---
title: "Time-domain specifications of 25/(s²+8s+25)"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:tr-tp-ts-mp-25-s2"
source: https://www.bctengineer.com/q/semester_3/control_systems/tr-tp-ts-mp-25-s2
site: BCT Engineering
---

# Time-domain specifications of 25/(s²+8s+25)

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2081 Baisakh, 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/tr-tp-ts-mp-25-s2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

For a second order system given by the transfer function G(s) = 25/(s² + 8s + 25), find the rise time, peak time, settling time and maximum overshoot.

## Short answer

tr = 0.833 s, tp = 1.047 s, ts = 1 s (2%), Mp = 1.52%.

## Answer

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

- $\omega_n^2=25 \Rightarrow \omega_n = 5$ rad/s
- $2\zeta\omega_n = 8 \Rightarrow \zeta = 0.8$ (underdamped)
- $\omega_d = \omega_n\sqrt{1-\zeta^2} = 5\times0.6 = 3$ rad/s
- $\theta = \cos^{-1}(0.8) = 36.87^\circ = 0.6435$ rad

$$
\begin{aligned}
t_r &= \frac{\pi-\theta}{\omega_d} = \frac{3.1416-0.6435}{3} = 0.833\ \text{s} \\
t_p &= \frac{\pi}{\omega_d} = \frac{3.1416}{3} = 1.047\ \text{s} \\
t_s &= \frac{4}{\zeta\omega_n} = \frac{4}{0.8\times5} = 1\ \text{s}\quad(2\%\ \text{criterion}) \\
M_p &= e^{-\zeta\pi/\sqrt{1-\zeta^2}} = e^{-0.8\pi/0.6} = e^{-4.189} = 0.0152
\end{aligned}
$$

With the 5% criterion, $t_s = \frac{3}{\zeta\omega_n} = 0.75$ s.

**Answer:** $t_r = 0.833$ s, $t_p = 1.047$ s, $t_s = 1$ s (2%), $M_p = 1.52\%$.

## Check yourself

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

1. For 25/(s^2+8s+25), zeta and wn are
   A. zeta=0.8, wn=5
   B. zeta=1.6, wn=5
   C. zeta=0.8, wn=25
   D. zeta=0.4, wn=5
   Answer: A. wn^2=25 and 2*zeta*wn=8. (Key point: Parameters)

2. wd equals
   A. 4 rad/s
   B. 5 rad/s
   C. 3 rad/s
   D. 0.6 rad/s
   Answer: C. wd = 5*sqrt(1-0.64) = 3. (Key point: Damped frequency)

3. Peak time tp equals
   A. 0.833 s
   B. 1.047 s
   C. 1 s
   D. 1.57 s
   Answer: B. tp = pi/wd = 1.047 s. (Key point: Peak time)

4. With the 2% criterion, ts and Mp are
   A. 0.75 s, 1.52%
   B. 1 s, 15.2%
   C. 1 s, 1.52%
   D. 1.25 s, 1.52%
   Answer: C. ts = 4/(0.8*5) = 1 s; Mp = e^(-4.189) = 1.52%. (Key point: Settling and overshoot)

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