Obtain Fourier series of the following full wave rectified sine wave shown in the figure,

Solution:

$ \boldsymbol{a}_0=\frac{2}{T} \int_0^{\frac{T}{2}} x(t) d t=\frac{2}{1} \int_0^{\frac{1}{2}} x(t) d t=\left[2 \int_0^{\frac{1}{2}} \sin \pi t d t\right]=2\left[-\frac{\cos \pi t}{\pi}\right]_0^{\frac{1}{2}}=-\frac{2}{\pi}\left[\cos \frac{\pi}{2}-\cos 0\right]=\frac{\mathbf{2}}{\boldsymbol{\pi}}\\ $

$ \boldsymbol{a}_{\boldsymbol{n}}=\frac{4}{T} \int_0^{\frac{T}{2}} x(t) \cos n \Omega_0 t d t=\frac{4}{1} \\\int_0^{\frac{1}{2}} \sin \pi t \cos n 2 \pi t d t=2 \int_0^{\frac{1}{2}}[\sin ((1+2 n) \pi t)+\sin ((1-2 n) \pi t)] d t $

$ \begin{aligned} = & 2\left[-\frac{\cos ((1+2 n) \pi t)}{(1+2 n) \pi}-\frac{\cos ((1-2 n) \pi t)}{(1-2 n) \pi}\right]_0^{\frac{1}{2}} \\\\ & =\frac{2}{\pi}\left[-\frac{\cos \left((1+2 n) \frac{\pi}{2}\right)}{1+2 n}-\frac{\cos \left((1-2 n) \frac{\pi}{2}\right)}{1-2 n}+\frac{1}{1+2 n}+\frac{1}{1-2 n}\right] \\\\ & =\frac{2}{\pi}\left[\frac{1}{1+2 n}+\frac{1}{1-2 n}\right]=\frac{2}{\pi}\left[\frac{1-2 n+1+2 n}{1-4 n^2}\right]=\frac{\mathbf{4}}{\boldsymbol{\pi}\left(\mathbf{1}-\mathbf{4 n}^2\right)}\\ \end{aligned}\\ $

$ \begin{gathered} 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 \\\\ \therefore x(t)=\frac{2}{\pi}+\sum_{n=1}^{\infty} \frac{4}{\pi\left(1-4 n^2\right)} \cos n 2 \pi t \end{gathered}\\ $