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
rs
r
s
:

rs≡|
Γ
v
|ei(
θ
r
−2βs)
r
s
Γ
v
θ
r
2
β
s

(1)
The vector
rs
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
2βs
2
β
s
as we move down the line. For every
rs
r
s
there is a corresponding
Zs
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
Zs
Zs
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

rs
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

rs
r
s
. What we would like to be able to do, is find

Zs
Z
0
Z
s
Z
0
, given an

rs
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),

rs
r
s
simply rotates around on the complex
plane. Given one

r
s
0
r
s
0
it is

*easy* to find another

rs
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 Figure 2, "A" represents an impedance of
Zs
Z
0
=4+2i
Z
s
Z
0
4
2
. 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
rs
r
s
plane instead. That way, if I knew one impedance (say
Z0
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
rs
r
s
around by
2βs
2
β
s
, 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 the above
equation and re-write 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:

∀φ,−π2≤φ≤π2:s=d(1−itanφ)
φ
2
φ
π
2
s
d
1
φ

(6)
Now we ask ourselves the question: what is the inverse of s?

1s=1d11−jtanφ
1
s
1
d
1
1
j
φ

(7)
We can substitute for

tanφ
φ
:

1s=1d11−isinφcosφ=1dcosφcosφ−isinφ
1
s
1
d
1
1
φ
φ
1
d
φ
φ
φ

(8)
And then, since

cosφ−isinφ=e−(iφ)
φ
φ
φ
1s=1dcosφe−(iφ)=1dcosφeiφ
1
s
1
d
φ
φ
1
d
φ
φ

(9)
A careful look at Figure 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′=seiΦ
s
s
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."*