There is a way that we can make things a good bit easier for ourselves however. The only drawback is that we have to do some complex analysis first, and look at a bilinear transform! Let's do one more substitution, and define another complex vector, which we can call r(s):
rs≡
Γ
v
ej(
θ
r
−2βs)
r
s
Γ
v
j
θ
r
2
β
s
(1)
The vector r(s) is just the rotating part of the crank diagram which we have been looking at. It has a magnitude equal to that of the reflection coefficient, and it rotates around at a rate
2βs2βs as we move down the line. For every r(s) there is a corresponding Z(s) which is given by:
Zs=
Z
0
1+rs1−rs
Z
s
Z
0
1
r
s
1
r
s
(2)
Now, it turns out to be easier if we talk about a normalized impedance, which we get by dividing Z(s) by
Z
0
Z
0
.
Zs
Z
0
=1+rs1−rs
Z
s
Z
0
1
r
s
1
r
s
(3)
which we can solve for
rs
r
s
rs=Zs
Z
0
−1Zs
Z
0
+1
r
s
Z
s
Z
0
1
Z
s
Z
0
1
(4)
This relationship is called a
bilinear transform. For every r(s) that we can imagine, there is one and only one
Zs/
Z
0
Z
s
/
Z
0
and for every
Zs/
Z
0
Z
s
/
Z
0
there is one and only one r(s). What we would like to be able to do, is find
Zs/
Z
0
Z
s
/
Z
0
, given an r(s). The reason for this should be readily apparent. Whereas, as we move along in s,
Zs/
Z
0
Z
s
/
Z
0
behaves in a most difficult manner (dividing one phasor by another), r(s) simply rotates around on the complex plane. Given one
r
s
0
r
s
0
it is
easy to find another r(s). We just rotate around!
We shall find the required relationship in a graphical manner. Suppose I have a complex plane, representing
Zs/
Z
0
Z
s
/
Z
0
And then suppose I have some point "A" on that plane and I want to know what impedance it represents. I just read along the two axes, and find that, for the example in Fig.2, that "A" represents an impedance of
Zs/
Z
0
=4+2j
Z
s
/
Z
0
4
2
j
. What I would like to do would be to get a grid similar to that on the
Zs/
Z
0
Z
s
/
Z
0
plane, but on the r(s) plane instead. That way, if I knew one impedance (say
Z(0)/
Z
0
=
Z
L
/
Z
0
Z(0)
/
Z
0
Z
L
/
Z
0
then I could find any other impedance, at any other s, by simply rotating r(s) around by 2βs2βs, and then reading of the new
Zs/
Z
0
Z
s
/
Z
0
from the grid I had developed. This is what we shall attempt to do.
Let's start with Equation 4 and rewrite it as:
rs=Zs
Z
0
+1−2Zs
Z
0
+1=1+2Zs
Z
0
+1
r
s
Z
s
Z
0
1
2
Z
s
Z
0
1
1

2
Z
s
Z
0
1
(5)
In order to use Equation 5, we are going to have to interpret it in a way which might seem a little odd to you. The way we will read the equation is to say: "Take
Zs/
Z
0
Z
s
/
Z
0
and add 1 to it. Invert what you get, and multiply by 2. Then add 1 to the result." Simple isn't it? The only hard part we have in doing this is inverting
Zs/
Z
0
+1
Z
s
/
Z
0
1
. This, it turns out, is pretty easy once we learn one very important fact.
The one fact about algebra on the complex plane that we need is as follows. Consider a vertical line, s, on the complex plane, located a distance d away from the imaginary axis. There are a lot of ways we could express the line s, but we will choose one which will turn out to be convenient for us. Let's let:
s=d(1−jtanφ)..........forπ2≤φ≤π2
s
d
1
j
φ
..........
for

π
2
≤
φ
≤
π
2
(6)
Now we ask ourselves the question: what is the inverse of s?
1s=1d
11−jtanφ
1
s
1
d
1
1
j
φ
(7)
We can substitute for
tanφ
φ
:
1s=1d
11−jsinφcosφ
=1d
cosφcosφ−jsinφ
1
s
1
d
1
1
j
φ
φ
1
d
φ
φ
j
φ
(8)
And then, since
cosφ−jsinφ=ejφ
φ
j
φ

j
φ
1s=1d
cosφejφ
=1d(cosφejφ)
1
s
1
d
φ

j
φ
1
d
(
φ
j
φ
)
(9)
A careful look at Fig.4 should allow you to convince yourself that Equation 9 is an equation for a circle on the complex plane, with a diameter = 1/d. If s is not parallel to the imaginary axis, but rather has its perpendicular to the origin at some angle ΦΦ, to make a line
s
'
s
'
. Since
s
'
=sejΦ
s
'
s
j
Φ
, taking 1/s simply will give us a circle with a diameter of 1/d, which has been rotated by an angle ΦΦ from the real axis. And so we come to the one fact we have to keep in mind: "The inverse of a straight line on the complex plane is a circle, whose diameter is the inverse of the distance between the line and the origin."