University of Illinois at Chicago ECE 423, Dr. D. Erricolo, Lecture 19 University of Illinois at Chicago ECE 423, Dr. D. Erricolo, Lecture 19 Frequency.

University of Illinois at Chicago

ECE 423, Dr. D. Erricolo, Lecture 19



Frequency analysis of three-conductor lines

We have already found that the transmission line equations for a three-conductor system are, in time domain:

For a sinusoidal steady-state excitation the above equations become:

t

tzILtzIR

z

tzV

),(),(

),(

t

tzVCtzVG

z

tzI

),(),(

),(

)(ˆˆ)(ˆ

)(ˆˆ)(ˆ

zVYdz

zId

zIZdz

zVd

(1)

(2)

(3)

(4)





where we have introduced the voltage and current phasors

and the per-unit length impedance and admittance matrices:

)(ˆ

)(ˆ)(ˆ

)(ˆ

)(ˆ)(ˆ

zI

zIzI

zV

zVzV

R

G

R

G

CjGY

LjRZ

ˆ

ˆ

(5)

(6)

(7)

(8)





Equations (3) and (4) are coupled first-order differential equations.In order to obtain uncoupled equations we can differentiate each equation with respect to z and substitute into the other to obtain

When (9) and (10) are obtained, it is important to keep the order of theproducts and since these do not commute in general.

A general solution is obtained by either solving (9) or(10). We willconsider the solution of (10).

YZ ˆˆ ZY ˆˆ

(9)

(10))(ˆˆˆ)(ˆˆ)(ˆ

)(ˆˆˆ)(ˆˆ)(ˆ

2

2

2

2

2

2

2

2

zIZYdz

zVdY

dz

zId

zVYZdz

zIdZ

dz

zVd





First we introduce a transformation into nodal currents by setting

And substituting into (10), which yields

If the transformation can diagonalize we obtain

And the modal equations (12) become uncoupled:

mITI ˆˆˆ

T

ZY ˆˆT

)(ˆˆˆˆˆ)(ˆ 1

2

2

zITZYTdz

zIdm

m

)(ˆ)(ˆ

)(ˆ)(ˆ

22

2

22

2

zIdz

zId

zIdz

zId

mRRmR

mGGmG

2

221

ˆ0

0ˆˆˆˆˆˆ

R

GTZYT

(11)

(12)

(13)

(14)

(15)





2Gˆ ,Zˆ YThe eigenvalues of and are referred to as propagation

constants and are used to form the solution:

2ˆR

mRz

mRz

mR

mGz

mGz

mG

IeIezI

IeIezI

RR

GG

ˆˆ)(ˆ

ˆˆ)(ˆ

ˆˆ

ˆˆ

Note that 1) the form of the solution is the same as for two-conductor lines; 2) the constants

mRmRmGmG IIII ˆ,ˆ,ˆ,ˆ

Equations (16) and (17) may be cast as: mG

zmG

zm IeIezI GG ˆˆ)(ˆ ˆˆ

where

z

zz

R

G

e

ee

ˆ

ˆˆ

0

0

mR

mGm

I

II

ˆ

ˆˆ

(16)

(17)

(18)

(19) (20)





Once a solution is found in terms of the nodal currents, the actualcurrents are found from:

mz

mz

m IeIeTITI GG ˆˆˆˆˆˆ ˆˆ

The voltage solution is obtained from:

m

zm

z IeIeTYdz

zIdYzV GG ˆˆˆˆˆ)(ˆˆ)(ˆ ˆˆ11

Observe that, since , then21ˆˆˆˆˆ

TZYT

ˆˆˆˆˆˆ 11 TYTZ

Hence (22) may be rewritten

m

zm

z IeIeTTTZzV GG ˆˆˆˆˆˆˆ)(ˆ ˆˆ11

(21)

(22)

(23)

(24)





The terms inside square brackets in the previous equation is the characteristic impedance matrix:

11 ˆˆˆˆˆ TTZZ c Let us now see how we can solve for the unknown constants . Weneed to add some additional constraints. For this purpose we look at the transmission line as a four-port circuit:

Three-conductor line

SR

NER NER VV ˆ)0(ˆ +

-

+

-

)0(GV)0(ˆ

GI)0(ˆ

RI

SV

FER VLV ˆ)(ˆ +

- - -

FER

FER

+

)(ˆ LVG

)(ˆ LIG)(ˆ LI R

z0zLz

This allows us to write:

Figure 1

)0(ˆˆˆ)0(ˆ IZVV ss )(ˆˆ)(ˆ LIZLV L(26) and (27)

mI

(25)





The previous quantities correspond to:

FE

LL

NE

ss

ss R

RZ

R

RZ

VV

0

0ˆ,0

0ˆ,0

ˆˆ

Evaluating the voltage, given by (22), at yields:0z

mm

mmc

IITI

IITZV

ˆˆˆ)0(ˆ

ˆˆˆˆ)0(ˆ

When (29) and (30) are combined with (26) we obtain:

mmcmmss IITZIITZV ˆˆˆˆˆˆˆˆˆ

Similarly, evaluating the voltage at yields:Lz

mL

mL

mL

mL

c

IeIeTLI

IeIeTZLV

ˆˆˆ)(ˆ

ˆˆˆˆ)(ˆ

ˆˆ

ˆˆ

When (32) and (33) are combined with (27) we obtain:

mL

mL

cmL

mL

L IeIeTZIeIeTZ ˆˆˆˆˆˆˆˆ

(28)

(29)

(30)

(31)

(32)

(33)

(34)





Equations (31) and (34) are now combined to give an algebraic matrixequation that determines

mI

0

ˆ

ˆ

ˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆ

s

m

m

LLc

LLc

scsc V

I

I

eTZZeTZZ

TZZTZZ

After the previous equation is solved current and voltage along the lineare computed using equations (21) and (22).

Looking at the line as a four-port network may be particularly useful ifwe are interested only in the values of voltages and currents at the endpoints of the line. For this purpose we need the chain parameter matrix

)0(ˆ

)0(ˆ

ˆˆ

ˆˆ

)(ˆ

)(ˆ

2221

1211

I

V

LI

LV

(35)

(36)





The parameters are computed from (29)-(32), and their expressionsare:

ij

1ˆˆ

22

11ˆˆ

21

1ˆˆ1

12

1ˆˆ1

11

ˆˆ2

1ˆ

ˆˆˆˆ2

1ˆ

ˆˆˆˆ2

1ˆ

ˆˆˆˆ2

1ˆ

TeeT

YTeeT

TeeTY

YTeeTY

LL

LL

LL

LL

Finally, another representation, obtained substituting (26)-(27) into (36), for the terminal currents is:

)0(ˆˆˆˆˆˆ)(ˆ

ˆˆˆˆ)0(ˆˆˆˆˆˆˆˆˆ

212221

112121221112

IZVLI

VZIZZZZ

ss

sLsLLs

(37)

(38)

(39)

(40)

(41)

(42)





An exact solution (lossless line immersed in homogeneous medium)

The solution previously found does not provide insights into the mechanism of crosstalk; hence here we consider an analytical solutionthat is obtained by making the following assumptions: 1) three-conductor line; 2) lossless; 3) homogenous medium

Lossless implies: CjYLjZGR ˆ,ˆ,0,0

Homogeneous implies: ICL from which

IZY 2ˆˆ

And we also observe that is diagonal so that . The propagation constant matrix is:

ZY ˆˆ IT ˆˆ

IvjIj ˆ

(43)

(44)





The elements of the chain parameter matrix reduce to:

IL

CL

LLjCLjv

LL

LLjLLjv

IL

cosˆ

sinsinˆ

sinsinˆ

cosˆ

22

21

12

11

When (45)-(48) are introduced into the representation (41) for theterminal current we obtain:

sL

sLLs

VLCjZL

LIL

ILjZLCjZL

LZZL

ˆˆsincos

)0(ˆˆˆsinˆˆcos

(45)

(46)

(47)

(48)

(49)





The expression for may be obtained upon substitution of (45)-(48)into (42) or by reciprocity. Applying reciprocity we obtain:

sLsLs VLILjZLCjZL

LZZL ˆ)(ˆˆˆsinˆˆcos

(50)

When (49) and (50) are solved for the terminal voltages due to crosstalkamong the wires one finds:

DC

DC

GLG

mFENE

FENE

GLGmFENE

NENER

VSk

LjCLcj

RR

RR

ISk

LjCLlj

RR

RSVV

ˆ1

1

2

ˆ1

2

Denˆ)0(ˆ

2

2

DCDC Gm

FENE

FENEGm

FENE

FEFER VLcj

RR

RRILlj

RR

RSVLV ˆˆ

Denˆ)(ˆ

(51)

(52)

)(ˆ LI





The various parameters that appear in (51) and (52) are:

)()1)(1(

)1)(1(1Den 2222

RGLGSGLRSR

SRLGLRSGRG CSjkSC

LC cos

L

LS

sin

1,))((

kcccc

c

ll

lk

mRmG

m

RG

m

Ls

LsmG

Ls

GG RR

RRLcc

RR

Ll

)(

FENE

FENEmR

FENE

RR RR

RRLcc

RR

Ll

)(





sLs

LG V

RR

RV

DC

ˆˆ

sLs

G VRR

IDC

ˆ1ˆ

21ˆ kvlcc

lZ G

mG

GCG

21ˆ kvlcc

lZ R

mR

RCR





Among the previous parameters, k is referred to as the coupling coefficient between the generator and receptor circuits; and when the two circuits are weakly coupled.

1k1k

are the characteristic impedances of each circuit in the CRCG ZZ ˆandˆ

presence of the other circuit.

RG and are the time constants of the circuit.

DCDC GG IV ˆandˆ are the voltage and current of the generator circuit for DC

excitation, respectively.

The remaining terms give the ratio of the termination resistance to thecharacteristic impedance of the line:

When one of the previous ratios is smaller or greater than one, the corresponding termination impedance is referred to as being a low-impedance load or high-impedance load, respectively.

CRFELRCRNESRCGLLGCGSSG ZRZRZRZR /,/,/,/

University of Illinois at Chicago ECE 423, Dr. D. Erricolo, Lecture 19 University of Illinois at Chicago ECE 423, Dr. D. Erricolo, Lecture 19 Frequency.

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