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suppose $y(n)=ax(n-1)+bx(n-2)+\dots$ ($y$ is the output and $x$ the input). What happens if I want to solve $x(n)$ from $y(n)$?

Z transform: $$Y(z)=G(z)X(z)\tag{1}$$ then $$X(z)=\frac{1}{G(z)}Y(z)\tag{2}$$

What are the properties of $1/G(z)$ ? If $(1)$ is causal what is the status of the inverse $(2)$? The roles of the poles and zeros have changed.

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As you've pointed out, inversion leads to poles at locations of the zeros of the original transfer function and vice versa. Assuming that $G(z)$ is causal and stable (i.e., it has all its poles inside the unit circle), we have to distinguish $3$ cases:

  1. $G(z)$ has at least some zeros outside the unit circle. This means its inverse has some poles outside the unit circle, and consequently, it cannot be causal and stable. If $G(z)$ has no zeros on the unit circle, there exists a stable impulse response corresponding to $1/G(z)$ but it cannot be causal. This is because a transfer function does not uniquely determine an impulse response. We can get different impulse responses corresponding to different regions of convergence of $1/G(z)$.

  2. $G(z)$ has some of its zeros on the unit circle. No stable inverse exists because $1/G(z)$ has poles on the unit circle.

  3. $G(z)$ has all its zeros inside the unit circle, i.e., it is a minimum-phase system. Consequently, $1/G(z)$ also has all its poles and zeros inside the unit circle (i.e., it is minimum-phase), and can be implemented by a causal and stable system.

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  • $\begingroup$ I see, let me explain further...G(z) has only one pole at z=0 and the magnitudes of all the zeros lie outside the unit circle (nothing on) - when I interpret the poles of (1/G(z)) in closed form - is it the anti-causal interpretation (a)^n u(-n-1)? Since y(n) is calculated in terms of past x(n) - but now x(n) is in terms of past/future y(n)? $\endgroup$
    – Michelle
    Commented Dec 13, 2017 at 7:43
  • $\begingroup$ @Michelle: If $1/G(z)$ has all its poles outside the unit circle, you get a stable anti-causal filter. $\endgroup$
    – Matt L.
    Commented Dec 13, 2017 at 7:54
  • $\begingroup$ Relief! The world is at piece again! Thanx very much. $\endgroup$
    – Michelle
    Commented Dec 13, 2017 at 7:58
  • $\begingroup$ Is it correct to say that a system is BIBO stable if (and only if) G(z) 's poles are all contained within the unit circle (not on) and all 1/G(z) 's poles are outside the unit circle? $\endgroup$
    – Michelle
    Commented Dec 13, 2017 at 12:30
  • $\begingroup$ @Michelle: No, BIBO stability just means that the output is bounded for any bounded input signal. For a rational transfer function that means that there must be no poles on the unit circle. If you further require causality, all poles must be inside the unit circle. BIBO stability says nothing about the zeros of the transfer function, and, consequently, nothing about the poles of the inverse system. $\endgroup$
    – Matt L.
    Commented Dec 13, 2017 at 12:49
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If G(z) is causal and minimum phase than 1/G(z) is causal and minimum phase as well.

Non-minimum phase systems have a non-causal inverse. Simple example: a 2-tap delay has in inverse that is "-2" taps of delay, i.e. it is non-causal.

In addition you need $|G(z)|> 0$, otherwise you have division by zero problems. In practice you need $|G(z)|> \epsilon$, where $\epsilon$ is a suitable small positive number, to avoid numerical and signal to noise problems.

This corresponds to the simple fact that whenever $|G(z)| = 0$ for some z, information is lost in the system and can therefore not be recovered.

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  • $\begingroup$ What you said about $|G(z)|$ being non-zero is only valid for $|z|=1$. A system with $G(z)=0$ for $|z|<1$ can perfectly be inverted by a causal and stable system. $\endgroup$
    – Matt L.
    Commented Dec 12, 2017 at 12:53

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