How can I find the inverse z transform of 1/(z-a)?

Question

I need to take the inverse z transform of $X(z) = \dfrac{.5}{(z-1)(z-.5)}$. I've used partial fractions to split this into $X(z) = \dfrac{1}{z-.5} - \dfrac{1}{z-1}$

But Here I'm stuck. This isn't in a table, and I'm not sure how to solve it. I've found pages with similar problems, e.g. $\dfrac{1}{(z-1)^2}$, but they claim that "this is simple" and just stop once they've handled the square. How would I go about deriving the inverse Z transform of $\dfrac{1}{z-a}$?

Answer 1

Perhaps the "inverse z-transform" in signal processing is known as "Laurent series" in mathmatics? In other words: find coefficients $a_n$ so that $$ \frac{1/2}{(z-1)(z-1/2)} = \sum_{n=-\infty}^{+\infty} a_n z^{-n} $$ I guess this series is supposed to be valid near $\infty$? If so, Maple says $$ \frac{1}{2}\;z^{-2}+\frac{3}{4}\;z^{-3}+\frac{7}{8}\;z^{-4}+\frac{15}{16}\,z^{-5}+\dots $$ Yes, it can be done by using the partial fraction expansion found in the question. Each of those series expansions is a geometric series. Then add the two series.

Answer 2

My professor responded to an e-mail and reminded me of a trick from class ... If you do the partial fraction expansion with $\dfrac{F(z)}{z}$ then you get a third fraction: $\dfrac{C}{z}$. Then multiply through by z and you get z in the numerator to match the tables of common transforms.