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Evaluate the Z transform of $x[n] = n^3$ where the signal is two sided.

I have tried using the basic definition of the Z transform ie.,

$$X(z) \triangleq \sum_\limits{n=-\infty}^{+\infty} x[n] \, z^{-n} $$

But what I get obviously is an infinite unbounded sum of sequence. Does that mean Z transform does not exist? IS there any alternate methods to obtain the transform? Please help.

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  • $\begingroup$ so who gave you that assignment? $\endgroup$ – robert bristow-johnson Oct 2 '18 at 0:59
  • $\begingroup$ Hi, Robert it was a question that I found in "Modern Control Engineering",K Ogata $\endgroup$ – Shehin A U Oct 2 '18 at 1:18
  • $\begingroup$ with $n^3$ getting bigger and bigger, without bound, i don't know how K Ogata can expect the summation to converge to a finite value no matter what $z$ is. $\endgroup$ – robert bristow-johnson Oct 2 '18 at 1:56
  • $\begingroup$ But in the case of Laplace Transform, we are able to converge the transform evenif the signal is unbounded. That is, $x(t) = t^2$ or in general $x(t)=t^n$ has an existing Laplace transform. I think similar is the case for z transform. Somehow it converges :( $\endgroup$ – Shehin A U Oct 2 '18 at 4:54
  • $\begingroup$ not if it's double-sided. if the $z^{-n}$ doesn't get you, the $z^{+n}$ will. $\endgroup$ – robert bristow-johnson Oct 2 '18 at 5:05
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[My path was erroneous, I made a confusion, but kept a modified answer for one-sided $\mathcal{Z}$-transforms]

With $\mathcal{Z}$-transforms, formulae with powers of $n$ often aim at making you handle basic operations on $\mathcal{Z}$-transforms: linearity, time-product/differentiation, integration, time-shift, see for instance: Properties of Z-Transform:

$$ \mathcal{Z}(nx[n]) = -z\frac{d}{dz}X(z)\,,$$ and $$ \mathcal{Z}(x[n-n_0]) = z^{-n_0}X(z)\,.$$

Hint: if your case were a one-sided $\mathcal{Z}$-transform, one could have focused on derivation, and on the simple geometric series:

$$\sum_{n=0}^{+\infty}z^{-n} = \frac{z}{z-1}\,.$$

As of today, as pointed out by Matt and Robert, I could not make sense of the $$\sum_{n=-\infty}^{+\infty}z^{-n}\,,$$ and I can see but a zero radius of convergence. For what it's worth, I will anyway try to add a common trick to try to derive something. One can use a stabilization with some $0 < a <1$:

$$\sum_{n=-\infty}^{+\infty}a^{|n|}z^{-n}\,,$$ with the secret thought that, if $a$ is just close to one, we could get almost what we wanted. I may have made mistakes again, so far this yields the formal form:

$$ \frac{(1-a^2)z}{(z-a)(1-az)}\,.$$

This still have a singular ROC, but one can play with $\mathcal{Z}$-transform properties.

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    $\begingroup$ That was so helpful, Laurent. But still two doubts vex me: 1) The question is $x(n)=n^3$ may be two sided. Can we apply those properties you have stated to this two sided signal? 2) The basic geometric series using the summation is valid only if the lower limit of the summation is '0', right? $\endgroup$ – Shehin A U Oct 2 '18 at 7:38
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    $\begingroup$ Does your last formula really make sense (with the given lower summation limit)? $\endgroup$ – Matt L. Oct 2 '18 at 17:02
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    $\begingroup$ That's such a polite to ask, and it is non-sense. OK for a one-sided version, correcting $\endgroup$ – Laurent Duval Oct 2 '18 at 19:40

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