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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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal 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
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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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
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
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
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
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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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
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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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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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
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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
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
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
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
MathematicsJournal of
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Mathematical Problems in Engineering
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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
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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OptimizationJournal of
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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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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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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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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
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
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
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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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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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
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
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