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Terminal Relations
1.0 Introduction
Our work in Chapter 3 resulted in ability to
compute two parameters:
 z in Ω/m
 y in mhos/m
We can use these parameters to compute the
series impedance and shunt admittance of
any length of line: even a 1 cm line, a 1 mm
line, or even an infinitesimal length of line.
From this notion, then, we see that the series
impedance and shunt admittance are in fact
distributed along the entire length of the line
and not simply lumped at a single location.
As a result, we refer to z and y as distributed
parameters. In these notes, we will derive
the so-called “long-line” transmission line
model.
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2.0 Model development
Consider the per-phase diagram in Fig. 1.
I1
I+dI
I
•••
I2
•••
zdx
dI
V1
V+dV
ydx
•••
V
V2
•••
dx
x
l
Fig. 1
In Fig. 1, V and I are phasors. We will
develop the variation of these phasors with
respect to position, but do not forget that the
phasors represent time varying quantities.
Note the differential length dx and the
corresponding small amounts of series
impedance zdx and shunt admittance ydx.
Let’s write KVL equation around inner loop:
V  dV  Izdx  V
 dV  Izdx
2
(1)
dV
 Iz
dx

(2)
(Same as eq. 4.2a in text)
We can also write a KCL equation at the
middle node:
I  dI  (V  dV ) ydx  I
(3)

dI  (V  dV ) ydx  Vydx  ydVdx (4)
In the right-hand-side of eq. (4), if the first
term Vydx, which contains dx, is small, then
the second term, ydVdx, which contains a
product of two infinitesimal numbers dVdx,
is really small. So let’s assume ydVdx=0.
Therefore:
dI  Vydx
(5)
dI
 Vy
 dx
(6)
(Same as eq. 4.2b in text).
Now differentiate eq. (6) wrspt x. This gives
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d 2 I dV

y
2
dx
dx
(7)
Now substitute eq. (2)
dV
 Iz
dx
(2)
into eq. (7) to get:
d 2I
 Izy
2
dx
(8)
Define the propagation constant:
  zy (units are 1/meters)
Note that γ is in general a complex number.
Then eq. (8) becomes:
d 2I
2

I

dx 2
(9)
(Same as eq. 4.4 in text).
Likewise, we could also differentiate eq. (2)
wrspt x, then substitute in eq. (6),
dI
 Vy
dx
(6)
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to get:
d 2V
2

V

dx 2
(10)
(Same as eq. 4.3 in text).
The text provides solution to eq. (10).
We will do it for eq. (9).
Taking the LaPlace transform of (10):
s 2 I ( s )  sI (0)  I(0)  I ( s ) 2 (11)
Notation above:
I(x) is x-domain expression of phasor I.
I(s) is Laplace transform of I(x).
Solve eq. (11) for I(s):
s 2 I ( s )  I ( s ) 2  sI (0)  I(0)
2
2

I
(
s
)
s



sI
(
0
)

I
(0)



sI (0)  I(0)
I ( s) 

s2   2


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(12)
The characteristic equation for this system
identifies the roots of the denominator of eq.
(12) as:
s
2

 2  0
(13)
Factoring eq. (13) results in:
s   s     0
(14)
And so the roots are s=-γ, +γ.
This means that the x-domain equation for
the voltage I(x) will have the form:
I ( x )  c1ex  c2e x
(15)
Now let’s do some tricky manipulation:
 e  x e x e   x e   x 

I ( x )  c1 



2
2
2 
 2
 e  x e  x e x e x 
 (16)
 c2 



2
2
2 
 2
Now distribute the coefficients:
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e x
e x
e  x
e  x
I ( x )  c1
 c1
 c1
 c1
2
2
2
2
e  x
e  x
e x
e x (17)
 c2
 c2
 c2
 c2
2
2
2
2
Reorganize the terms:
e x
e  x
e x
e  x
I ( x )  c1
 c1
 c1
 c1
2
2
2
2
e x
e  x
e  x
e x (18)
 c2
 c2
 c2
 c2
2
2
2
2
Factor out the coefficients from pairs of
terms:
 e  x e  x 
 e x e   x 
  c1 

I ( x )  c1 


2 
2 
 2
 2
 e x e  x 
 e  x e x 
  c2 
 (19)
 c2 


2 
2 
 2
 2
Let’s multiply inside and outside of last term
by -1. Also, combine fractions to get:
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 e x  e  x 
 e x  e  x 
  c1 

I ( x )  c1 
2
2




 e x  e  x 
 e  x  e  x 
  c2 
 (20)
 c2 
2
2




Note that each “column” of eq. (20) contains
a common factor. Factoring, we obtain:
 e  x  e  x 

I ( x )  c1  c2 
2


 e x  e  x 
(21)

 c1  c2 
2


Now define C1=c1+c2 and C2=c1-c2 and use
in eq. (21) to get:
 e x  e x 
 e x  e x 
  C 2 

I ( x )  C1 
2
2



 (22)
The expressions within the parentheses of
eq. (22) are the hyperbolic cosine and sine,
respectively, and eq. (22) can be re-written
as:
I ( x)  C1 coshx  C2 sinh x
(23)
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To determine C1 and C2, we use the
“boundary” condition, obtained from Fig. 1,
that when x=0, I=I2 and V=V2. We also
know that cosh and sinh have the following
characteristics:
Fig. 2
Applying this information to eq. (23), we
obtain:
I (0)  C1 cosh(0)  C2 sinh( 0)
(24)
 I 2  C1
(25)
How to obtain C2? Need to find another
boundary condition….
Recalling eq. (6):
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dI
 Vy
dx
or, as a function of x,
dI ( x )
 V ( x) y
dx
(26)
We can evaluate this at x=0, which gives:
dI ( x )
 V ( 0) y
(27)
dx x  0
But from Fig. 1, V(0)=V2, therefore
dI ( x )
 V2 y
dx x  0
(28)
Now differentiate eq. (23) to get:
dI ( x ) d
 C1 coshx  C 2 sinh x  (29)
dx
dx
We can derive the derivatives of the
hyperbolic functions using their exponential
expressions. Or we can look them up in
tables, finding:
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d
du( x )
cosh u( x ) 
sinh u( x )
dx
dx
d
du( x )
sinh u( x) 
cosh u( x )
dx
dx
Alternatively, we note that we are trying to
get their derivatives at x=0. Inspecting the
slopes of the plots of Fig. 2 at x=0, one can
conclude that:
d
C1 cosh x  C2 sinh x  x  0
dx
 C1 (  0)  C2 (  1)
(30)
Equating eqs. (28) and (30), we get:
dI ( x )
 V2 y  C 2
dx x  0

C2 
(31)
V2 y

Recalling  
(32)
zy , eq. (32) becomes:
V2 y
C2 
zy
(33)
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Multiplying top and bottom of (33) by
we obtain:
V2 y
C2 
zy
y
y

V2 y y
y z
 V2
y,
y
z (34)
Define: characteristic impedance of the line
ZC 
z
y (units are ohms)
(35)
Substituting eq. (35) into (34) we get:
V2
C2 
ZC
(36)
Substituting eq. (25) and (36) into (23):
V2
I ( x )  I 2 cosh x 
sinh x
(37)
ZC
Same as second equation in (4.9) of text.
The other equation in (4.9), which the text
derives, is:
V ( x )  V2 cosh x  ZC I 2 sinh x
(38)
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Equations (37) and (38) give the voltage and
current anywhere on a transmission line if
we know the receiving end voltage and
current.
In particular, we can obtain the voltage and
current at the sending end, i.e., x=l, to be:
V2
I ( l )  I1  I 2 cosh l 
sinh l
ZC
V (l )  V1  V2 cosh l  ZC I 2 sinh l
Same as eqs. (4.10) in text.
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