# Absolutely summable signals

Given an absolutely summable signal $x[n]$, the $z$-transform $X^z(z)$ is rational with a pole at $z=0.5$.

Given the following the statements:

• $x[n]$ has a finite support in the time domain.
• $x[n]$ is a left sided signal.
• $x[n]$ is a right sided signal.
• $x[n]$ is a two sided signal.

Can anyone please tell which statements are true and explain why or why not?

• I know I'm a nitpicker, but systems have transfer functions, signals don't. But, that's just semantics; in the end, all we care about is that $X$ is the $z$-Transform of $x$. – Marcus Müller Jul 17 '18 at 17:01
• You are all right, edited :). – Sama Assi Jul 17 '18 at 17:08
• These kind of questions should be worded carefully to remove undecidable ambiguity as much as possible and to clearly underline the range of decidable unknowns about the quantity... – Fat32 Jul 18 '18 at 13:56

Finite support signals cannot have poles (other than $z=0$ or $z=\infty$).
• Since the signal is absolutely summable we can conclude that z=1 is inside the ROC, so if we choose for example the signal $x[n]=\frac{1}{2}^{|n|}$ it would be two sided with a suitable ROC no? – Sama Assi Jul 17 '18 at 17:22
• Hi: the z-transform ( which is really just the closed form of the infinite series ) of your example is $\frac{1}{1- 0.5 z}$ if you sum from $0$ to $\infty$.. If you sum in the other direction, -n, the infinite series doesn't converge. So, I'm not absolutely sure but I think that's why it's called right sided. Someone hopefully will confirm. – mark leeds Jul 17 '18 at 17:28