# Prove the following property of All-Pass Filters

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?

• 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..." Sep 10, 2021 at 8:57
• The same question was also answered here. Sep 10, 2021 at 9:51
• Yeah...I saw that....but the elegant solution posted by @ZR Han below perfectly answered my question.... Sep 10, 2021 at 11:30

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 $$z=0$$.

• Thanks a lot, @ZR Han!!!!!!!! Sep 10, 2021 at 9:40
• @Orpheus Glad to help :) Sep 10, 2021 at 9:42
• I think there is a small correction....in the second equation I think the numerator should be $1-dz-d^{*}z^{*}+|d|^2|z|^2$ Sep 10, 2021 at 13:25
• @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. Sep 10, 2021 at 13:53
• ...noted...thanks a million! Sep 10, 2021 at 14:41