(So far, 13 people got it right out of 25 for a success rate of 52%)

Here is the transfer function for a causal, stable discrete-time system:

And from that we can plot the poles and zeros on the z-plane as follows (the dashed line indicates the unit circle):

Pick one:
This is a minimum phase system!
This is NOT a minimum phase system!

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Comment by July 23, 2024

I remember this same question from my skool daze.  It's because the extra z factor in the denominator is a pure delay and that puts in the pole at z=0 and increases the phase without changing the magnitude response.  If you remove that extra z factor (and pole at z=0), you get a minimum-phase filter.  But a pure delay cannot be minimum phase.

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Comment by July 23, 2024

My issue is with the rule "if all its poles and zeros are inside the unit circle" - we don't normally consider infinite poles and zeros as "poles and zeros." Do we?

I think everywhere this rule is given it should also be clarified that infinite poles or zeros must be considered.

But you're right, the simple fact there is an extra unit delay means it can't be minimum-phase, but then you have to argue based on phase and not "poles and zeros."

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Comment by July 23, 2024

This may help perhaps, at least from an independent knowledgeable source. Julius Orion Smith explicitly mentions poles at infinity. It seems to all add up in terms of purpose and meaning of "minimum phase": https://www.dsprelated.com/freebooks/filters/Defin...

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Comment by July 24, 2024

I propose a new criterion that I believe is simpler to evaluate:

A rational system function

$H(z) = \frac{B(z)}{A(z)}$

is minimum-phase if and only if its poles and zeros are within the region

$0 < |z| < 1$.

This way any poles or zeros at 0, which correspond to zeros or poles at infinity, respectively, and which therefore would result in a non-minimum phase system (or inverse system), are excluded.

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Comment by July 24, 2024

I don't think that will work. That condition where a pole at the origin for a causal stable system corresponds to a zero at infinity only applies to all-pass systems, and specifically the unit delay. We can have a system with a zero at infinity and a pole arbitrarily anywhere within the unit circle which results in a mixed phase system.

A counter-case of the proposed criterion is any other causal stable system with more poles than zeros in the unit circle, for example:

$$H(z) = \frac{z-0.9}{(z-0.7)(z-0.5)}$$

This system has two zeros at z=0.9 and z=infinity (because in the limit at z approaches infinity, H(z) approaches zero), and two poles at z=0.7 and z=0.5.

As for any new criterion, I don't think anything stated by Proakis/Manolakis is incorrect. I propose simply clarifying (if it is a point of confusion) that the criterion that already exists applies to all poles and zeros, not just the finite ones. (And to observe that all systems have the same number of poles and zeros when we include poles and zeros at infinity).  So for the case of causal stable systems, if there are more finite poles than finite zeros, it's not minimum phase. If there are more finite zeros than finite poles, it's not causal. If there are the same number of finite poles and finite zeros, and they are all inside the unit circle, it is "minimum phase" with the practical application that that system has a stable, causal (and min phase) inverse, and it will have the least possible time delay for the given magnitude response.

Am I capturing that all correctly?

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Comment by July 25, 2024

Dan, you are right, that won't work! And it won't work for the reason you state.

Also let me confirm for the record that (of course) Proakis/Manolakis are correct. I had to re-read section 3.3.1 (fourth edition), in which they confirm "poles and zeros" mean both finite and infinite poles and zeros.

This section is separated from that on minimum-phase (5.5.2) and I had forgotten (as is blatantly and embarassingly obvious by now) about infinite poles and zeros.

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Comment by July 25, 2024

Oh thanks for letting me know Randy! I'm glad they said it explicitly; I have the third edition and could not find that (I'll have to look again in that other section), but I do think it's a very good clarification. That said, do you think this made for a good, thought provoking quiz question? Hopefully it causes no embarrassment to anyone.

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Comment by July 25, 2024

You are most welcome, Dan.

Chapter 3 is "The z-Transform and Its Application to the Analysis of LTI Systems", subsection 3 is "Rational z-Transforms", and sub-subsection 3.3.1 is "Poles and Zeros."

Well you caused me to re-read the entire sections on z-transforms, region of convergence, etc., so yeah, it was a great question IMO.

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Comment by August 15, 2024

Dan/All,

I recently ran across this definition of minimum-phase signal in Appendix 13A, p.474, of Probability, Random Variables, and Stochastic Processes, Athanasios~Papoulis, 1991, Third, WCB/McGraw-Hill:

A function

$H(z) = \sum_{n=0}^{\infty} h_n z^{-n}$

is called minimum-phase if it is analytic and its inverse $1/H(z)$ is also analytic for $|z| \ge 1$.

I believe he means analytic in the complex variables sense, i.e., that it satisfies the Cauchy-Riemann conditions for such functions.

It troubles me that this is only so for $|z| \ge 1$, but the definition may be worth some discussion.

--Randy

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Comment by August 16, 2024

So this definition is consistent with not having any poles or zeros for $|z|>1$, right? If there are any poles or zeroes then it isn’t analytic.

It helps me to see this intuitively to consider a FIR system with a single zero at $z=r$ with $|r|<1$. For causality it will necessarily have a pole at the origin. It’s reverse is $z^{-1}H(1/z^*)$ (with that extra delay just to be causal otherwise it would be the pure time reverse filter and conjugated). It’s reverse has the exact same magnitude response but will have a much larger phase deviation (only because of that pesky delay we had to add for it to be causal, otherwise the phase is nearly the same, just conjugated!). We note too that any time we add a zero outside the unit circle that the frequency response when plotted on the complex plane will enclose the origin while if they are all inside the unit circle they won’t enclose the origin at all—- if you graph that it gives great insight into how that point leads to “minimum phase”!

For the graphical description of this last point, see https://dsp.stackexchange.com/a/75013/21048

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Comment by July 22, 2024

If this system is not minimum phase, then Proakis/Manolakis have it wrong (p.357, Digital Signal Processing: Principles, Algorithms, and Applications, John~G.~Proakis and Dimitris~G.~Manolakis, 2007, fourth, Prentice Hall):

Specifically, an IIR system with system function

$H(z) = \frac{B(z)}{A(z)}$

is called minimum phase if all its poles and zeros are inside the unit circle.

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