How can I find inverse z transform of $$X(z)=\frac{5}{(z-2)^{2}}$$ ?

It is known that x[n] is causal.

EDIT: Here is what I have done. Since signal x(n) is causal, convergence of z transform of that signal will be outside of circle with radius r:

enter image description here

We have in bracket sum which represents z transform of signal:

enter image description here

But I don't know what to do next.

  • $\begingroup$ wow, two poles outside the unit circle. this should be interesting. $\endgroup$ – robert bristow-johnson Apr 24 '15 at 0:42
  • $\begingroup$ even wolfram knows the answer... $\endgroup$ – jojek Apr 24 '15 at 1:04
  • $\begingroup$ Thanks jojek, but I'm interested in how to get final solution. Wolfram can't show me that :) $\endgroup$ – etf Apr 24 '15 at 11:46

You know the geometric series

$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r},\quad |r|<1\tag{1}$$

Due to convergence of (1) (for $|r|<1$) you can take the derivative with respect to $r$ by taking the element-wise derivatives of the left-hand side. Equating the derivatives of both sides of (1) gives

$$\sum_{n=0}^{\infty}nr^{n-1}=\sum_{n=1}^{\infty}nr^{n-1}=\sum_{n=0}^{\infty}(n+1)r^n=\frac{1}{(1-r)^2},\quad |r|<1\tag{2}$$

With $r=az^{-1}$ you get the following $\mathcal{Z}$-transform relation:

$$(n+1)a^nu[n]\Longleftrightarrow \frac{1}{(1-az^{-1})^2}=\frac{z^2}{(z-a)^2}\tag{3}$$

where $u[n]$ is the unit step sequence. With the time-shifting property of the $\mathcal{Z}$-transform (see here) you get the following pair:

$$(n-1)a^{n-2}u[n-2]\Longleftrightarrow \frac{1}{(z-a)^2}\tag{4}$$

With (4) you immediately get

$$\frac{5}{(z-2)^2}\Longleftrightarrow 5(n-1)2^{n-2}u[n-2]$$

Of course, all of this is only valid inside the region of convergence, i.e. for $|z|>a=2$.

| improve this answer | |
  • $\begingroup$ hey Matt, looks like etf believes in the stick more than the carrot. at least you got a check mark for your effort. $\endgroup$ – robert bristow-johnson Apr 25 '15 at 20:14
  • $\begingroup$ @robertbristow-johnson: I guess the OP wanted to see how to get the result. And that's indeed not shown in many basic DSP texts. $\endgroup$ – Matt L. Apr 26 '15 at 9:17

note, from wikipedia

$$ \mathcal{Z}\left\{ n a^n u[n] \right\} = \frac{az^{-1} }{ (1-a z^{-1})^2 } $$

see if you can make your problem fit that form. (gonna get a funky result because $a=2$.)

| improve this answer | |
  • $\begingroup$ Hi. I didn't solved it yet... $\endgroup$ – etf Apr 24 '15 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.