--- title: "FT of unit impulse and rectangular pulse" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:ft-impulse-rectangular-pulse" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-impulse-rectangular-pulse site: BCT Engineering --- # FT of unit impulse and rectangular pulse Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2070 Bhadra, 6 marks - Video: [Fourier Transform of Useful Functions (Unit impulse, Unit Step, Signum and Rectangular Function)](https://www.youtube.com/watch?v=_Sd3sXGUAdY) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-impulse-rectangular-pulse - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Find the Fourier transform of continuous time unit impulse and rectangular pulse. Discuss the result. ## Short answer δ(t) ↔ 1, a flat spectrum. Rectangular pulse of width τ: X(jω) = 2 sin(ωτ/2)/ω = τ sinc(ωτ/2π), a real sinc with zeros at multiples of 2π/τ. ## Answer #### FT of the unit impulse $\delta(t)$ $$ X(j\omega) = \int_{-\infty}^{\infty}\delta(t)\,e^{-j\omega t}\,dt = e^{-j\omega\cdot 0} = 1 $$ using the sifting property $\int\delta(t)f(t)dt = f(0)$. So $\delta(t) \leftrightarrow 1$. **Discussion:** The impulse has a flat spectrum: it contains all frequencies with equal amplitude and zero phase. This is why the response of an LTI system to $\delta(t)$ (the impulse response) gives its full frequency response $H(j\omega)$. #### FT of a rectangular pulse Let $x(t) = 1$ for $|t| < \tau/2$ and $0$ otherwise (width $\tau$, height 1). $$ \begin{aligned} X(j\omega) &= \int_{-\tau/2}^{\tau/2}e^{-j\omega t}dt = \frac{e^{j\omega\tau/2} - e^{-j\omega\tau/2}}{j\omega} \\ &= \frac{2\sin(\omega\tau/2)}{\omega} = \tau\,\frac{\sin(\omega\tau/2)}{\omega\tau/2} \end{aligned} $$ **Discussion:** - The spectrum is a real sinc function, peak value $\tau$ (the pulse area) at $\omega = 0$. - Zeros at $\omega = \pm\frac{2\pi}{\tau}, \pm\frac{4\pi}{\tau}, \dots$; most energy lies in the main lobe $|\omega| < 2\pi/\tau$. - The spectrum extends to infinity: a time-limited signal is not band-limited. - Narrower pulse means wider spectrum. As $\tau \to 0$ with area kept at 1 (height $1/\tau$), the pulse becomes $\delta(t)$ and the sinc flattens to 1, which matches the first result. ```text delta(t) -> 1 rect -> sinc ^ 1 +--------- 1 +--+ tau ^ | | | | .|. ---+---> t ---+-----> w -----+--+-- . . | . . 0 0 -t/2 t/2 -----'--+--'---> w -2pi/t 2pi/t ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the FT of δ(t)? A. δ(ω) B. 1/(jω) C. 1 D. 2π Answer: C. By sifting, ∫δ(t)e^(-jωt)dt = 1. (Key point: Impulse FT) 2. What does δ(t) <-> 1 show? A. Only DC content B. Only high frequencies C. Spectrum zero D. Equal amplitude at all frequencies (flat spectrum) Answer: D. The impulse contains all frequencies equally. (Key point: Meaning) 3. What is the FT of a unit-height pulse of width τ? A. 2 sin(ωτ/2)/ω B. sin(ωτ)/ω C. 2 sin(ωτ)/ω D. τ sin(ωτ/2) Answer: A. Peak value is τ at ω = 0. (Key point: Rect pulse) 4. Where are the first zeros of the rect-pulse spectrum? A. ω = ±π/τ B. ω = ±2π/τ C. ω = ±τ D. ω = ±2πτ Answer: B. sin(ωτ/2) = 0 at ωτ/2 = π. 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