0
$\begingroup$

I am looking for the inverse $\mathcal Z$-transform of the following:

$$ \frac{1}{1-\frac 12 z^{-1}}+\frac{1}{1+\frac 13 z^{-1}} $$

When the region of convergence is $z > 1/3$. I have found the $\mathcal Z$-transform for when $z > 1/2$ as a right sided signal:

$$ \left(\frac{1}{2}\right)^n u[n] + \left(-\frac{1}{3}\right)^n u[n] $$

I can't seem to find a inverse $\mathcal Z$-transform when the ROC goes outwards from the inner pole. Does the inverse $\mathcal Z$-transform not exist for $z>1/3$?

$\endgroup$
0
$\begingroup$

There exist $3$ sequences with the given expression as their $\mathcal{Z}$-transform. These $3$ sequences correspond to $3$ different regions of convergence (ROCs):

  1. $|z|>\frac12$: right-sided
  2. $\frac13<|z|<\frac12$: two-sided
  3. $|z|<\frac13$: left-sided

It's important that you figure out both right-sided and left-sided inverse $\mathcal{Z}$-transforms of the basic transform

$$X(z)=\frac{1}{1+az^{-1}}$$

and their respective ROCs. Then it's easy to find all $3$ inverse transforms for the given example.

$\endgroup$
  • $\begingroup$ Yeah that's what I thought. I found the inverse z transforms for the 3 cases above, but was also asked to find it for an additional fourth case for lzl>1/3, but I suppose it does not exist. Thanks! $\endgroup$ – Mark Tian Feb 23 '18 at 7:20

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.