---
title: "Unit step response of 4/[s(s+2)]"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:unit-step-4-s-s2"
source: https://www.bctengineer.com/q/semester_3/control_systems/unit-step-4-s-s2
site: BCT Engineering
---

# Unit step response of 4/[s(s+2)]

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2065 Shrawan (old course), 8 marks
- Video: [Unit Step Response of an Underdamped Second Order System in Control Systems](https://www.youtube.com/watch?v=Bq_X4_ShARw) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/unit-step-4-s-s2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Find the unit step response of a unity feedback system whose open loop transfer function is given by G(s)H(s) = 4/[s(s+2)].

## Short answer

The closed-loop system is underdamped with ζ=0.5, ωn=2 rad/s. Unit step response: c(t) = 1 - 1.1547e^{-t} sin(1.732t+60°), with 16.3% overshoot and settling time 4 s.

## Answer

For unity feedback, the closed-loop transfer function is $T(s)=\frac{G(s)}{1+G(s)}$. We find $C(s)$ for $R(s)=1/s$ and take the inverse Laplace transform.

#### Closed-loop transfer function

$$
\begin{aligned}
\frac{C(s)}{R(s)} &= \frac{\frac{4}{s(s+2)}}{1+\frac{4}{s(s+2)}} = \frac{4}{s^2+2s+4}
\end{aligned}
$$

Compare with the standard form $\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$:

- $\omega_n^2 = 4 \Rightarrow \omega_n = 2$ rad/s
- $2\zeta\omega_n = 2 \Rightarrow \zeta = 0.5$ (underdamped, $0<\zeta<1$)
- $\omega_d = \omega_n\sqrt{1-\zeta^2} = 2\sqrt{0.75} = \sqrt{3} = 1.732$ rad/s
- $\zeta\omega_n = 1$, $\theta = \cos^{-1}\zeta = 60^\circ = \pi/3$ rad

#### Response by partial fractions

$$
\begin{aligned}
C(s) &= \frac{4}{s(s^2+2s+4)} = \frac{1}{s} - \frac{s+2}{s^2+2s+4} \\
&= \frac{1}{s} - \frac{(s+1)}{(s+1)^2+3} - \frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{(s+1)^2+3}
\end{aligned}
$$

Taking the inverse Laplace transform:

$$
\begin{aligned}
c(t) &= 1 - e^{-t}\left[\cos\sqrt{3}t + \frac{1}{\sqrt{3}}\sin\sqrt{3}t\right] \\
&= 1 - \frac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}}\sin(\omega_d t+\theta) \\
&= 1 - 1.1547\,e^{-t}\sin\left(1.732t + 60^\circ\right), \quad t\ge 0
\end{aligned}
$$

Check: at $t=0$, $c(0)=1-1.1547\sin 60^\circ = 1-1 = 0$; as $t\to\infty$, $c(t)\to 1$, so the steady-state error is zero (type-1 system with step input).

#### Time-domain specifications (for completeness)

| Quantity | Formula | Value |
|---|---|---|
| Rise time $t_r$ | $\frac{\pi-\theta}{\omega_d}$ | $\frac{2.094}{1.732}=1.209$ s |
| Peak time $t_p$ | $\frac{\pi}{\omega_d}$ | $1.814$ s |
| Peak overshoot $M_p$ | $e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ | $0.163$ = 16.3% |
| Settling time $t_s$ (2%) | $\frac{4}{\zeta\omega_n}$ | $4$ s |

```text
 c(t)
 1.163 |      _
   1.0 |----/---\_____________
       |   /       ''
       |  /
       | /
     0 |/_____________________ t
          tp=1.81 s
```

**Answer:** $c(t) = 1 - 1.1547\,e^{-t}\sin(1.732t+60^\circ)$; the response is underdamped ($\zeta=0.5$, $\omega_n=2$ rad/s) with 16.3% overshoot and settles to 1.

## Check yourself

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

1. For G = 4/[s(s+2)], C/R is
   A. 4/(s^2+2s)
   B. 4/(s^2+2s+4)
   C. 2/(s^2+2s+2)
   D. 4/(s^2+4s+4)
   Answer: B. G/(1+G) gives s^2+2s+4. (Key point: Closed loop)

2. wn and zeta are
   A. wn=4, zeta=0.25
   B. wn=2, zeta=1
   C. wn=2, zeta=0.5
   D. wn=1, zeta=2
   Answer: C. wn^2=4 and 2*zeta*wn=2 give zeta=0.5. (Key point: Parameters)

3. The unit step response is
   A. 1 - 1.1547 e^(-t) sin(1.732t + 60°)
   B. 1 - e^(-2t) sin(1.732t)
   C. 1 - 1.1547 e^(-2t) sin(2t+60°)
   D. 1 - e^(-t) cos(t)
   Answer: A. Using zeta*wn=1, wd=sqrt(3), theta=60°, 1/sqrt(1-zeta^2)=1.1547. (Key point: Step response)

4. The overshoot and 2% settling time are
   A. 16.3%, 4 s
   B. 16.3%, 2 s
   C. 4.3%, 4 s
   D. 50%, 1 s
   Answer: A. Mp = e^(-pi*0.5/0.866) = 16.3%, ts = 4/(0.5*2) = 4 s. (Key point: Overshoot and ts)

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