Transmission Matrix
Comments on Example 1, and
Lumped Circuit Equivalent
1.0 Introduction
In these notes, I want to write the equations
from last time in a matrix form. Then I will
say a few words about example 4.1. Then I
am going to skip over examples 4.2, 4.3,
section 4.2, then jump to Sections 4.3 and
4.4.
2.0 Transmission matrix (Section 4.3)
We jump to Section 4.3 because it is easy to
do so. Recall the equations developed from
last time relating the sending end voltage
and current (V1, I1) to receiving end voltage
and current (V2, I2) of a transmission line.
V1  V2 cosh l  ZC I 2 sinh l
(1)
V2
I1  I 2 coshl 
sinh l
(2)
ZC
1
These equations can be written in matrix
form:
Z C sinh l 
 cosh l
V
 1 
V2 

 I    1 sinh l
I 
cosh

l

(3)
 1
 2
 ZC

Reference Module D1 of your EE 303 notes,
Fig. D1.16, repeated here for convenience:
Z
I1
V1
I2
Y1
Y2
Z  V2 
V1   1  Y2 Z

 I  Y  Y  ZY Y 1  Y Z   I 
 1  1 2
1 2
1  2 
V2
Fig. D1.16: ABCD parameters for “” circuit [1]
So eq. (3) is the relationship that we call the
“ABCD” parameters for a two-port network:
V1   A B  V2 
 I   C D   I 
(4)
 2 
 1 
Clearly, in this case, we see that:
1
A  cosh l
C
sinh l
ZC
(5)
B  Z C sinh l
D  cosh l
2
The ABCD parameters are also referred to
as the transmission parameters, where the
following matrix is defined as the
transmission matrix:
 A B
T

(6)
C
D


With this notation, eq. (4) becomes:
V1 
V2 
I   T I 
(7)
 1
 2
Two comments about this:
a. It is easy to get the receiving end (V2, I2)
as a function of the sending end quantities
(V1, I1):
V
V2 
1  1 
I   T I 
(8)
 2
 1
The reason why this is easy is because
of the hyperbolic identity:
cosh 2 x  sinh 2 x  1
and therefore the determinate of T is
|T|=AD-BC=1
and therefore
3
 D  B
T 

(9)

C
A


b. It is easy to express sending end or
receiving end quantities as a function of
the other when there are multiple twoports chained together. Figure 1 illustrates
the case of two 2-ports chained together.
1
I1a
V1a
I2a
I1b
a
I2b
b
V2b
Fig. 1
If we know Ta and Tb, where
 Aa
Ta  
Ca
 Ab
Tb  
Cb
Ba 
Da 
Bb 
Db 
then it is easy to show that
V
V1 
V2 
V2 
1  1 
I   T I  ,
I   T I 
 1
 2
 2
 1
where
T 1  Ta1Tb1
T  TaTb ,
4
3.0 Comments on Example 4.1
The text provides three nice examples,
Examples 4.1-4.3 on pp 93-96. I will not go
through them (you can read for yourself in
the text) but rather I will just focus (for now)
on the first one and draw out what are some
conceptual implications of which you should
be aware. Please consider (look for, think
about, connect with) these concepts as you
review these examples on your own.
Example 4.1:
The main idea of this example is to give you
some practice in using the equations derived
last time and repeated here for convenience:
V1  V2 cosh l  ZC I 2 sinh l
(1)
V2
I 1  I 2 coshl 
sinh l
(2)
ZC
The example gives you information to
compute everything on the right hand side:
 The line is l=225 miles long.
5
V2 and I2 (the voltage and current phasors
on receiving end of the transmission line)
 y and z and therefore γ and ZC.
And from this, we compute V1 and I1 (the
voltage and current phasors on sending end
of the transmission line) using eqs. (1), (2).

Note it requires evaluation of the hyperbolic
functions. We illustrate details of this. With
γl=0.4688∟83.95° =0.0494+j0.466
ZC=387.3∟-6.05° Ω
and recalling that
e l  e   l
e l  e  l
cosh l 
; sinh l 
2
2
l
 l
l
 l
2
cosh

l

e

e
;
2
sinh

l

e

e

6
Evaluation of 2coshγl:
2 cosh l  el  e l
 e 0.0494 j 0.466  e 0.0494 j 0.466
 e 0.0494e j 0.466  e 0.0494e  j 0.466
 e 0.0494 (cos 0.466  j sin 0.466)
 e 0.0494 (cos 0.466  j sin 0.466)
1.0506(0.8934  j0.4493)
 0.9518(0.8934 - j0.4493)
 0.9386 + j0.4721  0.8503 - j0.4276
1.7889 + j0.0445  1.78951.425
Aside: In evaluating the sin and cos terms
above, the arguments (which come from
Im(γl)) are radians, so do not plug them into
your calculator as if they are in degrees
unless you first convert them to degrees.
Comment: Another useful relation [2, pg.
205] is that the hyperbolic sines and cosines
of the complex argument x=(+j)x may
be separated into real and imaginary parts by
use of the following:
7
cosh x  cosh(  j ) x  cosh x cos x  j sinh x sin x
sinh l  sinh(   j ) x  sinh x cos x  j cosh x sin x
Therefore:
1.78951.425
coshl 
 0.89481.425
2
which is about what the text gives at the
bottom of page 93.
Two other notes:
1. Eqs. (1) and (2) were derived for a perphase model and therefore we must use
line-neutral voltages.
2. The statement on page 94: “It is
convenient
to
pick
∟V2=0°,…”
Remember, we can always pick any one
quantity in a circuit to be the reference.
Applying eqs. (1) and (2), the text gets the
following result:
V1 =89,280∟19.39°
I1 =162.42∟14.76°
8
I want to work this problem using a lumped
parameter π-equivalent model, where
Z =z*l=(0.807∟77.9°)Ω/mile*225miles
=181.6∟77.9°
Y =y*l=(j5.38*10-6)Mhos/mile*225miles
=j.00121
We model half of Y, i.e., Y/2=j0.000605, on
the receiving end and on the sending end. So
our circuit appears as in Fig. 2.
IZ
I1
I2=184.1∟-18.195°
Z=181.6∟77.9°
V1
IY1
IY2
Y/2=
j0.000605
Y/2=
j0.000605
Load
V2=76,200∟0°
Fig. 2
Using this lumped parameter π-model, we
can make the following calculations:
IY2 =V2Y/2=76,200(j.000605)=j46.101
IZ =IY2+I2
=j46.101+184(cos18.195-jsin18.195)
=j46.101+174.8 - j57.454
=174.8-j11.353=175.1683∟-3.7161°
9
The impact of line charging on current into
the load is to make IZ almost 1.0 pf.
V1 =V2+IZZ
=76,200+175.1683∟-3.7161°(181.6∟77.9°)
=76,200+31,811∟74.1839°
=76,200+8670.1+j30,607=90,220∟19.83°
IY1=V1Y/2=90,220∟19.83°(j.000605)
=54.5831∟109.83°
I1 =IY1+IZ
=54.5831∟109.83°+175.1683∟-3.7161°
=161.3198∟14.35°
Compare results:
V1
I1
Distributed 89,280∟19.39° 162.42∟14.76°
parameter
model
90,220∟19.83° 161.3198∟14.35°
Lumped
parameter
π-model
Lumped parameter π-model is not bad!
10
Would it be better or worse if the line was
only 50 miles long?
We are interested to see if we can devise the
relation between the two models to see if we
can better understand the difference between
them.
4.0 Lumped circuit model
Our question is the following. Can we create
a lumped parameter π-model that gives
exactly what the distributed parameter
model gives? Our work here relates to that
of Section 4.4 in the text.
In other words, we want to find parameters
Y’ and Z’ that make the relation between
sending and receiving end quantities in Fig.
3 be exactly the same as the distributed
parameter model gives.
11
IZ
I1
I2
Z’
IY1
IY2
Y’/2
Y’/2
V1
V2
Fig. 3
Note that Y’, Z’ are NOT the same as Y, Z
as used in the example of Fig. 2 above.
 Y and Z are the lumped parameters
obtained from zl and yl, respectively. Our
example shows that these values do NOT
give the same results as the distributed
parameter model gives.
 Y’ and Z’ are the lumped parameters that
give exactly the same results as the
distributed parameter model gives.
We know that the distributed parameter
model relates sending and receiving lines
according to eqs. (4) and (5), repeated here
for convenience:
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V1   A B  V2 
 I   C D   I 
 2 
 1 
A  cosh l
(4)
1
C
sinh l
ZC
(5)
B  Z C sinh l
D  cosh l
So our plan of attack materializes at this
point. We must
1.Derive the ABCD-parameters for the
network of Fig. 3 in terms of Y’ and Z’.
2.Equate them to the appropriate expressions
in eq. (5).
3.Solve for Y’ and Z’.
We have actually done the equivalent of
Step 1 in the calculations for the Exercise 1
lumped parameter model done above
(except we used Z there and here we use Z’).
We can express this calculation here:
13
Voltage relation:
V1 =V2+drop across Z’.
=V2+IZZ’
=V2+(I2+IY2)Z’
=V2+(I2+V2Y’/2)Z’
=(1+Y’Z’/2)V2+Z’I2
Current relation:
I1 =IY1+IZ
=IY1+IY2+I2
=V1Y’/2+V2Y’/2+I2
Use voltage relation to eliminate V1, to get:
I1 =[(1+Y’Z’/2)V2+Z’I2]Y’/2+V2Y’/2+I2
=(Y’/2+Y’2Z’/4+Y’/2)V2+(1+Z’Y’/2)I2
=Y’(1+Y’Z’/4)V2+(1+Z’Y’/2)I2
From the voltage and current relations, we
infer the following expressions for A,B,C,D:
A=(1+Y’Z’/2)
C=Y’(1+Y’Z’/4)
B=Z’
D=1+Z’Y’/2
Equating these expressions with eqs. (5)
results in:
14
A  cosh l  1  Y ' Z ' / 2
B  Z C sinh l  Z '
1
C
sinh l  Y ' (1  Y ' Z ' / 4)
ZC
(10)
D  cosh l  1  Z 'Y ' / 2
Solving for Z’ is very easy (second
equation). Solving for Y’ is more difficult.
The text does both (pg 100), and also
provides a bit of manipulation that I will
avoid. The result is:
sinh l
Z '  Z C sinh l  Z
l
2
l
tanh( l / 2)
Y'
tanh  Y
ZC
2
l / 2
15
Note that the right-hand-side of expressing
Z’ and Y’ relate it to Z=zl and Y=yl. In
other words, the multipliers of Z and Y in
those expressions are “correction factors”
that need to be applied to Z and Y in order
for the lumped parameter model, when
based on zl and yl, to give results equal to
the distributed parameter model.
A useful exercise to prove the middle terms
equal the right-hand-side terms in the above
is to show that
ZC 
Z
Z
Z
Z



l
zyl
zlyl
ZY
ZY Z ZY
ZY Y Y Z
Z




ZY
ZY
Y
Y Y Y
Y
2
Y

...
Z C l / 2
[1] ISU EE 303 Class Notes: “Introduction to Distribution Systems.”
[2] O. Elgerd, “Electric Energy Systems Theory: An Introduction,” 2nd
edition, McGraw-Hill, 1982.
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