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

Short answer

For x(t)=e^{−a|t|}, the Fourier transform is X(ω)=2a/(a²+ω²). The spectrum is real, even, and positive, peaking at 2/a at ω=0.

Find the Fourier transform of the signal x(t) = e^(-a|t|), where (0 < a < ∞) is real-valued and |t| denotes the absolute value of (t).

x(t) = e^(-a|t|), a > 0, which is e^(at) for t < 0 and e^(-at) for t ≥ 0.

X(ω) = ∫_{-∞}^{0} e^(at) e^(-jωt) dt + ∫₀^∞ e^(-at) e^(-jωt) dt

First integral = ∫_{-∞}^0 e^((a - jω)t) dt = 1/(a - jω)

Second integral = ∫₀^∞ e^(-(a + jω)t) dt = 1/(a + jω)

X(ω) = 1/(a - jω) + 1/(a + jω) = [(a + jω) + (a - jω)] / (a² + ω²)

X(ω) = 2a / (a² + ω²)

The spectrum is real, even and positive (as the signal is real and even), with a peak 2/a at ω = 0 and falling to zero as ω → ∞.

✓ Answer checked by Claude Opus 5.5 (high effort)

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