If A(z) is an all-pass filter given by

$$A(z)=\frac{z^{-1}-d_1}{1-d_1 z^{-1}}$$ where $d_1$ is a real coefficient

Then prove that $$ |A(z)|\left\{ \begin{aligned} &<1 &|z|>1 \\ &=1 &z=1 \\ &>1 &|z|<1 \end{aligned} \right. $$ Note: Most of the proofs available for this property assume complex coefficients but it should be valid for real coefficients too right?

  • $\begingroup$ Anyways, the way forward seems to be writing out what $\lvert A(z)\rvert$ is (as in: how do you calculate the absolute of a complex value?). Or, you could first simply try the $z=0$ special case (after cancelling $z^{-1}$ from your fraction). Quite possibly doing that already helps you refine your approach to ask an easier question than "prove for me that..." $\endgroup$ Sep 10, 2021 at 8:57
  • $\begingroup$ The same question was also answered here. $\endgroup$
    – Matt L.
    Sep 10, 2021 at 9:51
  • $\begingroup$ Yeah...I saw that....but the elegant solution posted by @ZR Han below perfectly answered my question.... $\endgroup$
    – Orpheus
    Sep 10, 2021 at 11:30

1 Answer 1


Of course it's valid for real coefficients because real number is a subset of complex number and the proof doesn't make any assumption whether the coefficient is complex or real.

For a complex $d$ $$ \begin{aligned} |A(z)|^2 &= A(z)A^*(z) = \frac{1-d^*z}{z-d} \frac{1-dz^*}{z^*-d^*}\\ &=\frac{1-(d^*z+dz^*)+|d|^2|z|^2}{|z|^2 - (d^*z+dz^*) + |d|^2} \end{aligned} $$ We know that the numerator and denominator of the above equation are both non-negative. To compare it with 1, subtract denominator with numerator $$ 1 + |d|^2|z|^2 - |z|^2-|d|^2 = (1-|d|^2)+ (|d|^2-1)|z|^2 = (1-|d|^2)(1-|z|^2) $$

If $d$ is real, then $$ \begin{aligned} |A(z)|^2 &= A(z)A^*(z) = \frac{1-dz}{z-d} \frac{1-dz^*}{z^*-d}\\ &=\frac{1-d(z+z^*)+d^2|z|^2}{|z|^2 - d(z+z^*) + d^2} \end{aligned} $$ Numerator minus denominator equals to $$ 1 + d^2|z|^2 - |z|^2-d^2 = (1-d^2)+ (d^2-1)|z|^2 = (1-d^2)(1-|z|^2) $$

For a stable all-pass filter we have $|d|<1$, so the sign of subtraction is determined by $1-|z|^2$, i.e., if

  1. $|z| > 1$, $|A(z)| < 1$
  2. $|z| = 1$, $|A(z)| = 1$ (that's an all-pass!)
  3. $|z| < 1$, $|A(z)| > 1$

It seems there is something wrong with your question, it is easy to see that $|A(z)|=-1/d\neq 0$ when $z=0$.

  • $\begingroup$ Thanks a lot, @ZR Han!!!!!!!! $\endgroup$
    – Orpheus
    Sep 10, 2021 at 9:40
  • $\begingroup$ @Orpheus Glad to help :) $\endgroup$
    – ZR Han
    Sep 10, 2021 at 9:42
  • $\begingroup$ I think there is a small correction....in the second equation I think the numerator should be $1-dz-d^{*}z^{*}+|d|^2|z|^2$ $\endgroup$
    – Orpheus
    Sep 10, 2021 at 13:25
  • $\begingroup$ @Orpheus corrected. The second equation is correct. When the coefficient is complex, the transfer function of an all-pass filter should be $H(z)=(1-d^*z)/(z-d)$ so that the zero and pole are reciprocal. $\endgroup$
    – ZR Han
    Sep 10, 2021 at 13:53
  • $\begingroup$ ...noted...thanks a million! $\endgroup$
    – Orpheus
    Sep 10, 2021 at 14:41

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