Fourier transform of CT periodic signals

A periodic signal is not absolutely integrable, so its ordinary Fourier integral does not converge. Its Fourier transform is obtained from its Fourier series using impulses.

Step 1: Fourier series. Let $x(t)$ be periodic with period $T_0$, $\omega_0 = 2\pi/T_0$:

$$x(t) = \sum_{k=-\infty}^{\infty} a_k\,e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T_0}\int_{T_0} x(t)\,e^{-jk\omega_0 t}\,dt$$

Step 2: FT of one complex exponential. Consider $X(j\omega) = 2\pi\,\delta(\omega - \omega_0)$. Its inverse FT is

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

by the sifting property. Hence

$$e^{j\omega_0 t} \;\longleftrightarrow\; 2\pi\,\delta(\omega - \omega_0), \qquad e^{jk\omega_0 t} \;\longleftrightarrow\; 2\pi\,\delta(\omega - k\omega_0)$$

Step 3: apply linearity to the Fourier series term by term:

$$X(j\omega) = \sum_{k=-\infty}^{\infty} 2\pi\,a_k\,\delta(\omega - k\omega_0)$$

Result: The FT of a periodic signal is a train of impulses located at the harmonic frequencies $k\omega_0$, with area $2\pi a_k$.

Link with the FT of one period: If $X_0(j\omega)$ is the FT of one period of $x(t)$, then $a_k = \frac{1}{T_0}X_0(jk\omega_0)$, so $X(j\omega) = \omega_0\sum_k X_0(jk\omega_0)\delta(\omega-k\omega_0)$.

Examples:

• $\cos\omega_0 t$: $a_{\pm1} = \frac12$, so $X(j\omega) = \pi[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)]$.
• $\sin\omega_0 t$: $a_{\pm1} = \pm\frac{1}{2j}$, so $X(j\omega) = \frac{\pi}{j}[\delta(\omega-\omega_0) - \delta(\omega+\omega_0)]$.
• Impulse train $\sum_n\delta(t - nT)$: $a_k = \frac1T$, so $X(j\omega) = \frac{2\pi}{T}\sum_k\delta\left(\omega - \frac{2\pi k}{T}\right)$.

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