---
title: "Find M, B and K from step response"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:mbk-from-step-response"
source: https://www.bctengineer.com/q/semester_3/control_systems/mbk-from-step-response
site: BCT Engineering
---

# Find M, B and K from step response

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/mbk-from-step-response
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

For the system as in figure (a), the unit step response is as in figure (b), determine M, B and K. [Figure (a): mass M hanging from a fixed support through spring K, with a damper B from M to the fixed ground below; displacement x downward; input is a force on M. Figure (b): unit step response x(t) rises to a peak of 2.36 at t = 2 sec and settles to a final value of 2]

## Short answer

Final value 2 gives K = 0.5 N/m. Overshoot 0.18 yields damping ratio ζ ≈ 0.479. Peak time 2 s gives ωn ≈ 1.79 rad/s. Then M = K/ωn² ≈ 0.156 kg and B = 2ζωn M ≈ 0.268 N-s/m.

## Answer

For a force input the mass-spring-damper is a standard second order system. The final value gives $K$, the overshoot gives $\zeta$, and the peak time gives $\omega_n$.

#### Transfer function
Force balance on $M$: $F = M\ddot{x} + B\dot{x} + Kx$, so
$$
\frac{X(s)}{F(s)} = \frac{1}{Ms^2 + Bs + K} = \frac{1/M}{s^2 + \frac{B}{M}s + \frac{K}{M}}
$$
with $\omega_n^2 = K/M$ and $2\zeta\omega_n = B/M$.

#### Step 1: $K$ from the final value
For a unit step force, $F(s) = 1/s$:
$$
x(\infty) = \lim_{s\to 0} s\cdot\frac{1}{s}\cdot\frac{1}{Ms^2+Bs+K} = \frac{1}{K} = 2
\;\Rightarrow\; K = 0.5\ \text{N/m}
$$

#### Step 2: $\zeta$ from the overshoot
$$
M_p = \frac{2.36 - 2}{2} = 0.18
$$
$$
\zeta = \frac{-\ln 0.18}{\sqrt{\pi^2 + (\ln 0.18)^2}} = \frac{1.7148}{\sqrt{9.8696 + 2.9405}} = \frac{1.7148}{3.5791} = 0.4791
$$

#### Step 3: $\omega_n$ from the peak time
$$
t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} = 2
\;\Rightarrow\;
\omega_n = \frac{\pi}{2\sqrt{1 - 0.4791^2}} = \frac{3.1416}{2(0.8778)} = 1.790\ \text{rad/s}
$$

#### Step 4: $M$ and $B$
$$
\begin{aligned}
M &= \frac{K}{\omega_n^2} = \frac{0.5}{1.790^2} = \frac{0.5}{3.2025} = 0.156\ \text{kg} \\
B &= 2\zeta\omega_n M = 2(0.4791)(1.790)(0.156) = 0.268\ \text{N-s/m}
\end{aligned}
$$

**Answer:** $M \approx 0.156$ kg, $B \approx 0.268$ N-s/m, $K = 0.5$ N/m.

## Check yourself

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

1. What is X(s)/F(s) for the mass-spring-damper?
   A. 1/(Ms^2 + Bs + K)
   B. 1/(Ms^2 + Ks + B)
   C. K/(Ms^2 + Bs)
   D. 1/(Bs^2 + Ms + K)
   Answer: A. F = M x'' + B x' + K x gives X/F = 1/(Ms^2 + Bs + K). (Key point: Transfer function)

2. Final value of the unit step response is 2. What is K?
   A. K = 2 N/m
   B. K = 0.5 N/m, since x(inf) = 1/K
   C. K = 1.18 N/m
   D. K = 0.18 N/m
   Answer: B. The final value theorem gives x(inf) = 1/K = 2. (Key point: Finding K)

3. The peak is 2.36 and final value 2. What are Mp and zeta?
   A. Mp = 0.36 and zeta = 0.32
   B. Mp = 0.18 and zeta = 0.18
   C. Mp = 0.18 and zeta = 0.479
   D. Mp = 1.18 and zeta = 0.12
   Answer: C. Mp = (2.36 - 2)/2 = 0.18, then zeta = -ln 0.18/sqrt(pi^2 + ln^2 0.18) = 0.479. (Key point: Overshoot and zeta)

4. With wn = 1.79 rad/s and K = 0.5, how are M and B found?
   A. M = K wn^2 and B = 2 zeta wn
   B. M = wn^2/K and B = zeta wn
   C. M = 1/K and B = 2 zeta wn
   D. M = K/wn^2 = 0.156 kg and B = 2 zeta wn M = 0.268 N-s/m
   Answer: D. wn^2 = K/M gives M, and 2 zeta wn = B/M gives B. (Key point: M and B)

## Related in Control Systems

- [F-V and F-I analogy of mass-spring-damper system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-fi-m-k1-k2-b2) — Control Systems, chapter 2
- [Find M, D, K from 10 lb response](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amdk-from-response-10lb) — Control Systems, chapter 3
- [State model of a mass-spring-damper system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-model-mass-spring-damper) — Control Systems, chapter 8
- [Second order model from step response (peak 1.25)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asecond-order-from-step-125) — Control Systems, chapter 3
- [State space model of vertical two-mass system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-mech-m1-m2-k-b) — Control Systems, chapter 8
- [Vertical two-mass system transfer function X1/F](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-vertical-m1-m2-k1-b1) — Control Systems, chapter 2

## Related in other subjects

- [Spring-mass-damper instrument parameters](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Aspring-mass-damper-instrument-numerical) — Instrumentation and Measurement, chapter 4
- [Unit step response for h(t) = exp(-at)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aunit-step-response-exp) — Data Communication, chapter 2
- [Force method: frame with support settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-settlement-200-ei) — Theory of Structures II, chapter 2
- [Piston with spring: final pressure when volume doubles](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Apiston-spring-gas-k100-volume-doubles) — Fundamental of Thermodynamics & Heat Transfer, chapter 1