Convergence; coefficients of ±1.5 square wave

Condition of convergence (Dirichlet conditions)

The Fourier series of a periodic $x(t)$ converges if, over one period:

1. $x(t)$ is absolutely integrable: $\int_T|x(t)|dt < \infty$.
2. $x(t)$ has a finite number of maxima and minima.
3. $x(t)$ has a finite number of finite discontinuities.

Then the series equals $x(t)$ wherever $x(t)$ is continuous and equals $\frac{1}{2}[x(t^-) + x(t^+)]$ at a jump. (Alternatively, finite energy in one period gives convergence in the mean-square sense.)

Fourier series coefficients

Period $T = 2$ s, so $\omega_0 = 2\pi/T = \pi$ rad/s.

$$x(t) = \begin{cases} 1.5, & 0 \le t < 1 \\ -1.5, & 1 \le t < 2 \end{cases}$$

DC term:

$$a_0 = \frac{1}{2}\left[\int_0^1 1.5\,dt + \int_1^2(-1.5)\,dt\right] = \frac{1}{2}(1.5 - 1.5) = 0$$

For $k \ne 0$:

$$\begin{aligned} a_k &= \frac{1}{2}\left[\int_0^1 1.5\,e^{-jk\pi t}dt - \int_1^2 1.5\,e^{-jk\pi t}dt\right] \\ &= \frac{1.5}{2}\left[\frac{1 - e^{-jk\pi}}{jk\pi} - \frac{e^{-jk\pi} - e^{-j2k\pi}}{jk\pi}\right] \end{aligned}$$

Using $e^{-j2k\pi} = 1$ and $e^{-jk\pi} = (-1)^k$:

$$\begin{aligned} a_k &= \frac{0.75}{jk\pi}\left[1 - (-1)^k - (-1)^k + 1\right] = \frac{0.75 \times 2\left[1 - (-1)^k\right]}{jk\pi} \\ &= \frac{1.5\left[1 - (-1)^k\right]}{jk\pi} \end{aligned}$$

So

$$a_k = \begin{cases} \dfrac{3}{jk\pi} = -j\dfrac{3}{k\pi}, & k \text{ odd} \\ 0, & k \text{ even (including } k = 0) \end{cases}$$

Phase: $-90^\circ$ for positive odd $k$ and $+90^\circ$ for negative odd $k$ (the signal is odd, so the $a_k$ are purely imaginary).

Equivalent trigonometric form:

$$x(t) = \frac{6}{\pi}\left[\sin\pi t + \frac{1}{3}\sin 3\pi t + \frac{1}{5}\sin 5\pi t + \dots\right]$$

Answer: $a_0 = 0$; $a_k = \dfrac{3}{jk\pi}$ for odd $k$; $a_k = 0$ for even $k$.