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I came across this situation in my textbook:

scan

However I have no clue about how you can (starting from the stepresponse on the left), get the polynome equation on the right.

Could someone please explain?

EDIT: what about this case? http://postimg.org/image/726noo2k3/b8bf257

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    $\begingroup$ You should upload your images through stackexchange's imgur service (you can use the little image icon in the editor toolbar). Since you don't have enough rep the image won't automatically show up, but another user can then embed them in without having to reupload a copy of the image. $\endgroup$ – SleuthEye Jan 28 '17 at 14:11
  • $\begingroup$ I'm not clicking on any link to "postimg.org", sorry. Follow @SleuthEye's advise. $\endgroup$ – Marcus Müller Jan 28 '17 at 20:09
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The right-hand polynomial is the Z-transform $X(z)$ of the discrete-time function $x[n]$, and is given by

$$X(z)=\sum_{n=0}^{\infty}x[n]z^{-n}$$

(unilateral or bilateral does not matter here, since $x[n]$ is causal anyway). So, if you now insert $x[n]$ you get

$$X(z)=\sum_{n=0}^{\infty}z^{-n}=\sum_{n=0}^\infty\left(z^{-1}\right)^n$$

You can use formula for the geometric series, as long as $|z|>1$ (i.e. $|z|>1$ is the region of convergence of the Z-transform):

$$X(z)=\sum_{n=0}^{\infty}\left(z^{-1}\right)^n=\frac{1}{1-z^{-1}}=\frac{z}{z-1}.$$

For the second case, i.e. $x[n]=a^nu[n]$, you can write the following:

$$X(z)=\sum_{n=0}^{\infty}a^nz^{-n}=\sum_{n=0}^{\infty}(az^{-1})^n$$

Then, again applying the geometric series formula you get (given that the series converges, i.e. $|az^{-1}|<1$) $$X(z)=\frac{1}{1-az^{-1}}$$

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  • $\begingroup$ I think doing it that way makes it a little more difficult for that case: imgur.com/60vfFj4 No? $\endgroup$ – hihaho Jan 28 '17 at 13:45
  • $\begingroup$ Wrong link. You can directly add images to your original post. Also, why dont you ask the difficult thing, when this is actually what you want? $\endgroup$ – Maximilian Matthé Jan 28 '17 at 13:51
  • $\begingroup$ Sorry, imgur is bit a pain in the ass today: postimg.org/image/726noo2k3/b8bf257 Because I expected a totally different answer. $\endgroup$ – hihaho Jan 28 '17 at 13:57
  • $\begingroup$ For the exponential decaying function, the calculation is similar. $\endgroup$ – Maximilian Matthé Jan 28 '17 at 14:06

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