---
title: "Find K and T from step response (overshoot 0.254)"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:k-t-from-step-0254"
source: https://www.bctengineer.com/q/semester_3/control_systems/k-t-from-step-0254
site: BCT Engineering
---

# Find K and T from step response (overshoot 0.254)

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2078 Kartik, 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/k-t-from-step-0254
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

The system below in figure (a), when subjected to unit step input, gives the output response as shown in figure (b). Determine the value of K and T. [Figure (a): R(s) → summing point (+, −) → K/[s(1+sT)] → C(s), unity negative feedback. Figure (b): unit step response with final value 1, peak overshoot 0.254 above the final value at t = 3 sec]

## Short answer

From peak overshoot 0.254, damping ratio ζ=0.4. Peak time 3 s gives ωd=1.0472 rad/s, thus ωn=1.1425 rad/s. Using the closed-loop form ωn²=K/T and 2ζωn=1/T, T=1.094 s and K≈1.43.

## Answer

From the response curve, peak overshoot $M_p = 0.254$ (25.4%) and peak time $t_p = 3$ s. These give $\zeta$ and $\omega_n$, which are then matched with the closed-loop transfer function.

#### Closed-loop transfer function

$$
\frac{C(s)}{R(s)} = \frac{\frac{K}{s(1+sT)}}{1+\frac{K}{s(1+sT)}} = \frac{K}{Ts^2+s+K} = \frac{K/T}{s^2+\frac{1}{T}s+\frac{K}{T}}
$$

Comparing with $\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$:

$$
\omega_n^2 = \frac{K}{T}, \qquad 2\zeta\omega_n = \frac{1}{T}
$$

#### Damping ratio

$$
\begin{aligned}
e^{-\zeta\pi/\sqrt{1-\zeta^2}} &= 0.254 \\
\frac{\zeta\pi}{\sqrt{1-\zeta^2}} &= -\ln 0.254 = 1.3704 \\
\zeta &= \frac{1.3704}{\sqrt{\pi^2+1.3704^2}} = 0.400
\end{aligned}
$$

#### Natural frequency

$$
\begin{aligned}
t_p &= \frac{\pi}{\omega_d} = 3 \Rightarrow \omega_d = 1.0472\ \text{rad/s} \\
\omega_n &= \frac{\omega_d}{\sqrt{1-\zeta^2}} = \frac{1.0472}{\sqrt{1-0.16}} = 1.1425\ \text{rad/s}
\end{aligned}
$$

#### K and T

$$
\begin{aligned}
T &= \frac{1}{2\zeta\omega_n} = \frac{1}{2\times0.4\times1.1425} = 1.094\ \text{s} \\
K &= \omega_n^2 T = 1.3053\times1.094 = 1.428
\end{aligned}
$$

**Answer:** $\zeta = 0.4$, $\omega_n = 1.142$ rad/s, giving $T \approx 1.094$ s and $K \approx 1.43$.

## 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, C/R is
   A. K/(Ts^2+s+K)
   B. K/(Ts^2+s)
   C. K/(s^2+Ts+K)
   D. KT/(Ts^2+s+K)
   Answer: A. C/R = G/(1+G) = K/(s(1+sT)+K) = K/(Ts^2+s+K). (Key point: Closed-loop TF)

2. Comparing with the standard form gives
   A. wn^2 = K and 2*zeta*wn = T
   B. wn^2 = K/T and 2*zeta*wn = 1/T
   C. wn^2 = KT and 2*zeta*wn = T
   D. wn^2 = K/T and 2*zeta*wn = T
   Answer: B. Dividing by T: s^2 + (1/T)s + K/T, so wn^2 = K/T and 2*zeta*wn = 1/T. (Key point: Matching coefficients)

3. For Mp = 0.254, zeta is about
   A. 0.254
   B. 0.400
   C. 0.600
   D. 0.800
   Answer: B. zeta*pi/sqrt(1-zeta^2) = -ln(0.254) = 1.3704, giving zeta = 0.40. (Key point: Damping ratio from overshoot)

4. With tp = 3 s and zeta = 0.4, wn is about
   A. 1.047 rad/s
   B. 1.1425 rad/s
   C. 2.5 rad/s
   D. 0.419 rad/s
   Answer: B. wd = pi/3 = 1.0472, wn = wd/sqrt(1-0.16) = 1.1425 rad/s. (Key point: wn from peak time)

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