Final value theorem of z-transform

Statement

Let $x(k) = 0$ for $k < 0$, with z-transform $X(z)$. Suppose all poles of $X(z)$ lie inside the unit circle, except possibly a simple pole at $z = 1$. Then

$$\lim_{k\to\infty}x(k) = \lim_{z\to1}\left[(1 - z^{-1})X(z)\right] = \lim_{z\to1}\left[\frac{z-1}{z}X(z)\right]$$

This condition means $(1 - z^{-1})X(z)$ has no poles on or outside the unit circle, so that $x(k)$ actually settles to a final value.

Proof

Consider the sequences $x(k)$ and $x(k-1)$. By definition and by the shift theorem:

$$\mathcal{Z}[x(k)] = \sum_{k=0}^{\infty}x(k)z^{-k} = X(z), \qquad \mathcal{Z}[x(k-1)] = \sum_{k=0}^{\infty}x(k-1)z^{-k} = z^{-1}X(z)$$

Subtract the two:

$$\sum_{k=0}^{\infty}x(k)z^{-k} - \sum_{k=0}^{\infty}x(k-1)z^{-k} = X(z) - z^{-1}X(z) = (1 - z^{-1})X(z)$$

Let $z \to 1$. Then $z^{-k} \to 1$:

$$\lim_{z\to1}(1 - z^{-1})X(z) = \sum_{k=0}^{\infty}\left[x(k) - x(k-1)\right]$$

The right side is a telescoping sum. Write out the partial sum up to $k = n$:

$$\begin{aligned} \sum_{k=0}^{n}\left[x(k) - x(k-1)\right] &= [x(0) - x(-1)] + [x(1) - x(0)] + \dots + [x(n) - x(n-1)] \\ &= x(n) - x(-1) = x(n) \end{aligned}$$

since $x(-1) = 0$. Taking $n \to \infty$:

$$\lim_{z\to1}(1 - z^{-1})X(z) = \lim_{n\to\infty}x(n)$$

This proves the theorem.

Example

For $X(z) = \dfrac{0.6321z}{(z-1)(z-0.3679)}$, the only pole outside $|z| < 1$ is the simple pole at $z = 1$. So:

$$x(\infty) = \lim_{z\to1}\frac{z-1}{z}\cdot\frac{0.6321z}{(z-1)(z-0.3679)} = \frac{0.6321}{0.6321} = 1$$

Use: finding steady-state errors of digital control systems without inverting $X(z)$.