Fourier series of square wave and DFT

(a) Fourier series of the square wave

A periodic signal of period $T$ and $\omega_0 = 2\pi/T$ is written as

$$f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos n\omega_0 t + b_n\sin n\omega_0 t\right)$$

with $a_0 = \frac{2}{T}\int_0^T f\,dt$, $a_n = \frac{2}{T}\int_0^T f\cos n\omega_0 t\,dt$, $b_n = \frac{2}{T}\int_0^T f\sin n\omega_0 t\,dt$.

  +A   +---------+         +---------+
       |         |         |         |
   0 --+---------+---------+---------+---> t
       |         |         |         |
  -A   |         +---------+         +----
       0        T/2        T       3T/2

The wave has zero mean over a period, so $a_0 = 0$. With the origin at the rising edge it is an odd function (about $t=0$, over $-T/2 < t < T/2$), so all $a_n = 0$.

$$b_n = \frac{2}{T}\left[\int_0^{T/2} A\sin n\omega_0 t\,dt - \int_{T/2}^{T} A\sin n\omega_0 t\,dt\right]$$ $$\int_0^{T/2}\sin n\omega_0 t\,dt = \frac{1-\cos n\pi}{n\omega_0}, \qquad \int_{T/2}^{T}\sin n\omega_0 t\,dt = \frac{\cos n\pi - 1}{n\omega_0}$$ $$b_n = \frac{2A}{T}\cdot\frac{2(1-\cos n\pi)}{n\omega_0} = \frac{2A(1-\cos n\pi)}{n\pi}$$

So $b_n = 4A/(n\pi)$ for odd $n$ and $b_n = 0$ for even $n$. Therefore

$$f(t) = \frac{4A}{\pi}\left[\sin\omega_0 t + \frac{1}{3}\sin 3\omega_0 t + \frac{1}{5}\sin 5\omega_0 t + \cdots\right]$$

Only odd harmonics exist, and their amplitudes fall as $1/n$.

(b) 4-point DFT

$$X_k = \sum_{n=0}^{3} x_n e^{-j2\pi kn/4}, \qquad x = \{1, 3, 2, 0\}$$

Using $e^{-j\pi k n/2} = 1, -j, -1, j$ for $kn = 0, 1, 2, 3 \pmod 4$:

• $X_0 = 1+3+2+0 = 6$
• $X_1 = 1 + 3(-j) + 2(-1) + 0 = -1 - j3$
• $X_2 = 1 + 3(-1) + 2(1) + 0 = 0$
• $X_3 = 1 + 3(j) + 2(-1) + 0 = -1 + j3$

Answer: $X = \{6,\ -1-j3,\ 0,\ -1+j3\}$.

Why FFT

A direct DFT needs about $N^2$ complex multiplications. The FFT (Cooley-Tukey) repeatedly splits the sum into even and odd samples and needs only about $(N/2)\log_2 N$. For $N = 1024$ this is about 5120 instead of about 1,000,000, so spectra can be obtained quickly, even in real time.