ECE 546 Lecture 07 Multiconductors - University Of Illinoisemlab.uiuc.edu/ece546/Lect_07.pdf ·...

ECE 546 – Jose Schutt‐Aine 1

ECE 546Lecture ‐ 07

MulticonductorsSpring 2018

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]


TELGRAPHER’S EQUATION FOR NCOUPLED TRANSMISSION LINES

V and I are the line voltage and line current VECTORSrespectively (dimension n).

V ILz t

I VCz t

V3(z)

V2(z)

V1(z)

...

L, C

z=0 z=l


N-LINE SYSTEM

L and C are the inductance and capacitance MATRICES respectively

1

2

n

II

I

I

1

2

n

VV

V

V

11 12

21 22

nn

L LL L

L

L

11 12

21 22

nn

C CC C

C

C

V3(z)

V2(z)

V1(z)

...

L, C

z=0 z=l


N-LINE ANALYSIS

In general LC ≠ CL and LC and CL are not symmetric matrices.

GOAL: Diagonalize LC and CL which will result in atransformation on the variables V and I.

Diagonalize LC and CL is equivalent to finding the eigenvalues ofLC and CL.

Since LC and CL are adjoint, they must have the sameeigenvalues.

2 2

2 2

V VLCz t

2 2

2 2

I ICLz t


MODAL ANALYSIS

LC and CL are adjoint matrices. Find matrices E and H such that

is the diagonal eigenvalue matrix whose entries are the inverses ofthe modal velocities squared.

2 21

2 2

EV EVELCEz t

2 21

2 2

HI HIHCLHz t

1 1 2mELCE HCLH


MODAL ANALYSIS

Second-order differential equation in modal space

2 22

2 2m m

mV Vz t

2 22

2 2m m

mI Iz t

Vm = EV Im = HI

V3(z)

V2(z)

V1(z)

...

L, C

z=0 z=l


Vm and Im are the modal voltage and current vectors respectively.

EIGENVECTORS

1

2

m

m

m

mn

VV

V

V

1

2

m

m

m

mn

II

I

I

Im = HI

Vm = EV


EIGEN ANALYSIS

* The eigenvalues of A satisfy the relation |A - I| = 0 where Iis the unit matrix.

* A scalar is an eigenvalue of a matrix A if there exists a vector X such that AX = X. (i.e. for which multiplication by a matrix is equivalent to an elongation of the vector).

* The vector X which satisfies the above requirement isan eigenvector of A.


EIGEN ANALYSIS

Assume A is an n x n matrix with n distinct eigenvalues, then,

* A has n linearly independent eigenvectors.

* The n eigenvectors can be arranged into an n ´ n matrix E;the eigenvector matrix.

* Finding the eigenvalues of A is equivalent to diagonalizing A.

* Diagonalization is achieved by finding the eigenvector matrix

E such that EAE-1 is a diagonal matrix.


EIGEN ANALYSIS

For an n-line system, it can be shown that

* Each eigenvalue is associated with a mode; the propagationvelocity of that mode is the inverse of the eigenvalue.

* LC can be transformed into a diagonal matrix whose entriesare the eigenvalues.

* LC possesses n distinct eigenvalues (possibly degenerate).

* There exist n eigenvectors which are linearly independent.


EIGEN ANALYSIS

* Each normalized eigenvector represents the relative lineexcitation required to excite the associated mode.

* Each eigenvector is associated to an eigenvalue and thereforeto a particular mode.


E : Voltage eigenvector matrix

H : Current eigenvector matrix

EIGENVECTORS

11 12 13

21 22 23

31 32 33

e e eE e e e

e e e

11 12 13

21 22 23

31 32 33

h h hH h h h

h h h

1 2mELCE

1 2mHCLH


1

2

3

1 0 0

10 0

10 0

m

mm

m

v

v

v

1

2

3

1 0 0

10 0

10 0

m

mm

m

v

v

v

11 12 13

21 22 23

31 32 33

h h hH h h h

h h h

11 12 13

21 22 23

31 32 33

e e eE e e e

e e e

Eigenvalues and Eigenvectors1 2

mELCE gives

1 2mHCLH

gives


MATCHINGNETWORK

+++

---e31

e32

e33 MODE C

MATCHINGNETWORK

+++

---e11

e12

e13 MODE A

MATCHINGNETWORK

+++

---e21

e22

e23 MODE B

Modal Voltage Excitation

Voltage EigenvectorMatrix

11 12 13

21 22 23

31 32 33

e e eE e e e

e e e


Modal Current Excitation

Current EigenvectorMatrix

MATCHINGNETWORK

h11

h12

h13 MODE A

MATCHINGNETWORK

h21

h22

h23 MODE B

MATCHINGNETWORK

h31

h32

h33 MODE C

11 12 13

21 22 23

31 32 33

h h hH h h h

h h h


Line variables are recovered using

V E1Vm

I H1Im

EIGENVECTORS

Special case: For a two-line symmetric system, E and H are equal and independent of L and C. The eigenvectors are [1,1] and [1,-1]

E : voltage eigenvector matrix

H : current eigenvector matrix

E and H depend on both L and C


MODAL AND LINE IMPEDANCE MATRICES

Zm m1ELH1

By requiring Vm = ZmIm, we arrive at

Zm is a diagonal impedance matrix which relates modal voltagesto modal currents

By requiring V = ZcI, one can define a line impedance matrix Zc.Zc relates line voltages to line currents

Zm is diagonal. Zc contains nonzero off-diagonal elements. We can show that

Zc E1ZmH E1m1EL


1

2( )

m

m

mn

j uv

j uv

j uv

e

eW u

e

1

2

1

1

1

m

mm

mn

v

v

v


MATRIX CONSTRUCTION

siiL L ( )m

ijijL L

( )

1

[ ]n

mii s ij

j

C C C

( )[ ] mij ijC C

[ ]ii sR R [ ] mij ijR R

( )

1

[ ]n

mii s ij

j

G G G

( )[ ] mij ijG G


Vm = EV Im = HI

Vm (z) = W(z)A +W(- z)B

Im (z) = Zm- 1 W(z)A - W(- z)B[ ]

1 1m mZ ELH

1 1 1c m mZ E Z H E EL

Modal Variables

General solution in modal space is:

Modal impedance matrix

Line impedance matrix


N-Line Network

Zs : Source impedance matrix

ZL : Load impedance matrix

Vs : Source vector


Reflection Coefficients

Source reflection coefficient matrix

s [1n EZsL1E1m ]1[1n EZsL1E1m ]

Load reflection coefficient matrix

L [1n EZLL1E1m ]1[1n EZLL1E1m ]

* The reflection and transmission coefficient are n x nmatrices with off-diagonal elements. The off-diagonal elements account for the coupling between modes.

* Zs and ZL can be chosen so that the reflection coefficient matrices are zero


TERMINATION NETWORK

yp11

yp22

yp12

V1

V2

I1 = yp11V1 +yp12(V1-V2)

I2 = yp22V2 +yp12(V2-V1)

In general for a multiline system

I YVV ZI Z Y1

Note : yii ypii, yij = -yji , (i≠j)ij

1Also, z ijy


Termination Network Construction

To get impedance matrix Z• Get physical impedance values ypij• Calculate yij's from ypij's• Construct Y matrix• Invert Y matrix to obtain Z matrix

Remark: If ypij = 0 for all i≠j, then Y = Z-1 and zii = 1/yii


1) Get L and C matrices and calculate LC product

2) Get square root of eigenvalues and eigenvectors of LC matrix m

3) Arrange eigenvectors into the voltage eigenvector matrix E

4) Get square root of eigenvalues and eigenvectors of CL matrix m

5) Arrange eigenvectors into the current eigenvector matrix H

6) Invert matrices E, H, m.

7) Calculate the line impedance matrix Zc Zc = E-1m-1EL

Procedure for Multiconductor Solution


8) Construct source and load impedance matrices Zs(t) and ZL(t)

9) Construct source and load reflection coefficient matrices 1(t) and 2(t).

10) Construct source and load transmission coefficient matrices T(t), T2(t)

11) Calculate modal voltage sources g1(t) and g2(t)

12) Calculate modal voltage waves:

a1(t) = T1(t)g1(t) + 1(t)a2(t - m)

a2(t) = T2(t)g2(t) + 2(t)a1(t - m)

b1(t) = a2(t - m)

b2(t) = a1(t - m)

Procedure for Multiconductor Solution


)(

)()(

a

mod

22mod

11mod

mi

mnnei

mei

mei

ta

tata

t

)(

mi is the delay associated with mode i. mi = length/velocity of mode i. The modal volage wave vectors a1(t) and a2(t) need to be stored since they contain the history of the system.

13) Calculate total modal voltage vectors:

Vm1(t) = a1(t) + b1(t)Vm2(t) = a2(t) + b2(t)

14) Calculate line voltage vectors:

V1(t) = E-1Vm1(t)V2(t) = E-1Vm2(t)

Wave‐Shifting Solution


ALS04 ALS240Drive Line 1

Drive Line 2

z=0 z=l

Drive Line 3

Sense Line 4

Drive Line 5

Drive Line 6

Drive Line 7

ALS04

ALS04

ALS04

ALS04

ALS04

ALS240

ALS240

ALS240

ALS240

ALS240

7-Line Coupled-Microstrip System

Ls = 312 nH/m; Cs = 100 pF/m;

Lm = 85 nH/m; Cm = 12 pF/m.


20010000

1

2

3

4

5Drive line 3 at Near End

Time (ns)2001000

-1

0

1

2

3

4

5Drive Line 3 at Far End

Time (ns)

Drive Line 3


2001000-1

0

1

2Sense Line at Near End

Time (ns)2001000

-1

0

1

2Sense Line at Far End

Time (ns)

The picture can't be displayed.

Sense Line


Multiconductor Simulation


LOSSY COUPLED TRANSMISSION LINES

Solution is best found using a numerical approach(See References)

V IRI Lz t

I VGV Cz t

V3(z)

V2(z)

V1(z)

...

L, C

z=0 z=l


1 2 3

31 1 211 12 13

VI V VC C Cz t t t

32 1 221 22 23

VI V VC C Cz t t t

3 31 231 32 33

I VV VC C Cz t t t

31 1 211 12 13

IV I IL L Lz t t t

32 1 221 22 23

IV I IL L Lz t t t

3 31 231 32 33

V II IL L Lz t t t

Three‐Line Microstrip


11 13( )V IL Lz t

11 13( )I VC Cz t

Subtract (1c) from (1a) and (2c) from (2a), we get

This defines the Alpha mode with:

1 3V V V 1 3I I I and

The wave impedance of that mode is:

11 13

11 13

L LZC C

and the velocity is 11 13 11 13

1uL L C C

Three‐Line – Alpha Mode


In order to determine the next mode, assume that

1 2 3V V V V

1 2 3I I I I

31 211 21 31 12 22 32 13 23 33

V II IL L L L L L L L Lz t t t

31 211 21 31 12 22 32 13 23 33

I VV VC C C C C C C C Cz t t t

By reciprocity L32 = L23, L21 = L12, L13 = L31

By symmetry, L12 = L23

Also by approximation, L22 L11, L11+L13 L11

Three‐Line – Modal Decomposition


31 211 12 13 12 112

V II IL L L L Lz t t t

In order to balance the right-hand side into I, we need to have

12 11 2 11 12 13 2 11 12 22L L I L L L I L L I

212 122L L

or 2

Therefore the other two modes are defined as

The Beta mode with

Three‐Line – Modal Decomposition


The Beta mode with

1 2 32V V V V

1 2 32I I I I

The characteristic impedance of the Beta mode is:

11 12 13

11 12 13

22

L L LZC C C

and propagation velocity of the Beta mode is

11 12 13 11 12 13

1

2 2u

L L L C C C

Three‐Line – Beta Mode


The Delta mode is defined such that

1 2 32V V V V

1 2 32I I I I

The characteristic impedance of the Delta mode is

11 12 13

11 12 13

22

L L LZC C C

The propagation velocity of the Delta mode is:

11 12 13 11 12 13

1

2 2u

L L L C C C

Three‐Line – Delta Mode


1 0 1

1 2 1

1 2 1

E

1 2 3

Alpha modeBeta mode*

Delta mode*

Symmetric 3‐Line Microstrip

In summary: we have 3 modes for the 3-line system

*neglecting coupling between nonadjacent lines


1 2 3

r = 4.3

113 17 5( / ) 16 53 16

5 17 113C pF m

346 162 67( / ) 152 683 152

67 162 346L nH m

0.45 0.12 0.450.5 0 0.50.45 0.87 0.45

E

0.44 0.49 0.440.5 0 0.50.10 0.88 0.10

H

Coplanar Waveguide


73 0 0( ) 0 48 0

0 0 94mZ

56 23 8( ) 22 119 22

8 23 56cZ

0.15 0 0( / ) 0 0.17 0

0 0 0.18pv m ns

1 2 3

r = 4.3

Coplanar Waveguide


WS' S'WS

rh

2Sk

S W

( ) : Complete Elliptic Integral of the first kindK k

'( ) ( ')K k K k

2 1/ 2' (1 )k k

30 '( ) (ohm)( )1

2

ocpr

K kZK k

1/ 22

1cpr

v c

Coplanar Waveguide


S WW

rh

120 '( ) (ohm)( )1

2

ocsr

K kZK k

Coplanar Strips


Characteristic Microstrip Coplanar Wguide Coplanar strips

eff* ~6.5 ~5 ~5

Power handling High Medium Medium

Radiation loss Low Medium Medium

Unloaded Q High Medium Low or High

Dispersion Small Medium Medium

Mounting (shunt) Hard Easy Easy

Mounting (series) Easy Easy Easy

* Assuming r=10 and h=0.025 inch

Qualitative Comparison

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