---
title: "Fourier transform of e^(−2|t−1|)"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-shifted-two-sided-exponential"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-shifted-two-sided-exponential
site: BCT Engineering
---

# Fourier transform of e^(−2|t−1|)

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2072 Magh, 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-shifted-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^(−2|t−1|)

## Short answer

First find FT of e^{-2|t|} = 4/(4+ω^2). By time-shift property, x(t)=e^{-2|t−1|} ↔ X(jω) = 4 e^{-jω}/(4+ω^2).

## Answer

**Method:** Find the FT of $e^{-2|t|}$, then use the time-shifting property.

**Step 1: FT of $x_1(t) = e^{-2|t|}$.**

$$
\begin{aligned}
X_1(j\omega) &= \int_{-\infty}^{0} e^{2t}e^{-j\omega t}dt + \int_{0}^{\infty} e^{-2t}e^{-j\omega t}dt \\
&= \frac{1}{2 - j\omega} + \frac{1}{2 + j\omega} \\
&= \frac{(2+j\omega) + (2-j\omega)}{4 + \omega^2} = \frac{4}{4 + \omega^2}
\end{aligned}
$$

**Step 2: time shift.** $x(t) = x_1(t-1)$, and $x_1(t - t_0) \leftrightarrow e^{-j\omega t_0}X_1(j\omega)$ with $t_0 = 1$:

$$
X(j\omega) = \frac{4}{4 + \omega^2}\,e^{-j\omega}
$$

**Answer:**

$$
e^{-2|t-1|} \;\longleftrightarrow\; \frac{4\,e^{-j\omega}}{4 + \omega^2}
$$

- Magnitude: $|X(j\omega)| = \frac{4}{4+\omega^2}$ (same as the unshifted signal; peak 1 at $\omega=0$, equal to the area under $x(t)$).
- Phase: $\angle X(j\omega) = -\omega$ (linear phase due to a delay of 1 s).

```text
 x(t)                |X(jw)|
 1 ^    .            1 ^   .
   |   . .             |  . .
   | .     .           | .   .
 ..+'       '..     ...|'     '...
---+---+---+---> t  ---+---+---+---> w
   0   1   2          -2   0   2
```

## Check yourself

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

1. What is the method to find the FT of e^(-2|t-1|)?
   A. Apply frequency shifting with ω0 = 1
   B. Apply time scaling by 2
   C. Differentiate in time
   D. Find FT of e^(-2|t|), then apply time shifting with t0 = 1
   Answer: D. x(t) = x1(t - 1) where x1 = e^(-2|t|). (Key point: Method)

2. What is the FT of e^(-2|t|)?
   A. 4/(4 + ω^2)
   B. 2/(4 + ω^2)
   C. 1/(2 + jω)
   D. 4/(2 + ω^2)
   Answer: A. 1/(2-jω) + 1/(2+jω) = 4/(4+ω^2). (Key point: Base transform)

3. What is X(jω)?
   A. 4 e^(jω)/(4 + ω^2)
   B. 4 e^(-jω)/(4 + ω^2)
   C. 4 e^(-j2ω)/(4 + ω^2)
   D. 4/(4+ω^2) shifted to ω = 1
   Answer: B. A delay of 1 s multiplies by e^(-jω). (Key point: Final answer)

4. How do magnitude and phase compare with the unshifted signal?
   A. Magnitude halves, phase zero
   B. Magnitude same, phase +ω
   C. Magnitude same, phase is -ω
   D. Magnitude same, phase -2ω
   Answer: C. Delay by 1 s leaves |X| unchanged and adds -ω. (Key point: Phase and magnitude)

## Close variants set in other exams

- [DTFT of aⁿu[n]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-geometric-sequence) — Signals and Systems, chapter 3, asked 4×
- [Fourier transform of e^(−at)cos ω₀t](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-damped-cosine) — Signals and Systems, chapter 3, asked 4×
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