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
ν
|ei(
θ
r
−2βs)
r
s
Γ
ν
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

Figure 1. 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
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

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

ss,

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
42
. 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 impedence (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
ss, by simply rotating
rs
r
s
around by
2βs
2
β
s
, and then reading off 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 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, ss, on the complex
plane, located a distance dd away
from the imaginary axis Figure 3. There are a lot
of ways we could express the line
ss, but we will choose one which
will turn out to be convenient for us. Let's let:

s=d(1−itanφ)
∀
φ
:φ∈
−π2
π2
s
d
1
φ
φ
φ
2
2

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

1s=1d11−itanφ
1
s
1
d
1
1
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

1d
1
d
. If

ss 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
′
Figure 5. Since

s
′
=seiφ
s
′
s
φ
, taking

1s
1
s
simply will give us a circle with a diameter of

1d
1
d
, which has been rotated by an angle

φφ from the real axis

Figure 6. 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."*
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