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.

  • $\begingroup$ so who gave you that assignment? $\endgroup$ Oct 2, 2018 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, 2018 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. $\endgroup$ Oct 2, 2018 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, 2018 at 4:54
  • $\begingroup$ not if it's double-sided. if the $z^{-n}$ doesn't get you, the $z^{+n}$ will. $\endgroup$ Oct 2, 2018 at 5:05

1 Answer 1


[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.

  • 1
    $\begingroup$ 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, 2018 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, 2018 at 17:02
  • 2
    $\begingroup$ That's such a polite to ask, and it is non-sense. OK for a one-sided version, correcting $\endgroup$ Oct 2, 2018 at 19:40

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