---
tag: PH61T3B7W7
kind: statement
role: theorem
name: Inverse Z-transform
msc: [44A55, 30B10]
uses: [3DBAMXA80B, 1P8WX3EGHF, 0GXMPDNYX3, 06ZGPYTWK8]
about: [5FY0JQGHYZ]
sources:
  - url: https://en.wikipedia.org/wiki/Z-transform
    locator: Inverse Z-transform
---

## Statement

Let $x = (x_n)_{n \in \Z}$ be a sequence of complex numbers whose [[bilateral Z-transform|3DBAMXA80B]] $X(z) = \sum_{n \in \Z} x_n z^{-n}$ has a [[region of convergence|1P8WX3EGHF]] that contains a circle $\abs{z} = \rho$ with $\rho > 0$.
Then for every $n \in \Z$,
$$x_n = \frac{1}{2\pi i} \oint_{\abs{z} = \rho} X(z) \, z^{n-1} \, dz,$$
where the [[contour integral|06ZGPYTWK8]] runs once counterclockwise around the circle.
In particular, $x$ is determined by the function $X$ on any one such circle.

## 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|0GXMPDNYX3]] 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 $n \ge 0$.

## Explanations

### L1 human

The Z-transform packs a sequence into a function, and this formula unpacks it.
Each term of the sequence is recovered by an average of the function around a circle.
The circle must be chosen inside the region where the transform converges, since that region tells which sequence one started from.

### L2 human

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.