Inverse z transform

1. Apr 24, 2015

etf

Hi!
My task is to find discrete signal x(n), if z transform of that signal is $$X(z)=\frac{5}{(z-2)^{2}}$$. It is known that signal is causal. 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:

We have in bracket sum which represents z transform of signal:

But I don't know what to do next :(

Last edited: Apr 24, 2015
2. Apr 24, 2015

rude man

The standard approach is
X(z) = 5/(z-2)2
Take X(z)/z
Form partial fraction expansion.
One of the denominators will be (z-2)2 which has to be dealt with uniquely. My hint:
multiply lhs and rhs of X(z)/z by z, then let z → ∞.
Now you'd have all the numerator coefficients of X(z)/z.
Now multiply by z to get X(z). Invert term-by-term. The term with (z-2)2 will probably throw you.
Hint: Consider the theorem {n(n-1)(n-2) ... (n-m+1)/m!αmnu[n] ⇔ z/(z-α)(m+1).