Find the Fourier series representation of the sawtooth waveform described at the beginning of this section (see Figure 23.12).
This function is defined by f(t)=t,-\pi<t<\pi, \text { and has period } T=2 \pi \text {. } It is an odd function and hence a_n = 0 for all n. To find the b_n we must evaluate
\begin{aligned} b_n & =\frac{1}{\pi} \int_{-\pi}^\pi t \sin n t \mathrm{~d} t \\ & =\frac{1}{\pi}\left\{\left[\frac{-t \cos n t}{n}\right]_{-\pi}^\pi+\int_{-\pi}^\pi \frac{\cos n t}{n} \mathrm{~d} t\right\} \\ & =\frac{1}{n \pi}\{-\pi \cos n \pi-\pi \cos n \pi\} \end{aligned}since the last integral vanishes. Therefore,
b_n=-\frac{2}{n} \cos n \pi=-\frac{2}{n}(-1)^nWe conclude that b_1=2, b_2=-1, b_3=\frac{2}{3}, \ldots . Therefore f (t) has Fourier series
f(t)=2\left\{\sin t-\frac{1}{2} \sin 2 t+\frac{1}{3} \sin 3 t-\frac{1}{4} \sin 4 t+\frac{1}{5} \sin 5 t-\cdots\right\}