# z-plane representation of single order all-pass filter

Started by September 3, 2014
I'm used to the s-plane representation of an all-pass filter, where the pole(s) are on the left-side of the y-axis, and the zeroes are mirror imaged across the y-axis. It's fairly clear that as omega increases (increasing frequency) in s = j * omega, that the length of the vectors that go from the y-axis to the mirror imaged poles and zeroes will be the same length (per pole-zero pair).

However, I do not grasp how this is possible on the z-plane, unless the pole and zero are right on top of each other. Using the bilinear transformation, the s-plane right hand side maps to the exterior of the unit circle in the z-plane, so it would imply that a zero on the right hand side of the s-plane will be outside the unit circle on the z-plane. But since z = e^(j * omega) is traversing a circular path, how is it possible that there are any two points on the plane whose magnitude ratio stays constant?

Just looking directly at the z-transform, it seems clear. I am simply having a bit of difficulty grasping it on the z-plane itself.
Try looking at page 274-275 in "Discrete Time Signal Processing", 2nd ed, O&S.The digital allpass function is (5.90):Hap(z) = (z^-1 - a*) / (1 - az^-1)It is shown in (5.91) that it has unity magnitude. For Hap(z) to be a causal, the pole must be inside the unit circle and the zero at the conjugate reciprocal location.You can rearrange Hap(z) to: Hap(z) = -a* (z - 1/a*) / (z -a)The magnitude is then:
|Hap(z)| = |-a*| |z - 1/a*| / |z -a|So the vectors |z - 1/a*| and |z -a| should not be equal length but there is a constant factor between them.Since it was shown that |Hap(z)| = 1, it proves that the ratio between the distance from e^jw to the zero and the distance from e^jw to the pole, is a constant.Figure 5.21 is a nice example to look at. You can for instance try comparing the length of the vectors of zero/pole pair (2, ½) to e^j0, e^j(pi/2), e^j(pi). You will see that each time the ratio is ½.I hope this helps you.
/Kim
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To: a...
Date: Tue, 2 Sep 2014 14:25:20 -0700
Subject: [audiodsp] z-plane representation of single order all-pass filter

I'm used to the s-plane representation of an all-pass filter, where the pole(s) are on the left-side of the y-axis, and the zeroes are mirror imaged across the y-axis. It's fairly clear that as omega increases (increasing frequency) in s = j * omega, that the length of the vectors that go from the y-axis to the mirror imaged poles and zeroes will be the same length (per pole-zero pair).

However, I do not grasp how this is possible on the z-plane, unless the pole and zero are right on top of each other. Using the bilinear transformation, the s-plane right hand side maps to the exterior of the unit circle in the z-plane, so it would imply that a zero on the right hand side of the s-plane will be outside the unit circle on the z-plane. But since z = e^(j * omega) is traversing a circular path, how is it possible that there are any two points on the plane whose magnitude ratio stays constant?

Just looking directly at the z-transform, it seems clear. I am simply having a bit of difficulty grasping it on the z-plane itself.