---
title: "Overshoot and settling time from pole plot"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:mp-ts-from-pole-plot"
source: https://www.bctengineer.com/q/semester_3/control_systems/mp-ts-from-pole-plot
site: BCT Engineering
---

# Overshoot and settling time from pole plot

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2072 Chaitra, 6 marks
- Video: [Time Response Specifications: Control Systems 2.4](https://www.youtube.com/watch?v=RRhApGSTG58) — CircuitBread
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/mp-ts-from-pole-plot
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Open loop pole/zero plot of a unity feedback system is shown in figure below. Determine maximum overshoot and settling time for its step response. [Figure: s-plane with open-loop poles at s = −2 ± j1; no zeros]

## Short answer

Open-loop poles at -2±j1 give closed-loop real part -2, so settling time ts = 2 s. With K=1, ωn²=6, ζ=0.8165, maximum overshoot Mp ≈ 1.18%.

## Answer

The open-loop poles at $s = -2 \pm j1$ give
$$
G(s) = \frac{K}{(s+2-j)(s+2+j)} = \frac{K}{s^2 + 4s + 5}
$$
No gain is marked on the plot, so take $K = 1$ (and show the general result).

#### Closed-loop transfer function (unity feedback)
$$
\frac{C(s)}{R(s)} = \frac{K}{s^2 + 4s + 5 + K}
$$
Closed-loop poles: $s = -2 \pm j\sqrt{1 + K}$, so $\sigma = \zeta\omega_n = 2$ for every $K$.

For $K = 1$:
$$
\begin{aligned}
\omega_n^2 &= 6 \;\Rightarrow\; \omega_n = 2.449\ \text{rad/s} \\
2\zeta\omega_n &= 4 \;\Rightarrow\; \zeta = \frac{2}{2.449} = 0.8165
\end{aligned}
$$

#### Maximum overshoot
$$
M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = e^{-\pi\sigma/\omega_d} = e^{-2\pi/\sqrt{2}} = e^{-4.443} = 0.0118 = 1.18\%
$$
In general, $M_p = e^{-2\pi/\sqrt{1+K}}$.

#### Settling time (2% criterion)
$$
t_s = \frac{4}{\zeta\omega_n} = \frac{4}{2} = 2\ \text{s}
$$
This is the same for any $K$, because the real part of the closed-loop poles is always $-2$.

**Answer:** with $K = 1$: $M_p \approx 1.18\%$ and $t_s = 2$ s.

(If the plotted poles $-2 \pm j1$ are taken directly as the dominant poles, $\zeta = 2/\sqrt{5} = 0.894$, $M_p = e^{-2\pi} = 0.19\%$, and $t_s$ is still 2 s.)

## Check yourself

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

1. With G(s)=K/(s^2+4s+5) and unity feedback, what is the closed-loop characteristic equation?
   A. s^2+4s+5-K=0
   B. s^2+4s+5+K=0
   C. s^2+4s+K=0
   D. s^2+(4+K)s+5=0
   Answer: B. Closed loop is G/(1+G) = K/(s^2+4s+5+K), so only the constant term changes. (Key point: Closed-loop poles)

2. Why is the settling time the same for every K?
   A. Damping ratio never changes with K
   B. Natural frequency never changes with K
   C. The real part of the closed-loop poles stays at -2
   D. The overshoot is independent of K
   Answer: C. The s coefficient is fixed at 4, so sigma = zeta*wn = 2 for any K, and ts = 4/sigma = 2 s. (Key point: Real part of poles)

3. For K=1, wn^2=6 and zeta=0.8165. The maximum overshoot is about
   A. 19%
   B. 0.19%
   C. 1.18%
   D. 11.8%
   Answer: C. Mp = exp(-pi*zeta/sqrt(1-zeta^2)) = exp(-4.443) = 0.0118 = 1.18%. (Key point: Overshoot for K=1)

4. Which formula gives the 2% settling time used here?
   A. 4/(zeta*wn)
   B. pi/wd
   C. 3/(zeta*wn)
   D. 1/(zeta*wn)
   Answer: A. The 2% criterion uses ts = 4/(zeta*wn) = 4/2 = 2 s. (Key point: Settling time)

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