Is $h[0] \neq 0$ a sufficient condition for the invertibility of a discrete, LTI, causal system? Can we get to similar results (i.e. get to some other sufficient condition(s)) for noncausal or anticausal systems?

  • $\begingroup$ There are lots of posts on the invertibility of systems on this site. This is one of many examples. $\endgroup$
    – Matt L.
    Feb 1, 2022 at 11:43

2 Answers 2


Any maximum phase discrete causal LTI system does not have a stable, causal inverse since to be maximum phase implies the system will have zeros outside the unit circle, which become poles once inverted.

A simple counter example of a maximum phase system with $h[0] \neq 0$ is:

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

Where $h[0]=1$

From this explanation we can deduce a sufficient condition for invertibility for causal systems, and then from that for non-causal systems as well which I will leave as an exercise (hint review Region of Convergence in how to provide the answer for non-causal systems once the solution for causal systems is understood).

  • $\begingroup$ I see that for the inverse to be causal and stable, the poles and zeros of the first system need to be inside the unit circle. However, what happens if we allow the inverse to be possibly noncausal and unstable? Can we talk about an inverse in that case? If the system is anticausal (no point of h in the positives), I understand that the poles need to be outside the unit circle, but will the inverse need to necesserily be anticausal, and stable, too, to deduce anything about the zeros? $\endgroup$
    – Dkpink
    Feb 1, 2022 at 10:34
  • $\begingroup$ If the system is anti-causal AND stable then the poles need to be outside the unit circle. The question of invertibility is the same: Where are the zeros? If there are zeros inside the unit circle (for that case) then there will not be a stable, anti-causal inverse. In accessing stability of inverses, it is all about the zeros prior to being inverted. The poles can be anywhere. This is the same as considering the stability of any system prior to inverting; the zeros can be anywhere, and it is all about the poles. $\endgroup$ Feb 1, 2022 at 12:39

Here is one useful way to look at. You can split every system into a minimum phase and an all-pass system, i.e.

$$H(\omega) = H_{min}(\omega) \cdot H_{ap}(\omega) = H_{min}(\omega) \cdot e^{j\phi(\omega)}, \phi \in \mathbb{R}$$

A minimum phase system is invertible and also minimum phase, no problem here. Let's invert the allpass

$$H^{-1}_{ap}(\omega) = \frac{1}{e^{j\phi(\omega)}} = e^{-j\phi(\omega)} = H^*_{ap}(\omega)$$

So inversion of an all pass is the same as complex conjugation. That results in a conjugate time reversal of the impulse response.

$$h_{ap,inv}[n] = h^*_{ap}[-n]$$

So unless the original system is minimum phase, the inverse will be non-casual. In practice you can often handle this by just adding sufficient amount of bulk delay.

Things are more tricky if you have zeros on the unit circle, i.e. $H(\omega_1) = 0 $ for some frequencies. This filter will "destroy" information at the zero frequencies and it's not recoverable, simply because $|H^{-1}(\omega_1)| \rightarrow \infty $


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