---
title: "Final value theorem of z-transform"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:final-value-theorem"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/final-value-theorem
site: BCT Engineering
---

# Final value theorem of z-transform

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2075 Bhadra, 1+3 marks
- Video: [Initial and Final Value Theorem - Z Transform - Signals and Systems](https://www.youtube.com/watch?v=dYzftpt6ZLc) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/final-value-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

State and prove final value theorem of z-transform.

## Short answer

If (1-z^{-1})X(z) has all poles inside the unit circle (except possibly a simple pole at z=1), then the final value is lim_{k→∞} x(k) = lim_{z→1} (1-z^{-1})X(z). It finds steady-state values without inverse transform.

## Answer

#### 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 $X(z)$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which expression gives lim x(k) as k tends to infinity?
   A. lim as z->1 of (1 - z^{-1})X(z)
   B. lim as z->infinity of X(z)
   C. lim as z->0 of (1 - z^{-1})X(z)
   D. lim as z->1 of X(z)
   Answer: A. The final value theorem uses (1 - z^{-1})X(z) evaluated at z = 1. (Key point: Statement of the theorem)

2. When is the theorem valid?
   A. All poles of X(z) inside the unit circle except possibly a simple pole at z = 1
   B. Any pole may lie on the unit circle
   C. All poles must be at z = 1
   D. Only when X(z) has no poles
   Answer: A. Otherwise (1 - z^{-1})X(z) has poles on or outside the unit circle and x(k) does not settle. (Key point: Condition of validity)

3. In the proof, why does the sum of [x(k) - x(k-1)] reduce to x(n)?
   A. It telescopes, and x(-1) = 0
   B. The terms are all equal
   C. Each term is zero except the last
   D. The z^{-k} factors cancel
   Answer: A. Successive terms cancel, leaving x(n) - x(-1) = x(n). (Key point: Proof idea)

4. What does subtracting Z[x(k-1)] = z^{-1}X(z) from X(z) give?
   A. (1 - z^{-1})X(z)
   B. (1 + z^{-1})X(z)
   C. zX(z)
   D. X(z)/z
   Answer: A. X(z) - z^{-1}X(z) = (1 - z^{-1})X(z). (Key point: Subtracting shifted sequences)

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