We are given a transfer function of a causal LTI system and must determine if it is an all-pass filter:

$$ H(z) = \frac{1 + 4z^{-2}}{4 - z^{-2}} $$

To the best of my recollection to determine if a system is an all-pass filter, we must first find the poles and zeros and see if they are conjugate reciprocal pairs. So working on that and finding the poles and zeros, we find that...

the poles are at $z = \tfrac{1}{2}, -\tfrac{1}{2}$

the zeros are at $z = 2j, -2j$

With that being said the poles and zeros are no conjugate reciprocal pairs. That being said, are there any exceptions or other criteria that would make a system an all-pass filter?


1 Answer 1


In an all-pass filter we should have $$H(z)H^*(1/z^*)=c^2$$ which means that the poles and zeros of the system $H^*(1/z^*)$ cancel the zeros and poles of $H(z)$. As a result, an all-pass filter has conjugate-reciprocal pole-zero pairs. For this system:

enter image description here

$$H^*(1/z^*)=\frac{4z^2+1}{-z^2+4}$$ So $$H(z)H^*(1/z^*)\neq c^2$$ You can also see that the poles and zeros are not conjugate and reciprocal: $\frac{1}{\left(\pm2j\right)^*}\neq\pm\frac{1}{2}$

  • $\begingroup$ equality in the last line ? $\endgroup$
    – Fat32
    Apr 29, 2017 at 21:42
  • $\begingroup$ $\frac{1}{(2j)^*} = \frac{1}{2}$ is it right ? :-) $\endgroup$
    – Fat32
    Apr 29, 2017 at 22:14
  • $\begingroup$ Thank you very much for the in-depth analysis. It makes sense how the conjugate and reciprocal aren't equivalent after looking at your plot. $\endgroup$ Apr 29, 2017 at 23:20
  • $\begingroup$ You're welcome @AndreYonadam $\endgroup$
    – msm
    Apr 29, 2017 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.