电磁暂态仿真lecture4

Modeling Of Transmission Lines

4.1

Transmission Line Equations in Frequency Domain

In the frequency domain, the equations for an n conductor transmission line are:

¶V Ð = Z(w) I¶x

Ð ¶I = Y(w) Vwhere I(x) and V(x) are the voltage

phasors at a distance x along a

(4.1)

¶x

whereZ(w) = R(w) + j X(w)

Y(w) = G(w) + j B(w)

For most frequencies of interest we may approximate the complicated series expansions for Yand Z due to Carson [1] by the formulae due to Wedepohl et al [2] and Deri et al [3]:

where

Zii = Zii(int) + Zii(ext)

0.3565ririm

coth (0.7777 mri) + W/mZii(int) =

22pripr

i

(4.2a)

withand

m(w) = jwmori(4.2b)

jwm

Zii(ext) = o ln 2hir+de

2pirwherede = jwmoand

jwmo

ln Zij = 2p

(yiÐyj)2 + (hi+hj+2de)2(yiÐyj)2 + (hiÐhj)2 W/m

with de as before.(4.2c)Here ri is the conductor resistivity (W-m), r is the earth resistivity, ri is the conductor radius (orGMR for bundled conductors) for conductor i.

Ó A.M. Gole, University of Manitoba, 1998

page 39

conductor i(y, h)conductor jriiirj(yj, hj)yhiihjFig. 4.1: Variables describing the geometry

Similarly[Y] = (Yij)where

[Y] = PÐ1Pij = j2pweo

Ð1

ln (yiÐyj)2 + (hi+hj)2dijwhere

dij = (yiÐyj)2 + (hiÐhj)2 for i¹j= ri for i=j

4.2Admittance Matrix Model for One Conductor Lossless Line

Hereri=0, r=0. Assume a single conductor line [4]Eqn. 4.2 and 4.3

Zii(int) = 0, Zii(ext) = jwmo ln 2hi2pri = jwL

and

Pii = (j2peow)-1 ln 2hri

Yii =2jwpeoor

ln2h = jwC

riwhere L and C are inductance/capacitance per unit length.Thus:

Ð ¶V

¶x

= jwL I frequency domain form

Ð ¶I

¶x

= jwC Vor:

Ó A.M. Gole, University of Manitoba, 1998

(4.3)

page 40

Ð Ð

¶V

¶i = L ¶x¶t = C ¶v

dt

time domain form(4.4)

¶I¶x

¶ Alembert's solution to (4.4) is:

i(x, t) = f1(xÐct) + f2(x+ct)v(x, t) = Zf1(xÐct) Ð f2(x+ct)

where c = 1/LC (phase velocity), Z = L/C (characteristic impedances)Note on rearranging,

v(x, t) + Z i(x, t) = 2Zf1(xÐct)v(x, t) Ð Z i(x, t) = Ð2Zf2(x+ct)

(4.5)

(4.6)

From (4.6) it is clear that an observer moving at velocity c along the line will see the quantityv+Zi= constant, because for him, xÐct=constant. Let the observer leave m at time tÐt and

ikmterminal

--------------k

traveller at velocity Fig. 4.2

arrive at k at time t, where t=d/c, d being the line length.

Thus,

vm=(tÐt) + Zim,k(tÐt) = vk(t) + Z(Ðikm(t))

or rearranging

ikm(t) = 1 vk(t) + Ik(tÐt) Zwhere

Similarly,

Ik(tÐt) = Ð 1 vm(tÐt) Ð imk(tÐt)

Zimk(t) = 1 vm(t) + Im(tÐt)

ZIm(tÐt) = Ð 1 vk(tÐt) Ð ikm(tÐt)

Z

(4.7)

m

--------------imkterminal

where

Thus we have the following Norton representation of the T-line:

Ó A.M. Gole, University of Manitoba, 1998

page 41

ikm(t)+vk(t)Ðimk(t)

ZIk(tÐt)Im(tÐt)ZFig. 4.3: Norton Representation

Example: Plot v1(t), i12(t), v2(t), i21(t) if

L = 400 nH/m, C = 40 pf/m, d = 2´105mThus t = dLC = 800ms

0.1W±0Thus:

i12Js100A010sv1Ð+1 s100i211 s100+vL1 s100Ð10V+v1Ð12+v2Ð100W

I1(tÐt)

I2(tÐt)Note:where

Z = L = 100WCthen

I1(tÐt) = Ð 1 v2(tÐt) Ð i21(tÐt)

100I2(tÐt) = Ð 1 v1(tÐt) Ð i12(tÐt)

100 10.01 v1 = Js Ð I1(tÐt)

for the two networks

and 0.02 v2 = ÐI2(tÐt)

from which V1(t) and V2(t) may be evaluated and hence the currents:

i12 = v1 + I1(tÑt)

100

i21 = v2 + I2(tÑt)

100Listed below in FORTRAN program to solve for these voltages and currents, and typical resultsare shown in Figs. 4.4 through 4.6 for various terminations on the line.

Ó A.M. Gole, University of Manitoba, 1998

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C****************************************************************************C Solution of Simple Transmission line problem

C**************************************************************************** real delt,fintim,prtime,v1,I1(20),I2(20),i12,i21 real L,C,Js

C____________________________________________________________________C open datafile for reading and output file for answer: open(unit=10,file='inp_file2') open(unit=11,file='out_file2')

write(11,*) 'This is the solution of a simple T-line problem'C____________________________________________________________________C Reading the datafile:C

read (10,*) delt,fintim,prtime,tup read(10,*)L,C,d,Rs,Rl

C Z is the char. imp., Rs is the source res., and Rl the load res.C____________________________________________________________________C Initialization Z=sqrt(L/C) g1=1/Rs+1/Z g2=1/Rl+ 1/Z z1=1/g1 z2=1/g2 v1=0.0 v2=0.0 Do i =1,20 I1(i)=0.0 I2(i)=0.0 enddo

time=0.0

C____________________________________________________________________C Setting the loop on time: nstep=fintim/delt nprt = prtime/delt

ndelay = d*sqrt(L*C)/delt

C____________________________________________________________________C The numerical solution:C

Do 1 J=1,nstep+1C

C Setting the source current: if(time.LT.tup)then Js=0.0 else

Js=100.0 endif

Ó A.M. Gole, University of Manitoba, 1998

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C (Note: at this time the values of v1 i12,etc,on the right hand sideC are from the previous timestep (t-delt) as the soln for time=t hasC as yet not been found!)

C_____________________________________________________________________C Now the solution:

v1 = (Js-I1(ndelay))*z1 v2 = -I2(ndelay)*z2C

i12 = v1/Z+I1(ndelay) i21 = v2/Z+I2(ndelay)

C_____________________________________________________________________C Calculating the history term (for use ndelay timesteps later...).: Do i=1,19 ii=20-i

I1(ii+1)=I1(ii) I2(ii+1)=I2(ii) enddo

I1(1) = -1/Z*v2- i21 I2(1) = -1/Z*v1- i12

C Writing the soln. to an output file:C

if(mod(J-1,nprt).eq.0) write(11,1111) time, v1, i12,v2,i21, . I1(ndelay),I2(ndelay)1111 format(f8.5,6G12.5)

C_____________________________________________________________________C Updating time:C

time=time+delt1 continue

C.....Now loop back for the next timestep.

C_____________________________________________________________________ write(6,*)'Thats all..folks!' stop end

Ó A.M. Gole, University of Manitoba, 1998

page 44

Results: RL = 100W (Characteristic impedance)

Note:a)The 800ms delay due to travel time.

b)The lack of reflections due to characteristic impedance termination.

Ó A.M. Gole, University of Manitoba, 1998

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RL = 106W

Note:a)The 800ms delay between the sending and receiving end.

b)The re-inforcement of the voltage signal due to reflections to twice itsoriginal magnitude.

c)The period of the oscillation = 1.6 ms (2 travel times).

Ó A.M. Gole, University of Manitoba, 1998

page 46

RL = 0.1W

Note:a)The zero voltage on the short circuited receiving end.

b)The current shows the same delays and periodic steps seen earlier, but

is building up to V1 = v1 (actually for a true s/c termination it will build to

R0.2Winfinity).

Ó A.M. Gole, University of Manitoba, 1998

page 47

4.3Multiconductor Lossless Line

Some theorems required for this material are now given below:

Theorem 4.1: The eigenvector matrix T for a n´n square matrix A with n distinct eigenvaluesl1...ln, diagonalizes A, i.e.

TÐ1 A T = L (a diagonal matrix

where Lij = dijljThe eigenvector matrix is a matrix whose ith column is an eigenvector corresponding to theeigenvalue li. A proof of this is straightforward and may be looked up in a suitable book onlinear algebra.

If A has repeated eigenvalues a strictly diagonal form is sometimes not possible. Howeverin transmission line problems such a situation does not arise.

Example 4.2

Consider the matrixA = 1 0 which has eigenvalues l1 = 1, l2 = 2 (found by solving |AÐlI| = 0.

3 2To find the first eigenvector p we have Ap = l1por

1 0 p1 = 1 p1

p23 2p2

i.e.p1 = p1 and 3p1 + 2p2 = p2 or 3p1 = Ðp2. As the two equations are linearly dependent, we

may choose any value for p1 (as long as we don't obtain the singular vector 0 ), and then p2 =

0

1Ð3p1. Thus is a suitable eigenvector. Similarly for the eigenvalue l2 = 2:Ð3

1 0 p1 = 2 p1

p23 2p2

orp1 = 2p1 and 3p1 + 2p2 = 2p2 .

Thus p1 = 0, p2 is arbitrary. 0 is therefore one suitable eigenvector.1Thus T = 1 0 is a possible transformation matrix

Ð3 1withTÐ1 = 1 0 , we have

3 1

TÐ1 AT = 1 0 ¥ 1 0 ¥ 1 0 = 1 0

3 13 23 10 2

= l1 00 l2

Ó A.M. Gole, University of Manitoba, 1998

page 48

, which is diagonal.

Theorem 4.2 If Y and Z correspond to the per unit length admittance and impedance matrices ofa transmission line with zero resistivity above a lossless earth, then ZY=YZ is a diagonal matrix.Proof: From equation 4.2

Z = [Zij] where (for i¹j)Zij = wherelikewise

Djwmo

ln ij2pdijDij = (yiÐyj)2 + (hi+hj)2 , dij = (yiÐyj)2 + (hiÐhj)2Y = PÐ1 where P = [Pij]

DijandPij = 1 ln dij2pÎojwThus we findSimilarly

and thusor

Zij = Ðw2moÎoPijZii = Ðw2moÎoPiiZ = Ðw2moÎoPZY = Ðw2moÎoIi¹j

(4.7)

where I is the identity matrix. Thus ZY is diagonal and similarly so is YZ.Note: Since [Z] = jw(L), Y = jw(C) it follows that

[L][C] = [C][L] is diagonal as well.

Theorem 4.3 If [L][C] = [C][L] = L where L is diagonal and [L] and [C] are n´n nonsingularmatrices with distinct eigenvalues, then the matrix T which diagonalizes [L] also diagonalizes[C], i.e. if TÐ1 [L] T is diagonal, then so is TÐ1 [C] T.Proof: Let TÐ1 [L] T = LLHowever[L] [C] = L so [L] = LCÐ1ThusorThusInverting

TÐ1 [LCÐ1] T = LL

L TÐ1[CÐ1] = LL as L is diagonal.

TÐ1[CÐ1] T = LÐ1LL which is still diagonalTÐ1 C T = (LÐ1LL)Ð1 = LC which is diagonal.

Consider now a multiphase lossless transmission line, for which we readily derive the timedomain equations in a form analogous to the derivation of Eqn. 4.4.Thus:

Ó A.M. Gole, University of Manitoba, 1998

page 49

Ð

¶v = [L] didt¶x

(4.8)

¶iÐ = [C] dvdt¶x

The only difference between Eqns. 4.4 and 4.8 being that the latter is a vector-matrix equation.

Note that we cannot directly obtain a d'Alembert type solution for Eqn. 4.8 because each componentequation is coupled and involves voltages or currents from more than one phase. Our attempt isto transform Eqn. 4.8 into two sets of n decoupled equations which can then be solved by themethod of d'Alembert.

Consider the matrix T such that

TÐ1 [L] T = L' where L' is diagonal.

Then by Theorem 4.3, TÐ1 [C] T = C' is also diagonal.

Consider the transformation v = T v'. Then equation 4.8 becomes

Ð ¶Tv¶x

= [L] dTidtÐ ¶Ti¶x

= [C] dTvdtor

Ð ¶v' = TÐ1x

[L] T di'¶dtÐ ¶i'¶x = TÐ1 [C] T dv'dtor

Ð ¶v'¶x

= L' di'dtÐ ¶i'¶x

= C' dv'dtEqn. 4.9a is a collection of pairs of n uncoupled equations because L', C' are diagonal.

Ð ¶v'j¶ = L'j

di'ix

dtÐ ¶i' i=1,2,...nj

¶x

= C'j dv'dtThus we get the n solution pairs

Ó A.M. Gole, University of Manitoba, 1998

(4.9)

(4.9a)

(4.9b)

page 50

i'j = f1(xÐcjt + f2(x+cjt)v'j = Zj (f1(xÐcjt) Ð f2(x+cjt)(4.10)

in a manner analogous to equation 4.5.

Herecj = 1/LjCj , Zj = Lj/CjThe quantities v and i are called phase quantities and v' and i' are called modal quantities.We thus obtain in the manner analogous to Eqn. 4.7 the equivalent circuit shown in Fig. 4.7.i1km(t)+v'1k(t)Ði2km(t)+v'2k(t)ÐI'2km(tÐt'2)Z'2I'2mk(tÐt'2)Z'2I'1km(tÐt'1)Z'1I'1mk(tÐt'1)Z'2i1mk(t)+v1mk(t)Ði2mk(t)+v2mk(t)Ð+v'nk(t)ÐI'nkm(tÐt'n)Z'nI'nmk(tÐt'n)Z'n+vnmk(t)ÐFig. 4.7: Multiconductor line in Modal Domain

In Fig. 4.7 the history terms are analogous to the ones developed earlier, i.e. for the jth mode.

I'jkm(tÐZ'j) = Ð 1 v'jm (tÐt'j) Ð i'jmk(tÐt'j)

Z'jI'jmk(tÐZ'j) = Ð 1 v'jk (tÐt'j) Ð i'jkm(tÐt'j)

Z'j

where Z'j = Lj/Cj , t'j = d LjCj d being the length of the lineWhen included in an admittance type formulation the Z' (orY') should be transferred to the phasedomain for inclusion in the network's Y matrix, i.e. Zo = TÐ1Z'T or Yo = TÐ1 Y' T

Ó A.M. Gole, University of Manitoba, 1998

page 51

(4.11)

ExternalNetworkTTExternalNetworkModal characteristicimpedance transferredto phaseModal history terms

Fig. 4.8: Interfacing to the external network

The method of solving the system with a transmission line now includes the following steps:1)2)3)4)5)6)7)

Calculate the transformation matrix T.

Calculate the modal impedance matrix Z' or admittance matrix Y' (Note: Y'i = 1 ) from

Z'i

Eqn. 4.10.

Calculate the transferred characteristic impedance matrix Zo (or Yo) as seen by the externalnetworks.

Form the admittance formulations for these networks using the standard techniques.

Calculate the history injections from equations 4.11. Note, you need modal voltages andcurrents to calculate these, so at every timestep you must evaluate v' = TÐ1v, i' = TÐ1i.The actual injections into the external systems are [T]Imk and [T]Ikm respectively.

Calculate all voltages and thence currents in the external network and return to Step 5 andthen to the next timestep.

Ó A.M. Gole, University of Manitoba, 1998

page 52

Using transformation matrices Fe and Fi.Assignment 4:

conductor 2

Gmr1=0.0682mpi=025mconductor 120mp=0

m0 = 4p x 10-7 weber/A-mÎ0 = 8.854 x 10-12 coul/N-m2

The 2 conductor transmission line is shown above

0.1W10V0±0Characteristicimpedance load

12mGmr2=0.1077m

Terminate it with a characteristic impedance load and excite it with the 10V step shown. Showvoltages and current waveforms on each phase at the sending and receiving ends. Also plot themodel quantities.

Repeat with an open circuit termination.

Ó A.M. Gole, University of Manitoba, 1998

page 53

4.4Inclusion of Line Resistance

If the frequency dependence of line resistance is ignored, the line may be modelled as twolossless sections with lumped resistor values R/4 at each end and a lumped resistor value of R/2in the centre as shown in Fig. 4.9.

R/2 R/4R/4+vkÐ

R/4 d/2ln d/2R/4+vmÐ

d=line

Fig. 4.9: Simple model of lossy line.

Dommel [4] claims that the incremental accuracy obtained by modeling the line with a largernumber of sections is only marginal and not worth the effort. Fig. 4.10 shows the resultantcurrent source Ð admittance formulation for the line in Fig. 4.9.kikmZCI'kIlZCk'l R/2nm'mimkZCI'nI'mZCZC=LCFig. 4.10: Equivalent circuit of lossy line.

If the midpoint information is not required then the circuit in Fig. 4.10 can be collapsed to thatshown in Fig. 4.11 which is similar to the one in Fig. 4.3.

+vkÐikmZIkImZimk+vmÐFig. 4.11: Collapsing Fig. 4.10 to eliminate central node.

The impedances and history currents in Fig. 4.11 are given by the following equations as thereader may show:

Z = L + R , t = d LCC4 page 54

Ó A.M. Gole, University of Manitoba, 1998

Ik(tÐt) = 1+h ¥ Ð 1 em(tÐt) Ð imk(tÐt)

2Z +

1Ðh

¥ Ð 1 ek(tÐt) Ð ikm(tÐt)2Z

(4.12)

and similarly for Im with k, m interchanged.Z Ð 1 R4Here h =

Z + 1 R

4The question arises as to what value of R should be used. If earth resistivity is ingored only theterm Zii (int) is a function of frequency (see eqn. 4.2), and this model is most valid as the rest ofthe mutual impedance terms remain frequency independent (thereby keeping the conditionssimilar to the lossless core).

At dc, eqn. 4.2a for the internal resistance yieldsjwmori riZii(int) = Lim + coth 0.7777 2prw®0i= ripri

2

0.3565 rijwmo ¥ ri + ripri2(4.13)

which is the dc resistance of the conductor (per unit length).

However, at any other frequency Zii(int) is complex for ri¹0 and thus yields different resistance

values for different frequencies (the so called 'skin-effect'). We may thus choose a value for Rthat matches the line resistance at the frequency of most interest to our particular study.

For a multiconductor line we use a model such as in Figures 4.9 through 4.11 for eachmode of the line. The modes are decomposed and recombined from phase quantities in themanner of section 4.3. This results in a different resistance for each mode. The transformationmatrices T and TÐ1 are evaluated as in section 4.3 (without considering the resistive terms,otherwise they would be complex and frequency dependant!) The modal resistance matrix isthen the real (and diagonal!) part of TÐ1 [Z]T.

4.4Transmission Line Parameter Calculations for a Lossy Line

With the inclusion of earth resistance the line parameters do not appear in the convenientform [Z]=jw[L] and Y=jw[C] as discussed in section 4.2. Also it is no longer true that the matrixthat diagonizes Z will also diagonalize Y. We still wish to use the treatment discussed in section4.3 and 4.4, i.e., the assumption of a lossless line, but we should select a set of parameters thatgives us most accuracy where desired.

The two parts of eqn. 4.1 can be substituted into each other to form the equations:

Ó A.M. Gole, University of Manitoba, 1998

page 55

¶V¶x2¶I¶x2

2

2

= (ZY)V(4.14)

= (YZ)INow consider the Transformation TVV' = V ,m TII' = Iwhere TV is such that

TÐ1V (ZY) TV = LZY

TÐ1I (YZ) TI = LYZ

i.e., TV diagonilizes the first equation of the pair 4.14 and TI diagonlizes the second.

Note that (ZY)T = YTZT = (YZ) because Y and Z are symmetric.

Thus LZY = LYZ = L because symmetric matrices have the same eigenvalues.Thus let us call

A=ZY

AT=YZ

Now TÐ1V A TV = Lon transposing:

TÐ1TTTV A (TV) = L = L (from eqn. 4.15a)

Ð1

But this is the same as the second equation in 4.15 if we consider TT = TVI.TThus we may choose TI = (TÐ1V) to satisfy equation 4.15b.

T

(4.15)

(4.16)

Now consider

T-1V ZY TV = L

Ð1

HenceTÐ1 Z (T TIVI) Y TV = L

Ð1

orTÐ1V Z TI TI Y TV = L

Similarly by considering the second equation in 4.15,

Ð1

TÐ1T YTVVI ZTI =L

It can be shown that if PQ=QP=L the P and Q themselves must be diagonal. Thus,

Ð1Ð1

TI Y TV and TV Z Ti are diagonalTÐ1I YTV =Ymor

TÐ1V ZTi =Zm

(4.17a)

(4.17b)

(4.18)

Now Ym and Zm are not of the form jwC', jwL', but for small losses are approximately so. We

Ó A.M. Gole, University of Manitoba, 1998

page 56

thus select a frequency of interest wo and evaluate

C' = YmjwoZL' = m

jwo

TVO = TV (wo)TIO = TI (wo)

(4.19)

We now (follow an identical procedure to that explained in section 4.3) but, use matrices TVO andTIO instead of the matrix T and TÐ1 in the calculations between phase and mode quantities.4.6

Treatment of Frequency Dependence

As can be seen from eqns. 4.2 and 4.3, Z(w) and Y(w) are in general functions of frequency.Carson [1] has solved such equations for linear terminations using frequency domain solutionmethods, in an exact manner. Unfortunately modeling in the time domain has proved to be quitedifficult. The treatment below is due to Marti [5]. For a single conductor line with equations(4.1) (repeated here), we have:

¶VÐ = Z(w) I¶x

Z, Y = impedance/admittance per unit length (4.1)

¶IÐ = Y(w) V¶xwith I and V as phasors, ([5], [6]).length = lkmFigure 4.11. Transmission Line

Vk(w) = cosh[g(w)l]Vm(w) ÐZC(w) sinh[g(w)l]Im(w)Ik(w) =

1 sinh[g(w)l]Vm(w) Ð cosh[g(w)l]Im(w)ZC(w)

(4.20)

and conversely for Vm, Im

Ó A.M. Gole, University of Manitoba, 1998

page 57

Here

ZC(w) = Z = Y(R+jwL)(G+jwC)g(w) = ZY = (R+jwL) (G+jwC)As in eqn. (4.6) we form the functions

Fk(w) = Vk(w) + ZC(w) Ik(w)Bk(w) = Vk(w) Ð ZC(w) Ik(w)

and likewise for m, we have

Fm(w) = Vm(w) + ZC Im(w)Bm(w) = Vm(w) Ð ZC Im(w)

We can eliminate Vm, Im, Vk, Ik to get:

Bk(w) =A1(w)Fm(w)Bm(w) =A1(w)Fk(w)where A1(w) = eÐg(w)L =

1cosh[g(w)L] + sinh[g(w) l]

(4.23)(4.22)(4.21)

A1(w) is called the weighting function and a1 (Fourier inverse of A1(w) is called the impulseresponse function.

Note that eqns. 4.21 and 4.22 can be used to solve for the Transmission line. The schematicrepresentation of eqn. 4.21 (for Bk Bm only) is as in Fig. 4.12.

Ik+vkбBk±BmZCZCIm+vmÐFig. 4.12: Schematic representation of Eqn. 4.23 for Bk, Bm.

When the circuit of Fig. 4.12 is drawn in the time domain, it is as in Fig. 4.13.

ik(t)+vk(t)ÐZC±bk(t)±ZCbm(t)im(t)+vm(t)Ðpage 58

Ó A.M. Gole, University of Manitoba, 1998

Fig. 4.13: Time domain Form.

Now

bk(t) = FÐ1A1(w) ¥ FR(w)

¥

=

t

fm(tÐx)a1(x) dx

¥

(4.24)

bm(t) =

t

fk(tÐx)a1(x) dx

Note that the old values of fm(t) and fk in eqn. 4.24 are old (history) values and are known fromprevious timesteps.

ZC is approximated by a Foster I type realization (approximation) [5] of the true ZC; and is of theform

R1R2RnRo........C1

C2

Cn

Fig. 4.14: Foster I Realization of ZC

It is not required to do an infinite integral as suggested by eqn. 4.24. This is so because a1(t) hasthe form shown in Fig. 4.15 and only has any significant value between t=t and t=t+p (say).Thus only values of fm(tÐx) between x=t and x=t+p need be considered in the integration toobtain bk(t).

a1(t)tt+pt

Fig. 4.15: Graph of a1(t).

We have only shown the single conductor case. For the multiconductor case we also obtain

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page 59

frequency dependent transformation matrices. For transmission lines we can often ignore thefrequency dependence and equate TV and TI to those calculated for the most important frequencyof interest to us. The treatment for cables however requires the use of frequency dependentmatrices and is a topic of continuing research.References:[1][2][3][4][5]

Carson, J.R., \"Wave propagation in overhead Wires with ground return\Technical Journal, Vol. 5, pp. 539-554.

Wedepohl, L.M. and Wilcox, D.J., \"Transient Analysis of Underground Power TransmissionSystems\

Deri, A., et al., \"Homogenous and Multilayer Earth Return\3686-3693.

Dommel, H.W., \"Digital Computer Solution of Electromagnetic Transients in Single andMulti-Phase Networks\Marti, J.R., \"Accurate Modelling of Frequency Dependent Transmission Lines inElectromagnetic Transient Simulations\1982, pp. 147-155.

Wang, X., Mathur, R.M., \"Real Time Digital Simulator of the Electromagnetic Transientsof Transmission Lines with Frequency Dependence\York, 19, paper WM 122-3 PWRD.

[6]

Ó A.M. Gole, University of Manitoba, 1998

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