Bilateral Z-transform

Remarks

Example: let $x_n = 2^{-\abs{n}}$. The part $\sum_{n \ge 0} (2z)^{-n}$ converges for $\abs{z} > 1/2$ and the part $\sum_{n \ge 1} (z/2)^n$ converges for $\abs{z} < 2$. So $X(z) = \dfrac{2z}{2z - 1} + \dfrac{z}{2 - z}$ on the annulus $1/2 < \abs{z} < 2$. Example with the same formula: for $a \ne 0$, $x_n = a^n$ for $n \ge 0$ (and $0$ otherwise) gives $z/(z - a)$ on $\abs{z} > \abs{a}$, while $x_n = -a^n$ for $n \le -1$ (and $0$ otherwise) gives $z/(z - a)$ on $\abs{z} < \abs{a}$. Non-example: $x_n = 1$ for all $n$ has an empty region of convergence, so the transform is not a function. The notation $X(z) = \mathcal{Z}\{x\}(z)$ is standard; signal processing texts write $x[n]$ for $x_n$.