# How do I find the ROC of a system if it has no poles

The output of a system of discrete time $y[n]$ is corellated with the input $x[n]$ through the equation $y[n]$.

$$y[n] = \frac 13\big(x[n-1]+x[n]+x[n+1]\big)$$

It then asks me to find the system function, the ROC, the zeroes and the poles.

I found out that $$Y(z) = \frac 13\left(z^{-1}X(z)+X(z)+zX(z)\right)$$

I found out that the system function is : $$H(z) = \frac {Y(z)}{X(z)} = \frac 1{3z}(z^{2}+z+1)$$

The pole here is 0. How do I tell if this sequence is left/right/two sided to figure out the ROC?

This is an unsolved exercise given by my professor.

• I don't see how you arrive at the third equation from your second equation. And also note that a system always has poles if you include poles at zero and/or infinity. – Matt L. Jan 30 '18 at 21:27
• Thanks for your answer. It seems I have forgotten to do division. I fixed it so the pole is zero. How do I figure out the ROC now? – thelaw Jan 30 '18 at 21:32
• You have two poles, and the ROC is in between those poles. – Matt L. Jan 30 '18 at 21:33
• Is the second pole infinity? And if so is it plus or negative infinity? I don't understand how to tell if this sequence is left/right/two sided. – thelaw Jan 30 '18 at 21:35

Since your time domain equation is of finite length, its Z-transform sum converges for (almost) any given $z$. Hence, the ROC of your system function spans the entire complex plane except for
• $z=0$, which is included only if your function is anti-causal and contains only negative potencies $z^{-n}$ which evaluate as $\frac{1}{z^n} \to \infty$ for $z=0$, or
• $|z| = \infty$, which will only be in the ROC for purely causal functions. Causal components $z^n$ will approach infinite values when $z$ is $\pm\infty$ (or any complex number of infinite radius), hence the absolute value.
Since your function has both causal and anti-causal parts, the sequence converges for $0 < |z| < \infty$. The region is limited by the two poles at $z=0$ and $|z|=\infty$.
• @thelaw: How is $-\infty<|z|<0$ supposed to make sense??? I hoped it would be clear that the magnitude $|z|$ of a complex number satisfies $|z|\ge 0$. – Matt L. Jan 31 '18 at 7:52