Question 17.5: Calculate the Fourier series for the function in Fig. 17.17....

The function in Fig. 17.17 is half-wave odd symmetric, so that a_{0} = 0 = a_{n} . It is described over half the period as

f(t) = t,           -1 <  t  <  1

T =  4, ω_{0} = 2π/T  =  π/2.  Hence,

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

Instead of integrating f(t) from 0 to 2, it is more convenient to integrate from -1  to 1. Applying Eq. (17.15d),

\int{ t \sin at dt} = \frac{1}{a^{2}} \sin at – \frac{1}{a} t \cos at                                 (17.15d)

b_{n} = \frac{4}{4} \int^{1}_{-1}{t \sin \frac{nπ t}{2} dt} = [ \frac{\sin nπt/2}{n^{2} π^{2}/4} – \frac{t \cos n π t /2}{n π /2}] \mid^{1}_{-1}

= \frac{4}{n^{2} π^{2}} [ \sin \frac{nπ}{2}  –  \sin(- \frac{nπ}{2})] – \frac{2}{nπ} [ \cos \frac{nπ}{2} – \cos (-\frac{nπ}{2}) ]

= \frac{8}{n^{2}π^{2}} \sin \frac{nπ}{2}

since \sin (-x) = – \sin x   is an odd function, while \cos(-x) = \cos x   is an even function. Using the identities for \sin nπ /2    in Table 17.1,

b_{n} = \frac{8}{n^{2}π^{2}}  (-1)^{(n  –  1)/2} ,            n = odd = 1 , 3 ,  5 , . . .

Thus,

f(t) = \sum \limits_{n = 1, 3 , 5}^{∞}{b_{n} \sin \frac{nπ}{2} t }

where b_{n}   is given above.