---
title: "Overshoot, settling time and error for 50/[s(s+2)]"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:mp-ts-50-s-s2-sse"
source: https://www.bctengineer.com/q/semester_3/control_systems/mp-ts-50-s-s2-sse
site: BCT Engineering
---

# Overshoot, settling time and error for 50/[s(s+2)]

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2076 Chaitra, 8 marks
- Video: [Problem to calculate rise time, peak time , peak overshoot , Max Peak Overshoot and settling time |](https://www.youtube.com/watch?v=pKBcv8YPEa4) — Mathspedia
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/mp-ts-50-s-s2-sse
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

For a unity feedback system, the open loop transfer function is G(s) = 50/[s(s+2)]. With unit step input find maximum overshoot and settling time. Also determine static error coefficients and steady state error if the input to the system is r(t) = 2 + 4t + 6t², t ≥ 0.

## Short answer

Closed-loop characteristic s²+2s+50 gives ζ=0.1414, ωn=7.071; maximum overshoot Mp = e^{-πζ/√(1-ζ²)} ≈ 63.84%, 2% settling time ts ≈ 4 s. Type 1 system: Kp=∞, Kv=25, Ka=0; for input 2+4t+6t², steady-state error is infinite due to parabolic component.

## Answer

Find $\zeta$ and $\omega_n$ from the closed-loop characteristic equation for the transient part. Then use the error coefficients (the system is Type 1) for the steady state part.

#### Closed-loop transfer function
$$
\frac{C(s)}{R(s)} = \frac{50}{s^2 + 2s + 50}
$$
$$
\begin{aligned}
\omega_n^2 &= 50 \;\Rightarrow\; \omega_n = 7.071\ \text{rad/s} \\
2\zeta\omega_n &= 2 \;\Rightarrow\; \zeta = \frac{1}{7.071} = 0.1414
\end{aligned}
$$

#### Maximum overshoot
$$
M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = e^{-\pi(0.1414)/0.9899} = e^{-0.4488} = 0.6384 = 63.84\%
$$

#### Settling time (2% criterion)
$$
t_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.1414 \times 7.071} = \frac{4}{1} = 4\ \text{s}
$$
(With the 5% criterion, $t_s = 3/\zeta\omega_n = 3$ s.)

#### Static error coefficients
$G(s)$ has one pole at the origin, so it is **Type 1**.
$$
\begin{aligned}
K_p &= \lim_{s\to 0} G(s) = \infty \\
K_v &= \lim_{s\to 0} sG(s) = \frac{50}{2} = 25 \\
K_a &= \lim_{s\to 0} s^2G(s) = 0
\end{aligned}
$$

#### Steady state error for $r(t) = 2 + 4t + 6t^2$
Write $6t^2 = 12\cdot\dfrac{t^2}{2}$.
$$
\begin{aligned}
e_{ss} &= \frac{2}{1+K_p} + \frac{4}{K_v} + \frac{12}{K_a} \\
&= 0 + \frac{4}{25} + \frac{12}{0} = 0 + 0.16 + \infty = \infty
\end{aligned}
$$

| Input part | Error |
|---|---|
| Step 2 | 0 |
| Ramp 4t | 0.16 |
| Parabola 6t² | ∞ |

**Answer:** $M_p = 63.84\%$, $t_s = 4$ s (2%); $K_p = \infty$, $K_v = 25$, $K_a = 0$; $e_{ss} = \infty$, because a Type 1 system cannot follow a parabolic input.

## Check yourself

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

1. For G = 50/[s(s+2)] with unity feedback, what are zeta and wn?
   A. wn = 7.07 rad/s and zeta = 0.1414
   B. wn = 50 rad/s and zeta = 0.02
   C. wn = 7.07 rad/s and zeta = 0.707
   D. wn = 2 rad/s and zeta = 1
   Answer: A. s^2 + 2s + 50 gives wn^2 = 50 and 2 zeta wn = 2, so zeta = 1/7.07 = 0.1414. (Key point: zeta and wn)

2. What is the maximum overshoot?
   A. 14.14%
   B. 6.38%
   C. 63.84%
   D. 36.2%
   Answer: C. Mp = exp(-pi zeta/sqrt(1-zeta^2)) = exp(-0.4488) = 0.6384. (Key point: Max overshoot)

3. What is ts by the 2% criterion?
   A. 3/(zeta wn) = 3 s
   B. 4/(zeta wn) = 4 s
   C. 4/wn = 0.57 s
   D. pi/wd = 0.45 s
   Answer: B. zeta wn = 1, so ts = 4 s (2% criterion). (Key point: Settling time)

4. The input 2 + 4t + 6t^2 is applied to this Type 1 system. What is ess?
   A. Infinite, since a Type 1 system cannot follow a parabolic input (Ka = 0)
   B. 0.16, from the ramp term only
   C. Zero, since Kp is infinite
   D. 0.5, from 12/Kv
   Answer: A. Step gives 0, ramp gives 4/25 = 0.16, but the parabola gives 12/Ka = infinity. (Key point: Steady state error)

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