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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?

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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. $(2)$:

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

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  • $\begingroup$ 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?? $\endgroup$
    – engr
    Dec 24 '19 at 16:37
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    $\begingroup$ @engr: Please see my edited answer. $\endgroup$
    – Matt L.
    Dec 25 '19 at 9:02
  • $\begingroup$ 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? $\endgroup$
    – engr
    Dec 25 '19 at 16:50
  • $\begingroup$ @engr: The minus sign wouldn't make any difference for the ROC. $\endgroup$
    – Matt L.
    Dec 25 '19 at 17:26

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