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Fourier transform of e^(-a|t|)

Short answer

The continuous‑time signal x(t)=e^{−a|t|} has Fourier transform X(ω)=2a/(a²+ω²). Splitting the integral yields the sum of two terms, giving a real, even spectrum with maximum 2/a at zero frequency.

Find the Fourier Transform of the continuous time signal x(t) = e^(-a|t|), for a > 0.

x(t) = e^(-a|t|), a > 0.

X(ω) = ∫_{-∞}^{∞} e^(-a|t|) e^(-jωt) dt

Split at t = 0:

X(ω) = ∫{-∞}^{0} e^(at) e^(-jωt) dt + ∫₀^∞ e^(-at) e^(-jωt) dt = [ e^((a - jω)t) / (a - jω) ]{-∞}^{0} + [ -e^(-(a + jω)t) / (a + jω) ]₀^∞ = 1/(a - jω) + 1/(a + jω) = 2a / (a² + ω²)

X(ω) = 2a / (a² + ω²)

The result is real and even, because x(t) is real and even. Its peak is 2/a at ω = 0, and the half-peak value occurs at ω = ±a. The area under X(ω) is 2π x(0) = 2π.

✓ Answer checked by Claude Opus 5.5 (high effort)

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