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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 |
|---|---|---|
| 0 | 1/a | 0 |
| a | 1/(a√2) | -45° |
| ∞ | 0 | -90° |
| -a | 1/(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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