Fourier transform of e^(−a|t|)

The signal $x(t) = e^{-a|t|}$, $a > 0$, is a two-sided decaying exponential. It is real and even:

$$x(t) = \begin{cases} e^{-at}, & t \ge 0 \\ e^{at}, & t < 0 \end{cases}$$

Fourier transform

$$\begin{aligned} X(\omega) &= \int_{-\infty}^{\infty} e^{-a|t|}e^{-j\omega t}dt = \int_{-\infty}^{0} e^{at}e^{-j\omega t}dt + \int_{0}^{\infty} e^{-at}e^{-j\omega t}dt \\ &= \left[\frac{e^{(a-j\omega)t}}{a-j\omega}\right]_{-\infty}^{0} + \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty} \\ &= \frac{1}{a-j\omega} + \frac{1}{a+j\omega} = \frac{(a+j\omega) + (a-j\omega)}{a^2+\omega^2} \end{aligned}$$

Both limits at $\pm\infty$ vanish because $a > 0$.

$$\boxed{X(\omega) = \frac{2a}{a^2+\omega^2}}$$

Spectrum

$X(\omega)$ is real, positive and even, so:

• Magnitude: $|X(\omega)| = \frac{2a}{a^2+\omega^2}$. The peak is $X(0) = 2/a$ (equal to the area under $x(t)$). At $\omega = \pm a$ it falls to half the peak, $1/a$, and it decays as $1/\omega^2$.
• Phase: $\angle X(\omega) = 0$ for all $\omega$.

 x(t)                       X(w)
  1        *                 2/a       *
          / \                         / \
        /     \              1/a    /     \
   __ /         \ __            __/         \__
 ---------+---------- t     ----+----+----+----- w
          0                    -a    0    a

Observation: a larger $a$ means faster decay in time and a wider spectrum, which shows the inverse relation between time duration and bandwidth. The spectrum has a Lorentzian shape, and by duality $\frac{2a}{a^2+t^2} \leftrightarrow 2\pi e^{-a|\omega|}$.