There
is a way that we can
make things a good bit easier for ourselves however. The onlydrawback 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 cancall
:
The vector
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 itrotates around at a rate
as we move down the line. For every
there is a corresponding
which is given by:
The vector r(s)
Now, it turns out to be easier if we talk about a
normalized
impedance , which we get by dividing
by
.
which we can solve for
This relationship is called a
bilinear
transform . For every
that we can imagine, there is one and only one
and for every
there is one and only one
. What we would like to be able to do, is find
, given an
. The reason for this should be readily
apparent. Whereas, as we move along in
,
behaves in a most difficult manner (dividing one
phasor by another),
simply rotates around on the complex plane. Given one
it is
easy to find another
. We just rotate around!
We shall find the required relationship in a
graphical manner. Suppose I have a complex plane, representing
. And then suppose I have some point "A" on that plane
and I want to know what impedance it represents. I just readalong the two axes, and find that, for the example in
, "A" represents an impedance of
. What I would like to do would be to get a grid
similar to that on the
plane, but on the
plane instead. That way, if I knew one impedence (say
then I could find any other impedance, at any other
, by simply rotating
around by
, and then reading off the new
from the grid I had developed. This is what we shall
attempt to do.
The complex impedance plane
Let's start with
and re-write it as:
In order to use
, we are going to have to
interpret it in a way which might seem a little odd to you. Theway we will read the equation is to say: "Take
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 hardpart we have in doing this is inverting
. 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 avertical line,
, on the complex
plane, located a distance
away
from the imaginary axis
. There are a lot
of ways we could express the line
, but we will choose one which
will turn out to be convenient for us. Let's let:
A vertical line, s, a distance, d, away from the
imaginary axis
Now we ask ourselves the question: what is the inverse of s?
We can substitute for
:
And then, since
A plot of 1/s
A careful look at
should allow you to
convince yourself that
is an equation for
a circle on the complex plane, with a diameter
. If
is not parallel to
the imaginary axis, but rather has its perpendicular to theorigin at some angle
, to make a line
. Since
, taking
simply will give us a circle with a diameter of
, which has been rotated by an angle
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 diameteris the inverse of the distance between the line and the
origin."
