---
title: "Transfer function from overshoot and settling time"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:tf-from-mp25-ts6"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-from-mp25-ts6
site: BCT Engineering
---

# Transfer function from overshoot and settling time

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2078 Bhadra, 4 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/tf-from-mp25-ts6
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

A system has 25% overshoot and settling time of 6 seconds for a unit step input. Determine the transfer function and calculate peak time. Assume ess as 2%.

## Short answer

ζ = 0.404, ωn = 1.651 rad/s, T(s) = 2.727/(s²+1.333s+2.727), tp = 2.08 s.

## Answer

Assume a standard second-order system $\frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Find $\zeta$ from overshoot and $\omega_n$ from settling time.

#### Damping ratio from $M_p = 0.25$

$$
\begin{aligned}
M_p &= e^{-\zeta\pi/\sqrt{1-\zeta^2}} = 0.25 \\
\frac{\zeta\pi}{\sqrt{1-\zeta^2}} &= -\ln 0.25 = 1.3863 \\
\zeta &= \frac{1.3863}{\sqrt{\pi^2+1.3863^2}} = \frac{1.3863}{3.4339} = 0.4037
\end{aligned}
$$

#### Natural frequency from $t_s = 6$ s (2% band)

$$
\begin{aligned}
t_s &= \frac{4}{\zeta\omega_n} = 6 \Rightarrow \zeta\omega_n = 0.6667 \\
\omega_n &= \frac{0.6667}{0.4037} = 1.651\ \text{rad/s}
\end{aligned}
$$

#### Transfer function

$\omega_n^2 = 2.727$, $2\zeta\omega_n = 1.333$:

$$
\frac{C(s)}{R(s)} = \frac{2.727}{s^2 + 1.333s + 2.727}
$$

#### Peak time

$$
\begin{aligned}
\omega_d &= \omega_n\sqrt{1-\zeta^2} = 1.651\sqrt{1-0.4037^2} = 1.511\ \text{rad/s} \\
t_p &= \frac{\pi}{\omega_d} = \frac{3.1416}{1.511} = 2.079\ \text{s}
\end{aligned}
$$

**Answer:** $\zeta = 0.404$, $\omega_n = 1.651$ rad/s, $T(s) = \dfrac{2.727}{s^2+1.333s+2.727}$, peak time $t_p = 2.08$ s.

## Check yourself

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

1. Which relation gives zeta from Mp = 0.25?
   A. zeta*pi/sqrt(1-zeta^2) = -ln(0.25)
   B. zeta = Mp
   C. zeta = 4/(ts*wn)
   D. zeta*pi/sqrt(1-zeta^2) = ln(4)/ts
   Answer: A. Mp = exp(-zeta*pi/sqrt(1-zeta^2)), so taking -ln gives the relation, with result zeta = 0.404. (Key point: Damping ratio from overshoot)

2. With the 2% criterion ts = 6 s and zeta = 0.4037, wn is about
   A. 0.667 rad/s
   B. 1.651 rad/s
   C. 2.727 rad/s
   D. 1.511 rad/s
   Answer: B. ts = 4/(zeta*wn) gives zeta*wn = 0.667, so wn = 0.667/0.4037 = 1.651 rad/s. (Key point: Natural frequency from settling time)

3. The resulting transfer function is
   A. 1.651/(s^2+0.667s+1.651)
   B. 2.727/(s^2+1.333s+2.727)
   C. 2.727/(s^2+0.667s+2.727)
   D. 1.333/(s^2+2.727s+1.333)
   Answer: B. wn^2 = 2.727 and 2*zeta*wn = 1.333 in the standard second-order form. (Key point: Transfer function)

4. Peak time is found from
   A. tp = pi/wn
   B. tp = pi/(wn*sqrt(1-zeta^2)) = pi/wd = 2.08 s
   C. tp = 4/(zeta*wn)
   D. tp = 1.5 s from 0.25 overshoot directly
   Answer: B. tp = pi/wd with wd = 1.511 rad/s, giving about 2.08 s. (Key point: Peak time)

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