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 :

The vector r⁢s 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 2 β s as we move down the line. For every r⁢s r s there is a corresponding Z⁢s 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 Z⁢s Zs by Z 0 Z 0 .

which we can solve for r⁢s r s 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 Z s Z 0 and for every Z⁢s Z 0 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 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 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 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 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 Fig.2, that "A" represents an impedance of Z⁢s Z 0 =4+2⁢j Z s Z 0 4 2 j . What I would like to do would be to get a grid similar to that on the Z⁢s Z 0 Z s Z 0 plane, but on the r(s) plane instead. That way, if I knew one impedance (say Z(0) Z 0 = Z L Z 0 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 2βs2βs, and then reading of the new Z⁢s Z 0 Z s Z 0 from the grid I had developed. This is what we shall attempt to do.

Let's start with the above equation 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 Z⁢s 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 Z⁢s 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, 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:

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

We can substitute for tanφ φ : And then, since cosφ−j⁢sinφ=e−(j⁢φ) φ j φ j φ

A careful look at Fig.4 should allow you to convince yourself that Equation 9 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 ΦΦ, to make a line s ′ s . Since s′=s⁢ej⁢Φ s s j Φ , taking 1/s simply will give us a circle with a diameter of 1/d, which has been rotated by an angle ΦΦ 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."

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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 (Unknown) on Aug 15, 2002 12:00 am -0500.