1
$\begingroup$

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:

derivation of the square magnitude of PDF

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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/…). $\endgroup$ Dec 2 '20 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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