$$ |\alpha|^2 = \alpha \alpha^* $$

Then, why does:

$$ |H(z)|^2 = H(z) H(z^{-1}) $$

instead of:

$$ |H(z)|^2 = H(z) H^*(z) $$


It's important to understand that the equation


is only valid on the unit circle, i.e., for $z=e^{j\omega}$. For complex-valued systems the general form of $(1)$ is


For real-valued systems we have $H(z)=H^*(z^*)$ and, consequently, $(1)$ is valid for real-valued systems.

The reason why we use $(1)$ or $(2)$ as a generalization of $|H(e^{j\omega})|^2$ to the whole complex plane is that we want an analytic function that is actually a $\mathcal{Z}$-transform of some sequence. And the function $H(z)H^*(z)$ does not satisfy that requirement.

Note that the IDFT of $H^*(e^{j\omega})$ is $h^*[-n]$. So if we find the $\mathcal{Z}$-transform of $h^*[-n]$ we have what we need:


  • $\begingroup$ for "complex valued system" and "real valued system" are your referring to h[n]? $\endgroup$ – MrCasuality Jan 20 at 18:30
  • $\begingroup$ (1) is valid for real-valued systems, but only on unit circle... would the same apply for complex valued system, (2) is only valid on unit circle? $\endgroup$ – MrCasuality Jan 20 at 18:32
  • $\begingroup$ Yes, that's the meaning of real-valued and complex-valued systems. $\endgroup$ – Matt L. Jan 20 at 18:33
  • $\begingroup$ And yes to your other question. $\endgroup$ – Matt L. Jan 20 at 18:34
  • $\begingroup$ analytic requirement because the magnitude of $|H(z)|^2$ won't converge on a value because it has poles on the z-plane that will make the magnitude infinite... thus, we can only find magnitude on unit circle that doesn't have any poles for realizable system. $\endgroup$ – MrCasuality Jan 20 at 18:43

if h[n] is real, then H(z) is conjugate symmetric:

$$ H(z) = H^*(z^*) $$

Or equivalently:

$$ H^*(z) = H(z^*) $$

Applying definition of H(z) symmetry to magnitude:

$$|\alpha|^2=\alpha\ \alpha^*$$

$$|H(z)|^2= H(z) H^*(z)$$

$$|H(z)|^2= H(z) H(z^{*})$$

I'm at a lost for why $z^* = z^{-1}$.

For a stable system:

$$ |H(e^{j\omega})|^2 = \left[ H(z)\ H^*\left(\frac{1}{z^*}\right)\right]_{z=e^{j\omega}} $$

Then if H(z) is restricted to be real:

$$ |H(e^{j\omega})|^2 = \left[ H(z)\ H\left(z^{-1}\right)\right]_{z=e^{j\omega}} $$


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