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 2βs2βs 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)Z(s) by Z0Z0.

which we can solve for r(s)r(s) This relationship is called a bilinear transform. For every r(s)r(s) that we can imagine, there is one and only one Z(s)/Z0Z(s)/Z0 and for every Z(s)/Z0Z(s)/Z0 there is one and only one r(s)r(s). What we would like to be able to do, is find Z(s)/Z0Z(s)/Z0, given an r(s)r(s). The reason for this should be readily apparent. Whereas, as we move along in s, Z(s)/Z0Z(s)/Z0 behaves in a most difficult manner (dividing one phasor by another), r(s)r(s) simply rotates around on the complex plane. Given one r(s0)r(s0) it is easy to find another r(s)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)/Z0Z(s)/Z0 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, that “A” represents an impedance of Z(s)/Z0 = 4 + 2jZ(s)/Z0=4+2j. What I would like to do would be to get a grid similar to that on the Z(s)/Z0Z(s)/Z0 plane, but on the r(s)r(s) plane instead. That way, if I knew one impedance (say Z(0)/Z0 = ZL/Z0Z(0)/Z0=ZL/Z0, then I could find any other impedance, at any other ss, by simply rotating r(s)r(s) around by 2βs2βs, and then reading of the new Z(s)/Z0Z(s)/Z0 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 Z(s)/Z0Z(s)/Z0 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)/Z0 + 1Z(s)/Z0+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. 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 ss?

We can substitute for tan φ tanφ: And then, since cos φ − j sin φ = e−jφcosφ−j sinφ=e−jφ

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= 1/d1/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′ = sejΦs′=sejΦ, taking 1/s1/s simply will give us a circle with a diameter of 1/d1/d, which has been rotated by an angle −Φ−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.”

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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 16, 2001 12:00 am -0500.