Prove the following property of All-Pass Filters

Question

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?

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 $z=0$.