---
title: "Gain factor K1 raising overshoot from 25% to 50%"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:k1-overshoot-25-to-50"
source: https://www.bctengineer.com/q/semester_3/control_systems/k1-overshoot-25-to-50
site: BCT Engineering
---

# Gain factor K1 raising overshoot from 25% to 50%

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2067 Asar (old course), 8 marks; 2081 Bhadra, 8 marks; 2080 Bhadra, 8 marks; 2079 Bhadra, 6 marks (asked 4 times)
- 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/k1-overshoot-25-to-50
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

The open loop transfer function of a unity feedback system is given by G(s) = K/[s(1+sT)], where 'K' is the gain constant and 'T' is time constant. With the gain multiplied by a factor K1 the maximum overshoot of the system is increased from 25% to 50%. Determine K1.

## Short answer

Damping ratios for 25% and 50% overshoot are ζ1 = 0.4037 and ζ2 = 0.2155. Since ζ ∝ 1/√K, K1 = (ζ1/ζ2)^2 ≈ 3.51.

## Answer

The closed-loop system is second order; increasing the gain lowers the damping ratio and raises the overshoot.

#### Closed-loop characteristic equation
$$
\frac{C(s)}{R(s)} = \frac{K}{Ts^2 + s + K} = \frac{K/T}{s^2 + \frac{1}{T}s + \frac{K}{T}}
$$
Comparing with $s^2 + 2\zeta\omega_ns + \omega_n^2$:
$$
\omega_n = \sqrt{\frac{K}{T}}, \qquad 2\zeta\omega_n = \frac{1}{T} \;\Rightarrow\; \zeta = \frac{1}{2\sqrt{KT}}
$$
So $\zeta \propto 1/\sqrt{K}$ (for fixed $T$).

#### Damping ratio from overshoot
$$
M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} \;\Rightarrow\; \zeta = \frac{-\ln M_p}{\sqrt{\pi^2 + (\ln M_p)^2}}
$$
- For $M_p = 0.25$: $\ln 0.25 = -1.3863$
$$
\zeta_1 = \frac{1.3863}{\sqrt{9.8696 + 1.9218}} = \frac{1.3863}{3.4339} = 0.4037
$$
- For $M_p = 0.50$: $\ln 0.5 = -0.6931$
$$
\zeta_2 = \frac{0.6931}{\sqrt{9.8696 + 0.4805}} = \frac{0.6931}{3.2172} = 0.2155
$$

#### Gain factor
With gain $K$: $\zeta_1 = \dfrac{1}{2\sqrt{KT}}$. With gain $K_1K$: $\zeta_2 = \dfrac{1}{2\sqrt{K_1KT}}$. Dividing:
$$
\frac{\zeta_1}{\zeta_2} = \sqrt{K_1} \;\Rightarrow\; K_1 = \left(\frac{\zeta_1}{\zeta_2}\right)^2 = \left(\frac{0.4037}{0.2155}\right)^2 = (1.8738)^2
$$

**Answer:** $K_1 \approx 3.51$. The gain must be multiplied by about 3.51 to raise the peak overshoot from 25% to 50%.

## Check yourself

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

1. For G(s) = K/[s(1+sT)] with unity feedback, how does ζ depend on K?
   A. ζ = 1/(2 sqrt(KT)), so it falls as K rises
   B. ζ = sqrt(KT)/2, so it rises with K
   C. ζ is independent of K
   D. ζ = K/(2T)
   Answer: A. Comparing Ts^2 + s + K with the standard form gives ζ = 1/(2 sqrt(KT)). (Key point: Damping vs gain)

2. Which relation gives ζ from the peak overshoot Mp?
   A. ζ = -ln Mp / sqrt(π^2 + (ln Mp)^2)
   B. ζ = Mp/π
   C. ζ = ln Mp/π
   D. ζ = e^(-Mp)
   Answer: A. Inverting Mp = exp(-πζ/sqrt(1-ζ^2)) gives this expression. (Key point: Overshoot to damping)

3. What damping ratios correspond to 25% and 50% overshoot?
   A. 0.4037 and 0.2155
   B. 0.2155 and 0.4037
   C. 0.25 and 0.50
   D. 0.707 and 0.5
   Answer: A. Lower overshoot means higher damping: 25% gives 0.4037 and 50% gives 0.2155. (Key point: Damping values)

4. What is K1?
   A. (ζ1/ζ2)^2 ≈ 3.51
   B. ζ1/ζ2 ≈ 1.87
   C. (ζ2/ζ1)^2 ≈ 0.28
   D. 2 (overshoot doubles)
   Answer: A. Since ζ ∝ 1/sqrt(K), ζ1/ζ2 = sqrt(K1), so K1 = (0.4037/0.2155)^2 ≈ 3.51. (Key point: Gain factor)

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