Consider a discrete-time causal and stable LTI system $S_1$​. The inverse system $S_2$​ is defined as the system that takes the output of $S_1$​ as its input and provides the input of $S_1$​ as its output. What can we infer about $S_2$?

I believe that since $S_1$ is stable we know that its poles lie inside the unit circle, therefore the zeroes of $S_2$ will lie inside the unit circle. (So $S_2$ is min-phase If I'm not mistaken)

Other than that what can we say about $S_2$'s poles and causality status?

  • 1
    $\begingroup$ Could you go into more detail about what you don't understand after having read this answer to a very related question? $\endgroup$
    – Matt L.
    Commented May 23 at 15:13
  • 1
    $\begingroup$ You gotta worry about the zeros of $S_1$. $S_1$ must not only be stable (poles inside the unit circle), but must also be minimum-phase (zeros inside the unit circle), because the transfer function of $S_2$ is the reciprocal of the transfer function of $S_1$ and then the zeros of $S_1$ become the poles of $S_2$. $\endgroup$ Commented May 23 at 16:09
  • $\begingroup$ @MattL. what I don't understand is if we just know that $S_1$ is causal and stable (nothing more, we don't know the location of its zeroes), then what do we know about $S_2$ other than its zeroes being inside the unit circle? e.g. Is $S_2$ also causal? Will $S_2$'s poles be inside and/or outside the unit circle? $\endgroup$
    – Nyquist-er
    Commented May 23 at 18:26
  • 2
    $\begingroup$ @Nyquist-er: Causality and stability of $S_1$ only tell you something about its poles, not about its zeros. Consequently, you only have information about the zeros of $S_2$ (they're inside the unit circle). From the zeros alone you can't say anything about causality or stability. $\endgroup$
    – Matt L.
    Commented May 23 at 18:38
  • $\begingroup$ @MattL. Thank you so much! This is exactly the answer I was looking for. I didn't realize that causality was related to the poles. If it's not too much trouble, could you explain how causality is connected to the poles? $\endgroup$
    – Nyquist-er
    Commented May 23 at 19:52

1 Answer 1


If a given system with a rational transfer function is causal and stable, we know that its poles must lie inside the unit circle. Consequently, the inverse system, whose zeros are the poles of the original system and vice versa, has all its zeros inside the unit circle. This tells us nothing about stability or causality of the inverse system.

Stability and causality are determined by the poles. Note that the poles and zeros (and the gain constant) generally don't uniquely characterize a system. Dependent on the region of convergence (ROC) of the transfer function, we get different systems with the same algebraic expression of the transfer function.

E.g., a causal system has a ROC that lies outside the pole with the largest magnitude. If that pole lies outside the unit circle, the system is unstable. However, the same transfer function can represent a stable system if we can find a ROC that includes the unit circle, but that system is non-causal.

In general, we can distinguish the following three cases:

  1. The ROC is an annulus between two poles. The corresponding system has a two-sided impulse response (i.e., it is non-causal). If the unit circle is inside the ROC, the system is stable.

  2. The ROC lies outside the pole with the largest magnitude. The corresponding system is causal. It is only stable if that pole has a magnitude less than $1$.

  3. The ROC is inside a circle whose radius is determined by the pole with the smallest magnitude. The corresponding system is anti-causal (left-sided). It is only stable if all poles lie outside the unit circle.

If the inverse system is required to be causal and stable, all its poles must lie inside the unit circle. Consequently, not only the poles but also all zeros of the original system must lie inside the unit circle. Systems with all its poles and zeros inside the unit circle are called minimum-phase systems. Only minimum-phase systems have causal and stable inverses.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.