---
title: "Peak and rise time from 40% overshoot"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:mp40-ts4-tp-tr"
source: https://www.bctengineer.com/q/semester_3/control_systems/mp40-ts4-tp-tr
site: BCT Engineering
---

# Peak and rise time from 40% overshoot

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2075 Chaitra, 8 marks
- Video: [Question 2 | Numerical on 2nd Order System to Calculate Rise Time, Peak Time, Maximum Overshoot, etc](https://www.youtube.com/watch?v=hghatcbQdQc) — Guidance Line
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/mp40-ts4-tp-tr
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

A system has 40% overshoot and requires a settling time of 4 seconds when given a step input. Find peak time and rise time.

## Short answer

40% overshoot gives ζ ≈ 0.28; 4 s settling time (2% criterion) gives ζωn=1 so ωn≈3.57 rad/s. Damped frequency ωd≈3.429 rad/s; peak time tp=π/ωd≈0.916 s; rise time tr=(π-cos⁻¹ζ)/ωd≈0.541 s.

## Answer

Overshoot fixes $\zeta$. The settling time fixes $\zeta\omega_n$ (2% criterion, $t_s = 4/\zeta\omega_n$). Then $\omega_d$ gives $t_p$ and $t_r$.

#### Step 1: Damping ratio from $M_p = 40\%$
$$
\zeta = \frac{-\ln 0.40}{\sqrt{\pi^2 + (\ln 0.40)^2}} = \frac{0.9163}{\sqrt{9.8696 + 0.8396}} = \frac{0.9163}{3.2725} = 0.280
$$

#### Step 2: Natural frequency from $t_s = 4$ s (2% criterion)
$$
t_s = \frac{4}{\zeta\omega_n} = 4 \;\Rightarrow\; \zeta\omega_n = 1 \;\Rightarrow\; \omega_n = \frac{1}{0.280} = 3.571\ \text{rad/s}
$$

#### Step 3: Damped frequency
$$
\omega_d = \omega_n\sqrt{1-\zeta^2} = 3.571\sqrt{1 - 0.0784} = 3.571(0.960) = 3.429\ \text{rad/s}
$$

#### Step 4: Peak time
$$
t_p = \frac{\pi}{\omega_d} = \frac{3.1416}{3.429} = 0.916\ \text{s}
$$

#### Step 5: Rise time (0 to 100%)
$$
\theta = \cos^{-1}\zeta = \cos^{-1}(0.280) = 1.287\ \text{rad}
$$
$$
t_r = \frac{\pi - \theta}{\omega_d} = \frac{3.1416 - 1.287}{3.429} = \frac{1.8546}{3.429} = 0.541\ \text{s}
$$

**Answer:** $t_p \approx 0.916$ s and $t_r \approx 0.541$ s ($\zeta = 0.28$, $\omega_n = 3.57$ rad/s).

(With the 5% criterion, $t_s = 3/\zeta\omega_n$, you get $\omega_n = 2.679$ rad/s, $t_p = 1.222$ s and $t_r = 0.721$ s.)

## Check yourself

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

1. What is zeta for 40% overshoot?
   A. 0.40
   B. 0.6
   C. 0.14
   D. 0.28
   Answer: D. zeta = -ln 0.4/sqrt(pi^2 + ln^2 0.4) = 0.9163/3.2725 = 0.28. (Key point: Zeta from overshoot)

2. How does ts = 4 s help find wn?
   A. ts = 4/(zeta wn) gives zeta wn = 1, so wn = 3.57 rad/s
   B. ts = pi/wn gives wn = 0.78
   C. ts = 4 zeta wn gives wn = 3.57
   D. ts = 4/wn gives wn = 1
   Answer: A. With the 2% criterion zeta wn = 4/ts = 1, so wn = 1/0.28. (Key point: Settling time use)

3. What is the peak time?
   A. pi/wn = 0.88 s
   B. pi/wd = 0.916 s
   C. 4/(zeta wn) = 4 s
   D. pi/(zeta wd) = 3.3 s
   Answer: B. wd = wn sqrt(1 - zeta^2) = 3.429, so tp = 3.1416/3.429 = 0.916 s. (Key point: Peak time)

4. How is the 0 to 100% rise time computed?
   A. tr = theta/wd = 0.375 s
   B. tr = pi/wd = 0.916 s
   C. tr = (pi - theta)/wd with theta = cos^-1 zeta, giving 0.541 s
   D. tr = 2.2/wn = 0.62 s
   Answer: C. theta = cos^-1(0.28) = 1.287 rad, so tr = 1.8546/3.429 = 0.541 s. (Key point: Rise time)

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