---
title: "Derive maximum overshoot of second order system"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:mp-expression-derivation"
source: https://www.bctengineer.com/q/semester_3/control_systems/mp-expression-derivation
site: BCT Engineering
---

# Derive maximum overshoot of second order system

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2070 Chaitra, 7 marks
- Video: [Peak Overshoot of Second Order System: Basics, Definition, and Derivation](https://www.youtube.com/watch?v=7gYsYhj4ntM) — 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/mp-expression-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

For a second order system with G(s) = ωn²/[s(s + 2ξωn)] and H(s) = 1, find expression for maximum overshoot on its unit step response, where ωn is natural frequency of oscillation and ξ is damping ratio, at underdamped situation. [Figure: R(s) → summing point (+, −) → G(s) → C(s), feedback H(s)]

## Short answer

Maximum overshoot for a unit step is M_p = exp(−ζπ/√(1−ζ²)), occurring at t_p = π/(ω_n√(1−ζ²)). It depends only on damping ratio ζ.

## Answer

**Maximum (peak) overshoot** is the largest amount by which the response exceeds its final value, usually given as a percentage of the final value. To find it, derive $c(t)$, find the time of the first peak, and substitute.

#### Closed-loop transfer function
$$
\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{\omega_n^2}{s^2 + 2\xi\omega_n s + \omega_n^2}
$$
For $\xi < 1$ the poles are complex:
$$
s = -\xi\omega_n \pm j\omega_n\sqrt{1-\xi^2} = -\xi\omega_n \pm j\omega_d, \qquad \omega_d = \omega_n\sqrt{1-\xi^2}
$$

#### Output for a unit step
With $R(s) = 1/s$:
$$
C(s) = \frac{\omega_n^2}{s(s^2 + 2\xi\omega_n s + \omega_n^2)}
$$
Partial fractions:
$$
\begin{aligned}
C(s) &= \frac{1}{s} - \frac{s + 2\xi\omega_n}{s^2 + 2\xi\omega_n s + \omega_n^2} \\
&= \frac{1}{s} - \frac{s + \xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} - \frac{\xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2}
\end{aligned}
$$
Using $\mathcal{L}^{-1}\left\{\frac{s+a}{(s+a)^2+\omega^2}\right\} = e^{-at}\cos\omega t$ and $\mathcal{L}^{-1}\left\{\frac{\omega}{(s+a)^2+\omega^2}\right\} = e^{-at}\sin\omega t$:
$$
\begin{aligned}
c(t) &= 1 - e^{-\xi\omega_n t}\cos\omega_d t - \frac{\xi\omega_n}{\omega_d}e^{-\xi\omega_n t}\sin\omega_d t \\
&= 1 - e^{-\xi\omega_n t}\left[\cos\omega_d t + \frac{\xi}{\sqrt{1-\xi^2}}\sin\omega_d t\right]
\end{aligned}
$$
Put $\cos\theta = \xi$ and $\sin\theta = \sqrt{1-\xi^2}$, so that $\cos\omega_d t\,\sin\theta + \sin\omega_d t\,\cos\theta = \sin(\omega_d t + \theta)$:
$$
c(t) = 1 - \frac{e^{-\xi\omega_n t}}{\sqrt{1-\xi^2}}\sin(\omega_d t + \theta), \qquad \theta = \tan^{-1}\frac{\sqrt{1-\xi^2}}{\xi}
$$

#### Peak time
At a peak, $\dfrac{dc}{dt} = 0$. Differentiating $c(t)$:
$$
\frac{dc}{dt} = \frac{\omega_n}{\sqrt{1-\xi^2}}e^{-\xi\omega_n t}\sin\omega_d t = 0
$$
So $\sin\omega_d t = 0$, which gives $\omega_d t = n\pi$. The first peak (maximum overshoot) is at $n = 1$:
$$
t_p = \frac{\pi}{\omega_d} = \frac{\pi}{\omega_n\sqrt{1-\xi^2}}
$$

#### Maximum overshoot
Substitute $t_p$ in $c(t)$. Since $\omega_d t_p = \pi$, $\sin(\pi + \theta) = -\sin\theta = -\sqrt{1-\xi^2}$:
$$
\begin{aligned}
c(t_p) &= 1 - \frac{e^{-\xi\omega_n\pi/\omega_d}}{\sqrt{1-\xi^2}}\left(-\sqrt{1-\xi^2}\right) \\
&= 1 + e^{-\xi\pi/\sqrt{1-\xi^2}}
\end{aligned}
$$
With final value $c(\infty) = 1$:
$$
M_p = c(t_p) - c(\infty) = e^{-\xi\pi/\sqrt{1-\xi^2}}
$$
$$
\%M_p = 100\,e^{-\xi\pi/\sqrt{1-\xi^2}}
$$

#### Remarks
- $M_p$ depends only on $\xi$, not on $\omega_n$.
- When $\xi$ increases, $M_p$ decreases: e.g. $\xi = 0.5$ gives 16.3%, $\xi = 0.707$ gives 4.3%, and $\xi \ge 1$ gives no overshoot.

## Check yourself

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

1. How is the time of the first peak found?
   A. Set c(t) = 1
   B. Set dc/dt = 0, giving sin(wd*t)=0 and wd*t = pi
   C. Set e^(-xi*wn*t) = 0
   D. Set t = 4/(xi*wn)
   Answer: B. Peaks occur where the slope is zero; the first one is at wd*t = pi. (Key point: Peak condition)

2. Peak time is
   A. pi/(wn*sqrt(1-xi^2))
   B. pi/wn
   C. 2*pi/wd
   D. 4/(xi*wn)
   Answer: A. tp = pi/wd with wd = wn*sqrt(1-xi^2). (Key point: Peak time)

3. At t = tp, sin(wd*tp + theta) = sin(pi + theta) equals
   A. sin(theta) = sqrt(1-xi^2)
   B. -sin(theta) = -sqrt(1-xi^2)
   C. cos(theta) = xi
   D. 0
   Answer: B. sin(pi+theta) = -sin(theta), which cancels the sqrt(1-xi^2) in the denominator and makes the term positive. (Key point: Key substitution)

4. Maximum overshoot Mp is
   A. e^(-xi*pi/sqrt(1-xi^2))
   B. e^(-xi*wn*pi)
   C. 1 + e^(-xi*pi/sqrt(1-xi^2))
   D. e^(-pi/xi)
   Answer: A. c(tp) = 1 + e^(-xi*pi/sqrt(1-xi^2)); subtracting the final value 1 gives Mp. It depends only on xi. (Key point: Result)

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