Given that:

  • $H(z)$ has 4 poles maximum.
  • $H(z)$ has a pole at $z_1=a+bi$

Given that the impulse response $h[n]$ is:

  • Symmetric: $h[n] = h[-n]$
  • Real: $\forall$$n$ , $h[n]$$\in$$\mathbb{R}$

How we can conclude that the other poles are the inverse and the complex reflection of $a+bi$ ?

  • $\begingroup$ Can you explain what they mean by a "real and symmetric transfer function"? It looks like the corresponding impulse response is real-valued and symmetric. $\endgroup$ – Matt L. Jul 16 '18 at 9:22
  • $\begingroup$ I've edited the question. $\endgroup$ – Sama Assi Jul 16 '18 at 10:02

The transfer function $H(z)$ is the $\mathcal{Z}$-transform of the impulse response $h[n]$. If $h[n]$ is real-valued, i.e., if $h[n]=h^*[n]$ we have


So if $z_1$ is a pole of $H(z)$, then $z_1^*$ must also be a pole.

If furthermore $h[n]$ is symmetric, i.e., if $h[n]=h[-n]$ we have


which means that if $z_1$ is a pole, then $1/z_1$ must also be a pole.

In sum, if $h[n]$ is real-valued and symmetric you get for each complex-valued pole $z_1$ three additional poles at $z_1^*$, $1/z_1$, and $1/z_1^*$


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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