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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.

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  • $\begingroup$ 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
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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 $|z|>a=2$.

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  • $\begingroup$ 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
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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 $a=2$.)

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

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