I found, as rule of thumb, that a system is causal and stable when it poles lies inside the unit circle.

However, more generally we should argue with region of convergence here, like in this example of Wikipedia:


It's hard to see for me, how region of convergence and causality is related. Could somebody bring light into this?

I'm sorry for bringing less self-thoughts about this, I did not find many useful explanations about this topic.


2 Answers 2


There are actually TWO conditions for causality so let's investigate them separately.

But first keep this in mind:

  • For Stability AND Causality: All poles must be inside the unit circle.
  • For only Stability: Unit Circle must be in ROC.

A causal LTI system has an impulse response $h[n]$ that is zero for $n < 0$, and therefore it is a right-sided sequence. As such, if $x[n]$ is a right -sided sequence, and if the circle $|z| = r_0$ is in the ROC, then all finite values of $z$ for which $|z| > r_0$ will also be in the ROC.


$$\frac{1}{1-z^{-1}} \xrightarrow[]{\mathcal{Z}^{-1}} u[n] \quad \textbf{if } \; |z| > 1$$

Plot of $u[n]$:


However, if the ROC was $|z| < 1$ then its IZT would be $-u[-n-1]$ which looks like this:


which is clearly not causal as per the test of causality for LTI systems.

Let's consider stability for a moment and see why the unit circle must be in the ROC?


$$\frac{1}{1-\alpha z^{-1}} \xrightarrow[]{\mathcal{Z}^{-1}} \alpha^n u[n] \quad \textbf{if } \; |z| > |\alpha|$$

Now, if $|\alpha|$ was larger than $1$ would $\alpha^n u[n]$ be stable? Remember that the stability of a discrete-time LTI system is equivalent to its impulse response being absolutely summable. Hence, no it won't be stable.

Now consider the case where $|z| < |\alpha|$ then we have the impulse response:

$$h[n] = -\alpha^n u[-n-1]$$

Here what restriction would you impose to have an absolutely summable impulse response (hint: Look at the second figure above)?

Coming back to causality if the system if causal as described by its impulse response being zero for $n<0$ then we write the Z-Transform as:

$$H(z) = \sum_{n=0}^{\infty} x[n] z^{-n}$$

Notice that this power series does not include any positive powers of $z$. As such, a DT LTI system is causal if and only if the ROC of its system function, $H(z)$,is the exterior of a circle, including infinity. This condition is ONLY for causality and as such the system might be unstable and causal.

The next condition is that if If $H(z)$ is rational, then for the system to be causal, the ROC must be outside the outermost pole and infinity must be in the ROC. Equivalently, the $\lim_{z \rightarrow \infty} H(z)$ must be finite. This is equivalent to the numerator of $H(z)$ having degree no larger than the denominator when both are expressed as polynomials in $z$. These are the only two conditions for causality.

However, when we also bring stability into play (necessary in real life) then we must incorporate the conditions together. Meaning that our ROC must be outside of some circle and that the unit circle must in the ROC! How can we achieve that? By making sure that all poles must be inside the unit circle.

  • $\begingroup$ However, when we also bring stability into play (necessary in real life). Point of order: you can build stable systems with unstable components given the right feedback parameters. When you do, you can both model and measure the unstable components transfer functions. I have, in fact, done it, with real-world components. So I'm not sure where the "necessary" comes from. $\endgroup$
    – TimWescott
    Jun 25, 2023 at 21:48
  • 2
    $\begingroup$ I meant the overall system needs to stable in real life... $\endgroup$ Jun 26, 2023 at 7:59

Knowing the pole locations in the complex plane is generally not sufficient for deciding whether the corresponding system is causal or stable. We need to know the region of convergence to get the complete picture.

For a discrete-time system to be causal, the region of convergence (ROC) must be outside a circle centered at $z=0$. A non-causal system (i.e., a system with an impulse response with non-zero values at positive as well as at negative indices) has a ROC that is an annulus centered at the origin of the complex plane. Finally, for an anti-causal system (with an impulse response that is zero for positive indices), the ROC is inside a circle centered at the origin of the complex plane.

Furthermore, a system is stable if the unit circle is contained in its ROC.

Consequently, a causal and stable system must have an ROC that lies outside a circle centered at $z=0$ AND the unit circle must be inside the ROC. Hence, the ROC must be of the form $|z|>r$ with $r<1$ (such that it includes the unit circle $|z|=1$). Or, equivalently, all poles must lie inside the unit circle.

  • $\begingroup$ Now I am curious, what happens when the ROC is only the unit circle? $\endgroup$ Jun 25, 2023 at 12:55
  • 1
    $\begingroup$ @DanBoschen: I think in that case we can't really talk about a ROC because a region should be two-dimensional. I would say that in such a case the inverse Z-transform doesn't exist, just the IDTFT. Such a sequence is not causal and also not BIBO stable. An example is the impulse response of an ideal low pass filter. $\endgroup$
    – Matt L.
    Jun 25, 2023 at 13:18
  • $\begingroup$ Great answer- thanks $\endgroup$ Jun 25, 2023 at 23:57

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