The amplitude of the first odd harmonic of the signal shown below is,

Solution:

$ \begin{aligned}\\ b_1 & =\frac{2}{T_0} \int_{t_0}^{t_2+\mathrm{T}_0} x(t) \sin \omega_0 t d t \\\\ \omega_0 & =\frac{2 \pi}{2 \pi}=1 \\\\ & =\frac{2}{2 \pi} \int_0^{2 \pi} x(t) \sin t d t \\\\ & =\frac{1}{\pi}\left[\int_0^\pi V \sin t d t+\int_\pi^{2 \pi}-V \sin t d t\right] \\\\ & =\frac{V}{\pi}\left[\left.\operatorname{Cos} t\right|_0 ^\pi-\left.\operatorname{Cos} t\right|_\pi ^{2 \pi}\right]\\ \end{aligned} $

$ \begin{aligned} & =\frac{V}{\pi}[-2-2] \\\\ b_1 & =\frac{-4 V}{\pi} \\\\ \left|b_1\right| & =\frac{4 \mathrm{~V}}{\pi}\\ \end{aligned}\\ $