---
title: "Z-transform of ZOH with plant 1/(s(s+1))"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:z-transform-zoh-plant"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-zoh-plant
site: BCT Engineering
---

# Z-transform of ZOH with plant 1/(s(s+1))

Digital Control System · Chapter 3: Analysis of Discrete Time Control System

- Asked in: 2072 Magh, 4 marks
- Video: [Discrete control #3: Designing for the zero-order hold](https://www.youtube.com/watch?v=yetLPW9sQaI) — Brian Douglas
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-zoh-plant
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

Obtain z-transform of f(s)=ZOH·1/s(s+1).

## Short answer

Use F(z) = (1−z⁻¹)Z[1/(s²(s+1))] with partial fractions. The result is F(z) = [(T−1+e^{-T})z + (1−e^{-T}−Te^{-T})]/[(z−1)(z−e^{-T})]. For T = 1 s, F(z) = (0.3679z+0.2642)/[(z−1)(z−0.3679)].

## Answer

The ZOH transfer function is $G_h(s) = \dfrac{1-e^{-Ts}}{s}$. The factor $e^{-Ts}$ is a delay of one sampling period, which becomes $z^{-1}$. So the transform is $(1-z^{-1})$ times the z-transform of the plant divided by $s$:

$$
F(z) = \mathcal{Z}\left[\frac{1-e^{-Ts}}{s}\cdot\frac{1}{s(s+1)}\right] = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right]
$$

#### Partial fractions
$$
\frac{1}{s^2(s+1)} = \frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1}
$$

#### z-transform of each term
$$
\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}
$$

#### Multiply by $(1-z^{-1}) = (z-1)/z$
$$
\begin{aligned}
F(z) &= \frac{T}{z-1} - 1 + \frac{z-1}{z-e^{-T}} \\
&= \frac{T(z-e^{-T}) - (z-1)(z-e^{-T}) + (z-1)^2}{(z-1)(z-e^{-T})} \\
&= \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})}
\end{aligned}
$$

#### With $T = 1$ s
$e^{-1} = 0.3679$, so $T - 1 + e^{-T} = 0.3679$ and $1 - e^{-T} - Te^{-T} = 1 - 2(0.3679) = 0.2642$.

$$
F(z) = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)} = \frac{0.3679z^{-1} + 0.2642z^{-2}}{1 - 1.3679z^{-1} + 0.3679z^{-2}}
$$

**Answer:** $F(z) = \dfrac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})}$, which for $T = 1$ s is $\dfrac{0.3679z + 0.2642}{(z-1)(z-0.3679)}$.

## Check yourself

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

1. How is the ZOH factor (1 - e^(-Ts))/s handled when taking the z-transform?
   A. e^(-Ts) becomes z, giving (1 - z)
   B. e^(-Ts) becomes z^-1 (one-period delay), giving (1 - z^-1) times Z[plant/s]
   C. The ZOH is ignored
   D. It becomes 1/(1 - z^-1)
   Answer: B. e^(-Ts) is a delay of T, i.e. z^-1, so the transform is (1 - z^-1) Z[G(s)/s]. (Key point: ZOH factor)

2. For ZOH followed by 1/(s(s+1)), which function is z-transformed (before multiplying by 1 - z^-1)?
   A. 1/(s(s+1))
   B. 1/(s^2 (s+1))
   C. 1/(s^2 (s+1)^2)
   D. 1/(s+1)
   Answer: B. The ZOH contributes an extra 1/s, so we transform 1/(s^2(s+1)). (Key point: Function to transform)

3. The partial fraction expansion of 1/(s^2(s+1)) is
   A. 1/s^2 - 1/s + 1/(s+1)
   B. 1/s^2 + 1/s - 1/(s+1)
   C. 1/s - 1/(s+1)
   D. 1/s^2 - 1/(s+1)
   Answer: A. Combining 1/s^2 - 1/s + 1/(s+1) gives 1/(s^2(s+1)), as the residues confirm. (Key point: Partial fractions)

4. For T = 1 s, the result F(z) is
   A. (0.3679z + 0.2642) / ((z-1)(z-0.3679))
   B. (0.6321z + 0.2642) / ((z-1)(z-0.3679))
   C. (0.3679z + 0.3679) / (z-0.3679)
   D. (0.2642z + 0.3679) / ((z-1)(z-0.3679))
   Answer: A. T - 1 + e^-T = 0.3679 and 1 - e^-T - T e^-T = 0.2642, with poles at 1 and e^-1. (Key point: Numerical result)

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