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Bilinear Transform

Module by: Bill Wilson. E-mail the author

Summary: Introduction of bilinear transform.

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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):

rs| Γ v |ej( θ r 2βs) r s Γ v j θ r 2 β s
(1)

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 2βs2βs as we move down the line. For every r(s) there is a corresponding Z(s) which is given by:

Zs= Z 0 1+rs1rs Z s Z 0 1 r s 1 r s
(2)

Figure 1: The vector r(s).
Figure 1 (7.17.png)

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 0 .

Zs Z 0 =1+rs1rs Z s Z 0 1 r s 1 r s
(3)
which we can solve for rs r s
rs=Zs Z 0 1Zs Z 0 +1 r s Z s Z 0 1 Z s Z 0 1
(4)
This relationship is called a bilinear transform. For every 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 r(s). What we would like to be able to do, is find Zs/ 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, Zs/ 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 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 Fig.2, that "A" represents an impedance of Zs/ Z 0 =4+2j Z s / Z 0 4 2 j . 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 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 Zs/ Z 0 Z s / Z 0 from the grid I had developed. This is what we shall attempt to do.

Figure 2: The complex impedance plane
Figure 2 (7.18.png)

Let's start with Equation 4 and re-write it as:

rs=Zs Z 0 +12Zs Z 0 +1=1+-2Zs Z 0 +1 r s Z s Z 0 1 2 Z s Z 0 1 1 - 2 Z s Z 0 1
(5)

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, 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(1jtanφ)..........for-π2φπ2 s d 1 j φ .......... for - π 2 φ π 2
(6)

Figure 3: A Vertical line, s, a distance, d, away from the imaginary axis
Figure 3 (7.19.png)

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

1s=1d 11jtanφ 1 s 1 d 1 1 j φ
(7)
We can substitute for tanφ φ :
1s=1d 11jsinφcosφ =1d cosφcosφjsinφ 1 s 1 d 1 1 j φ φ 1 d φ φ j φ
(8)
And then, since cosφjsinφ=e-jφ φ j φ - j φ
1s=1d cosφe-jφ =1d(cosφejφ) 1 s 1 d φ - j φ 1 d ( φ j φ )
(9)

Figure 4: A plot of 1/s
Figure 4 (7.20.png)

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 ' =sejΦ 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."

Figure 5: The line s multiplied by ejΦ j Φ
Figure 5 (7.21.png)
Figure 6: Inverse of rotated line
Figure 6 (7.22.png)

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