Say I have two minimum-phase filters:

$$\frac{A(z)}{B(z)} \: \text{ and } \: \frac{C(z)}{D(z)}$$

That is, the roots of $A(z)$, $B(z)$, $C(z)$ and $D(z)$ are all in the stable region.

If add them together:

$$\frac{A(z)}{B(z)} + \frac{C(z)}{D(z)} = \frac{A(z)D(z) + C(z)B(z)}{B(z) D(z)}$$

The result is causal, and stable (because the roots of $B(z) D(z)$ are also in the stable region). But it might not be minimum-phase, because the numerator $A(z)D(z) + C(z)B(z)$ can (in general) have zeroes outside the stable region.

What conditions are required for this summed filter to also be minimum-phase?

EDIT: I found a sufficient but not necessary condition here, but that's not quite what I was hoping for


2 Answers 2


Just looking at it I can see it's related to control theory. If you make a contrived open-loop gain $$H(z) = \frac{A(z)D(z)}{B(z)C(z)}$$ and then wrap it with unity-gain feedback, you get $$G(z) = \frac{A(z)B(z)C(z)D(z)}{A(z)D(z) + B(z)C(z)}$$.

So any stability test that takes $H(z)$ as an open-loop gain will equally be a test to see if your sum is a minimum-phase system. If you have some parameter variation then you could use root-locus plots, Bode plot or Nyquist plot analysis, or robust control methods to determine the amount of variation that could be tolerated while keeping the system minimum phase.

As an example, if you let your transfer function be $$k\frac{A(z)}{B(z)} + \frac{C(z)}{D(z)}$$ then you could easily cast this as a root-locus problem, and with a bit of effort you could put it into Bode plot form or feed it into a robust control solution that's designed to work in the $z$ domain.

There's probably a more direct path from here to there than going through control theory, but my hammer says "control theory" on it, and by gosh your problem is looking a lot like a nail to me!


I don't think you will have much luck there. Minimum phase means that all the roots of all polynomials $A,B,C,D$ are inside the unit circle. That means that the product of two polynomials will also have their roots inside the unit circle so $A \cdot D$, $B \cdot C$, and $B \cdot D$ are all good.

However it's NOT good for $A \cdot D + B \cdot C$. The roots of a sum of polynomials is an iffy problem and there is no easy way to express them as the roots of the individual polynomials. See https://math.stackexchange.com/questions/1789320/roots-of-sum-of-two-polynomials-with-known-roots

I'm afraid that you need to slog through this one at a time on a case by case basis

  • $\begingroup$ So you're saying a complete answer might not exist, just a bunch of separate sufficient-but-not-necessary individual cases? $\endgroup$
    – cloudfeet
    Nov 30, 2020 at 20:17
  • $\begingroup$ What I'm saying that for each individual case you have to plug in the actual numbers, calculate the roots of the numerator polynomial and determine whether they are all inside the unit circle or not. $\endgroup$
    – Hilmar
    Dec 1, 2020 at 13:30
  • $\begingroup$ Yeah - there are definitely some general cases where we know the answer without doing that, as I linked in the question, but they're only a partial answer. $\endgroup$
    – cloudfeet
    Dec 2, 2020 at 18:41

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