Let's suppose I have to find the impulse response of a discrete time LTI system given a specified input and its output through the system. I think I'm going to get the $\mathcal Z$-transform of input-output and then as everyone knows $$H(z)=\frac{Y(z)}{X(z)}$$ The problem is: I want it in time domain, so I need to inverse transform $H(z)$. How will I choose the ROC (hence causality / noncausality)?

  • $\begingroup$ It's better to give a specific example. Anyway, when you have $H(z)$, you can determine the ROC by looking at the poles locations (in the complex plane) such that in the ROC their must be no poles. $\endgroup$ Jun 15, 2017 at 11:26

1 Answer 1


The function $H(z)$ without a specified ROC is generally not sufficient for uniquely determining the time domain sequence $h[n]$. You need more information about the desired $h[n]$. There are three types of ROCs:

  1. the ROC is an annulus ($R_1<|z|<R_2$): $h[n]$ is a two-sided sequence (non-causal)
  2. the ROC is defined by $|z|>R$: $h[n]$ is a right-sided sequence
  3. the ROC is defined by $|z|<R$: $h[n]$ is a left-sided sequence

If you're looking for a stable system, then the ROC needs to include the unit circle $|z|=1$. If that does not uniquely define the ROC, then you need to know if you're looking for a left-sided, a right-sided, or a two-sided sequence.

  • $\begingroup$ this is to say: one pair input/output isn't enough to determine impulse response (how the system really works). I am not very sure about it... isn't the system linear? $\endgroup$ Jun 15, 2017 at 13:57
  • $\begingroup$ @Surferonthefall: If you just have the Z-transforms (without ROC), then there's usually no unique solution. But if you know that the system is causal then the ROC is implied and you get a unique solution. $\endgroup$
    – Matt L.
    Jun 15, 2017 at 15:06
  • $\begingroup$ Since you have the input and output in the time domain, the region of convergence for both $Y(z)$ and $X(z)$ are known. $\endgroup$
    – msm
    Jun 16, 2017 at 17:03

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.