---
title: "Fourier transform of e^(−a|t|)"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-two-sided-exponential"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-two-sided-exponential
site: BCT Engineering
---

# Fourier transform of e^(−a|t|)

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2082 Chaitra, 5 marks; 2081 Chaitra, 5 marks; 2079 Asoj, 5 marks; 2079 Chaitra, 5 marks; 2078 Chaitra, 5 marks (asked 5 times)
- 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-two-sided-exponential
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Find the Fourier transform of the signal x(t) = e^(−a|t|), a > 0 and also plot the frequency spectrum.

## Short answer

X(ω) = 2a/(a²+ω²). The magnitude |X(ω)| = 2a/(a²+ω²) is even, peak 2/a at ω=0, half-maximum at ω=±a, and decays as 1/ω²; phase is zero for all ω.

## Answer

The signal $x(t) = e^{-a|t|}$, $a > 0$, is a two-sided decaying exponential. It is real and even:
$$
x(t) = \begin{cases} e^{-at}, & t \ge 0 \\ e^{at}, & t < 0 \end{cases}
$$

#### Fourier transform
$$
\begin{aligned}
X(\omega) &= \int_{-\infty}^{\infty} e^{-a|t|}e^{-j\omega t}dt = \int_{-\infty}^{0} e^{at}e^{-j\omega t}dt + \int_{0}^{\infty} e^{-at}e^{-j\omega t}dt \\
&= \left[\frac{e^{(a-j\omega)t}}{a-j\omega}\right]_{-\infty}^{0} + \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty} \\
&= \frac{1}{a-j\omega} + \frac{1}{a+j\omega} = \frac{(a+j\omega) + (a-j\omega)}{a^2+\omega^2}
\end{aligned}
$$
Both limits at $\pm\infty$ vanish because $a > 0$.
$$
\boxed{X(\omega) = \frac{2a}{a^2+\omega^2}}
$$

#### Spectrum
$X(\omega)$ is real, positive and even, so:
- **Magnitude:** $|X(\omega)| = \frac{2a}{a^2+\omega^2}$. The peak is $X(0) = 2/a$ (equal to the area under $x(t)$). At $\omega = \pm a$ it falls to half the peak, $1/a$, and it decays as $1/\omega^2$.
- **Phase:** $\angle X(\omega) = 0$ for all $\omega$.

| $\omega$ | 0 | ±a | ±2a | ±3a |
|---|---|---|---|---|
| $X(\omega)$ | 2/a | 1/a | 0.4/a | 0.2/a |

```text
 x(t)                       X(w)
  1        *                 2/a       *
          / \                         / \
        /     \              1/a    /     \
   __ /         \ __            __/         \__
 ---------+---------- t     ----+----+----+----- w
          0                    -a    0    a
```

**Observation:** a larger $a$ means faster decay in time and a wider spectrum, which shows the inverse relation between time duration and bandwidth. The spectrum has a Lorentzian shape, and by duality $\frac{2a}{a^2+t^2} \leftrightarrow 2\pi e^{-a|\omega|}$.

## Check yourself

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

1. What is X(ω)?
   A. 1/(a + jω)
   B. 2a/(a^2 + ω^2)
   C. 2/(a^2 + ω^2)
   D. a/(a^2 - ω^2)
   Answer: B. 1/(a-jω) + 1/(a+jω) = 2a/(a^2 + ω^2). (Key point: Result)

2. Why is the phase of X(ω) zero for all ω?
   A. X(ω) is real and positive because x(t) is real and even
   B. Because a > 0 only
   C. Because the signal is causal
   D. Because |X(ω)| is decreasing
   Answer: A. A real even signal has a real transform, and 2a/(a^2+ω^2) is positive. (Key point: Phase and symmetry)

3. What is the peak value X(0) and the value at ω = ±a?
   A. 2/a and 1/a
   B. 1/a and 1/(2a)
   C. 2a and a
   D. 2/a and 2/(a^2)
   Answer: A. X(0) = 2a/a^2 = 2/a, and X(a) = 2a/(2a^2) = 1/a, half the peak. (Key point: Spectrum values)

4. What happens to the spectrum when a increases?
   A. It becomes narrower
   B. It becomes wider because the signal decays faster in time
   C. It stays the same
   D. Its peak moves to ω = a^2
   Answer: B. Faster decay in time means wider bandwidth: duration and bandwidth are inversely related. (Key point: Time-bandwidth relation)

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