CTFT of e^(jω₀t)

$x(t) = e^{j\omega_0t}$ has $|x(t)| = 1$ for all $t$, so it is not absolutely integrable and the FT integral does not converge in the ordinary sense. We use an impulse in frequency.

Step 1: guess and verify. Take $X(j\omega) = 2\pi\,\delta(\omega - \omega_0)$ and find its inverse FT:

$$\begin{aligned} x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\,\delta(\omega - \omega_0)\,e^{j\omega t}\,d\omega \\ &= e^{j\omega_0t} \end{aligned}$$

by the sifting property. Since the FT is unique,

$$e^{j\omega_0t} \;\longleftrightarrow\; 2\pi\,\delta(\omega - \omega_0)$$

(Equivalently: $1 \leftrightarrow 2\pi\delta(\omega)$, then frequency shifting by $\omega_0$.)

          X(jw)
          |        ^ 2*pi
          |        |
----------+--------+-------> w
          0        w0

Discussion:

• The spectrum is a single impulse at $\omega = \omega_0$ with area $2\pi$: all the signal content is at one frequency.
• The spectrum is not symmetric about $\omega = 0$, because $x(t)$ is complex.
• The magnitude is constant forever, so the signal has infinite energy but finite power (1). It is a power signal, which is why an impulse appears instead of an ordinary function.
• This pair is the basis for the FT of all periodic signals: from $x(t) = \sum a_ke^{jk\omega_0t}$ we get $X(j\omega) = \sum 2\pi a_k\delta(\omega - k\omega_0)$. For example, $\cos\omega_0t \leftrightarrow \pi[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)]$.