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)


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.

| improve this answer | |
  • $\begingroup$ Would you mind explaining the reasoning behind ? Thanks for the answer! $\endgroup$ – J. Barbosa Jun 1 '16 at 10:50
  • $\begingroup$ @J.Barbosa: I've added an explanation to my answer. $\endgroup$ – Matt L. Jun 1 '16 at 11:39

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