Inverse Z-transform

Remarks

On the circle the series $\sum_k x_k z^{-k}$ converges absolutely and uniformly, so it can be integrated term by term after multiplication by $z^{n-1}$. The integral of $z^{n-1-k}$ around the circle is $2\pi i$ when $k = n$ and $0$ otherwise, which leaves $x_n$. This is the coefficient formula for the Laurent series of $X$ on its annulus. Example: $X(z) = z/(z - a)$ on $\abs{z} > \abs{a}$; for $n \ge 0$ the residue theorem gives $x_n = a^n$, and for $n < 0$ the integrand $z^n/(z - a)$ has total residue $0$ inside a large circle, so $x_n = 0$. The circle must lie in the region of convergence: the same rational function $z/(z - a)$ on $\abs{z} < \abs{a}$ gives $x_n = -a^n$ for $n \le -1$ and $x_n = 0$ for $n \ge 0$.

Going deeper

For a student

In practice, $X$ is often a rational function, and one inverts it by partial fractions and a table of pairs such as $a^n \leftrightarrow z/(z - a)$. The contour formula justifies this method and shows that the result is unique once the region of convergence is fixed. With $z = e^{i\omega}$ on the unit circle, the formula becomes the inverse of the discrete-time Fourier transform. It is the discrete analogue of the Bromwich inversion integral for the Laplace transform.