---
title: "Z-transform of X(s) = 1/(s(s+1))"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-s-s-plus-1"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-s-s-plus-1
site: BCT Engineering
---

# Z-transform of X(s) = 1/(s(s+1))

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2071 Magh, 4 marks
- Video: [Digital control 3: The Z-transform](https://www.youtube.com/watch?v=NtYne9Nnsi8) — Corné van Daalen
- 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-s-s-plus-1
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the z-transform of X(s)=1/s(s+1).

## Short answer

Using partial fractions, X(z)= (1-e^{-T})z/((z-1)(z-e^{-T})).

## Answer

Expand $X(s)$ in partial fractions, take the z-transform of each sampled time function, and combine.

#### Step 1: Partial fractions

$$
X(s) = \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}
$$

#### Step 2: Time function and its samples

$$
x(t) = 1 - e^{-t}, \quad t \ge 0 \;\Rightarrow\; x(kT) = 1 - e^{-kT}
$$

#### Step 3: Z-transform of each term

| $X(s)$ | $x(kT)$ | $X(z)$ |
|---|---|---|
| $\dfrac{1}{s}$ | $1$ | $\dfrac{1}{1-z^{-1}}$ |
| $\dfrac{1}{s+1}$ | $e^{-kT}$ | $\dfrac{1}{1-e^{-T}z^{-1}}$ |

$$
\begin{aligned}
X(z) &= \frac{1}{1 - z^{-1}} - \frac{1}{1 - e^{-T}z^{-1}} \\
&= \frac{(1 - e^{-T}z^{-1}) - (1 - z^{-1})}{(1-z^{-1})(1-e^{-T}z^{-1})} \\
&= \frac{(1 - e^{-T})z^{-1}}{(1-z^{-1})(1-e^{-T}z^{-1})}
\end{aligned}
$$

In positive powers of $z$:

$$
X(z) = \frac{(1 - e^{-T})z}{(z-1)(z - e^{-T})}
$$

#### Numerical form for $T = 1$ s

$e^{-1} = 0.3679$, so

$$
X(z) = \frac{0.6321z}{(z-1)(z-0.3679)} = \frac{0.6321z}{z^2 - 1.3679z + 0.3679}
$$

#### Check

- Initial value: $\lim_{z\to\infty}X(z) = 0 = x(0)$.
- Final value: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1-e^{-T}}{1-e^{-T}} = 1 = x(\infty)$.

**Answer:** $X(z) = \dfrac{(1-e^{-T})z^{-1}}{(1-z^{-1})(1-e^{-T}z^{-1})} = \dfrac{(1-e^{-T})z}{(z-1)(z-e^{-T})}$.

## Check yourself

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

1. How is X(s) = 1/(s(s+1)) decomposed?
   A. 1/s - 1/(s+1)
   B. 1/s + 1/(s+1)
   C. 1/(s+1) - 1/s
   D. 2/s - 1/(s+1)
   Answer: A. A = 1 at s=0 and B = -1 at s=-1, so x(t) = 1 - e^{-t}. (Key point: Partial fractions)

2. What is X(z) for this X(s)?
   A. (1-e^{-T})z/((z-1)(z-e^{-T}))
   B. (1+e^{-T})z/((z-1)(z-e^{-T}))
   C. z/((z-1)(z-e^{-T}))
   D. (1-e^{-T})/((z-1)(z-e^{-T}))
   Answer: A. Subtracting z/(z-1) - z/(z-e^{-T}) gives numerator (1-e^{-T})z. (Key point: Final X(z))

3. Which pair of checks does X(z) = 0.6321z/((z-1)(z-0.3679)) satisfy?
   A. x(0) = 0 and x(infinity) = 1
   B. x(0) = 1 and x(infinity) = 0
   C. x(0) = 0 and x(infinity) = 0
   D. x(0) = 1 and x(infinity) = 1
   Answer: A. X(z) tends to 0 as z tends to infinity, and the final value theorem gives 0.6321/0.6321 = 1, matching 1 - e^{-kT}. (Key point: Checks with initial and final value)

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