# Inverse $\mathcal Z$-transform when region of convergence goes outwards from the inner pole?

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 $z>1/3$?

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.

• 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! – Mark Tian Feb 23 '18 at 7:20