Bilinear Transform2.132000/09/232007/08/14 09:50:30.480 GMT-5BillWilsonwlw@madriver.netBillWilsonwlw@madriver.netElizabethGregoryelizabeth.gregory@gmail.comJeffreyMSilvermanJSilverman@astro.berkeley.eduGerardWysockigerardw@rice.edubilinear transformIntroduction of bilinear transform.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:
rsΓνθr2βsThe vector
rs 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 as we move down the line. For every
rs there is a corresponding
Zs which is given by:
ZsZ01rs1rs
Now, it turns out to be easier if we talk about a normalized
impedance, which we get by dividing
Zs by
Z0.
ZsZ01rs1rswhich we can solve for
rsrsZsZ01ZsZ01
This relationship is called a bilinear
transform. For every
rs that we can imagine, there is one and only one
ZsZ0 and for every
ZsZ0 there is one and only one
rs. What we would like to be able to do, is find
ZsZ0, given an
rs. The reason for this should be readily
apparent. Whereas, as we move along in s,
ZsZ0 behaves in a most difficult manner (dividing one
phasor by another),
rs simply rotates around on the complex plane. Given one
rs0 it is easy to find another
rs. We just rotate around!
We shall find the required relationship in a
graphical manner. Suppose I have a complex plane, representing
ZsZ0. 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 , "A" represents an impedance of
ZsZ042. What I would like to do would be to get a grid
similar to that on the
ZsZ0plane, but on the
rs plane instead. That way, if I knew one impedence (say
Z0Z0ZLZ0 then I could find any other impedance, at any other
s, by simply rotating
rsaround by
2βs, and then reading off the new
ZsZ0 from the grid I had developed. This is what we shall
attempt to do.
Let's start with and re-write it as:
rsZsZ012ZsZ011-2ZsZ01
In order to use , 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
ZsZ0 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
ZsZ01. 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:
sd1φφφ22
Now we ask ourselves the question: what is the inverse of s?
1s1d11φ
We can substitute for
φ:
1s1d11φφ1dφφφ
And then, since
φφφ1s1dφφ1dφφ
A careful look at should allow you to
convince yourself that is an equation for
a circle on the complex plane, with a diameter
1d. If s is not parallel to
the imaginary axis, but rather has its perpendicular to the
origin at some angle φ, to make a line
s′. Since
s′sφ, taking
1s simply will give us a circle with a diameter of
1d, which has been rotated by an angle
φ 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."