--- title: "First-order system impulse and step response" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:first-order-system-response" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/first-order-system-response site: BCT Engineering --- # First-order system impulse and step response Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2069 Bhadra, 5 marks - Video: [Unit Step Response of First-Order Systems: Concepts, Derivation, and Graphical Understanding](https://www.youtube.com/watch?v=9_1nprBoEbo) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/first-order-system-response - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question Derive the expression for impulse response and step response of first order continuous time system described by the differential equation τ dy(t)/dt + y(t) = x(t). ## Short answer Using Fourier transforms on tau y'(t)+y(t)=x(t) gives H(w)=1/(1+jw tau). The impulse response is h(t)=(1/tau)e^(-t/tau)u(t); integrating it gives the step response s(t)=(1-e^(-t/tau))u(t), which reaches 63.2% at t=tau. ## Answer **Given:** a first-order system with time constant $\tau$: $$ \tau\frac{dy(t)}{dt} + y(t) = x(t) $$ (e.g. an RC low pass circuit with $\tau = RC$). Assume the system is initially at rest. #### Impulse response Take the Fourier transform, using $\frac{d}{dt} \leftrightarrow j\omega$: $$ \begin{aligned} (j\omega\tau + 1)\,Y(\omega) &= X(\omega) \\ H(\omega) = \frac{Y(\omega)}{X(\omega)} &= \frac{1}{1 + j\omega\tau} = \frac{1/\tau}{1/\tau + j\omega} \end{aligned} $$ Using the pair $e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}$ with $a = 1/\tau$: $$ h(t) = \frac{1}{\tau}e^{-t/\tau}u(t) $$ **Check:** for $t > 0$, $\tau h' + h = -\frac{1}{\tau}e^{-t/\tau} + \frac{1}{\tau}e^{-t/\tau} = 0$, and the jump $h(0^+) = 1/\tau$ makes $\tau\,dh/dt$ contain $\delta(t)$, matching the input. #### Step response $$ \begin{aligned} s(t) &= \int_{-\infty}^{t} h(\lambda)\,d\lambda = \int_0^{t}\frac{1}{\tau}e^{-\lambda/\tau}d\lambda \\ &= \left[-e^{-\lambda/\tau}\right]_0^{t} \\ &= \left(1 - e^{-t/\tau}\right)u(t) \end{aligned} $$ **Features:** - $s(\tau) = 1 - e^{-1} = 0.632$: the output reaches 63.2% of the final value in one time constant. - It reaches about 98% after $4\tau$; final value is 1. - Smaller $\tau$ means a faster response (wider bandwidth $1/\tau$ rad/s). ```text h(t) s(t) 1/tau|\ 1 | .-------- | \ 0.632| .-' | `-. | / | `---___ | / 0 +--------------> t 0 +/-------------> t 0 tau 0 tau ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is H(w) for tau dy/dt + y = x? A. 1/(1 + j w tau) B. 1/(1 - j w tau) C. tau/(1 + j w) D. 1 + j w tau Answer: A. Fourier transform gives (j w tau + 1)Y = X. (Key point: Frequency response) 2. What is h(t)? A. (1 - e^(-t/tau)) u(t) B. e^(-t/tau) u(t) C. (1/tau) e^(-t/tau) u(t) D. tau e^(-t/tau) u(t) Answer: C. With a = 1/tau, H = (1/tau)/(a + jw) inverts to (1/tau) e^(-t/tau) u(t). (Key point: Impulse response) 3. How is s(t) obtained and what is it? A. Differentiate h(t), giving -(1/tau^2) e^(-t/tau) B. Integrate h(t), giving (1 - e^(-t/tau)) u(t) C. Square h(t), giving e^(-2t/tau) D. Integrate h(t), giving e^(-t/tau) u(t) Answer: B. The integral of (1/tau) e^(-lambda/tau) from 0 to t is 1 - e^(-t/tau). (Key point: Step response) 4. What is s(tau)? A. 0.368, i.e. 36.8% of the final value B. 1 C. 0.632, i.e. 63.2% of the final value D. 0.5 Answer: C. s(tau) = 1 - e^-1 = 0.632; smaller tau gives a faster response. 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