How can I find the inverse z transform of 1/(z-a)? 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 $\dfrac{1}{z-a}$?
 A: 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.
A: 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.
A: I completely understand your question; I was looking for the answer to this online, and found nothing.
The answer to this actually stems from the time-shift property of the z-transform, which states that $x(n-n_0) = (z^{-n_0})X(z)$.  You can rewrite $\frac{1}{z-a}$ as
$z^{-1}  \frac{z}{z-a}$, which is in the form of the time-shifting formula.  Therefore, the inverse z-transform of $\frac{1}{z-a}$ is as follows:
$a^{n-1}u(n-1)$, $|z| > |a|$
-or-
$-a^{n-1}u(-n+(a-1))$
Hope this helps.
