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Fourier transform of e^(-at)u(t) and spectra

Find the Fourier transform of the single sided exponential function x(t) = e^(-at)u(t) (a>0). Plot its magnitude and phase spectra.

x(t) = e^(-at) u(t), a > 0

X(ω) = ∫ x(t) e^(-jωt) dt = ∫₀^∞ e^(-at) e^(-jωt) dt = ∫₀^∞ e^(-(a + jω)t) dt

= [ -e^(-(a + jω)t) / (a + jω) ]₀^∞ = 1/(a + jω)

The integral converges since a > 0 (the signal is absolutely integrable).

X(ω) = 1/(a + jω)

Magnitude and phase

|X(ω)| = 1 / √(a² + ω²)

φ(ω) = ∠X(ω) = -tan⁻¹(ω/a)

ω|X|φ
01/a0
a1/(a√2)-45°
∞0-90°
-a1/(a√2)+45°
Magnitude spectrum |X(w)|
            1/a
          .-'-.
        .'  |  '.      1/(a*sqrt2) at w = -a and w = +a
     _.'    |    '._
 ---'-------+-------'---> w
      -a    0    a

Phase spectrum (degrees)
 +90 -._
        '-._           +45 at w = -a
   0 -------'-._---------> w
                '-._   -45 at w = +a
 -90                '-
      -a    0    a

The magnitude spectrum is even and bell-shaped, with the peak 1/a at ω = 0 and a half-power (3 dB) point at ω = a. The phase spectrum is odd: it falls from +90° (as ω → -∞) through 0 at ω = 0 to -90° (as ω → +∞). This is the response of a first-order RC low-pass filter with a = 1/RC.

✓ Answer checked by Claude Opus 5.5 (high effort)

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