$X(z) = \displaystyle \frac{1}{z}{\left(1-\frac{z^2}{4}\right)\left(1+\frac{1}{z}\right)\left(1-z\right)}$

Using partial fractions expansion i came up to this:

$\displaystyle \frac{1}{X(z)} = \frac{-\frac{z}{3}}{1-\frac{z}{2}} + \frac{\frac{z}{3}}{1+\frac{z}{2}} + \frac{\frac{2z}{3}}{1-z}$

I tried but i can't go on any further.


You just need to multiply the terms and sort by powers of $z$. If you do so you should get


So the work is in multiplying out the terms, not in actually figuring out the inverse $\mathcal{Z}$-transform, because from $(1)$ you can immediately write down $x[n]$:


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  • 1
    $\begingroup$ i like the text stylization: "$\mathcal{Z}$-transform". $\endgroup$ – robert bristow-johnson Aug 30 '15 at 21:28

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