Question 17.3: Find the Fourier series expansion of f(t) given in Fig. 17.......

The function f(t)  is an odd function. Hence a_{0} = 0 = a_{n} .   The period is T = 4 ,  and ω_{0} = 2 π /T = π/2 ,   so that

b_{n} = \frac{4}{T} \int^{T/2}_{0}{f(t) \sin n ω_{o} t} dt

= \frac{4}{4} [ \int^{1}_{0}{1 \sin \frac{nπ}{2} t  dt }  +  \int^{2}_{1}{0 \sin \frac{nπ}{2} t  dt} ]

= – \frac{2}{nπ} \cos \frac{nπ t}{2} \mid^{1}_{0} = \frac{2}{nπ} (1  –  \cos \frac{nπ}{2})

Hence,

f(t) = \frac{2}{π} \sum\limits_{n = 1}^{∞}{\frac{1}{n} (1  –  \cos \frac{nπ}{2})  \sin \frac{nπ}{2} t}

which is a Fourier sine series.