Fourier transform of e^(−at)cos ω₀t

For the Fourier transform to exist, the signal is taken as causal: $x(t) = e^{-at}\cos(\omega_0 t)\,u(t)$ with $a > 0$. Without $u(t)$ the signal grows without bound as $t \to -\infty$, and it has no transform.

Method: known pair plus frequency shifting

The known pair is

$$e^{-at}u(t) \xleftrightarrow{F} \frac{1}{a + j\omega}$$

Write the cosine using Euler's formula:

$$x(t) = \frac{1}{2}e^{-at}u(t)e^{j\omega_0 t} + \frac{1}{2}e^{-at}u(t)e^{-j\omega_0 t}$$

The frequency-shift property, $e^{\pm j\omega_0 t}g(t) \leftrightarrow G(\omega \mp \omega_0)$, gives

$$\begin{aligned} X(\omega) &= \frac{1}{2}\left[\frac{1}{a + j(\omega - \omega_0)} + \frac{1}{a + j(\omega + \omega_0)}\right] \\ &= \frac{1}{2}\cdot\frac{2(a + j\omega)}{(a + j\omega - j\omega_0)(a + j\omega + j\omega_0)} \\ &= \frac{a + j\omega}{(a + j\omega)^2 + \omega_0^2} \end{aligned}$$

The second line uses a common denominator: the numerators add to $2a + 2j\omega$. The last step uses $(p - jq)(p + jq) = p^2 + q^2$ with $p = a + j\omega$ and $q = \omega_0$.

$$\boxed{X(\omega) = \frac{a + j\omega}{(a + j\omega)^2 + \omega_0^2}}$$

Check by direct integration

$$\begin{aligned} X(\omega) &= \int_0^{\infty}e^{-at}\frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2}e^{-j\omega t}dt \\ &= \frac{1}{2}\left[\frac{1}{a + j(\omega - \omega_0)} + \frac{1}{a + j(\omega + \omega_0)}\right] \end{aligned}$$

This is the same expression. A numerical check with $a = 1.3$, $\omega_0 = 2$ and $\omega = 0.9$ gives $0.288 + j0.046$ from both the integral and the formula.

Spectrum

$|X(\omega)|$ has peaks near $\omega = \pm\omega_0$, of height about $\frac{1}{2a}$ when $a \ll \omega_0$. The width of each peak is set by $a$. This is the spectrum of a damped oscillation, like the impulse response of a second-order underdamped system.