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An s-Plane to z-Plane Mapping Example

Rick LyonsSeptember 24, 201610 comments

While surfing around the Internet recently I encountered the 's-plane to z-plane mapping' diagram shown in Figure 1. At first I thought the diagram was neat because it's a good example of the old English idiom: "A picture is worth a thousand words." However, as I continued to look at Figure 1 I began to detect what I believe are errors in the diagram.

Reader, please take a few moments to see if you detect any errors in Figure 1.

            Figure 1: One (incorrect) Internet 's-plane to z-plane
                           mapping' diagram.

I have drawn what I think is a corrected version of Figure 1. Given the various loci of points in Figure 1's s-plane, my version of a correct 's-plane to z-plane mapping' diagram shown in Figure 2.

           Figure 2: Corrected 's-plane to z-plane mapping' diagram.

Hopefully my Figure 2 is worth 1001 words. If there are any errors in that figure I hope a perceptive reader lets me know.



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Comment by Tim WescottSeptember 23, 2016
Just eyeballing things -- aren't you both wrong on the constant damping factor line? The angle of departure from z = 1 should be the same as the angle of departure from s = 0. In fact, for any exact mapping -- and therefore any decent approximate mapping -- the behavior of the poles close to z = 1 should just be a scaled and offset version of the behavior of the poles close to s = 0.
[ - ]
Comment by Rick LyonsSeptember 23, 2016
You are mistaken. The constant damping factor locus of points (circular light purple curve) on the z-plane is correct. That line starts at a frequency of zero and the frequency increases in the negative-frequency direction. As the frequency goes more and more negative the line's magnitude (e^-0.5) does not change.

Your phrase " The angle of departure from z = 1" makes no sense to me because the constant damping factor locus of points (circular light purple curve) on the z-plane does NOT intersect the z = 1 point.
[ - ]
Comment by yatesSeptember 28, 2016
Tim, I agree with Rick on the constant damping factor line.
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Comment by yatesSeptember 28, 2016
Rick, are you sure the mapping the original diagram used was z = e^S and not z = e^S* (conjugate)? You didn't provide a reference to the original diagram so I can't check.

--Randy
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Comment by Rick LyonsOctober 2, 2016
Hi Randy. OK, I'm back at my office. I double checked the web page that presented the original mapping diagram. That original mapping diagram was indeed based on the definition z = e^s.
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Comment by Rick LyonsSeptember 28, 2016
Hi Randy. I'm away from my office for a few days. I can't remember the reference off the top of my head, but I'll get back to you regarding your question.
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Comment by itxsMarch 19, 2021

Hi. Thanks to useful picture. But the bilinear transform is conformal transform, so angle between dark-purple and blue / dark-purple and green must be 45 degrees, as in source picture.

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Comment by Rick LyonsMarch 19, 2021

I'm sorry itxs. I don't understand what "angle between dark-purple and blue / dark-purple and green must be 45 degrees, as in source picture" means.

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Comment by andyp71September 27, 2023

Hi, I am not itxs, but let me clarify a bit:

This means that bilinear transform is a conformal map, so angles between lines should be preserved. Angle formed by the intersecting dark-purple and blue lines in the red point on the right should be equal to 45 degrees, as on the left picture.

One more thing concerns me a bit. Bilinear transform should map lines into lines or circles and circles - into circles or lines. Dark purple line does not meet this requirement.

Hope this helps.


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Comment by sdohertyDecember 2, 2020

Great diagram. I teach a course in digital control and students tend to struggle with s to z-plane mapping. Picture definitely worth 100 words. Especially like the units of the jw in terms of pi which seems more understandable that units of ws. Thanks for sharing

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