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] \\\\ & =\frac{1}{2}\left[\left[\frac{\sin n \Omega_0 t}{n \Omega_0}\right]_{-1}^1-\left[\frac{\sin n \Omega_0 t}{n \Omega_0}\right]_1^3\right]=\frac{1}{2}\left[\left[\frac{\sin n \frac{\pi}{2} t}{n \frac{\pi}{2}}\right]_{-1}^1-\left[\frac{\sin n \frac{\pi}{2} t}{n \frac{\pi}{2}}\right]_1^3\right]\\ \end{aligned}\\ $

$\begin{aligned}\\ =\frac{1}{2}\left(\frac{2}{n \pi}\right)[\sin n & \left.\frac{\pi}{2}-\left(\sin n \frac{\pi}{2}(-1)\right)-\left(\sin n \frac{\pi}{2}(3)-\sin n \frac{\pi}{2}\right)\right] \\\\ & =\left[\frac{1}{n \pi}\right]\left[\sin n \frac{\pi}{2}+\sin n \frac{\pi}{2}-\sin 3 n \frac{\pi}{2}+\sin n \frac{\pi}{2}\right]=\frac{1}{n \pi}\left[3 \sin \frac{n \pi}{2}-\sin \left(2 n \pi-\frac{n \pi}{2}\right)\right] \\\\ & =\frac{1}{n \pi}\left[3 \sin \frac{n \pi}{2}-\left(-\sin n \frac{\pi}{2}\right)\right]=\frac{\mathbf{4}}{n \pi}\left[\sin n \frac{\pi}{2}\right]\\ \end{aligned}$

Evaluation of, $b_n$

$\boldsymbol{b}_{\boldsymbol{n}}=\frac{2}{T} \int_{t_0}^{t_0+T} x(t) \sin n \Omega_0 t d t=\frac{2}{4}\left[\int_{-1}^1 \sin n \Omega_0 t d t+\int_1^3-\sin n \Omega_0 t d t\right]\\ $

$\begin{aligned}\\ & =\frac{1}{2}\left[\left[\frac{-\cos n \Omega_0 t}{n \Omega_0}\right]_{-1}^1-\left[\frac{-\cos n \Omega_0 t}{n \Omega_0}\right]_1^3\right]=\frac{1}{2}\left[\left[\frac{-\cos n \frac{\pi}{2} t}{n \frac{\pi}{2}}\right]_{-1}^1+\left[\frac{\cos n \frac{\pi}{2} t}{n \frac{\pi}{2}}\right]_1^3\right] \\\\ & =\frac{1}{2}\left[\frac{-2}{n \pi}\left(\cos n \frac{\pi}{2}-\cos n \frac{\pi}{2}(-1)\right)+\frac{2}{n \pi}\left(\cos n \frac{\pi}{2}(3)-\cos n \frac{\pi}{2}\right)\right] \\\\ & =\frac{1}{2}\left[0+\frac{2}{n \pi}\left(\cos \left(2 n \pi-\frac{n \pi}{2}\right)-\cos n \frac{\pi}{2}\right)\right]=\left[\frac{1}{n \pi}\left(\cos n \frac{\pi}{2}-\cos n \frac{\pi}{2}\right)\right]=\mathbf{0} \end{aligned}\\ $

Trigonometric Fourier series,

$\begin{aligned} x(t)=a_0+\sum_{n=1}^{\infty} & a_n \cos n \Omega_0 t+\sum_{n=1}^{\infty} b_n \sin n \Omega_0 t\\ \\ & =\sum_{n=1}^{\infty} \frac{4}{n \pi} \sin \left(\frac{n \pi}{2}\right) \cos n \Omega_0 t=\sum_{n=1}^{\infty} \frac{4}{n \pi} \sin \left(\frac{n \pi}{2}\right) \cos n \frac{\pi}{2} t\\ \end{aligned}\\ $