---
title: "Fourier transform of a ramp pulse"
subject: "Data Communication"
chapter: 3
key: "semester_5/data_communication:ft-ramp-signal"
source: https://www.bctengineer.com/q/semester_5/data_communication/ft-ramp-signal
site: BCT Engineering
---

# Fourier transform of a ramp pulse

Data Communication · Chapter 3: Signal Analysis

- Asked in: 2070 Asar, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/ft-ramp-signal
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-3

## Question

Determine the Fourier transform of the following signal, x(t). [Figure: x(t) rises linearly from 0 at t = 0 to 1 at t = 1, then drops to 0; zero elsewhere]

## Short answer

Using integration by parts, the Fourier transform of the unit ramp from t=0 to 1 is X(ω)=[(1+jω)e^{−jω}−1]/ω². At ω=0, the limit gives X(0)=1/2, the area under the ramp.

## Answer

Reading the figure: x(t) = t for 0 ≤ t ≤ 1 and 0 elsewhere.

X(ω) = ∫₀^1 t e^(-jωt) dt

Integrate by parts (u = t, dv = e^(-jωt) dt, so v = e^(-jωt)/(-jω)):

= [ t e^(-jωt) / (-jω) ]₀^1 + (1/(jω)) ∫₀^1 e^(-jωt) dt
= -e^(-jω)/(jω) + (1/(jω)) · [e^(-jωt)/(-jω)]₀^1
= -e^(-jω)/(jω) + (e^(-jω) - 1)/ω²

Since -1/(jω) = j/ω:

**X(ω) = [ (1 + jω) e^(-jω) - 1 ] / ω²**

In rectangular form, using e^(-jω) = cos ω - j sin ω:

X(ω) = [ (cos ω + ω sin ω - 1) + j(ω cos ω - sin ω) ] / ω²

|X(ω)| = √(2 + ω² - 2cos ω - 2ω sin ω) / ω²

At ω = 0: X(0) = ∫₀^1 t dt = **1/2** (the area under the ramp). The expression gives this as a limit (verified numerically).

## Check yourself

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

1. The integral is
   A. ∫ e^(-jωt) dt
   B. ∫0^1 t dt
   C. ∫0^1 t e^(-jωt) dt
   D. ∫ t^2 e^(-jωt)
   Answer: C. x=t on 0..1. (Key point: Setup)

2. The integral is solved by
   A. Substitution only
   B. Duality
   C. Laplace
   D. Integration by parts
   Answer: D. u=t. (Key point: Method)

3. X(ω) =
   A. [(1+jω)e^(-jω) - 1]/ω^2
   B. 1/(jω)
   C. e^(-jω)/ω
   D. (sin ω)/ω
   Answer: A. Derived result. (Key point: Result)

4. X(0) equals
   A. 1
   B. 1/2
   C. 0
   D. Infinity
   Answer: B. Area under ramp. (Key point: X(0))

## Related in Data Communication

- [Fourier transform of a DC signal](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aft-dc-signal) — Data Communication, chapter 3
- [Fourier transform of e^(-a|t|)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aft-exp-abs) — Data Communication, chapter 3
- [Fourier transform of e^(-a|t|)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aft-exp-abs-ct) — Data Communication, chapter 3
- [Why the Fourier transform is needed](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Awhy-fourier-transform-needed) — Data Communication, chapter 3, asked 2×
- [Properties of the Fourier transform](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afourier-transform-properties) — Data Communication, chapter 3, asked 2×
- [Fourier transform of e^(-at)u(t) and spectra](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aft-exp-causal-spectra) — Data Communication, chapter 3, asked 2×

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