FT of periodic signals; pulse example

FT of a continuous-time periodic signal

Let $x(t)$ be periodic with period $T_0$ and $\omega_0 = 2\pi/T_0$. Its exponential Fourier series is

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

The inverse FT of $2\pi\delta(\omega - k\omega_0)$ is $\frac{1}{2\pi}\int 2\pi\delta(\omega-k\omega_0)e^{j\omega t}d\omega = e^{jk\omega_0 t}$, so $e^{jk\omega_0 t} \leftrightarrow 2\pi\delta(\omega - k\omega_0)$. By linearity,

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

Example: periodic pulse

Assumption: In one period $T_0$, $x(t) = 1$ for $0 < t < T$ and $0$ for $T < t < T_0$ (with $T < T_0$), repeated every $T_0$.

Fourier coefficients:

$$\begin{aligned} a_k &= \frac{1}{T_0}\int_0^{T} e^{-jk\omega_0 t}dt = \frac{1}{T_0}\cdot\frac{1 - e^{-jk\omega_0 T}}{jk\omega_0} \\ &= \frac{1}{T_0}\cdot\frac{e^{-jk\omega_0 T/2}\left(2j\sin(k\omega_0 T/2)\right)}{jk\omega_0} \\ &= \frac{\sin(k\omega_0 T/2)}{k\pi}\,e^{-jk\omega_0 T/2}, \quad k \ne 0 \end{aligned}$$

(using $T_0\omega_0 = 2\pi$), and $a_0 = \frac{T}{T_0}$ (the average value, also the limit of the formula as $k\to0$).

Fourier transform:

$$X(j\omega) = \frac{2\pi T}{T_0}\delta(\omega) + \sum_{k\ne0}\frac{2\sin(k\omega_0 T/2)}{k}\,e^{-jk\omega_0 T/2}\,\delta(\omega - k\omega_0)$$

Interpretation:

• The spectrum is a set of impulses at $\omega = k\omega_0$. Their strengths follow a sinc-shaped envelope $\frac{2\sin(\omega T/2)}{\omega}\cdot\omega_0$, which is $\omega_0$ times the FT of one pulse.
• The factor $e^{-jk\omega_0 T/2}$ is a linear phase because the pulse is centred at $t = T/2$ rather than at $t=0$.
• Example: duty cycle $T/T_0 = 1/2$ gives $a_0 = 1/2$ and $a_k = 0$ for even $k \ne 0$ (only odd harmonics).

 x(t)
 1 +---+     +---+     +---+
   |   |     |   |     |   |
---+---+-----+---+-----+---+---> t
   0   T    T0  T0+T  2T0