Find the trigonometric Fourier series for the periodic signal x(t) as shown in Figure,,

Solution:

Evaluation of, $a_0$

$ \begin{gathered} a_0=\frac{1}{T} \int_{t_0}^{t_0+T} x(t) d t=\frac{1}{4}\left[\int_{-1}^1 1 d t+\int_1^3-1 d t\right]=\frac{1}\\{4}\left[[t]_{-1}^1-1[t]_1^3\right]=\frac{1}{4}[(1-(-1))-(3-1)] \\\\ =\frac{1}{4}[2-2]=\mathbf{0}\\ \end{gathered}\\ $

Evaluation of ,$a_n$

$ \begin{aligned} \boldsymbol{a}_{\boldsymbol{n}} & =\frac{2}{T} \int_{t_0}^{t_{0+T}} x(t) \cos n \Omega_0 t d t=\frac{2}\\{4}\left[\int_{-1}^1 \cos n \Omega_0 t d t+\int_1^3(-1) \cos n \Omega_0 t d t\right] \\\\ …