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What is the inverse Z-transform of $\frac{1}{(1-z^{-1})^2}$?

Title says it all. I have a one line solution but can't work out how to get there from tables or first principals.

Thanks!

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    Is this what it means? Find the coefficients $x_n$ in closed form: $$ \frac{1}{(1-z^{-1})^2} = \sum_{n=0}^\infty x_n z^{-n} $$ And if that's not it, tell us what it is!2012-01-05

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