0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $x$. $\endgroup$ – Marcus Müller Jul 17 '18 at 17:01
  • $\begingroup$ You are all right, edited :). $\endgroup$ – Sama Assi Jul 17 '18 at 17:08
  • $\begingroup$ 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... $\endgroup$ – Fat32 Jul 18 '18 at 13:56
2
$\begingroup$

Absolute summability implies that the Fourier transform converges uniformly, so this means that the unit circle is contained in the region of convergence (ROC).

Now, the pole at 0.5 is the limit of the ROC, so the ROC is from 0.5 outwards (in order to include the unit circle). This means that the signal is right-sided.

Finite support signals cannot have poles (other than $z=0$ or $z=\infty$).

$\endgroup$
  • 1
    $\begingroup$ But it also can be two sided right? $\endgroup$ – Sama Assi Jul 17 '18 at 17:09
  • $\begingroup$ @SamaAssi: No, it can't be two-sided because for that you'd need another pole outside the unit circle. For two-sided signals the ROC is an annulus centered at the origin of the complex plane. $\endgroup$ – Matt L. Jul 17 '18 at 17:12
  • $\begingroup$ 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? $\endgroup$ – Sama Assi Jul 17 '18 at 17:22
  • $\begingroup$ 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. $\endgroup$ – mark leeds Jul 17 '18 at 17:28
  • 2
    $\begingroup$ @SamaAssi: Yes, a two-sided signal can be absolutely summable; my comment was about your question where you have a single pole. Your example of a two-sided signal has two poles, but it is indeed abolutely summable. As long as the unit circle is inside the ROC the signal is absolutely summable, no matter if it's right-sided, left-sided, or two-sided. $\endgroup$ – Matt L. Jul 17 '18 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.