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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 $a+bi$ ?

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  • $\begingroup$ 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
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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

$$H(z)=\sum_{n=-\infty}^{\infty}h[n]z^{-n}=\sum_{n=-\infty}^{\infty}h^*[n]z^{-n}=\left[H(z^*)\right]^*\tag{1}$$

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

$$H(z)=\sum_{n=-\infty}^{\infty}h[n]z^{-n}=\sum_{n=-\infty}^{\infty}h[-n]z^{-n}=\sum_{n=-\infty}^{\infty}h[n]z^{n}=H\left(\frac{1}{z}\right)\tag{2}$$

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 $1/z_1^*$

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