Product of a minimum phase signal times its inverse

Question

I got a problem stated as H(z) is the z trasform of h[n], which is a real, stable, even, two sided sequence. H(z)=G(z)*G(z^-1), where g[n] is a causal, minimum phase signal. Shouldn't the product of G(z) and its inverse equal to 1, the identity system? Did I miss out something?

Answer 1

What you missed is the fact that generally $$G(z^{-1})\neq G^{-1}(z)$$ So the inverse of the system $G(z)$ is $G^{-1}(z)$ since clearly $G(z) \cdot G^{-1}(z)=G(z)\cdot \frac{1}{G(z)}=1$ which is the identity system.

However, $G(z^{-1})$ means all the poles and zeros in $G(z)$ are inverted. So for example if $G(z)=\frac{z-z_0}{z-p_0}$ (with a pole $p_0$ and a zero $z_0$), then $G(z^{-1})=\frac{z^{-1}-z_0}{z^{-1}-p_0}$, whose poles and zeros are $\frac{1}{p_0}$ and $\frac{1}{z_0}$.