---
title: "Second-order step response and damping"
subject: "Instrumentation and Measurement"
chapter: 4
key: "semester_4/instrumentation_and_measurement:second-order-step-response-damping"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/second-order-step-response-damping
site: BCT Engineering
---

# Second-order step response and damping

Instrumentation and Measurement · Chapter 4: Dynamic Response of Measurement System

- Asked in: year not given, 8 marks
- Video: [unit step response second order system](https://www.youtube.com/watch?v=KwUlz5MZuXM) — techgurukula
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/second-order-step-response-damping
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-4

## Question

Derive the response of an underdamped second-order measurement system to a unit step input. Sketch and compare the responses for underdamped, critically damped and overdamped cases, and explain why a damping ratio of about 0.7 is preferred in instruments.

## Short answer

Underdamped (ζ<1) response has overshoot y = 1 - e^(-ζωn t) sin(ωd t + φ)/√(1-ζ²). Critical damping (ζ=1) gives fastest non-oscillatory response. Overdamped (ζ>1) is slow. Damping ratio ≈0.7 yields only 4.6% overshoot and near-minimum settling time with good bandwidth.

## Answer

The second-order model in standard form is

$$
\frac{1}{\omega_n^2}\ddot{q}_o + \frac{2\zeta}{\omega_n}\dot{q}_o + q_o = K q_i
$$

For a step input $q_i = q_s$, $q_o(0) = \dot q_o(0) = 0$. Let $y = q_o/(Kq_s)$, so $y$ tends to 1.

#### Underdamped case ($0 < \zeta < 1$)
The characteristic roots are $s = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}$. With damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$, the solution is $y = 1 + e^{-\zeta\omega_n t}(A\cos\omega_d t + B\sin\omega_d t)$. The conditions $y(0) = 0$ and $\dot y(0) = 0$ give $A = -1$ and $B = -\zeta\omega_n/\omega_d$:

$$
y(t) = 1 - \frac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}}\sin\left(\omega_d t + \phi\right), \qquad \phi = \cos^{-1}\zeta
$$

Setting $\dot y = 0$ gives peak time $t_p = \pi/\omega_d$, and the peak overshoot

$$
M_p = \exp\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}\right)
$$

#### Other cases
- **Critically damped ($\zeta = 1$):** $y = 1 - (1 + \omega_n t)e^{-\omega_n t}$. Fastest approach to the final value without overshoot.
- **Overdamped ($\zeta > 1$):** $y = 1 - \dfrac{e^{-\zeta\omega_n t}}{2\sqrt{\zeta^2-1}}\left[\dfrac{e^{\omega_n\sqrt{\zeta^2-1}\,t}}{\zeta - \sqrt{\zeta^2-1}} - \dfrac{e^{-\omega_n\sqrt{\zeta^2-1}\,t}}{\zeta + \sqrt{\zeta^2-1}}\right]$. No overshoot but slow.

```text
 y
 1.4|  .-.            zeta = 0.2
 1.0|-/---`-.__.---- zeta = 0.7
    | /  _.--------- zeta = 1
    |/ .'  ........  zeta = 2 (slow)
    +----------------> w_n t
```

| Case | $\zeta$ | Overshoot | Speed |
|---|---|---|---|
| Underdamped | $<1$ | Yes (large for small $\zeta$) | Fast, oscillatory |
| Critical | $=1$ | None | Fastest non-oscillatory |
| Overdamped | $>1$ | None | Sluggish |

#### Why $\zeta \approx 0.7$
At $\zeta = 0.7$ the overshoot is only 4.6%, the 2% settling time is near its minimum, and the amplitude response is flat within 1% up to about $0.4\,\omega_n$ with a phase lag nearly proportional to frequency (little distortion). Hence most instruments use $\zeta = 0.6$ to $0.7$.

## Check yourself

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

1. For an underdamped system, the damped frequency is
   A. ωn sqrt(1-ζ^2)
   B. ωn / sqrt(1-ζ^2)
   C. ζ ωn
   D. ωn (1-ζ)
   Answer: A. ωd = ωn sqrt(1-ζ^2); the roots are -ζωn ± jωd. (Key point: Damped frequency)

2. Peak overshoot for the underdamped step response is
   A. exp(-πζ/sqrt(1-ζ^2))
   B. exp(πζ/sqrt(1-ζ^2))
   C. 1 - ζ
   D. ζ/sqrt(1-ζ^2)
   Answer: A. Mp = exp(-πζ/sqrt(1-ζ^2)), which is large for small ζ. (Key point: Overshoot)

3. Which is true of critical damping compared with overdamping?
   A. Critical gives fastest approach without overshoot; overdamped is sluggish
   B. Critical overshoots; overdamped does not
   C. Both overshoot
   D. Overdamped is the fastest
   Answer: A. At ζ = 1 there is no overshoot and the response is the fastest non-oscillatory; ζ > 1 is slow. (Key point: Critical vs over damping)

4. Why is ζ of about 0.7 preferred in instruments?
   A. Zero overshoot
   B. Small overshoot (4.6%), near-minimum settling time and flat amplitude response with nearly linear phase
   C. Slowest response for stability
   D. Infinite bandwidth
   Answer: B. At 0.7 overshoot is only 4.6%, settling is near minimum and distortion is small. (Key point: Why 0.7)

## Related in Instrumentation and Measurement

- [Second-order transducer frequency and step response](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Asecond-order-transducer-numerical) — Instrumentation and Measurement, chapter 4
- [Step response of a first-order system](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Afirst-order-step-response-time-constant) — Instrumentation and Measurement, chapter 4
- [Characteristics and response of zero-order system](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Azero-order-system-characteristics) — Instrumentation and Measurement, chapter 4
- [Models of zero, first and second order systems](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Azero-first-second-order-models) — Instrumentation and Measurement, chapter 4

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