TITLE: Inverse $Z$ transform of $\frac{1}{z-a}$
QUESTION [0 upvotes]: I don't really get what's happening here and I haven't been able to find a single example on how to get the inverse $Z$-transform of $\frac{1}{z-a}$.
Can anyone show the way?

REPLY [2 votes]: That's the more elementary transform, it's a consequence of the geometric sum:
$$ \sum_{k=0}^\infty b^k=\frac{1}{1-b} \hspace{1cm} |b|<1$$
Hence
$$\frac{1}{z-a} =z^{-1} \frac{1}{1-a z^{-1}} =z^{-1}\sum_{k=0}^\infty (a z^{-1})^k = \sum_{t=1}^\infty a^{t-1}z^{-t} \tag{1}$$
(this converges when $|a/z|<1$)
But also
$$\frac{1}{z-a}=-a^{-1}\frac{1}{1-z/a}=-a^{-1}\sum_{k=0}^\infty (z/a)^k = \sum_{t=-\infty}^0 -a^{t-1} z^{-t} \tag{2}$$
(this converges when $|z/a|<1$)
Then the antitransform is given by $(1)$ (causal: $x[n]=a^{n-1} u[n-1]$) in the region $|z|>|a|$ ("outside the circle"). In the other ROC ($|z|<|a|$, "inside the circle") we get the anticausal sequence $x[n]=-a^{n-1} u[-n]$.
(Remember: a $X(z)$ does not univocally defines the antitransform - you need both $X(z)$ and the ROC to obtain $x[n]$)
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