FS of unit-amplitude rectangular pulses

Take a periodic train of rectangular pulses with unity amplitude, period $T$ and pulse width $2T_1$, centred at $t = 0$:

$$x(t) = \begin{cases} 1, & |t| < T_1 \\ 0, & T_1 < |t| < T/2 \end{cases} \qquad x(t+T) = x(t), \quad \omega_0 = \frac{2\pi}{T}$$

 x(t)
  1     ___       ___       ___
       |   |     |   |     |   |
 ______|   |_____|   |_____|   |___ t
     -T1   T1  T-T1  T+T1
       (width 2T1, period T)

Fourier series coefficients

DC term (the average value):

$$a_0 = \frac{1}{T}\int_{-T_1}^{T_1} dt = \frac{2T_1}{T}$$

For $k \ne 0$:

$$\begin{aligned} a_k &= \frac{1}{T}\int_{-T_1}^{T_1} e^{-jk\omega_0 t}dt = \frac{1}{T}\cdot\frac{e^{-jk\omega_0 T_1} - e^{jk\omega_0 T_1}}{-jk\omega_0} \\ &= \frac{2}{k\omega_0 T}\sin(k\omega_0 T_1) = \frac{\sin(k\omega_0 T_1)}{k\pi} \end{aligned}$$

This can also be written $a_k = \frac{2T_1}{T}\,\mathrm{sinc}\!\left(\frac{2kT_1}{T}\right)$, with $\mathrm{sinc}(x) = \frac{\sin \pi x}{\pi x}$.

Example: T = 4T₁ (duty cycle 50%)

$a_k = \frac{\sin(k\pi/2)}{k\pi}$: $a_0 = 0.5$, $a_{\pm1} = 1/\pi \approx 0.318$, $a_{\pm2} = 0$, $a_{\pm3} = -1/(3\pi) \approx -0.106$, $a_{\pm5} \approx 0.064$.

Magnitude spectrum

 |a_k|
 0.5              |
 0.32          |  |  |
 0.11       |  |  |  |  |
 0.06    |  |  |  |  |  |  |
 --------+--+--+--+--+--+--+--- k
        -5 -3 -1  0  1  3  5
     (|a_k| = 0 for even k != 0)

The lines sit at $k\omega_0$ under a sinc-shaped envelope. A wider pulse ($T_1$ larger) gives a narrower envelope. A longer period ($T$ larger) gives more closely spaced lines, and as $T \to \infty$ the scaled coefficients $Ta_k$ merge into the continuous Fourier transform $2\sin(\omega T_1)/\omega$.