsed thenode and point voltages with respect to common-ground119874 (Figure 1) are reproduced faithfully and without delaybut with geometrically progressive amplitude attenuationregardless of the impedance 119885

1

It is clear that relations (7)ndash(9) offer many other pos-sibilities for the selection of 119885

119892 119885119871 1198851 and 119885

2that lead

to generation of versatile ladders realizing different types offrequency selective networks An interesting one seems tobe the common-ground RLC ladder realizing (a) pulse delaywith independently selected Elmorersquos delay and rise timesandor (b) true delay for either pulsed or analog frequencylimited input signals These topics will be considered in thefollowing section

3 Elmorersquos Delay and Rise Times for SelectedTypes of Common-Ground Uniform RLCLadders

Consider the ladder in Figure 1 having 1198851= 119877 + 119871 sdot 119904 119884

2=

11198852= 119866 + 119862 sdot 119904 and 119885

119892= 119885119871= (119871119862)

12 Suppose that

it holds the relation 120572 = 119877119871 = 119866119862 similar to the oneassociated with per-unit-length parameters of distortionlesstransmission line [18 19] Let us define the auxiliary parameter120573 = 1(119871 sdot 119862)

12 and let us recast the network voltagetransmittances describing the transfer of emf E to the ladderpoints with voltages 119880

0 1198801 119880

119873(Figure 1)mdashin the form

suitable for calculating of Elmorersquos delay and rise times [7](Appendix B) Firstly we should observe that the followingholds

1198851

1198852

= (119904 + 120572

120573)

2

1198851

119885119871

=119885119871

119885119888

sdot sinh (120591) = 119904 + 120572

120573

119885119888

119885119871

sdot sinh (120591) = 119904 + 120572

120573sdot [(

119904 + 120572

120573)

2

+ 2]

1

119885119892+ 119885119871

sdot (119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (120591)

=119904 + 120572

2 sdot 120573sdot [(

119904 + 120572

120573)

2

+ 3]

(24)

and then by using (8) and the Properties 2 and 3 let us expressthe point voltage transmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873 minus 1)

and 119879119873(119904) = 119880

119873119864 in the following form

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [

2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587) ])

minus1

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 7: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 7

1198801015840119898

119864= (sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (1 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119898) sdot 120591]

sinh [(119873 + 1) sdot 120591]

1198801015840

119898

1198800

= (2 sdot sinh [(119873 minus 119898) sdot 120591])

times ( cosh (119873 sdot 120591) sdot sinh (120591)

+ (2 +1198851

1198852

) sdot sinh (119873 sdot 120591))

minus1

=2 sdot sinh [(119873 minus 119898) sdot 120591]

sinh[(119873+1) sdot 120591]+sinh (119873 sdot 120591)

119898 = 0119873 minus 1

(19)

Case F (119885119892= 119885119871= 1198851+ 2 sdot 119885

2) From (8) and (9) after using

relations 119885119888= 1198852sdot sinh(120591) 119885

1= 119885119888 sdot tanh(1205912) cosh(120591) =

1 + 11988511198852 and 119885

119888sdot sinh(120591)(119885

1+ 21198852) = 119885

11198852 it readily

follows that

119868119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (2 sdot (1198851+ 2 sdot 119885

2)

sdot [ cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

2 sdot (1198851+ 2 sdot 119885

2) sdot sinh [(119873 + 1) sdot 120591]

119868119899

1198680

= ( cosh [(119873 minus 119899) sdot 120591]

+1198851+ 2 sdot 119885

2

119885119888

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +1198851+ 2 sdot 119885

2

119885119888

sdot sinh (119873 sdot 120591))

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]

sinh [(119873 + 1) sdot 120591] + sinh (119873 sdot 120591)

119880119899

119864= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times(2 sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh [(119873 minus 119899 + 1) sdot 120591] minus sinh [(119873 minus 119899) sdot 120591]

2 sdot sinh [(119873 + 1) sdot 120591]

119880119899

1198800

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times (cosh (119873 sdot 120591) +119885119888

(1198851+ 2 sdot 119885

2)sdot sinh (119873 sdot 120591))

minus1

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119880119899

119868119899

= ( cosh [(119873 minus 119899) sdot 120591]

+119885119888

1198851+ 2 sdot 119885

2

sdot sinh [(119873 minus 119899) sdot 120591])

times ( sinh [(119873 minus 119899) sdot 120591 ]

+119885119888

(1198851+ 2 sdot 119885

2)sdot cosh [(119873 minus 119899) sdot 120591 ])

minus1

sdot 119885119888

=cosh [(119873 minus 119899 + 1) sdot 120591] + cosh [(119873 minus 119899) sdot 120591]

sinh [(119873 minus 119899 + 1) sdot 120591] + sinh [(119873 minus 119899) sdot 120591]sdot 119885119888

119899 = 0119873

(20)1198801015840119898

119864= (cosh [(119873 minus 119898) sdot 120591])

times ((1198851

1198852

+ 2)

sdot [cosh (119873 sdot 120591) +1198851+ 1198852

119885119888

sdot sinh (119873 sdot 120591)])

minus1

=sinh (120591) sdot cosh [(119873 minus 119898) sdot 120591]

[1 + cosh (120591)] sdot sinh [(119873 + 1) sdot 120591]

8 Mathematical Problems in Engineering

1198801015840119898

1198800

=2

(11988511198852) + 2

sdotcosh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)+(119885119888 (1198851+ 2 sdot 119885

2)) sdot sinh (119873sdot120591)

=2 sdot cosh [(119873 minus 119898) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(21)

Case G (119885119892= 119885119871= 119885119888= 119896 sdot 119885

1(119896 gt 1)) From (7)

it is easily obtained that 1198852= [(119896

2minus 1)2] sdot 119885

1 sinh(120591) =

2 sdot119896(1198962minus1) cosh(120591) = (1198962+1)(1198962minus1) exp(120591) = (119896+1)(119896minus1) and tanh(1205912) = 1119896 And from (8) and (9) it follows that

119868119899

119864=

1

2 sdot 119896 sdot 1198851

sdot (119896 minus 1

119896 + 1)

119899

119868119899

1198680

=119880119899

1198800

= (119896 minus 1

119896 + 1)

119899

119880119899

119864=1

2sdot (

119896 minus 1

119896 + 1)

119899

119880119899

119868119899

= 119885119888= 119896 sdot 119885

1 119899 = 0119873

(22)

1198801015840119898

119864=119896 minus 1

2 sdot 119896sdot (

119896 minus 1

119896 + 1)

119898

1198801015840119898

1198800

=119896 minus 1

119896sdot (

119896 minus 1

119896 + 1)

119898

119898 = 0119873 minus 1

(23)

From (22) and (23) it can be seen that even whenemf e [with Laplace transform 119864 = 119864(119904)] is pulsed thenode and point voltages with respect to common-ground119874 (Figure 1) are reproduced faithfully and without delaybut with geometrically progressive amplitude attenuationregardless of the impedance 119885

1

It is clear that relations (7)ndash(9) offer many other pos-sibilities for the selection of 119885

119892 119885119871 1198851 and 119885

2that lead

to generation of versatile ladders realizing different types offrequency selective networks An interesting one seems tobe the common-ground RLC ladder realizing (a) pulse delaywith independently selected Elmorersquos delay and rise timesandor (b) true delay for either pulsed or analog frequencylimited input signals These topics will be considered in thefollowing section

3 Elmorersquos Delay and Rise Times for SelectedTypes of Common-Ground Uniform RLCLadders

Consider the ladder in Figure 1 having 1198851= 119877 + 119871 sdot 119904 119884

2=

11198852= 119866 + 119862 sdot 119904 and 119885

119892= 119885119871= (119871119862)

12 Suppose that

it holds the relation 120572 = 119877119871 = 119866119862 similar to the oneassociated with per-unit-length parameters of distortionlesstransmission line [18 19] Let us define the auxiliary parameter120573 = 1(119871 sdot 119862)

12 and let us recast the network voltagetransmittances describing the transfer of emf E to the ladderpoints with voltages 119880

0 1198801 119880

119873(Figure 1)mdashin the form

suitable for calculating of Elmorersquos delay and rise times [7](Appendix B) Firstly we should observe that the followingholds

1198851

1198852

= (119904 + 120572

120573)

2

1198851

119885119871

=119885119871

119885119888

sdot sinh (120591) = 119904 + 120572

120573

119885119888

119885119871

sdot sinh (120591) = 119904 + 120572

120573sdot [(

119904 + 120572

120573)

2

+ 2]

1

119885119892+ 119885119871

sdot (119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (120591)

=119904 + 120572

2 sdot 120573sdot [(

119904 + 120572

120573)

2

+ 3]

(24)

and then by using (8) and the Properties 2 and 3 let us expressthe point voltage transmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873 minus 1)

and 119879119873(119904) = 119880

119873119864 in the following form

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [

2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587) ])

minus1

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 8: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

8 Mathematical Problems in Engineering

1198801015840119898

1198800

=2

(11988511198852) + 2

sdotcosh [(119873 minus 119898) sdot 120591]

cosh (119873 sdot 120591)+(119885119888 (1198851+ 2 sdot 119885

2)) sdot sinh (119873sdot120591)

=2 sdot cosh [(119873 minus 119898) sdot 120591]

cosh [(119873 + 1) sdot 120591] + cosh (119873 sdot 120591)

119898 = 0119873 minus 1

(21)

Case G (119885119892= 119885119871= 119885119888= 119896 sdot 119885

1(119896 gt 1)) From (7)

it is easily obtained that 1198852= [(119896

2minus 1)2] sdot 119885

1 sinh(120591) =

2 sdot119896(1198962minus1) cosh(120591) = (1198962+1)(1198962minus1) exp(120591) = (119896+1)(119896minus1) and tanh(1205912) = 1119896 And from (8) and (9) it follows that

119868119899

119864=

1

2 sdot 119896 sdot 1198851

sdot (119896 minus 1

119896 + 1)

119899

119868119899

1198680

=119880119899

1198800

= (119896 minus 1

119896 + 1)

119899

119880119899

119864=1

2sdot (

119896 minus 1

119896 + 1)

119899

119880119899

119868119899

= 119885119888= 119896 sdot 119885

1 119899 = 0119873

(22)

1198801015840119898

119864=119896 minus 1

2 sdot 119896sdot (

119896 minus 1

119896 + 1)

119898

1198801015840119898

1198800

=119896 minus 1

119896sdot (

119896 minus 1

119896 + 1)

119898

119898 = 0119873 minus 1

(23)

From (22) and (23) it can be seen that even whenemf e [with Laplace transform 119864 = 119864(119904)] is pulsed thenode and point voltages with respect to common-ground119874 (Figure 1) are reproduced faithfully and without delaybut with geometrically progressive amplitude attenuationregardless of the impedance 119885

1

It is clear that relations (7)ndash(9) offer many other pos-sibilities for the selection of 119885

119892 119885119871 1198851 and 119885

2that lead

to generation of versatile ladders realizing different types offrequency selective networks An interesting one seems tobe the common-ground RLC ladder realizing (a) pulse delaywith independently selected Elmorersquos delay and rise timesandor (b) true delay for either pulsed or analog frequencylimited input signals These topics will be considered in thefollowing section

3 Elmorersquos Delay and Rise Times for SelectedTypes of Common-Ground Uniform RLCLadders

Consider the ladder in Figure 1 having 1198851= 119877 + 119871 sdot 119904 119884

2=

11198852= 119866 + 119862 sdot 119904 and 119885

119892= 119885119871= (119871119862)

12 Suppose that

it holds the relation 120572 = 119877119871 = 119866119862 similar to the oneassociated with per-unit-length parameters of distortionlesstransmission line [18 19] Let us define the auxiliary parameter120573 = 1(119871 sdot 119862)

12 and let us recast the network voltagetransmittances describing the transfer of emf E to the ladderpoints with voltages 119880

0 1198801 119880

119873(Figure 1)mdashin the form

suitable for calculating of Elmorersquos delay and rise times [7](Appendix B) Firstly we should observe that the followingholds

1198851

1198852

= (119904 + 120572

120573)

2

1198851

119885119871

=119885119871

119885119888

sdot sinh (120591) = 119904 + 120572

120573

119885119888

119885119871

sdot sinh (120591) = 119904 + 120572

120573sdot [(

119904 + 120572

120573)

2

+ 2]

1

119885119892+ 119885119871

sdot (119885119888+119885119892sdot 119885119871

119885119888

) sdot sinh (120591)

=119904 + 120572

2 sdot 120573sdot [(

119904 + 120572

120573)

2

+ 3]

(24)

and then by using (8) and the Properties 2 and 3 let us expressthe point voltage transmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873 minus 1)

and 119879119873(119904) = 119880

119873119864 in the following form

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2

+2 sdot 1205732sdot sin2 [

2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587) ])

minus1

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 9: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 9

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ( (119904 + 120572) sdot [(119904 + 120572)2+ 3 sdot 120573

2]

sdot

119873minus1

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2sdot sin2 ( 119894 sdot 120587

2 sdot 119873)] + 2 sdot 120573

sdot

119873

prod119895=1

[(119904 + 120572)2

+2 sdot 1205732sdot sin2 (

2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(25)

As the calculation example of Elmorersquos times let us supposefor this type of ladder that the specified parameters are119873 = 6119877 = 50 [Ω] 119871 = 10 [mH] 119866 = 5 [mS] and 119862 = 1 [120583F] Wefirstly calculate 119885

119871= 119885119892= (119871119862)

12= 100 [Ω] 120572 = 5 sdot 103

[1s] and 120573 = 104 [1s] and then we recast 119879119899(119904) = 119880

119899E

(119899 = 0119873 minus 1) and 119879119873(119904) = 119880

119873E (25) in the following form

119879119899(119904) = (

1205732

2)

119899

sdot (1198600(119873 119899) + 119860

1(119873 119899) sdot 119904 + 119860

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198602119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198612119873+1

(119873) sdot 1199042119873+1

)minus1

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573) (1198610(119873) + 119861

1(119873) sdot 119904 + 119861

2(119873)

sdot 1199042+ sdot sdot sdot + 119861

2119873+1(119873) sdot 119904

2119873+1)minus1

(26)

where all ldquoArdquo and ldquoBrdquo coefficients are positive To applyElmorersquos definitions of delay and rise times (Appendix B) forall points in the ladder with zero initial conditions (Figure 1)and excited at 119905 = 0 with the step-voltage e with amplitude119864 [V] wemust do the following

(i) Firstly check that the point voltages 119906119896= 119906119896(t) (119896 =

0119873) have no overshoots or eventually if they arepresent overshoots must be less than 5 of thosevoltages steady state values [7] For example if wehave assumed for ladder with 119873 = 6 sections thatE = 10 [V] then its voltage step-responses 119906

119896=

119906119896(119905)(119896 = 0 6) and the node response 1199061015840

0as well

obtained through pspice simulation and depicted inFigures 2 and 3 in the interval 119905 isin [0 750] [120583s]

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500123

4

5678

9

10

Volta

ges (

V)

Time (120583s)

u0

u2

u1

u3

u0998400

Figure 2 A set of voltage responses at the selected points of theconsidered ladder (and at the119872

0node)

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750Time (120583s)

080

160

240320400480560

640720

800

Volta

ges (

mV

)

u3

u4

u5u6

Figure 3 Another set of voltage responses at the selected points ofthe considered ladder

reveal that Elmorersquos definitions cannot be appliedfor responses u

0and 1199061015840

0 since the occurrence of

overshoots whereas for all other point voltages 119906119896=

119906119896(t) (119896 = 1 6) those definitions are applicable

(ii) Secondly calculate the coefficients A0(Nn) A

1(Nn)

A2(N n) B

0(N) B

1(N) and B

2(N) so as to determine

the parameters a1(Nn) a

2(Nn) b

1(N) and b

2(N)

of the normalized transfer functions 119879119899(s)T119899(0)

(119899 = 0119873 minus 1) and T119873(s)T119873(0) obtained from (26)

according to the relations

119879119899(119904)

119879119899(0)

= (1 + 1198861(119873 119899) sdot 119904 + 119886

2(119873 119899) sdot 119904

2

+ sdot sdot sdot + 1198862119873minus2119899+1

(119873 119899) sdot 1199042119873minus2119899+1

)

times (1 + 1198871(119873) sdot 119904 + 119887

2(119873) sdot 119904

2

+ sdot sdot sdot + 1198872119873+1

(119873) sdot 1199042119873+1

)minus1

119879119899(0) = (

1205732

2)

119899

sdot1198600(119873 119899)

1198610(119873)

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 10: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

10 Mathematical Problems in Engineering

Table 1 Delay and rise times of the point voltages in the considered common-ground uniform RLC ladder

n 0 1 2 3 4 5 61198861[ms] 0813 0680 0547 0415 0286 0158 0

1198862[120583s] 0323 0225 0145 0082 0037 0010 0

1198871[ms] 0804 0804 0804 0804 0804 0804 0804

1198872[120583s] 0315 0315 0315 0315 0315 0315 0315

119879119863[ms]1 Not applicable 0124 0257 0388 0518 0647 0804

119879119877[ms]1 Not applicable 0157 0200 0229 0250 0286 0326

119879119863119888

[ms]2 Not applicable 0115 0257 0395 0528 0658 0817119879119877119888[ms]2 Not applicable 0190 0252 0295 0324 0349 0373

119906119899(infin) [V]3 5999 2999 1498 0747 0370 0178 0075

1The times calculated according to Elmorersquos definitions2The times obtained through pspice simulation according to the classical definitions of 119879119863119888 (ldquo50rdquo delay time) and 119879119877119888 (ldquo10ndash90rdquo rise time)3All data are related to the ladder with zero initial conditions

119886119894=

119860119894(119873 119899)

1198600(119873 119899)

[119894 = 0 (2119873 minus 2119899 + 1) 119899 = 0 (119873 minus 1)]

119887119895=

119861119895(119873)

1198610(119873)

[119895 = 0 (2119873 + 1)]

119879119873(119904)

119879119873(0)

= (1) (1 + 1198871(119873) sdot 119904

+1198872(119873) sdot 119904

2+ + 119887

2119873+1(119873) sdot 119904

2119873+1)minus1

119879119873(0) =

1205732119873+1

2119873minus1 sdot 1198610(119873)

(27)

Calculation of coefficients 1198600(N n)119860

1(N n) 119860

2(N

n) 1198610(N) 119861

1(N) and 119861

2(N) in (26) might be a

tedious task especially when N is large since thesecoefficients are produced from (25) as cumbersomeexpressions which cannot be put in the closed formSo we must resort to making of a numerical applica-tion (say in MATHCAD) for automatic calculation of1198600(N n) 119860

1(N n) 119860

2(N n) 119861

0(N) 119861

1(N) 119861

2(N)

1198861(N n) 119886

2(N n) 119887

1(N) 119887

2(N) and Elmorersquos times

for any given set of input parameters 119873 119899 119877 119871 119862 119866provided that the condition 119877119871 = 119866119862 is satisfied

(iii) Calculate Elmorersquos delay time (119879119863) and rise time (119879

119877)

of any point step-voltage response having no over-shoot according to definitions (B5) in Appendix Band by using (27)

119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

119879119877(119873 119899)

= radic2 sdot 120587 sdot 11988721(119873) minus 1198862

1(119873 119899) + 2 sdot [119886

2(119873 119899) minus 119887

2(119873)]

(28)

For RLC ladder having 119873 = 6 119877 = 50 [Ω] 119871 = 10 [mH]119866 = 5 [mS] 119862 = 1 [120583F]mdashexcited at 119905 = 0 by step emf 119890 withamplitude E = 10 [V] the obtained results are summarizedin Table 1

Another interesting ladder in Figure 1 is the onewith1198851=

119877+119871 sdot 119904 1198842= 1119885

2= 119866+119862 sdot 119904119885

119892= 0[Ω] 119885

119871= (119871119862)

12 and120572 = 119877119871 = 119866119862 Again let it be 120573 = 1(119871 sdot 119862)

12 By usingof (8) and Properties 2 and 3 the point voltage transmittances119879119899(119904) = 119880

119899E (119899 = 0119873 minus 1) and 119879

119873(119904) = 119880

119873E are obtained

as follows

119879119899(119904) = (

1205732

2)

119899

sdot ((119904 + 120572) sdot

119873minus119899

prod119894=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [ 119894 sdot 120587

2 sdot (119873 minus 119899)]

+ 120573 sdot

119873minus119899

prod119895=1

(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 [2 sdot 119895 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 11: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 11

119879119873(119904) = (

1205732

2)

119873

sdot (2 sdot 120573)

times ((119904 + 120572) sdot

119873

prod119894=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 ( 119894 sdot 120587

2 sdot 119873)]

+ 120573 sdot

119873

prod119895=1

[(119904 + 120572)2+ 2 sdot 120573

2

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

(29)

In this case Elmorersquos delay and rise times can be cal-culated in the similar way as it has been done in (25) Thespecified set of parameters 119877 119871 119862 119866119873 119899 provided that thevoltages of selected ladder points andor nodes satisfy thecondition (i) Elmorersquos times can be calculated by using (26)ndash(29) but further consideration of this point will be left to thereader

Also an interesting ladder is the one with 1198851= 1198772 119884

2=

11198852= 119862 sdot 119904 119885

119892= 0 [Ω] and 119885

119871rarr infin [Ω](Figure 1)

which resembles to a delay line [13] but it does not trulybehave like it and rather may be used for generation ofdelayed time markers with Elmorersquos times expressible in theclosed form To see this let us produce the point voltagetransmittances 119879

119899(119904) = 119880

119899E (119899 = 0119873) and the node voltage

transmittances 1198791015840119898(119904) = 1198801015840

119898119864(119898 = 0119873 minus 1) by using (8)

(9) and Property 2 or Case A (10)

119879119899(119904) =

cosh [(119873 minus 119899) sdot 120591]

cosh (119873 sdot 120591)= (

1

119877 sdot 119862)119899

sdot (

119873minus119899

prod119894=1

119904 +4

119877 sdot 119862

sdot sin2 [ 2 sdot 119894 minus 1

4 sdot (119873 minus 119899)sdot 120587])

times (

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

119899 = 0119873 minus 1

(30)

119879119873(119904) = (

1

119877 sdot 119862)119873

sdot (2)(

119873

prod119895=1

[119904 +4

119877 sdot 119862

sdot sin2 (2 sdot 119895 minus 1

4 sdot 119873sdot 120587)])

minus1

1198791015840

119898(119904) =

2 (119877 sdot 119862)

119904 + (4 (119877 sdot 119862))sdot (119879119898+ 119879119898+1

) 119898 = 0119873 minus 1

(31)

Observe that if 119904 rarr 0 then also 120591 rarr 0 and from(30) and (31) it follows that 119880

119899119864 rarr 1 (119899 = 1119873) and

1198801015840119898119864 997888rarr 1 (119898 = 0119873 minus 1) This is in obvious physical

agreement with the network behaviour in DC operatingregime The previous conclusions could be also formallyverified by using Remark 1 (Appendix A) in calculation of119879119899(119899 = 1119873) and 1198791015840

119898(119898 = 0119873 minus 1) for 119904 = 0 If

excitation e(t) of the network with zero initial conditions isstep voltage at 119905 = 0 with amplitude 119864 then 119864 = 119864(119904) =

L119890(119905) = 119864119904 From (30) (31) and Remark 1 we canalso easily see that 119906

119896(0) = lim

119904rarrinfin[119864 sdot 119879119896(119904)] = 0 [V] and

119906119896(infin) = lim

119904rarr0[119864 sdot 119879119896(119904)] = 119864 [V] (119896 = 1119873) In this case

it can be shown that (i) the node and the point voltages arestrictly monotone in 119905 [11] so that Elmorersquos definitions can beapplied leading to (ii) closed-form expressions of delay andrise times for node voltages [15] To illustrate the point (i)let us consider the network in Figure 4 whereon capacitancevoltages are denoted with 119909

119894= 119909119894(119905) (119894 = 1119873) Let us

introduce notation 120594 = 119905119877119862 119909119894(119905) = 120585

119894(120594) x(119905) =

[1199091(119905) 1199092(119905) sdot sdot sdot 119909

119873(119905)]

T 120585(120594) = [1205851(120594) 1205852(120594) sdot sdot sdot 120585

119873(120594)]

T

(ldquoTrdquo-operation of matrix transposition) and 119890(119905) = 120576(120594)For the ladder in Figure 4 the following system of state-spaceequations can be written in 120594-domain

119889120585 (120594)

119889120594= A sdot 120585 (120594) + B sdot 120576 (120594)

A =

[[[[[

[

minus3 1 0 sdot sdot sdot 0

1 minus2 1 sdot sdot sdot 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 sdot sdot sdot 1 minus2 1

0 sdot sdot sdot 0 1 minus1

]]]]]

]

(119873 times 119873 real symmetric tridiagonal matrix)

(32)

B = [2 0 sdot sdot sdot 0]T (119873 times 1 column vector) 120585(0) =

[0 0 sdot sdot sdot 0]T [V] (119873 times 1 column vector of state initial

conditions)Since the real symmetric119873times119873 regular tridiagonalmatrix

A is hyperdominant [17] then it is also positive definiteand both similar and congruent to diagonal matrix D =

diag(1198891 1198892 119889

119873) [20] where 119889

119894gt 0 (119894 = 1119873) So there

exists an orthogonal matrix Q [20] such that minusQminus1 sdot A sdot

Q = minusQT sdot A sdot Q = = D If we introduce the coordinatetransformation 120585(120594) = Q sdot 120577(120594) then the vector differentialequation in 120585(120594) = [1205851(120594) 120585

2(120594) sdot sdot sdot 120585

119873(120594)] (32) takes on

the following coordinate decoupled form

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Mathematical PhysicsAdvances in

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

12 Mathematical Problems in Engineering

e = u0 x1 = u0998400

u1 u2C C C

x2 = u1998400

xN = uNminus1998400

uNuNminus1

R2 R2 R2 R2 R2 R2middot middot middot

middot middot middot

Figure 4 The common-ground uniform integrating RC ladder with N sections

d120577 (120594)d120594

= Qminus1 sdot A sdotQ sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

= minusD sdot 120577 (120594) +Qminus1 sdot B sdot 120576 (120594)

(33)120577(0) = [0 0 sdot sdot sdot 0]

T (119873 times 1 column vector of the ldquotrans-formedrdquo state-space initial conditions)

For119873 = 8 the matricesQ andDwith entries rounded upto 3 decimal places are being obtained as

Q =

[[[[[[[[[[

[

minus0497 0478 minus0441 0386 minus0317 0235 minus0145 0049

0478 minus0317 0049 0235 minus0441 0497 minus0386 0145

minus0441 0049 0386 minus0478 0145 0317 minus0497 0235

0386 0235 minus0478 minus0049 0497 minus0145 minus0441 0317

minus0317 minus0441 0145 0497 0049 minus0478 minus0235 0386

0235 0497 0317 minus0145 minus0478 minus0386 0049 0441

minus0145 minus0386 minus0497 minus0441 minus0235 0049 0317 0478

0049 0145 0235 0317 0386 0441 0478 0497

]]]]]]]]]]

]

(34)

D =

[[[[[[[[[[

[

3961 0 0 0 0 0 0 0

0 3663 0 0 0 0 0 0

0 0 3111 0 0 0 0 0

0 0 0 2390 0 0 0 0

0 0 0 0 1610 0 0 0

0 0 0 0 0 0888 0 0

0 0 0 0 0 0 0337 0

0 0 0 0 0 0 0 0038

]]]]]]]]]]

]

(35)

Now if we suppose that the excitation 119890(119905) = 120576(120594) is the stepvoltage at 119905 = 0 with amplitude 119864 = 1 [V] then the uniquesolution of the vector differential equation (32) is producedin following form

120585 (120594)

= Q sdot

[[[[[[[[[[

[

1 minus 119890minus1198891sdot120594

1198891

0 sdot sdot sdot 0

01 minus 119890minus1198892sdot120594

1198892

sdot sdot sdot 0

0 sdot sdot sdot d 0

0 sdot sdot sdot 01 minus 119890minus119889119873sdot120594

119889119873

]]]]]]]]]]

]

sdotQminus1 sdot B wherefrom (as it was expected)

we obtain that

x (infin) = 120585 (infin) = minusAminus1 sdot B

= [1 1 sdot sdot sdot 1]T[V] (119873 times 1 column vector)

(36)

The point voltages are obtained according to relations 119906119895(120594) =

[120585119895(120594) + 120585

119895+1(120594)]2 (119895 = 1119873 minus 1) and finally for 119873 = 8

we can produce by using (36) the closed form solutions ofpoint and node voltages (in [V]) as ldquo120585rdquo and ldquo119906rdquo functions ofnormalized (ie dimensionless) ldquotimerdquo 120594 = 119905119877119862

1205851(120594) = 1 minus 0125 sdot 119890

minus3961sdot120594minus 0125 sdot 119890

minus3663sdot120594

minus 0125 sdot 119890minus3111sdot120594

minus 0125 sdot 119890minus2390sdot120594

minus 0125 sdot 119890minus1610sdot120594

minus 0125 sdot 119890minus0888sdot120594

minus 0125 sdot 119890minus0337sdot120594

minus 0125 sdot 119890minus0038sdot120594

1199061(120594) = 1 minus 0002 sdot 119890

minus3961sdot120594minus 0021 sdot 119890

minus3663sdot120594

minus 0555 sdot 119890minus3111sdot120594

minus 0100 sdot 119890minus2390sdot120594

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 13: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 13

minus 0149 sdot 119890minus1610sdot120594

minus 0194 sdot 119890minus0888sdot120594

minus 0229 sdot 119890minus0337sdot120594

minus 0247 sdot 119890minus0038sdot120594

1205852(120594) = 1 + 0120 sdot 119890

minus3961sdot120594+ 0083 sdot 119890

minus3663sdot120594

+ 00139 sdot 119890minus3111sdot120594

minus 0076 sdot 119890minus2390sdot120594

minus 0174 sdot 119890minus1610sdot120594

minus 0264 sdot 119890minus0888sdot120594

minus 0333 sdot 119890minus0337sdot120594

minus 0370 sdot 119890minus0038sdot120594

1199062(120594) = 1 + 0005 sdot 119890

minus3961sdot120594+ 0035 sdot 119890

minus3663sdot120594

+ 0061 sdot 119890minus3111sdot120594

+ 0039 sdot 119890minus2390sdot120594

minus 00584 sdot 119890minus1610sdot120594

minus 0216 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0485 minus 119890minus0038sdot120594

1205853(120594) = 1 minus 0110 sdot 119890

minus3961sdot120594minus 00128 sdot 119890

minus3663sdot120594

+ 0109 sdot 119890minus3111sdot120594

+ 0155 sdot 119890minus2390sdot120594

+ 00572 sdot 119890minus1610sdot120594

minus 0168 sdot 119890minus0888sdot120594

minus 0428 sdot 119890minus0337sdot120594

minus 0601 sdot 119890minus0038sdot120594

1199063(120594) = 1 minus 0006 sdot 119890

minus3961sdot120594minus 00372 sdot 119890

minus3663sdot120594

minus 0013 sdot 119890minus3111sdot120594

+ 0085 sdot 119890minus2390sdot120594

+ 0126 sdot 119890minus1610sdot120594

minus 0045 sdot 119890minus0888sdot120594

minus 0404 sdot 119890minus0337sdot120594

minus 0705 sdot 119890minus0038sdot120594

1205854(120594) = 1 + 0097 sdot 119890

minus3961sdot120594minus 00616 sdot 119890

minus3663sdot120594

minus 0135 sdot 119890minus3111sdot120594

+ 0016 sdot 119890minus2390sdot120594

+ 0196 sdot 119890minus1610sdot120594

+ 00770 sdot 119890minus0888sdot120594

minus 0380 sdot 119890minus0337sdot120594

minus 0809 sdot 119890minus0038sdot120594

1199064(120594) = 1 + 0009 sdot 119890

minus3961sdot120594+ 0026 sdot 119890

minus3663sdot120594

minus 0047 sdot 119890minus3111sdot120594

minus 0072 sdot 119890minus2390sdot120594

+ 0107 sdot 119890minus1610sdot120594

+ 0165 sdot 119890minus0888sdot120594

minus 0291 sdot 119890minus0337sdot120594

minus 0897 sdot 119890minus0038sdot120594

1205855(120594) = 1 minus 00780 sdot 119890

minus3961sdot120594+ 0115 sdot 119890

minus3663sdot120594

+ 0041 sdot 119890minus3111sdot120594

minus 0160 sdot 119890minus2390sdot120594

+ 00193 sdot 119890minus1610sdot120594

+ 0253 sdot 119890minus0888sdot120594

minus 0203 sdot 119890minus0337sdot120594

minus 0986 sdot 119890minus0038sdot120594

1199065(120594) = 1 minus 0009 sdot 119890

minus3961sdot120594minus 0004 sdot 119890

minus3663sdot120594

+ 0065 sdot 119890minus3111sdot120594

minus 0056 sdot 119890minus2390sdot120594

minus 0084 sdot 119890minus1610sdot120594

+ 0229 sdot 119890minus0888sdot120594

minus 0080 sdot 119890minus0337sdot120594

minus 1055 sdot 119890minus0038sdot120594

1205856(120594) = 1 + 0059 sdot 119890

minus3961sdot120594minus 0123 sdot 119890

minus3663sdot120594

+ 00890 sdot 119890minus3111sdot120594

+ 00470 sdot 119890minus2390sdot120594

minus 0188 sdot 119890minus1610sdot120594

+ 0205 sdot 119890minus0888sdot120594

+ 0042 sdot 119890minus0337sdot120594

minus 1125 sdot 119890minus0038sdot120594

1199066(120594) = 1 + 0011 sdot 119890

minus3961sdot120594minus 0011 sdot 119890

minus3663sdot120594

minus 0026 sdot 119890minus3111sdot120594

+ 0094 sdot 119890minus2390sdot120594

minus 0140 sdot 119890minus1610sdot120594

+ 0089 sdot 119890minus0888sdot120594

+ 0157 sdot 119890minus0337sdot120594

minus 1172 sdot 119890minus0038sdot120594

1205857(120594) = 1 minus 0036 sdot 119890

minus3961sdot120594+ 0101 sdot 119890

minus3663sdot120594

minus 0141 sdot 119890minus3111sdot120594

+ 0142 sdot 119890minus2390sdot120594

minus 00929 sdot 119890minus1610sdot120594

minus 00260 sdot 119890minus0888sdot120594

+ 0273 sdot 119890minus0337sdot120594

minus 1220 sdot 119890minus0038sdot120594

1199067(120594) = 1 minus 0012 sdot 119890

minus3961sdot120594+ 0031 sdot 119890

minus3663sdot120594

minus 0037 sdot 119890minus3111sdot120594

+ 0020 sdot 119890minus2390sdot120594

+ 0029 sdot 119890minus1610sdot120594

minus 0130 sdot 119890minus0888sdot120594

+ 0342 sdot 119890minus0337sdot120594

minus 1245 sdot 119890minus0038sdot120594

1205858(120594) = 71 + 0012 sdot 119890

minus3961sdot120594minus 00380 sdot 119890

minus3663sdot120594

+ 0067 sdot 119890minus3111sdot120594

minus 0102 sdot 119890minus2390sdot120594

+ 0152 sdot 119890minus1610sdot120594

minus 0234 sdot 119890minus0888sdot120594

+ 0412 sdot 119890minus0337sdot120594

minus 1270 sdot 119890minus0038sdot120594

(37)All ldquo120585rdquo and ldquo119906rdquo functions of 120594 are depicted in Figures 5

and 6 in the ldquotimerdquo interval 120594 isin [0 120]Assume thatΩ

0= 4(119877 sdot119862) is normalizing frequency and

recast (30) and (31) in the following form by using Remark 1(Appendix A)119879119899(119904)

= (

119873minus119899

prod119894=1

1 +119904

Ω0sdot sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

)

times (

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

119899 = 0119873 minus 1

(38)

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 14: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

14 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100 110 120minus01

0010203040506070809

1

120594

1205851(120594)

u1(120594)

1205852(120594)

u2(120594)

1205853(120594)

1205854(120594)u3(120594)

u4(120594)

Figure 5 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

0 10 20 30 40 50 60 70 80 90 100 120110minus01

0010203040506070809

1

u6(120594)1205857(120594)

u7(120594)

1205858(120594)

1205856(120594)

1205855(120594)

u5(120594)

120594

Figure 6 120585(120594) and 119906(120594) functions for the specified nodes and pointsof ladder in Figure 4

119879119873(119904)

=(1)times(

119873

prod119895=1

[1 +119904

Ω0sdot sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

])

minus1

1198791015840

119898(s) =

119879119898+ 119879119898+1

2 sdot (1 + (119904Ω0)) 119898 = 0119873 minus 1

(39)

The coefficients 1198861(119873 119899) 119886

2(119873 119899) 119887

1(119873) and 119887

2(119873) (27)

necessary for calculation of 119879119863(119873 119899) and 119879

119877(119873 119899) according

to Elmorersquos definitions (28) are obtained for 119879119899(119904) (119899 =

1119873 minus 1) (38) in the following form (see Corollary 3 inAppendix A)

1198861(119873 119899) =

1

Ω0

sdot

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=119877 sdot 119862

2sdot (119873 minus 119899)

2

1198871(119873) =

1

Ω0

sdot

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)

=119877 sdot 119862

2sdot 1198732

(40)

1198862(119873 119899) =

1

2 sdot Ω20

sdot

[

119873minus119899

sum119894=1

1

sin2 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]]

2

minus

119873minus119899

sum119894=1

1

sin4 [((2 sdot 119894 minus 1) (4 sdot (119873 minus 119899))) sdot 120587]

=(119877 sdot 119862)

2

24sdot (119873 minus 119899)

2sdot [(119873 minus 119899)

2minus 1]

(41)

1198872(119873) =

1

2 sdot Ω20

sdot

[

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732minus 1)

(42)

provided that 1198861(119873119873) = 119886

2(119873119873) = 0 and finaly

Elmorersquos delay and rise times for point voltages 119906119899(119899 = 1119873)

(Figure 4) are calculated in the closed form according to therelations119879119863(119873 119899) = 119887

1(119873) minus 119886

1(119873 119899)

=119877 sdot 119862

2sdot [1198732minus (119873 minus 119899)

2]

(43)

119879119877(119873 119899)

= radic2sdot120587sdot11988721(119873)minus1198862

1(119873 119899)+2sdot[119886

2(119873 119899)minus119887

2(119873)]

= 119877 sdot 119862 sdot radic120587

6

sdot radic[1198732 minus (119873 minus 119899)2] sdot 2 sdot [1198732 + (119873 minus 119899)

2] + 1

(44)

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 15: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 15

whereas Elmorersquos delay times T1015840119863(Nm) and rise times T1015840

119877(N

m) for node voltages u1015840119898(119898 = 0119873 minus 1) (Figure 1) are

calculated by using (38) and (39)

1198791015840

119863(119873119898) = 119887

1015840

1(119873) minus 119886

1015840

1(119873119898)

1198791015840

119877(119873119898)

= radic2 sdot 120587 sdot 119887101584021(119873)minus11988610158402

1(119873119898)+2 sdot [1198861015840

2(119873119898)minus1198871015840

2(119873)]

(45)

where

1198861015840

1(119873119898) =

119877 sdot 119862

4sdot [(119873 minus 119898)

2+ (119873 minus 119898 minus 1)

2]

1198871015840

1(119873) =

1

Ω0

sdot [

[

1 +

119873

sum119895=1

1

sin2 (((2 sdot 119895 minus 1) (4 sdot 119873)) sdot 120587)]

]

=119877 sdot 119862

4sdot (21198732+ 1)

(46)

1198861015840

2(119873119898) =

(119877 sdot 119862)2

48

times (119873 minus 119898)2sdot [(119873 minus 119898)

2minus 1]

+ (119873 minus 119898 minus 1)2sdot [(119873 minus 119898 minus 1)

2minus 1]

(47)

1198871015840

2(119873) =

1

2 sdot Ω20

sdot

[1 +

119873

sum119894=1

1

sin2 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)]

2

minus 1 minus

119873

sum119894=1

1

sin4 (((2 sdot 119894 minus 1) (4 sdot 119873)) sdot 120587)

=(119877 sdot 119862)

2

24sdot 1198732sdot (1198732+ 2)

(48)

From (43) it follows that 119879119863(N N)prop 1198732 and from (44) that

119879119877(N n) gt 2 sdot 119879

119863(N n) (119899 = 1119873) which does not qualify

this type of ladder for using as pulse delay line in its ownright If in the ladder on Figure 4 119873 = 6 119877 = 50 [Ω] and119862 = 2 [nF] and e(t) is periodic pulse train with amplitude10 [V] duty-cycle 50 and period 100 [120583s] then excitatione(t) and node voltages 119909

2(119905) = 1199061015840

1(t) 1199095(119905) = 1199061015840

4(t) and point

voltageu3(t)mdashobtained by pspice simulation are depicted on

Figure 7 in interval 119905 isin [0 120][120583s]

0 10 20 30 40 50 60 70 80 90 100 120110Time (120583s)

0

123

456789

10

Volta

ges (

V)

e(t)

x2(t) = u1998400(t)

u3(t)

x5(t) = u4998400(t)

Figure 7 Some of the pspice simulation results for six segmentintegrating ladder in Figure 4 with 119877 = 50 [Ω] 119862 = 2 [nF]and pulse-train excitation e(t) with amplitude 10 [V] period 119879 =

100 [120583s] and duty-cycle 50

E

C C C C R

L L LL

R R R RU

R R R R

U0998400 U1

998400 UNminus2998400

middot middot middot

middot middot middotN capacitors

Figure 8 Common-ground uniform ladder which may play therole of artificial delay line (ADL) for sufficiently large N

Finally consider that the ladder in Figure 8 is analogue tothe ladder in Figure 1 Assume 120572 = 119877119871 = 1(119877 sdot 119862) and

119885119892= 1198851=1

2sdot119877 sdot 119871 sdot 119904

119877 + 119871 sdot 119904=119877

2sdot

119904

119904 + 120572

Z2=

1

C sdot s

119885119871= 1198851+

1

1 + 119877 sdot 119862 sdot 119904

=119877

2sdot119904 + 2120572

119904 + 120572(119885119892+ 119885119871= 119877)

(49)

In this case from (7) we obtain

119885119888= radic119885

1sdot (1198851+ 2 sdot 119885

2) =

119877

2sdot119904 + 2120572

119904 + 120572= 119885119871

sinh (120591) =119885119888

1198852

=119904 sdot (119904 + 2120572)

2120572 sdot (119904 + 120572)

cosh (120591) = 1 +1198851

1198852

=(119904 + 120572)

2+ 1205722

2120572 sdot (119904 + 120572)

(50)

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 16: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

16 Mathematical Problems in Engineering

From (50) it is found that eminus120591 = 120572(s + 120572) and from (9) itfollows that

1198791015840

119898(119904) =

1198801015840

119898

119864=119885119871minus 1198851

119885119871+ 1198851

sdot 119890minus119898sdot120591

=120572

119904 + 120572sdot 119890minus119898sdot120591

= (120572

119904 + 120572)119898+1

=1

(1 + (119904120572))119898+1

(119898 = 0119873 minus 2)

(51)

119879 (119904) =119880

119864=1198801015840119873minus2

119864sdot

119880

1198801015840119873minus2

= 1198791015840

119873minus2sdot

120572

119904 + 120572

=1

(1 + (119904120572))119873minus1

sdot120572

119904 + 120572=

1

(1 + (119904120572))119873

(52)

Relation (52) could have been produced in a less formalway by observing that the ladder in Figure 8 is a constantresistance network when 119877119871 = 1(119877 sdot 119862) since theinput impedances (denoted by dashed arrows) seen from thepertinent pairs of nodes are all equal to R If the ladder hasno initial conditions and if it is excited by unit-step voltage119890(119905) at 119905 = 0 119864 = 119864(119904) = L[119890(119905)] = 1119904 then for overallstep response 119906(119905) = Lminus1[119880(119904)] it holds that (a) 119906(0) =

lim119904rarrinfin

119904 sdot 119880(119904) = 0 [V] (b) 119906(infin) = lim119904rarr0

119904 sdot 119880(119904) = 1 [V]and (c) u(t) is a monotone increasing function in t (ie u(t)has no overshoots) since we have

0 le 119906 (119905) = Lminus1[(

120572

119904 + 120572)119873

sdot1

119904]

= 1 minus 119890minus120572sdot119905

sdot

119873minus1

sum119896=0

(120572 sdot 119905)119896

119896le 1

119889119906 (119905)

119889119905=120572119873 sdot 119890minus120572sdot119905 sdot 119905119873minus1

(119873 minus 1)gt 0 119905 ge 0

(53)

From (52) we obtain the ldquoardquo and ldquobrdquo coefficients necessary forcalculation of Elmorersquos delay time 119879

119863and rise time 119879

119877(B5)

for the network in Figure 8

1198861= 1198862= 0

1198871=119873

120572 119887

2=119873 sdot (119873 minus 1)

2 sdot 1205722

119879119863= 1198871=119873

120572=119871

119877sdot 119873 = 119873 sdot radic119871 sdot 119862

119879119877=1

120572sdot radic2120587 sdot 119873

= 119879119863sdot radic

2120587

119873= radic2120587 sdot 119873 sdot 119871 sdot 119862

(54)

The procedure for determining the ladder parameterswhen both Elmorersquos times 119879

119863and 119879

119877are specified consists

0 2 4 6 8 10 12 14 16 18 20Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

u9998400

(12951 m 9)

(4834362 120583 5)

(9833841 120583 5)

(7262636 120583 1000 m)

Figure 9 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05[mH] and 119862 = 5 [120583F] The output is 119906(119905)

of the following three steps (a) firstly we calculate 119873 =

ant[2120587 sdot (119879119863119879119877)2] + 1 and then from (54) we calculate the

actual rise time 119879Ract = 119879119863sdot (2120587119873)

12lt 119879119877(119879119863remains

unchanged) (b) assume C and (c) calculate 119877 = (119879119863119873)119862

and 119871 = (119879119863119873)2C When solely Elmorersquos delay time 119879

119863is

specified and 119879119877is left unspecified we arbitrarily select some

reasonably great N so as to produce the sufficiently small 119879119877

of the output u for the unit-step input and then follow thesteps (b) and (c)

But in the case when N is small and we want to realizeElmorersquos delay time 119879

119863for pulse train with period T and

duty cycle 120575 it can be shown that one of the followingtwo conditions must be satisfied in order to prevent severedistortion of transmitted ldquopulsesrdquo (i) 119879

119863le (1 minus 120575) sdot 1198792

when 120575 ge 12 or (ii) 119879119863

le 120575 sdot 1198792 when 120575 lt 12 Forthe ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 = 05

[mH] and 119862 = 5 [120583F] excited by the pulse train e(t) withamplitude 10 [V] period119879 = 10 [ms] and duty-cycle 120575 = 06the obtained results of pspice simulation in the time interval119905 isin [0 20] [ms] are depicted in Figure 9 whereon are plottedthe excitation e(t) voltage 1199061015840

9(t) of the 10th capacitor and

the voltage u(t) of the 20th capacitor Elmorersquos delay and risetimes for voltage 1199061015840

119898minus1of the mth capacitor (119898 = 1119873 minus 1)

(Figure 8) are respectively 119898 sdot radic119871 sdot 119862 and radic2120587 sdot 119898 sdot 119871 sdot 119862according to (54) For example Elmorersquos delay time of 1199061015840

9(119905)

is 10 sdot radic119871 sdot 119862 = 05 [ms] and of u(t) is 20 sdot radic119871 sdot 119862 = 1 [ms]Both these values are slightly different from those obtainedaccording to the classical definition of delay time and theyare denoted in Figure 9 for comparison In this case Elmorersquosrise time for u obtained from (54) is119879

119877= radic12058710 [ms]asymp 5605

[120583s] whereas the classical rise time obtained from Figure 9 isslightly different 119879

119877119888asymp 5689[120583s]

Consider now the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that the condition (i) isviolated by supposing that119879 = 2 [ms] and 120575 = 06The resultsof pspice simulation in the time interval 119905 isin [0 5] [ms] aredepicted in Figure 10 and obviously exhibit a severe distortionof ldquopulsesrdquo u being transmitted to the end of the ladder

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 17: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 17

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 10 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 06 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

0 500120583 1 m 15 m 2 m 25 m 3 m 35 m 4 m 45 m 5 mTime (s)

0

123

456789

10

Volta

ges (

V)

e

u

Figure 11 Realization of Elmorersquos delay time 119879119863= 1 [ms] for pulse-

train input e(t) with amplitude 10 [V] period 119879 = 2 [ms] and duty-cycle 120575 = 04 by ladder in Figure 8 with 119873 = 20 119877 = 10 [Ω] 119871 =

05 [mH] and 119862 = 5 [120583F] The output is u(t)

Consider again the same ladder excited by the pulse traine(t) with amplitude 10 [V] provided that condition (ii) isviolated by supposing that 119879 = 2 [ms] and 120575 = 04 Theresults of pspice simulation are depicted in Figure 11 in thetime interval 119905 isin [0 5] [ms] and obviously exhibit a severedistortion of ldquopulsesrdquo u being transmitted to the end of theladder

And finally consider the ladder in Figure 8 which shouldrealize Elmorersquos delay time 119879

119863= 5 [ms] and rise time

119879119877= 5605 [120583s] Let the excitation e(t) be the pulse-train

with amplitude 10 [V] period 119879 = 20 [ms] and duty-cycle 120575 = 05 To avoid a severe distortion of output pulsesu(t) it is sufficient to follow the previously formulated steps(a) divide (c) So according to the step (a) we firstly calculate119873 = 500 Then in the step (b) we assume that say 119862 =

1 [120583F] and finally in the step (c) we calculate 119877 = 10

[Ω] and 119871 = 100 [120583H] The time variations of e(t) andu(t) obtained by pspice simulation are depicted in Figure 12

0 4 8 12 16 20 24 28 32 36 40Time (ms)

0

123

456789

10

Volta

ges (

V)

e

u

(52828m 9)

(50086m 5)

(47078m 1)

Figure 12 Realization of Elmorersquos delay time 119879119863= 5 [ms] and rise-

time119879119877= 5605 [120583s] for pulse-train input 119890(119905)with amplitude 10 [V]

period 119879 = 20 [ms] and duty-cycle 120575 = 05 by ladder in Figure 8with 119873 = 500 119877 = 10 [Ω] 119871 = 100 [120583H] and 119862 = 1 [120583F] Theoutput is u(t)

in the time interval 119905 isin [0 40] [ms] Therefrom we findthe classical delay and rise times of the output voltage u(t)119879119863119888

asymp 5 [ms] and 119879119877119888

asymp 575 [120583s] respectively Both thesetimes are slightly different from the pertinent Elmorersquos times(54) whose obvious advantage is that are explicitly calculableas functions of 119873 119877 119871 119862 parameters of common-grounduniform RLC ladder (Figure 8)

In the previous examples we have investigated Elmorersquostimes of voltages at points andor nodes in several character-istic types of common-ground uniform RLC ladders excitedeither by step or by pulse-train emfs In themost general caseconsider the realization of physical delay time 119879

119863for arbitrary

excitation e(t) (ie true delay time) by using the RLC ladderin Figure 8 (true delay line) If we presume for this networkthat119873 rarr infin then from (52) and (54) it readily follows that

lim119873rarrinfin

119879 (119904) = lim119873rarrinfin

1

(1 + (119904120572))119873

= lim119873rarrinfin

[(1 +119904 sdot 119879119863

119873)

119873(119904sdot119879119863)

]

minus119904sdot119879119863

= 119890minus119904sdot119879119863

(55)

so that the overall network response becomes 119906(119905) =

Lminus1[119880(119904)] = Lminus1[119890minus119904sdot119879119863 sdot 119864(119904)] = 119890(119905 minus 119879119863) (ie u(t) is the

exact replica of the excitation e(t) delayed by the time119879119863) Let

us firstly introduce the notation

119911 =119904 sdot 119879119863

119873= (120590 + 119895 sdot 120596) sdot

119879119863

119873

=120590 sdot 119879119863

119873+ 119895 sdot

120596 sdot 119879119863

119873

Re 119911 =120590 sdot 119879119863

119873

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 18: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

18 Mathematical Problems in Engineering

01 001 000100001

01001000100001

258681

260872

263064

265255

267446

269638

271829

|f(z)|

RezImz

1eminus05 1eminus05

Figure 13 The plot of function |119891(119911)| in the 119911-region of interest

Im 119911 =120596 sdot 119879119863

119873

119891 (119911) = (1 + 119911)1119911

(56)

and then investigate the conditions under which the functionf (z) becomes close to the number e It is obvious from (55)that these conditions relate to certain restrictions in the spanof N 120590 and 120596 Observe that 119891(119911) rarr 119890 when 119911 rarr 0 Aftersome manipulation from (56) we easily obtain

1003816100381610038161003816119891 (119911)1003816100381610038161003816 = 119890(Im119911(Re2119911+Im2119911))sdot119886 tan[Im119911(1+Re119911)]

sdot ([1 + Re 119911]2

+ Im2 119911)Re119911(2sdot[Re2119911+Im2119911])

Arg [119891 (119911)] = Re 119911Re2 119911 + Im2 119911

sdot 119886 tan [ Im 119911

1 + Re 119911]

minusIm (119911)

2 sdot [Re2 119911 + Im2 119911]

sdot ln ([1 + Re 119911]2 + Im2 119911)

(57)

where Arg[f (z)] is expressed in radians and should bemultiplyed by 180120587 to be expressed in [deg]

The function |119891(119911)| is depicted in Figure 13 From theset of its associated numerical data it can be seen that forRe119911 = 0 we have |119891(119911)| = 27182 when 0 le Im119911 le

2120587 sdot 10minus3 and |119891(119911)| rarr 119890 when Im119911 rarr 0 In other wordsfor |119891(119911)| to be close to e it is sufficient in this case to providethat 2120587 sdot119891max sdot 119879119863119873 le 2120587 sdot 10minus3 where 119891max is the maximumfrequency in the spectrumof the signal e(t) being transmittedthrough the RLC ladder in Figure 8 having almost true timedelay equal to Elmorersquos delay time 119879

119863 or 119873 ge 119891max sdot [119879119863]

where [119879119863] is delay time 119879

119863expressed in [ms]

The function Arg[119891(119911)] [deg] is depicted in Figure 14From the set of its associated numerical data it can be seen

minus251114

minus209262

minus16741

minus125557

minus0837048

minus0418524

0

01 001 000100001

01001000100001

arg[f(z)]

(deg

)

RezImz

1eminus05 1eminus05

Figure 14The plot of function Arg[119891(119911)] in the z-region of interest

minus25m

minus3eminus018

25m

50m

75m

100m

125m

Art

ifici

al d

elay

-line

abso

lute

erro

rae

r(t)=u(t)minuse

02468

10

minus10

minus8

minus6

minus4

minus2

0 10 20 30 40 50 60 70 80 90 100Time (ms)

(5m 1120367m)

(17505m minus39471m)

(22707m 25039eminus048)

(62495m 10)

(67495m 99961)

(42505m 39471m)

(67505 m minus39471m)

CPU time asymp 54 (s)

aer

(92505 m 39471m)

e

u

(tminus

TD) (

V)

Volta

gese

(t)

andu(t)

(V)

Figure 15 PSPICE verification of delay time 119879119863

= 5 [ms] forplusmn10 [V]20 [Hz] sine-wave input e(t)

that for Re119911 = 0 we have minus01799964 lt Arg[119891(119911)] [deg]le 0 when 0 le Im119911 le 2120587 sdot 10minus3 and Arg[119891(119911)] rarr

0 when Im119911 rarr 0 For Arg[119891(119911)] le 0 to be close to 0 sayto be |Arg[119891(119911)]| lt 018 [deg] it is sufficient in this case alsoto provide that 2120587sdot119891max sdot119879119863119873 le 2120587sdot10minus3or 119873 ge 119891max sdot [119879119863]

For properly selected N (ie 119879119863119873 le 1(1000 sdot 119891max))

the ladder parameters are found according to the previouslyformulated steps (b) divide (c) arbitrarily select C and calculate119877 = (119879

119863N )C and 119871 = (119879

119863119873)2119862

As a final example consider the ladder in Figure 8 withparameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] 119871 =

100 [120583H] and the Elmorersquos delay time 119879119863

= 119873 sdot (119871119877) =

119873 sdot radic119871 sdot 119862 = 5 [ms] Since condition 119891max sdot 119879119863119873 le 10minus3

is satisfied for 119891max le 100 [Hz] then this ladder may be usedas true delay line for every input signal e(t) with maximumspectral frequency up to 119891max with output u(t) being almostthe replica of e(t) with no attenuation and virtually constantdelay 119879

119863equal to Elmorersquos delay time The results for this

ladder excited by various input signals 119890(119905) produced by usingof PSPICE simulation are depicted in Figures 15 16 17 18 and19 There on it may be noticed that the overall output voltageof this ladder as of artificial delay-line (ADL) is representedby119906(119905) asymp 119890(119905minus119879

119863) with absolute error aer(119905) = 119906(119905)minus119890(119905minus119879

119863)

In the paper [12] it is proved that by using distor-tionless transmission line with resistive load equal to the

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

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Stochastic AnalysisInternational Journal of

Page 19: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 19

0 10 20 30 40 50 60 70 80 90 100Time (ms)

02468

10

25m

50m

75m

Art

ifici

al d

elay

-line

abso

lute

erro

r

minus10

minus8

minus6

minus4

minus2

minus100m

minus75m

minus50m

minus25m

minus1eminus017

(46288m 43577n) (458621m 67711m)

(648621m 211398m)(552021 m 49975)(502021m 50101)

(918421m 643787m)

(561221m minus131795m)

CPU time asymp 57 (s)

aer

euVo

ltage

se(t)

andu(t)

(V)

aer(t)=u(t)minuse (tminus

TD) (

V)

Figure 16 PSPICE verification of delay time 119879119863= 5 [ms] for input

wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide 100

[Hz] and linearly increasing amplitude 0 divide 10 [V]

0 5 10 15 20 25 30 35 40Time (ms)

0

5

10

15

Volta

ges (

V)

minus15

minus10

minus5

(27014m 136422) (77014m 136419)

e

u

Figure 17 PSPICE results for transfer with delay 119879119863

= 5 [ms]of CAM input wave e(t) with carrier plusmn10 [V]100 [Hz] modulationindex 119898 = 05 and the intelligence frequency 50 [Hz] within ADLin Figure 8

0 10 20 30 40 50 60 80 9070 100Time (ms)

0

3

7

10

Volta

ges (

V)

minus10

minus7

minus3

e u

(313552m 99913)

(361552m 99838)

Figure 18 PSPICE results for transfer with delay 119879119863= 5 [ms] of

FM input wave e(t) with carrier plusmn10 [V]50 [Hz] modulation index119898 = 2 and intelligence frequency 25 [Hz] within ADL in Figure 8

0 20 40 60 80 100 120 160 180140 200Time (ms)

0

3

78

10

Volta

ges (

V)

minus10

minus7

minus8

minus3minus2

2

minus5

5(241891m 2333)

(291891m 23321)

(1241287m 2333)

(1291287m 23321)

e

u

Figure 19 PSPICE results for transfer with delay 119879119863= 5 [ms] of

input wave e(t) with period 119879 = 01 [s] linear frequency chirp 0 divide100 [Hz] and linearly increasing amplitude 0 divide 10 [V] within ADLin Figure 8

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

990991992993994995996997998999

1000A

mpl

itude

of t

he

(159m 9997503 m) (318149m 999 m)(100m 9901791 m)

outp

ut v

olta

geu

(mV

)

Figure 20 The amplitude-frequency characteristics of the consid-ered ladder in the frequency range 119891 isin [0 100] [Hz]

line characteristic impedance radic11987110158401198621015840 (1198711015840 and 1198621015840 are per-unit-length inductance and capacitance of line resp) evena relativelly small signal delay 119879

119863cannot be realistically

produced For example signal delay 119879119863

= 5 [ms] can berealized with lossless line having the constant parameters1198711015840= 15 [120583Hm] 1198621015840 = 18 [pFm] and load resistance

(11987110158401198621015840)12

= 500radic3 [Ω] at the distance 119889 = 119879119863sdot

(1198711015840 sdot 1198621015840)minus12

asymp 96225 [km] from the line sending endbut when distortionless line is lossy and terminated withcharacteristic impedance (11987110158401198621015840)12 then besides the timedelay 119889 sdot (1198711015840 sdot 1198621015840)

12 at distance d the signal attenuationfactor exp[119889 sdot (1198771015840 sdot 1198661015840)] (1198771015840 and 1198661015840 are the per-unit-lengthresistance and conductance of the lossy line resp) mustbe also taken into account [18] Realization of pulse delay119879119863

= 5 [ms] with lumped RLC network in Figure 8 isverified in Figure 12 and seems to be the more comfortableapproach than using lengthy distributed parameter networksIn general realization of any time delay 119879

119863for signal with

maximum spectral frequency 119891max can be accomplished byusing the uniform ladder in Figure 8 with length N selectedaccording to the condition 119879

119863119873 le 1(1000 sdot 119891max) The

ladder parameters are calculated according to relations 119877 =

(119879119863119873)119862 and 119871 = (119879

119863119873)2119862 where 119862 may be selected

arbitrarily

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 20: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

20 Mathematical Problems in Engineering

0 10 20 30 40 50 60 70 80 90 100Frequency (Hz)

minus180

minus144

minus108

minus72

minus36

0

Phas

e of t

he o

utpu

tvo

ltage

u(d

eg)

(10 minus18)(100 minus1799976)

Figure 21 The phase-frequency characteristics of the consideredladder in the frequency range 119891 isin [0 100] [Hz]

Finally in Figures 20 and 21 the results of PSPICE AC-analysis are depicted with 1 [V] amplitude AC-input in theentire frequency range [0 100] [Hz] for network in Figure 8with parameters 119873 = 500 119862 = 1 [120583F] 119877 = 10 [Ω] and 119871 =

100 [120583H] In Figure 20 we see that the amplitude-frequencycharacteristics of the network are almost flat and in Figure 21we see that its phase-frequency characteristics are virtuallylinear with slope minus18 [degHz] Hence we conclude that thenetwork in Figure 8 may satisfactorily realize the time delay119879119863for any input signal e(t) with spectrum in range 0 divide 100

[Hz]

4 Conclusions

In the paper are derived general closed-form expressions fornetwork functions of common-ground uniform and pas-sive ladders with complex double terminations are derivedmaking distinction in analysis between the signal transfer toladder nodes and to its points The obtained results are thensimplified for ladders with seven specific pairs of complexdouble terminations Elmorersquos delay and rise times calculatedfor specific important types of RLC ladders indicated slightdeviation fromdelay and rise times values obtained accordingto their classical definitions

For the common-ground integrating RC ladder with stepinput Elmorersquos delay and rise times are produced in closedform both for ladder nodes and ladder points It has beenshown that this type of ladder could not be recommended aspulse delay line in its own right since its Elmorersquos rise-times ofpoint voltages are not less than the twice of their delay times

And finally in this paper we have proposed a specifictype of common-ground uniform RLC ladder amenable forapplication as delay line both for pulsed and analog inputsignals For this type of ladder Elmorersquos delay and rise timesrelating to the node voltages are produced in the closed formwhich offers a possibility for realization of (a) the pulse delayline with arbitrarily specified Elmorersquos delay and rise timesand (b) the true delay line both for pulsed and analog inputsignals with arbitrarily selected delay time For both cases(a) and (b) in the paper a procedure for the determinationof ladder length and calculation of all its RLC parametersis specified Recall that the ladder network topology ispreferable one in the network synthesis since it has very lowsensitivity with respect to variations of RLC parameters Theobtained results are illustratedwith several practical examplesand are verified through pspice simulation

Appendices

A

Property 1 If 119911 isin C (C-set of complex numbers) thencos(119899 sdot 119911) is the 119899th order trigonometric polynomial in cos(z)and sin(119899 sdot 119911) is the product of sin(z) and the (119899 minus 1)th ordertrigonometric polynomial in cos(z)

Proof Let 119895 = radicminus1 and let us make the following binomialexpansion

cos (119899 sdot 119911) + 119895 sdot sin (119899 sdot 119911)

= [cos (119911) + 119895 sdot sin (119911)]119899

=

119899

sum119896=0

(119899

119896) sdot (119895)

119896

sdot cos119899minus119896 (119911) sdot sin119896 (119911)

=

119901le1198992

sum119901=0

(minus1)119901(119899

2119901) sdot cos119899minus2119901 (119911) sdot sin2119901 (119911) + 119895

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119902minus1 (119911) sdot sin2119902+1 (119911)

(A1)

From (A1) it readily follows that

cos (119899 sdot 119911) =

119901le1198992

sum119901=0

(minus1)119901(119899

2119901)

sdot cos119899minus2119901 (119911) sdot [1 minus cos2 (119911)]119901

sin (119899 sdot 119911) = sin (119911)

sdot

119902le(119899minus1)2

sum119902=0

(minus1)119902(

119899

2119902 + 1)

sdot cos119899minus2119901minus1 (119911) sdot [1 minus cos2 (119911)]119902

(A2)

Property 2 If 119911 isin C then the following finite productexpansion holds

cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)] (A3)

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 21: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 21

Proof According to Property 1 cos(119899 sdot 119911) is the 119899th orderpolynomial in cos(z) Then we have

cos (119899 sdot 119911) = 119860 sdot

119899

prod119894=1

[cos (119911) minus 120585119894] 119860 is a constant (119860 = 0)

120585i = cos(2119894 minus 12119899

sdot 120587) 119894 = 1 119899

(A4)

Now consider the complex equation 1199112119899 = minus1with roots 119911119896=

expplusmn119895sdot[(2119896minus1)2119899]sdot120587 (119896 = 1 119899)Then it holds the followingfactorization

1199112119899+ 1 =

119899

prod119896=1

(119911 minus 119890119895sdot((2119896minus1)2119899)sdot120587

)

sdot (119911 minus 119890minus119895sdot((2119896minus1)2119899)sdot120587

)

=

119899

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(2119896 minus 1

2119899sdot 120587) + 1]

(A5)

For 119911 = 1 from (A5) we obtain the identityprod119899119896=119894

sin(((2119896 minus1)4119899) sdot 120587) = radic22119899 Then for 119911 = 0 from (A4) it firstlyfollows that 1119860 = prod

119899

119894=1[1 minus cos(((2119894 minus 1)2119899) sdot 120587)] = 2

119899sdot

[prod119899

119894=1sin(((2119894 minus 1)4119899) sdot 120587)]2 = 2119899 sdot (222119899) = 21minus119899 or 119860 =

2119899minus1 and then upon substitution 119911 rarr 119895 sdot 119911 also in (A4) thefollowing implication is obtained

cos (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cos (119911) minus cos(2119894 minus 12119899

sdot 120587)]

119911rarr119895sdot119911

997904rArr cosh (119899 sdot 119911) = 2119899minus1

sdot

119899

prod119894=1

[cosh (119911) minus cos(2119894 minus 12119899

sdot 120587)]

(A6)

Property 3 If 119911 isin C then the following finite productexpansion holds

sinh (119899 sdot 119911) = 2119899minus1

sdot sinh (119911)

sdot

119899minus1

prod119896=1

[cosh (119911) minus cos(119896 sdot 120587119899

)] (A7)

Proof According to Property 1 sin(119899 sdot 119911) is the product ofsin(z) and a polynomial in cos(z) of order 119899minus1Then we have

Sin (119899 sdot 119911) = 119861 sdot sin (119911)

sdot

119899minus1

prod119894=1

[cos (119911) minus 120577119894] 119861 is a constant (119861 = 0)

120577119894= cos(119894 sdot 120587

119899) 119894 = 1 119899 minus 1

(A8)

Now consider the complex equation 1199112119899 = 1 with roots

119911 = plusmn1 and 119911119896= exp[plusmn119895 sdot (119896 sdot 120587119899)] (119896 = 1 119899 minus 1) Then

the following factorization holds

1199112119899minus 1 = (119911

2minus 1) sdot

119899minus1

prod119896=1

(119911 minus 119890119895sdot(119896sdot120587119899)

) sdot (119911 minus 119890minus119895sdot(119896sdot120587119899)

)

= (1199112minus 1) sdot

119899minus1

prod119896=1

[1199112minus 2 sdot 119911 sdot cos(119896 sdot 120587

119899) + 1]

(A9)

wherefrom for 119911 = 1 we obtain the identity prod119899119896=1

sin(119896 sdot1205872119899) = radic1198992

119899minus1Then for 119911 = 0 from (A8) it firstly followstaht 119899119861 = prod119899minus1

119894=1[1minuscos(119894sdot120587119899)] = 2119899minus1sdot[prod119899

119894=1sin(119894 sdot 1205872119899)]2 =

2119899minus1

sdot (11989922119899minus2

) = 119899 sdot 21minus119899 or 119861 = 2

119899minus1 and then uponsubstitution 119911 rarr 119895 sdot119911 also in (A8) the following implicationis obtained

sin (119899 sdot 119911) = 2119899minus1 sdot sin (119911) sdot119899minus1

prod119894=1

[cos (119911) minus cos(119894 sdot 120587119899)]119911rarr119895sdot119911

997904rArr

sinh (119899 sdot 119911) = 2119899minus1 sdot sinh (119911) sdot119899minus1

prod119894=1

[cosh (119911) minus cos(119894 sdot 120587119899)]

(A10)

Remark 1 From proofs of Properties 2 and 3 we explicitlyemphasize the following two identities having been obtainedin passing

119899

prod119896=1

sin(2119896 minus 14119899

sdot 120587) =radic2

2119899

119899

prod119896=1

sin(119896 sdot 1205872119899

) =radic119899

2119899minus1

(A11)

Property 4 If 119909 isin R (R-set of real numbers) then thefollowing identity holds

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899) = sin (119899 sdot 119909) (A12)

Proof Theroots of the polynomial 1199112119899minus2sdot[cos(2sdot119899sdot119909)]sdot119911119899+1 =0 are exp[119895 sdot (plusmn2 sdot119909+2sdot119896 sdot120587119899)] (119896 = 0 119899 minus 1) and they can bewritten in the following form 119911

119896= exp[119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)]

and 119911lowast119896= exp[minus119895 sdot (2 sdot 119909 + 2 sdot 119896 sdot 120587119899)] = exp119895 sdot [minus2 sdot 119909 + 2 sdot

(119899 minus 119896) sdot 120587119899] (119896 = 0 119899 minus 1) The considered polynomial hasthe following factorization

1199112119899minus 2 sdot [cos (2 sdot 119899 sdot 119909)] sdot 119911119899 + 1

=

119899minus1

prod119896=0

(119911 minus 119911119896) sdot (119911 minus 119911

lowast

119896)

=

119899minus1

prod119896=0

[1199112minus 2 sdot cos(2 sdot 119909 + 2 sdot 119896 sdot 120587

119899) sdot 119911 + 1]

(A13)

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 22: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

22 Mathematical Problems in Engineering

wherefrom for 119911 = 1 it follows that

|sin (119899 sdot 119909)| = 2119899minus1 sdot1003816100381610038161003816100381610038161003816100381610038161003816

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

1003816100381610038161003816100381610038161003816100381610038161003816

(A14)

It is easy to see that if sin(119899 sdot 119909) ge 0 hArr 119909 isin [2 sdot 119898 sdot 120587 2 sdot

119898 sdot +120587119899] 119898 isin N (N is the set of natural numbers) thensin(119909 + 119896 sdot 120587119899) ge 0 for 119896 = 0 119899 minus 1 And the opposite ifsin(119899 sdot119909) lt 0 [hArr 119909 isin (2sdot119898sdot120587+120587119899 2 sdot119898sdot120587+2120587119899) 119898 isin

N] then for 119896 = 0 119899 minus 2 we see that it holds sin(119909+119896sdot120587119899) gt

0 and for 119896 = 119899minus1 it holds sin(119909+119896sdot120587119899) lt 0 Hence we haveproved that modulus signs can be lifted on both sides of therelation (A14) and therefrom the relation (A12) follows

Corollary 2 At its regular points the function 119891(119909) (x isin R)

is equal to zero

119891 (119909) =

119899minus1

sum119896=0

1

sin2 (119909 + (119896 sdot 120587119899))

minus1198992

sin2 (119899 sdot 119909)

(A15)

Proof By differentiating (A12) in x we easily obtain that inregular points of f (x) it holds that

2119899minus1

sdot

119899minus1

prod119896=0

sin(119909 + 119896 sdot 120587

119899)

sdot [

119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899)] = 119899 sdot cos (119899 sdot 119909)

or119899minus1

sum119896=0

cotan(119909 + 119896 sdot 120587

119899) = 119899 sdot cotan (119899 sdot 119909)

(A16)

and by differentiating in x the second relation in (A16)(A15) immediately follows Observe that the statement of thiscorollary also holds if in (A14) k runs from 1 to n (insteadfrom 0 to 119899 minus 1)

Corollary 3 The following two identities hold

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

119899

sum119896=1

1

sin4 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)=4

3sdot 1198992sdot (2 sdot 119899

2+ 1)

(A17)

Proof If in (A15) of Corollary 3 we make the substitution119899 rarr 2 sdot 119899 then let k runs from 1 to 2 sdot 119899 and finally select

119909 = minus120587(4 sdot 119899) the relation (A15) is being transformed intothe following form

2119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 4 sdot 119899

2

Since the 119896th and the (2119899minus119896+1)th entries are equal forall119896= 12119899997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888997888rarr

119899

sum119896=1

1

sin2 (((2 sdot 119896 minus 1) (4 sdot 119899)) sdot 120587)= 2 sdot 119899

2

(A18)

and thus with the first of relations in (A17) is proved Nowsuppose that in (A15) k runs from 1 to n and differentiatethis relation twice As a result after some manipulations it isobtained that

119899

sum119896=1

1

sin4 (119909 + (119896 sdot 120587119899))

=1198992

sin2 (119899 sdot 119909)sdot [

1198992

sin2 (119899 sdot 119909)+2

3sdot (1 minus 119899

2)]

(A19)

If in (A19) we make the substitution 119899 rarr 2 sdot 119899 let k runsfrom 1 to 2 sdot 119899 select 119909 = minus120587(4 sdot 119899) and finally use the samearguments as in (A18) we immediately obtain the second ofrelations in (A17)

B

Let T(s) be the voltage transfer function of linear lumpedelectrical network with zero initial conditions and let T](s)be its normalized form T(s)T(0)

119879 (119904) = 119879 (0) sdot1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

119879] (119904) =1 + 1198861sdot 119904 + 119886

2sdot 1199042 + sdot sdot sdot + 119886

119894sdot 119904119894

1 + 1198871sdot 119904 + 119887

2sdot 1199042 + sdot sdot sdot + 119887

119895sdot 119904119895

(B1)

where usually it holds that 119895 gt 119894 since for most lumpedphysical systems the number of poles in transfer functionis larger than the number of zeros The normalized unit-step response 119906](119905) of this network and its normalizedimpulse response ℎ](119905) are related through relations ℎ](119905) =119889119906](119905)119889119905 = Lminus1119879](119904) and 119906](119905) = Lminus1119879](119904)119904 In theabsence of the initial conditions [hArr 119906](0) = 0] we have119906](infin) = 119879](0) = 1 Convenient definitions of delay time(119879119863) and rise time (119879

119877) of a network (or other physical

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 23: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Mathematical Problems in Engineering 23

systems) applicable only when its step response ismonotonic(nonovershooting) originate from Elmore [7]

119879119863= intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905

119879119877= radic2120587 sdot int

infin

0

(119905 minus 119879119863)2

sdot ℎ] (119905) sdot 119889119905

= radic2120587 sdot radicintinfin

0

1199052 sdot ℎ] (119905) sdot 119889119905 minus 1198792

119863

intinfin

0

ℎ] (119905) sdot 119889119905 = 119906] (infin) minus 119906] (0)

= 119879] (0) minus 119879] (infin) = 1

(B2)

In many practical situations it has been noticed that whenthe unit-step response of network or system hasaovershootless than 5 of its steady-state unity value (B2) will still beholding and the obtained results will be in close agreementwith those produced by using the conventional definitions ofdelay and rise times [7] Since 119879](119904) = Lℎ](119905) then fromElmorersquos definitions (B2) it is obtained that

119879] (119904) = intinfin

0

ℎ] (119905) sdot 119890minus119904sdot119905

sdot 119889119905

= intinfin

0

ℎ] (119905) sdot

infin

sum119896=0

(minus1)119896sdot(119904 sdot 119905)119896

119896sdot 119889119905

= intinfin

0

ℎ] (119905) sdot 119889119905 minus 119904 sdot intinfin

0

119905 sdot ℎ] (119905) sdot 119889119905 +1199042

2

sdot intinfin

0

1199052sdot ℎ] (119905) sdot 119889119905 minus sdot sdot sdot

= 1 minus 119904 sdot 119879119863+1199042

2sdot (

1198792

119877

2120587+ 1198792

119863) minus sdot sdot sdot

(B3)

and after the application of the long division on 119879](119904) in (B1)it follows that

119879] (119904) = 1 minus (1198871minus 1198861) sdot 119904

+ (1198872

1minus 1198861sdot 1198871+ 1198862minus 1198872) sdot 1199042+ sdot sdot sdot

(B4)

Finally from (B3) and (B4) we easily identify Elmorersquos timesas functions of only the coefficients 119886

1 1198862 1198871 and 119887

2[19]

119879119863= 1198871minus 1198861

119879119877= radic2 sdot 120587 [1198872

1minus 11988621+ 2 sdot (119886

2minus 1198872)]

(B5)

Acknowledgment

This work is supported in part by the Serbian Ministry ofScience andTechnologicalDevelopment throughProjects TR32048 and III 41006

References

[1] C A Mead and L A Conway Introduction to VLSI SystemsAddison Wesley Reading Mass USA 1980

[2] P Penfield and J Rubinstein ldquoSignal delay in RC tree networksrdquoin Proceedings of the 2nd Caltech Conference on Very Large ScaleIntegration pp 269ndash283 1981

[3] J Rubsinstein P Penfield Jr and M A Horowitz ldquoSignal delayin RC tree networksrdquo IEEE Transactions on Computer-AidedDesign of Integrated Circuits and Systems vol 2 no 3 pp 202ndash211 1983

[4] M A Horowitz ldquoTiming models for MOS pass networksrdquo inProceedings of International Symposium on Circuits and Systemspp 198ndash201 1983

[5] T-M Lin and C A Mead ldquoSignal delay in general RCnetworksrdquo IEEE Transactions on Computer-Aided Design ofIntegrated Circuits and Systems vol 3 no 4 pp 331ndash349 1984

[6] J L Wyatt Jr ldquoSignal delay in RC mesh networksrdquo IEEETransactions on Circuits and Systems vol 32 no 5 pp 507ndash5101985

[7] W C Elmore ldquoThe transient response of damped linear net-works with particular regard to wideband amplifiersrdquo Journalof Applied Physics vol 19 no 1 pp 55ndash63 1948

[8] P K Chan ldquoAn extension of Elmorersquos delayrdquo IEEE Transactionson Circuits and Systems vol 33 no 11 pp 1147ndash1149 1986

[9] P K Chan ldquoAn extension of Elmorersquos delay and its applicationfor timing analysis of MOS pass transistor networksrdquo IEEETransactions on Circuits and Systems vol 33 no 11 pp 1149ndash1152 1986

[10] R Gupta B Tutuianu and L T Pileggi ldquoThe elmore delayas a bound for rc trees with generalized input signalsrdquo IEEETransactions on Computer-Aided Design of Integrated Circuitsand Systems vol 16 no 1 pp 95ndash104 1997

[11] P Penfield Jr and J Rubinstein ldquoSignal delay in RC networksrdquoin Proceedings of the 18th Design Automation Conference (DACrsquo81) pp 613ndash617 July 1981

[12] D B Kandic B D Reljin and I S Reljin ldquoOn modellingof two-wire transmission lines with uniform passive laddersrdquoMathematical Problems in Engineering vol 2012 Article ID351894 42 pages 2012

[13] F T Ulaby and M M Maharbiz Circuits National Technologyand Science Press Allendale NJ USA 2009

[14] T Lamdan ldquoCalculation of rsquoElmorersquo delay for RC laddernetworksrdquo Proceedings of the IEE vol 123 no 5 pp 411ndash4121976

[15] S C Dutta Roy ldquoCalculation of Elmore-defined delay and risetimes of RC ladder networksrdquo Proceedings of the IEE vol 124no 11 pp 1001ndash1002 1977

[16] D B Kandic and B D Reljin ldquoVoltage transfer-functions andrise- and delay-times of homogeneous ladder networksrdquo inProceedings of the 10th International Symposium on TheoreticalElectrical Engineering pp 151ndash155 Magdeburg Germany 1999

[17] L Weinberg Network Analysis and Synthesis McGraw-HillNew York NY USA 1961

[18] YHKuTransient Circuit Analysis DVanNostrandNewYorkNY USA 1961

[19] M S Ghausi and J J Kelly Introduction to Distributed-Parameter Networks with Application to Integrated CircuitsHolt Reinhart and Winston New York NY USA 1968

[20] B Noble and JWDanielApplied Linear Algebra Prentice-HallEnglewood Cliffs NJ USA 2nd edition 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 24: Research Article Electrical Network Functions of …downloads.hindawi.com/journals/mpe/2013/691740.pdfResearch Article Electrical Network Functions of Common-Ground Uniform Passive

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Recommended