Unfortunately, since we don't know what value
the phasor
V
+
V
+
has, these equations do not do us a whole lot of good!
One way to deal with this is to simply divide this equation into this equation. That gets
rid of
V
+
V
+
and the
eiβs
β
s
and so we now come up with a *new*
variable, which we shall call line impedance,
Zs
Z
s
.

Zs≡VsIs=
Z
0
1+
Γ
ν
e-2iβs1−
Γ
ν
e-2iβs
Z
s
V
s
I
s
Z
0
1
Γ
ν
-2
β
s
1
Γ
ν
-2
β
s

(1)
Zs
Z
s
represents the ratio of the total voltage to the total
current anywhere on the line. Thus, if we have a line

LL long, terminated with a load
impedance

Z
L
Z
L
, which gives rise to a terminal reflection coefficient

Γ
ν
Γ
ν
, then if we substitute

Γ
ν
Γ
ν
and

LL into

Equation 1, the

ZL
Z
L
which we calculate will be the "apparent" impedance
which we would see looking into the input terminals to the line!

There are several ways in which we can look at
Equation 1. One is to try to put it into a more
tractable form, that we might be able to use to find
Zs
Z
s
, given some line impedance
Z
0
Z
0
, a load impedance
Z
L
Z
L
and a distance, ss away
from the load. We can start out by multiplying top and bottom by
eiβs
β
s
, substituting in for
Γ
ν
Γ
ν
, and then multiplying top and bottom by
Z
L
+
Z
0
Z
L
Z
0
.

Zs=
Z
0
(
Z
L
+
Z
0
)eiβs−
Z
L
e−(iβs)(
Z
L
+
Z
0
)eiβs−(
Z
L
−
Z
0
)e−(iβs)
Z
s
Z
0
Z
L
Z
0
β
s
Z
L
Z
0
β
s
Z
L
Z
0
β
s
Z
L
Z
0
β
s

(2)
Next, we use Euler's relation, and substitute

cosβs±isinβs
±
β
s
β
s
for the exponential. Lots of things will cancel out,
and if we do the math carefully, we end up with

Zs=
Z
0
Z
L
+i
Z
0
tanβs
Z
0
+i
Z
L
tanβs
Z
s
Z
0
Z
L
Z
0
β
s
Z
0
Z
L
β
s

(3)
For some people, this equation is more satisfying than

Equation 1, but for me, both are about equally opaque in
terms if seeing how

Zs
Z
s
is going to behave with various loads, as we move down
the line towards the generator.

Equation 3
*does* have the nice property that it is easy
to calculate, and hence could be put into MATLAB or a
programmable calculator. (In fact you could program

Equation 1 just as well for that matter.) You could specify
a certain set of conditions and easily find

Zs
Z
s
, but you would not get much insight into how a
transmission line actually behaves.