Let's say I have a $\mathcal Z$-transform that represents some transfer function and its has some ROC.

  • My question is how do I know if this system is causal?

I know that if the ROC contains the unit circle in the complex plane the system is stable and it has DTFT but I don't know how I can evaluate the causality. This also applies to the continuous domain, I believe the system is causal if the ROC is right sided, for example, $\Re(\sigma)>1$ but I also don't know why. (All of this for LTI systems)


1 Answer 1


If the ROC is outside a circle in the complex $z$-plane ($|z|>a$), then the corresponding system is causal. If it is inside a circle ($|z|<a$), the system is anti-causal. If the ROC is a ring ($a<|z|<b$), the impulse response of the corresponding system is two-sided, i.e., the system is non-causal. In all cases, if the ROC contains the unit circle, the system is stable.

Consider the causal sequence $x[n]=a^nu[n]$ (where $u[n]$ is the discrete-time unit step sequence). The bilateral $\mathcal{Z}$-transform is defined by


Since $x[n]$ is causal we get

$$X(z)=\sum_{n=0}^{\infty}a^nz^{-n}=\sum_{n=0}^{\infty}\left(\frac{a}{z}\right)^n=\frac{1}{1-\left(\frac{a}{z}\right)}=\frac{z}{z-a},\qquad \left|\frac{a}{z}\right|<1\tag{2}$$

where I've used the formula for the geometric series. Note that the series only converges for $|a|<|z|$, i.e., outside a circle in the complex plane.

In a completely analogous manner we get for an anti-causal sequence such as $x[n]=a^nu[-n]$

$$X(z)=\sum_{n=-\infty}^{0}a^nz^{-n}=\sum_{n=0}^{\infty}a^{-n}z^n=\sum_{n=0}^{\infty}\left(\frac{z}{a}\right)^n=\frac{1}{1-\left(\frac{z}{a}\right)}=\frac{a}{a-z},\qquad \left|\frac{z}{a}\right|<1\tag{3}$$

which converges for $|z|<|a|$, i.e., inside a circle in the complex plane.

Finally, a two-sided sequence can be split up into a causal and an anti-causal sequence, with an ROC that is the overlap of the two individual ROCs, which, if it is not empty, is a ring in the complex plane.

  • $\begingroup$ Would you mind explaining the reasoning behind ? Thanks for the answer! $\endgroup$
    – J. Barbosa
    Commented Jun 1, 2016 at 10:50
  • $\begingroup$ @J.Barbosa: I've added an explanation to my answer. $\endgroup$
    – Matt L.
    Commented Jun 1, 2016 at 11:39

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.