--- title: "Underdamped second order unit step response derivation" subject: "Control Systems" chapter: 3 key: "semester_3/control_systems:second-order-step-derivation" source: https://www.bctengineer.com/q/semester_3/control_systems/second-order-step-derivation site: BCT Engineering --- # Underdamped second order unit step response derivation Control Systems · Chapter 3: System Transfer Function and Responses - Asked in: 2071 Shrawan, 4 marks - Video: [unit step response second order system under damped derivation](https://www.youtube.com/watch?v=19eCOPJdNKs) — techgurukula - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/second-order-step-derivation - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3 ## Question For an open loop transfer function with unity feedback G(s) = ωn²/[s(s + 2ξωn)] where ξ < 1, derive an expression for output when unit step input is applied. ## Short answer For G(s)=ω_n²/[s(s+2ζω_n)] with ζ<1, the unit step response is c(t)=1 – (e^{-ζω_n t}/√(1−ζ²)) sin(ω_d t+θ), where ω_d = ω_n√(1−ζ²) and θ = tan⁻¹(√(1−ζ²)/ζ). ## Answer For $\xi < 1$ (underdamped) the unit step response is a damped sinusoid that settles at 1. #### Closed-loop transfer function $$ \frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{\omega_n^2}{s^2 + 2\xi\omega_n s + \omega_n^2} $$ For $\xi < 1$ the poles are complex: $$ s = -\xi\omega_n \pm j\omega_n\sqrt{1-\xi^2} = -\xi\omega_n \pm j\omega_d, \qquad \omega_d = \omega_n\sqrt{1-\xi^2} $$ #### Output for a unit step With $R(s) = 1/s$: $$ C(s) = \frac{\omega_n^2}{s(s^2 + 2\xi\omega_n s + \omega_n^2)} $$ Partial fractions: $$ \begin{aligned} C(s) &= \frac{1}{s} - \frac{s + 2\xi\omega_n}{s^2 + 2\xi\omega_n s + \omega_n^2} \\ &= \frac{1}{s} - \frac{s + \xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} - \frac{\xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} \end{aligned} $$ Using $\mathcal{L}^{-1}\left\{\frac{s+a}{(s+a)^2+\omega^2}\right\} = e^{-at}\cos\omega t$ and $\mathcal{L}^{-1}\left\{\frac{\omega}{(s+a)^2+\omega^2}\right\} = e^{-at}\sin\omega t$: $$ \begin{aligned} c(t) &= 1 - e^{-\xi\omega_n t}\cos\omega_d t - \frac{\xi\omega_n}{\omega_d}e^{-\xi\omega_n t}\sin\omega_d t \\ &= 1 - e^{-\xi\omega_n t}\left[\cos\omega_d t + \frac{\xi}{\sqrt{1-\xi^2}}\sin\omega_d t\right] \end{aligned} $$ Put $\cos\theta = \xi$ and $\sin\theta = \sqrt{1-\xi^2}$, so that $\cos\omega_d t\,\sin\theta + \sin\omega_d t\,\cos\theta = \sin(\omega_d t + \theta)$: $$ c(t) = 1 - \frac{e^{-\xi\omega_n t}}{\sqrt{1-\xi^2}}\sin(\omega_d t + \theta), \qquad \theta = \tan^{-1}\frac{\sqrt{1-\xi^2}}{\xi} $$ The response is a sine wave at $\omega_d$ that decays inside the envelopes $1 \pm e^{-\xi\omega_n t}/\sqrt{1-\xi^2}$ and settles at the final value 1, so the steady state error is zero. ```text c(t) | _ | / \ _ 1|--/---\---/-\--------- final value | / \_/ |/ +------------------------ t ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For unity feedback with G = wn^2/[s(s+2*xi*wn)], C/R is A. wn^2/(s^2+2*xi*wn*s) B. wn^2/(s^2+2*xi*wn*s+wn^2) C. 1/(s^2+2*xi*wn*s+wn^2) D. wn^2/(s+2*xi*wn) Answer: B. G/(1+G) gives denominator s^2+2*xi*wn*s+wn^2. (Key point: Closed-loop TF) 2. For xi<1, the closed-loop poles are A. -xi*wn +/- j*wn*sqrt(1-xi^2) B. +/- j*wn C. -wn and -xi*wn D. -xi*wn +/- wn*sqrt(xi^2-1) Answer: A. Complex poles with real part -xi*wn and imaginary part wd = wn*sqrt(1-xi^2). (Key point: Pole location) 3. The unit step response for xi<1 is A. 1 - e^(-xi*wn*t)*sin(wd*t) B. 1 - e^(-wn*t)*cos(wd*t) C. 1 - [e^(-xi*wn*t)/sqrt(1-xi^2)]*sin(wd*t + theta) D. 1 + e^(-xi*wn*t)*sin(wd*t) Answer: C. Partial fractions and inverse Laplace give this damped sinusoid, with theta = atan(sqrt(1-xi^2)/xi). (Key point: Step response) 4. What is the steady state error for this step response? A. Zero, since c(t) settles at 1 B. Equal to xi C. Infinite D. 1/wn Answer: A. The exponential term decays, so c(t) approaches 1. (Key point: Final value) ## Related in Control Systems - [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 - [Closed vs open loop step response analytically](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Acl-vs-ol-step-response) — Control Systems, chapter 1, asked 2× - [Unit step response of 4/[s(s+2)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Aunit-step-4-s-s2) — Control Systems, chapter 3 - [Derive maximum overshoot of second order system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amp-expression-derivation) — Control Systems, chapter 3 - [Derivative feedback in second order time response](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Aderivative-feedback-second-order) — Control Systems, chapter 7 - [Transfer function with G3 feedforward, H1 and unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2g3-h1-unity-2070o) — Control Systems, chapter 3 ## Related in other subjects - [Second-order step response and damping](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Asecond-order-step-response-damping) — Instrumentation and Measurement, chapter 4 - [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 - [Digital PI design, plant e^(−5s)/(s+0.4), step response](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Api-design-dead-time-5s-step) — Digital Control System, chapter 4 - [Controller design with 8 samples per cycle, step response and Kv](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adesign-8-samples-step-kv) — Digital Control System, chapter 4