Bilinear Transform **new** 2003/06/16 16:39:08.275 GMT-5 2003/06/17 01:44:37.790 GMT-5 Bill Wilson wlw@rice.edu Bill Wilson wlw@rice.edu Elizabeth Gregory lizzardg@rice.edu Jeffrey Silverman jsilv@rice.edu bilinear transform Introduction 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 r s : r s Γ ν θ r 2 β s 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 2 β s as we move down the line. For every r s there is a corresponding Z s which is given by: Z s Z 0 1 r s 1 r s

The Vector r(s)

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 s Z 0 1 r s 1 r s which we can solve for r s r s Z s Z 0 1 Z s Z 0 1 This relationship is called a bilinear transform. For every r s that we can imagine, there is one and only one Z s Z 0 and for every Z s Z 0 there is one and only one r s . What we would like to be able to do, is find Z s Z 0 , given an r s . The reason for this should be readily apparent. Whereas, as we move along in s, 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 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 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 , "A" represents an impedance of Z s Z 0 42 . What I would like to do would be to get a grid similar to that on the Z s Z 0 plane, but on the r s plane instead. That way, if I knew one impedence (say 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 2 β s , and then reading off the new Z s Z 0 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: r s Z s Z 0 1 2 Z s Z 0 1 1 -2 Z s Z 0 1 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 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 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 φ φ φ 2 2

A Vertical Line, s, a Distance, d, Away From the Imaginary Axis

Now we ask ourselves the question: what is the inverse of s? 1 s 1 d 1 1 φ We can substitute for φ : 1 s 1 d 1 1 φ φ 1 d φ φ φ And then, since φ φ φ 1 s 1 d φ φ 1 d φ φ

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 1 d . 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 ′ s φ , taking 1 s simply will give us a circle with a diameter of 1 d , 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."

The Line s' The line s multiplied by φ

Inverse of a Rotated Line