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Fourier transform of a ramp pulse

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.

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]

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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