So far, the transmission lines we have looked at have been
"ideal". That is they have been lossless and
dispersionless. Lest you leave the course with a false idea of
how things really work, we should go back
to our model and try to get things adjusted just a bit.
As you can probably imagine, a real transmission line is going
to have some series resistance, associated with the real losses
in the copper wire. There may also be some shunt conductance,
if the insulating material holding the two conductors has some
leakage current. We will need to include these effects along
with the distributed inductance and capacitance which we have
already talked about. Fixing up the model accordingly, we now
draw a section of line
Δx
Δ
x
long as shown in Figure 1. Taking the
same voltage loop and current sum that we did back in the
discussion of transmission
lines, we come up with the following version of the
telegrapher's equations.
∂∂xVxt=-RIxt-L∂∂tIxt
x
V
x
t
R
I
x
t
L
t
I
x
t
(1)
and
∂∂xIxt=-GVxt+C∂∂tVxt
x
I
x
t
G
V
x
t
C
t
V
x
t
(2)
Clearly, we would like to simplify things if we can. Let's
again make a sinusoidal time excitation assumption, and let
Ixt
I
x
t
and
Vxt
V
x
t
become phasors. Since the time variation is now
represented by a simple
ⅇⅈωL
ω
L
the time derivatives become just
ⅈω
ω
. We have
∂∂xVx=-R+ⅈωLIx
x
V
x
R
ω
L
I
x
(3)
and
∂∂xIx=-G+ⅈωCVx
x
I
x
G
ω
C
V
x
(4)
The way to get a solution is, of course, just like we have
always done. Take the derivative with respect to
xx of Equation 3
∂2∂x2Vx=-R+ⅈωL∂∂xIx
x
2
V
x
R
ω
L
x
I
x
(5)
and then plug in
Equation 4
∂2∂x2Vx=R+ⅈωLG+ⅈωCVx
x
2
V
x
R
ω
L
G
ω
C
V
x
(6)
The obvious solution to this (See how easy this gets after
you've done it once or twice) is
Vx=
V
0
ⅇ±γx
V
x
V
0
±
γ
x
(7)
with
γ=R+ⅈωLG+ⅈωC
γ
R
ω
L
G
ω
C
(8)
This number, γγ is called
the complex propagation constant. Obviously, in
general, it will have both a real and an imaginary part:
γ=α+ⅈβ
γ
α
β
(9)
and we have
Vx=
V
0
ⅇ±α+ⅈβx
V
x
V
0
±
α
β
x
(10)
Let's choose the minus sign in the exponent, and write the two
terms as a product.
Vx=
V
0
ⅇ-αxⅇ-ⅈβx
V
x
V
0
α
x
β
x
(11)
We see we have something similar to what we had before, but with
just a minor difference. The
ⅇ-ⅈβx
β
x
term is the propagating term which tells us how the
phase angle of the phasor changes as we move along the line, and
acts just like the
ββ term we
had before. Thus
β=2πλ
β
2
λ
(12)
and
ν
p
=ωβ
ν
p
ω
β
(13)
The αα is called the
attenuation coefficient, and obviously, the
ⅇ-αx
α
x
term in Equation 11 causes the
amplitude of the wave to decrease as it moves down the line.
Figure 2 is a sketch of what a wave
would look like if it is both propagating down the transmission
line and also being attenuated. In a distance
1α
1
α
the amplitude of the propagating wave has fallen to
ⅇ-1
-1
of the value it had when it started.
Let's take the minus sign solution in Equation 7 and substitute back into Equation 3
∂∂xVx=-γ
V
0
ⅇ-γx=-R+ⅈωLIx
x
V
x
γ
V
0
γ
x
R
ω
L
I
x
(14)
From which we get
Ix=γR+ⅈωL
V
0
ⅇ-γx=R+ⅈωLG+ⅈωCR+ⅈωLVx=G+ⅈωCR+ⅈωLVx
I
x
γ
R
ω
L
V
0
γ
x
R
ω
L
G
ω
C
R
ω
L
V
x
G
ω
C
R
ω
L
V
x
(15)
Thus we can say
Vx=
Z
0
Ix
V
x
Z
0
I
x
(16)
where
Z
0
=R+ⅈωLG+ⅈωC=
R
0
+ⅈ
X
0
Z
0
R
ω
L
G
ω
C
R
0
X
0
(17)
In general, in order to find
αα,
ββ,
R
0
R
0
, and
X
0
X
0
, we would have to find the square root given in Equation 8 and Equation 17 for specific values of
RR,
LL,
GG, and
CC. On the other hand, we could
maybe come up with some reasonable approximations which might
suffice for cases of real interest. Obviously, if a line is very
lossy, we would not be very interested in using it, and so
except in some very special cases where an extremely lossy line
is unavoidable (usually having to do with signals at very high
frequencies) we might see if we can find a low loss
approximation.
