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I am newbie to $\mathcal Z$-transform, I searched to find the magnitude of a function in $z$-domain, but I couldn't find anything, for example when we have $$ H(z) = \frac{z-3}{z-0.5} $$ How do you compute $\lvert H(z)\rvert^2$? I know that $\lvert H(z)\rvert^2 = H(z)\cdot H\left(z^{-1}\right)$, is it right? when I calculate it for my example: $$ H(z)\cdot H\left(z^{-1}\right) = \frac{z-3}{z-0.5}\cdot\frac{z^{-1}-3}{z^{-1}-0.5} = \frac{10-3z-3z^{-1}}{1.25-0.5z-0.5z^{-1}} $$ but it is not a numerical value like Fourier domain.

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  • $\begingroup$ you need to evaluate at $z$ and get the absolute value of that expression $\endgroup$ – percusse Nov 30 '16 at 21:39
  • $\begingroup$ @percusse Sorry, I didn't understand, I did evaluate at z, but what do you mean by getting absolute value of expression? can you help me with the example above? $\endgroup$ – user137927 Dec 1 '16 at 8:20
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The $z$-transform can be evaluated at any point on the complex plane that is also in the ROC of the $z$-transform. To find the magnitude of $H(z)$, you can find the magnitude of numerator and denumerator separately, and then divide the results.

Let's say $H(z)=\frac{A(z)}{B(z)}$ So

$$|H(z)|=\frac{|A(z)|}{|B(z)|}$$ and let's assume $A(z)=z-a$ and $B(z)=z-b$ . $$|A(z)|=|z-a|$$ Let $z=re^{j\phi}$. So $$\begin{align} |A(z)|&=|re^{j\phi}-a|\\&=|(r\cos(\phi)-a)+j r\sin(\phi)|\\ &=\sqrt{(r\cos(\phi)-a)^2+r^2\sin^2(\phi)}\\ &=\sqrt{r^2(\sin^2(\phi)+\cos^2(\phi))+a^2-2ar\cos(\phi)}\\ &=\sqrt{r^2+a^2-2ar\cos(\phi)} \end{align}$$ Similarly $$|B(z)|=\sqrt{r^2+b^2-2br\cos(\phi)}$$ and

$$|H(z)|=\frac{\sqrt{r^2+a^2-2ar\cos(\phi)}}{\sqrt{r^2+b^2-2br\cos(\phi)}}$$

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  • $\begingroup$ $|H(z)|$ is a function of $z$ and hence, a function of $r$ and $\phi$. You cannot find them, except for a given $|H(z)|$. To grasp the idea try plotting this assuming some values for $r$ and $\phi$. $\endgroup$ – msm Dec 1 '16 at 11:59

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