# Squared magnitude of System Function H(z)

If:

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

Then, why does:

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

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

It's important to understand that the equation

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

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

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

For real-valued systems we have $$H(z)=H^*(z^*)$$ and, consequently, $$(1)$$ is valid for real-valued systems. Note that both $$(1)$$ and $$(2)$$ are only valid on the unit circle.

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:

$$\sum_nh^*[-n]z^{-n}=\left(\sum_nh[n](z^*)^{n}\right)^*=H^*\left(\frac{1}{z^*}\right)\tag{3}$$

• for "complex valued system" and "real valued system" are your referring to h[n]? Jan 20 '19 at 18:30
• (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? Jan 20 '19 at 18:32
• Yes, that's the meaning of real-valued and complex-valued systems. Jan 20 '19 at 18:33
• And yes to your other question. Jan 20 '19 at 18:34
• 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. Jan 20 '19 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:

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