Fourier transform of cos ω₀t

Step 1: write the cosine with exponentials (Euler).

$$x(t) = \cos\omega_0 t = \frac{1}{2}\left(e^{j\omega_0t} + e^{-j\omega_0t}\right)$$

Step 2: FT of a complex exponential. The inverse FT of $2\pi\delta(\omega - \omega_0)$ is

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

so $e^{j\omega_0t} \leftrightarrow 2\pi\delta(\omega-\omega_0)$ and $e^{-j\omega_0t} \leftrightarrow 2\pi\delta(\omega+\omega_0)$.

Step 3: linearity.

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

Answer: $\cos\omega_0t \leftrightarrow \pi[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)]$

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

Remarks:

• The everlasting sinusoid is not absolutely integrable, so its FT exists only in the generalised sense, with impulses.
• All its power is at the single frequency $\omega_0$ (and $-\omega_0$ in the two-sided spectrum).
• In terms of $f$: $\cos 2\pi f_0t \leftrightarrow \frac{1}{2}[\delta(f-f_0) + \delta(f+f_0)]$. Similarly $\sin\omega_0t \leftrightarrow \frac{\pi}{j}[\delta(\omega-\omega_0) - \delta(\omega+\omega_0)]$.