Z-transform

Remarks

Example: for $x_n = a^n$ when $n \ge 0$ and $x_n = 0$ when $n < 0$, with $a \in \C$, the series is geometric with ratio $a z^{-1}$. It converges absolutely exactly when $\abs{z} > \abs{a}$, and then $X(z) = z/(z - a)$. Non-example: for $x_n = 1$ for all $n \in \Z$, the series $\sum_{n \in \Z} z^{-n}$ converges for no $z$. The part with $n \ge 0$ needs $\abs{z} > 1$, and the part with $n < 0$ needs $\abs{z} < 1$, so the sequence has no Z-transform as a function. The sum over all $n \in \Z$ is the bilateral form, and the sum over $n \ge 0$ is the unilateral form. A function $X(z)$ alone does not determine $x$: the set of convergence is part of the data. Substituting $z = e^{i\omega}$ gives the discrete-time Fourier transform, when the unit circle lies in the set of convergence.