---
title: "FT of e^(−at)u(t); amplitude and phase"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-one-sided-exponential"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-one-sided-exponential
site: BCT Engineering
---

# FT of e^(−at)u(t); amplitude and phase

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2076 Bhadra, 6+2 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-one-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^(−at)u(t), where a is real constant. Also, draw its amplitude and phase spectra.

## Short answer

For a > 0, the Fourier transform is 1/(a + jω), with magnitude 1/sqrt(a² + ω²) and phase -arctan(ω/a). The amplitude spectrum is even, the phase spectrum is odd, and the 3 dB bandwidth is ω = a.

## Answer

For the transform to exist, $a > 0$ (otherwise $e^{-at}u(t)$ grows and is not absolutely integrable).

#### Fourier transform

$$
\begin{aligned}
X(j\omega) &= \int_{-\infty}^{\infty} e^{-at}u(t)\,e^{-j\omega t}\,dt = \int_{0}^{\infty} e^{-(a+j\omega)t}\,dt \\
&= \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty} \\
&= 0 - \frac{1}{-(a+j\omega)} = \frac{1}{a+j\omega}, \quad a > 0
\end{aligned}
$$

The upper limit is zero because $|e^{-(a+j\omega)t}| = e^{-at} \to 0$ as $t \to \infty$.

**Magnitude and phase:**

$$
|X(j\omega)| = \frac{1}{\sqrt{a^2+\omega^2}}, \qquad \angle X(j\omega) = -\tan^{-1}\left(\frac{\omega}{a}\right)
$$

| $\omega$ | $\lvert X\rvert$ | Phase |
|---|---|---|
| $0$ | $1/a$ | $0$ |
| $\pm a$ | $\frac{1}{\sqrt2\,a} = 0.707/a$ | $\mp 45^\circ$ |
| $\to \pm\infty$ | $\to 0$ | $\to \mp 90^\circ$ |

#### Amplitude and phase spectra

```text
 |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 amplitude spectrum is even and the phase spectrum is odd, as expected for a real signal.
- The spectrum is low-pass, with 3 dB bandwidth $\omega = a$ rad/s. Faster decay (larger $a$) gives a wider bandwidth.

## Check yourself

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

1. Why must a > 0 for the FT of e^(-at)u(t)?
   A. Otherwise the signal is non-causal
   B. Otherwise u(t) is undefined
   C. Otherwise the FT is purely imaginary
   D. Otherwise the signal grows and is not absolutely integrable
   Answer: D. For a ≤ 0 the signal does not decay, so the Fourier integral does not converge. (Key point: Existence)

2. What is X(jω)?
   A. 1/(a + jω)
   B. 1/(a - jω)
   C. a/(a^2 + ω^2)
   D. 1/(a^2 + ω^2)
   Answer: A. ∫0^∞ e^(-(a+jω)t) dt = 1/(a+jω), the upper limit vanishing because e^(-at) → 0. (Key point: Transform)

3. What are |X| and the phase at ω = a?
   A. 1/a and 0°
   B. 0.707/a and -45°
   C. 0.5/a and -90°
   D. 1/(2a) and +45°
   Answer: B. |X| = 1/sqrt(a^2 + a^2) = 0.707/a and phase = -tan^-1(1) = -45°. (Key point: Spectra values)

4. Which symmetry holds for this real signal's spectra?
   A. Magnitude odd, phase even
   B. Both even
   C. Magnitude even, phase odd
   D. Both odd
   Answer: C. For any real x(t), |X(jω)| is even and ∠X(jω) is odd. (Key point: Symmetry)

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