# Squared magnitude of the Z-transform

I am basically new to the $$z$$-transform and there are some points regarding its square magnitude that I do not understand.

Basically I do not understand how in slide 4 of PDF, they arrive at the following expression of $$|H(z)|^2$$: $$P(z)=|H(z)|^2=H(z)\ast H^*(1/z^*)$$ If we start with $$p(n)=h(n)\ast h(-n)$$ but directly take the $$z$$-transform of the above expression (without substituting $$h(n)$$ by $$h^*(-n)$$ as they do in PDF) we arrive at: $$P(z)=H(z)\ast H(1/z)$$ where I am using the following properties of the $$z$$-transform: $$x[n] \longleftrightarrow X(z)$$ $$x[-n] \longleftrightarrow X(1/z)$$ However, it is has been said before (StackExchange) that this expression ($$P(z)=H(z)\ast H(1/z)$$) is only valid for the unit circle, however I fail to see in which part of the derivation I made the imposition that $$|z|=1$$.

For reference I attach a capture of the lines I am referring to in PDF:

I understand your confusion. What we are looking for is a function of $$z$$ which equals the squared magnitude $$|H(e^{j\omega})|^2$$ of a Fourier transform $$H(e^{j\omega})=\mathcal{F}\{h[n]\}$$ when evaluated on the unit circle $$|z|=1$$. Note that the inverse discrete-time Fourier transform of $$|H(e^{j\omega})|^2$$ is given by the convolution $$h[n]\star h^*[-n]$$, where $$^*$$ denotes complex conjugation. Since the $$\mathcal{Z}$$-transform of $$h^*[-n]$$ is given by $$H^*(1/z^*)$$ we see that the $$\mathcal{Z}$$-transform of $$h[n]\star h^*[-n]$$ is given by

$$\mathcal{Z}\big\{h[n]\star h^*[-n]\big\}=H(z)H^*\left(\frac{1}{z^*}\right)\tag{1}$$

If $$h[n]$$ is real-valued we obviously have $$h^*[-n]=h[-n]$$, and $$(1)$$ is equivalent to

$$\mathcal{Z}\big\{h[n]\star h[-n]\big\}=H(z)H\left(\frac{1}{z}\right)\tag{2}$$

It's important to understand that the equations

$$\big|H(z)\big|^2=H(z)H^*\left(\frac{1}{z^*}\right)\tag{3}$$

and, for real-valued $$h[n]$$,

$$\big|H(z)\big|^2=H(z)H\left(\frac{1}{z}\right)\tag{4}$$

are only valid on the unit circle $$|z|=1$$.

In general we of course have

$$\big|H(z)\big|^2=H(z)H^*(z)\tag{5}$$

but $$(5)$$ is no valid $$\mathcal{Z}$$-transform of any time domain sequence, because it is not analytic. On the other hand, $$(3)$$ and $$(4)$$ are valid $$\mathcal{Z}$$-transforms, namely of the sequence $$h[n]\star h^*[-n]$$. On the unit circle they equal the squared magnitude of the Fourier transform of $$h[n]$$, i.e.,

$$H(z)H^*\left(\frac{1}{z^*}\right)=\big|H(e^{j\omega})\big|^2,\qquad |z|=1\tag{6}$$

Let's look at a simple example. With $$h[n]=\delta[n]+\delta[n-1]$$, we have $$H(z)=1+z^{-1}$$. Since $$h[n]$$ is real-valued we can use Eq. $$(4)$$:

$$H(z)H\left(\frac{1}{z}\right)=(1+z^{-1})(1+z)=z+2+z^{-1}\tag{7}$$

The actual squared magnitude of $$H(z)$$ is

$$\big|H(z)\big|^2=|1+z^{-1}|^2=1+2\textrm{Re}\{z^{-1}\}+|z|^{-2}\tag{8}$$

On the unit circle $$z=e^{j\omega}$$ both $$(7)$$ and $$(8)$$ equal

$$\big|H(e^{j\omega})\big|^2=2+2\cos(\omega)\tag{9}$$

However, the inverse $$\mathcal{Z}$$-transform of $$(7)$$ equals the inverse Fourier transform of $$|H(e^{j\omega})|^2$$, whereas $$(8)$$ is no $$\mathcal{Z}$$-transform.

• Thank you very much @Matt L. that was really helpful and quick. Just to clarify my understanding I have made a new post (I did not want to put it here as a comment because it is rather long and involves some lines of maths):(dsp.stackexchange.com/questions/71738/…). Dec 2 '20 at 9:36