# Magnitude of function in $z$ domain

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.

• you need to evaluate at $z$ and get the absolute value of that expression – percusse Nov 30 '16 at 21:39
• @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? – user137927 Dec 1 '16 at 8:20

## 1 Answer

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

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

• $|H(z)|$ 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 $\phi$. – msm Dec 1 '16 at 11:59