---
title: "Overshoot, settling and peak time of rotational system"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:step-torque-j1-b1-k1"
source: https://www.bctengineer.com/q/semester_3/control_systems/step-torque-j1-b1-k1
site: BCT Engineering
---

# Overshoot, settling and peak time of rotational system

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2067 Asar (old course), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/step-torque-j1-b1-k1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

A step torque T(t) is applied in a system shown in figure below. Find the percent overshoot, settling time and peak time for output θ2(t). [Figure: torque T(t) applied to inertia 1 kg-m² (angle θ1(t)); inertia connected through viscous damper 1 N-m-s/rad to a point of angle θ2(t), which is connected through a torsional spring 1 N-m/rad to the fixed wall]

## Short answer

The closed-loop transfer function is 1/(s²+s+1) giving ωn=1 rad/s and ζ=0.5. Percent overshoot = 16.3%, settling time (2%) = 8 s, and peak time = 3.63 s.

## Answer

Write the torque equations at each angle, eliminate $\theta_1$ to get $\Theta_2(s)/T(s)$, then compare with the standard second order form.

#### Equations of motion
At $\theta_1$ (inertia $J = 1$, damper $D = 1$ between $\theta_1$ and $\theta_2$):
$$
J\ddot{\theta}_1 + D(\dot{\theta}_1 - \dot{\theta}_2) = T
\;\Rightarrow\;
(s^2 + s)\Theta_1 - s\Theta_2 = T(s)
$$
At $\theta_2$ (no inertia; damper and spring $K = 1$ to the wall):
$$
D(\dot{\theta}_2 - \dot{\theta}_1) + K\theta_2 = 0
\;\Rightarrow\;
-s\Theta_1 + (s + 1)\Theta_2 = 0
$$

#### Transfer function
From the second equation, $\Theta_1 = \dfrac{s+1}{s}\Theta_2$. Substituting in the first:
$$
\begin{aligned}
(s^2 + s)\frac{s+1}{s}\Theta_2 - s\Theta_2 &= T \\
\left[(s+1)^2 - s\right]\Theta_2 &= T \\
\frac{\Theta_2(s)}{T(s)} &= \frac{1}{s^2 + s + 1}
\end{aligned}
$$

#### Parameters
$$
\omega_n^2 = 1 \Rightarrow \omega_n = 1\ \text{rad/s}, \qquad 2\zeta\omega_n = 1 \Rightarrow \zeta = 0.5
$$

#### Percent overshoot
$$
\%OS = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 100\,e^{-\pi(0.5)/0.866} = 100\,e^{-1.814} = 16.3\%
$$

#### Settling time (2% criterion)
$$
t_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.5 \times 1} = 8\ \text{s}
$$

#### Peak time
$$
t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} = \frac{3.1416}{0.866} = 3.63\ \text{s}
$$

**Answer:** $\%OS = 16.3\%$, $t_s = 8$ s, $t_p = 3.63$ s.

## Check yourself

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

1. Which equation applies at the inertia node theta1?
   A. J*theta1'' + D*(theta1' - theta2') = T
   B. J*theta1'' + K*theta1 = T
   C. D*theta2' + K*theta2 = T
   D. J*theta1'' = T
   Answer: A. The inertia is driven by T and loaded by the damper to theta2. (Key point: Torque equation at theta1)

2. Eliminating theta1 gives Theta2/T equal to
   A. 1/(s^2+s)
   B. 1/(s^2+s+1)
   C. 1/(s^2+2s+1)
   D. s/(s^2+s+1)
   Answer: B. [(s+1)^2 - s]Theta2 = T, i.e. s^2+s+1. (Key point: Transfer function)

3. From s^2+s+1: wn and zeta are
   A. wn=1, zeta=1
   B. wn=2, zeta=0.25
   C. wn=1, zeta=0.5
   D. wn=0.5, zeta=1
   Answer: C. wn^2 = 1 and 2*zeta*wn = 1 give zeta = 0.5. (Key point: Parameters)

4. With zeta=0.5, wn=1: %OS, ts, tp are
   A. 16.3%, 8 s, 3.63 s
   B. 4.3%, 8 s, 3.14 s
   C. 16.3%, 4 s, 3.63 s
   D. 16.3%, 8 s, 3.14 s
   Answer: A. %OS = 16.3%, ts = 4/0.5 = 8 s, tp = pi/0.866 = 3.63 s. (Key point: Results)

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