I'm working through and learning about Inverse $\mathcal Z$-transforms right now, and I'm getting caught up on trying to find the Inverse $\mathcal Z$-transform of the following: $$\sum_{k=1}^P\frac{1}{z^{60k+1}-az^{60k}}$$ I'm not really sure how to proceed. I'm trying to manipulate the equation by pulling out $z^{30k}$, but it seems to not be going anywhere. If anyone can point me in the right direction, I'd appreciate it.
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$\begingroup$ Off-topic, but I just realized you deleted the other question that I had answered. You should know that while you have the right to delete your own post, it is something inappropriate and rude when someone has already answered your question. This is because [by doing so] you also delete other people's post which does not belong to you. $\endgroup$– msmNov 7, 2016 at 3:28
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$\begingroup$ Actually I un-deleted that question since for me it seemed quite relevant + it had an answer. $\endgroup$– jojeck ♦Nov 8, 2016 at 10:39
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$\begingroup$ Thanks @jojek. I don't expect any response from this user, to be honest. It was just to prevent it being repeated (hopefully) for others... $\endgroup$– msmNov 8, 2016 at 12:38
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1 Answer
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Multiply the top and bottom of the fraction by $z^{-60k}$ in each term, and you are left with each term being:
$$ z^{-60k}\frac{1}{z-a} = z^{-60k-1}\frac{1}{1-az^{-1}} $$
The first factor delays the output by $60k +1$ samples, and the remaining factor is a simple one pole transfer function, giving:
$$ x[n] = \sum_{k=1}^{P} a^{n - 60k - 1} u[n - 60k - 1] $$
