--- title: "FT of exp(−a|t|)u(t)" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:ft-exp-abs-step" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-exp-abs-step site: BCT Engineering --- # FT of exp(−a|t|)u(t) Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2075 Baisakh, 5 marks - Video: [Fourier Transform of Basic Signals (Exponential Signals)](https://www.youtube.com/watch?v=WjOlJ2lPZOg) — Neso Academy - 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-exp-abs-step - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Find and sketch the Fourier Transform of exponential signal: x(t) = exp(−a|t|) u(t), a > 0. ## Short answer Because of u(t), the signal equals e^(-at)u(t) for a > 0, so its Fourier transform is 1/(a + jω). Magnitude is 1/sqrt(a² + ω²), phase is -arctan(ω/a), giving an even low-pass magnitude and odd phase. ## Answer Because of $u(t)$, the signal is zero for $t<0$, and for $t>0$ we have $|t| = t$. So $$ x(t) = e^{-a|t|}u(t) = e^{-at}u(t), \quad a > 0 $$ **Fourier transform:** $$ \begin{aligned} X(j\omega) &= \int_{0}^{\infty} e^{-at}\,e^{-j\omega t}\,dt = \int_0^\infty e^{-(a+j\omega)t}dt \\ &= \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_0^\infty = \frac{1}{a+j\omega} \end{aligned} $$ **Magnitude and phase:** $$ |X(j\omega)| = \frac{1}{\sqrt{a^2+\omega^2}}, \qquad \angle X(j\omega) = -\tan^{-1}\frac{\omega}{a} $$ | $\omega$ | $\lvert X\rvert$ | Phase | |---|---|---| | $0$ | $1/a$ | $0^\circ$ | | $\pm a$ | $0.707/a$ | $\mp45^\circ$ | | $\pm\infty$ | $0$ | $\mp90^\circ$ | **Sketches:** ```text x(t) 1 ^. | . | ' . | ' . . _ _ ---+---------------> t 0 |X(jw)| 1/a * * | * 0.707/a *.........* (half-power points, w = +/- a) * | * * | * --------------+--------------> w -a 0 a phase +90 | ...... +45 | . 0 +--------.---------> w | 0 . -45 | . -90 | ...... ``` The magnitude is even, the phase is odd, and the spectrum is low-pass with half-power frequency $\omega = a$. (Note: if the signal were $e^{-a|t|}$ without $u(t)$, the answer would be $\frac{2a}{a^2+\omega^2}$, which is real and even.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does e^(-a|t|)u(t) reduce to? A. e^(-a|t|) for all t B. e^(at)u(-t) C. 2e^(-at)u(t) D. e^(-at)u(t) Answer: D. u(t) zeroes t < 0, and for t > 0, |t| = t. (Key point: Simplifying signal) 2. What is X(jω)? A. 1/(a + jω) B. 2a/(a^2 + ω^2) C. 1/(a - jω) D. a/(a + jω) Answer: A. The integral of e^(-(a+jω)t) over 0 to ∞ gives 1/(a+jω). (Key point: Transform) 3. What is the phase of X(jω)? A. tan^-1(ω/a) B. -tan^-1(ω/a) C. -2 tan^-1(ω/a) D. 0 Answer: B. Phase of 1/(a+jω) is the negative of the angle of a+jω. (Key point: Phase) 4. How would the result differ for e^(-a|t|) without u(t)? A. 1/(a + jω), the same B. a/(a^2 + ω^2) C. 2a/(a^2 + ω^2), real and even D. 2/(a + jω) Answer: C. The two-sided exponential is real and even, so its FT is real and even: 1/(a+jω) + 1/(a-jω). 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