FT of e^(−at)u(t); amplitude and phase

For the transform to exist, $a > 0$ (otherwise $e^{-at}u(t)$ grows and is not absolutely integrable).

Fourier transform

$$\begin{aligned} X(j\omega) &= \int_{-\infty}^{\infty} e^{-at}u(t)\,e^{-j\omega t}\,dt = \int_{0}^{\infty} e^{-(a+j\omega)t}\,dt \\ &= \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty} \\ &= 0 - \frac{1}{-(a+j\omega)} = \frac{1}{a+j\omega}, \quad a > 0 \end{aligned}$$

The upper limit is zero because $|e^{-(a+j\omega)t}| = e^{-at} \to 0$ as $t \to \infty$.

Magnitude and phase:

$$|X(j\omega)| = \frac{1}{\sqrt{a^2+\omega^2}}, \qquad \angle X(j\omega) = -\tan^{-1}\left(\frac{\omega}{a}\right)$$

Amplitude and phase spectra

 |X(jw)|
  1/a         *
            * | *
0.707/a  *.........*   (half-power points, w = +/- a)
      *       |       *
  *           |           *
--------------+--------------> w
        -a    0    a

 phase
 +90 | ......
 +45 |       .
   0 +--------.---------> w
     |        0 .
 -45 |            .
 -90 |              ......

• The amplitude spectrum is even and the phase spectrum is odd, as expected for a real signal.
• The spectrum is low-pass, with 3 dB bandwidth $\omega = a$ rad/s. Faster decay (larger $a$) gives a wider bandwidth.