# How we can find ROC(Region of convergence) given a signal in z domain?

I read that ROC is a region,which is a set of values where z transform is defined ,that is it converges

Lets say i have a discrete time time signal $$x[n]=n^2 u(n)$$ and i want to find its ROC(Region of convergence),how can i do that?

You need to determine the values of the complex variable $$z$$ for which the series

$$\sum_{n=1}^{\infty}n^2z^{-n}\tag{1}$$

converges. So you need to know a few things about infinite series.

For this case a simple test is the ratio test. You take the ratio of two successive elements of the series and compute the limit:

$$L=\lim_{n\to\infty}\left|\frac{(n+1)^2z^{-(n+1)}}{n^2z^{-n}}\right|=\lim_{n\to\infty}\left|\frac{(n+1)^2}{n^2}\right|\left|z^{-1}\right|=\frac{1}{|z|}\tag{2}$$

The series converges absolutely for $$L<1$$, i.e., for $$1/|z|<1$$, or $$|z|>1$$. So the ROC is the region $$|z|>1$$.

Note that the ROC would not change if we used any other power of $$n$$ in $$(1)$$.

EDIT:

A step for step explanation of Eq. $$(2)$$:

\begin{align}L&=\lim_{n\to\infty}\left|\frac{(n+1)^2z^{-(n+1)}}{n^2z^{-n}}\right|\\&=\lim_{n\to\infty}\left|\frac{(n+1)^2}{n^2}\right|\left|\frac{z^{-(n+1)}}{z^{-n}}\right|\\&=\lim_{n\to\infty}\left|\frac{(n+1)^2}{n^2}\right|\left|z^{-1}\right|\\&=\frac{1}{|z|}\lim_{n\to\infty}\left|\frac{(n+1)^2}{n^2}\right|\\&=\frac{1}{|z|}\lim_{n\to\infty}\left(\frac{n^2+2n+1}{n^2}\right)\\&=\frac{1}{|z|}\lim_{n\to\infty}\left(1+\frac{2}{n}+\frac{1}{n^2}\right)\\&=\frac{1}{|z|}\left(1+\underbrace{\lim_{n\to\infty}\frac{2}{n}}_{0}+\underbrace{\lim_{n\to\infty}\frac{1}{n^2}}_{0}\right)\\&=\frac{1}{|z|}\end{align}

• Please kindly elaborate eq(2) right side ,how you reach/arrived at 1/|z|,especillay considering how you removed limit and put limit equal to 1??
– engr
Dec 24 '19 at 16:37
• @engr: Please see my edited answer. Dec 25 '19 at 9:02
• What if we had a minus sign in eq(1) before n2, ROC will be same that is |z|>1 or will it be opposite |z|<1?
– engr
Dec 25 '19 at 16:50
• @engr: The minus sign wouldn't make any difference for the ROC. Dec 25 '19 at 17:26