Skip to content Skip to navigation


You are here: Home » Content » Bilinear Transform


Recently Viewed

This feature requires Javascript to be enabled.


(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.

Bilinear Transform

Module by: Bill Wilson. E-mail the author

Summary: Introduction of bilinear transform.

Note: You are viewing an old version of this document. The latest version is available here.

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

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

Figure 1: The vector r(s).
Figure 1 (717.png)

Now, it turns out to be easier if we talk about a normalized impedance, which we get by dividing Zs Zs by Z 0 Z 0 .

Zs Z 0 =1+rs1rs Z s Z 0 1 r s 1 r s
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
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 (718.png)

Let's start with the above equation 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

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 2 φ π 2

Figure 3: A Vertical line, s, a distance, d, away from the imaginary axis
Figure 3 (719.png)

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

1s=1d 11jtanφ 1 s 1 d 1 1 j φ
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
And then, since cosφjsinφ=e(jφ) φ j φ j φ
1s=1d cosφe(jφ) =1d cosφ ejφ 1 s 1 d φ j φ 1 d φ j φ

Figure 4: A plot of 1/s
Figure 4 (720.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 (721.png)
Figure 6: Inverse of rotated line
Figure 6 (722.png)

Content actions

Download module as:

Add module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens


A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks