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 :

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: 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 . which we can solve for rs r s 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: 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:

Now we ask ourselves the question: what is the inverse of s? We can substitute for tanφ φ : And then, since cosφ−ⅈsinφ=ⅇ-ⅈφ φ φ φ 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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This work is licensed by Bill Wilson under a Creative Commons Attribution License (CC-BY 1.0), and is an Open Educational Resource.

Last edited by Raymond Wagner on Aug 14, 2007 9:50 am GMT-5.