Question 23.16: Obtain the expressions for the Fourier coefficients ao, an a......
Obtain the expressions for the Fourier coefficients a_o, a_n and b_n in Equations (23.3), (23.4) and (23.5).
\begin{aligned} & a_0=\frac{2}{T} \int_0^T f(t) \mathrm{d} t \quad (23.3)\\ & a_n=\frac{2}{T} \int_0^T f(t) \cos \frac{2 n \pi t}{T} \mathrm{~d} t \quad \text { for } n \text { a positive integer }\quad (23.4) \\ & b_n=\frac{2}{T} \int_0^T f(t) \sin \frac{2 n \pi t}{T} \mathrm{~d} t \quad \text { for } n \text { a positive integer }\quad (23.5) \\ & \end{aligned}Assume that f (t) can be expressed in the form
f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n \cos \frac{2 n \pi t}{T}+b_n \sin \frac{2 n \pi t}{T}\right) (23.6)
Multiplying Equation (23.6) through by \cos \frac{2 m \pi t}{T} and integrating from 0 to T we find
\begin{aligned} \int_0^T f(t) \cos \frac{2 m \pi t}{T} \mathrm{~d} t= & \int_0^T \frac{a_0}{2} \cos \frac{2 m \pi t}{T} \mathrm{~d} t \\ & +\int_0^T \sum_{n=1}^{\infty}\left(a_n \cos \frac{2 n \pi t}{T}+b_n \sin \frac{2 n \pi t}{T}\right) \cos \frac{2 m \pi t}{T} \mathrm{~d} t \end{aligned}If we now assume that it is legitimate to interchange the order of integration and summation we obtain
\begin{aligned} \int_0^T f(t) \cos \frac{2 m \pi t}{T} \mathrm{~d} t= & \int_0^T \frac{a_0}{2} \cos \frac{2 m \pi t}{T} \mathrm{~d} t \\ & +\sum_{n=1}^{\infty} \int_0^T\left(a_n \cos \frac{2 n \pi t}{T}+b_n \sin \frac{2 n \pi t}{T}\right) \cos \frac{2 m \pi t}{T} \mathrm{~d} t \end{aligned}The first integral on the r.h.s. is easily shown to be zero unless m = 0. Furthermore, we can use the previously found orthogonality properties (Table 23.1) to show that the rest of the integrals on the r.h.s. vanish except for the case when n = m, in which case the r.h.s. reduces to \frac{a_m T}{2}. Consequently,
a_m=\frac{2}{T} \int_0^T f(t) \cos \frac{2 m \pi t}{T} \mathrm{~d} t \quad \text { for all positive integers } mas required. When m = 0 all terms on the r.h.s. except the first vanish and we obtain
\begin{aligned} \int_0^T f(t) \mathrm{d} t & =\int_0^T \frac{a_0}{2} \mathrm{~d} t \\ & =\frac{a_0 T}{2} \end{aligned}so that
a_0=\frac{2}{T} \int_0^T f(t) \mathrm{d} tTo obtain the formula for the bn multiply Equation (23.6) through by \sin \frac{2 m \pi t}{T} and integrate from 0 to T.
\begin{aligned} \int_0^T f(t) \sin \frac{2 m \pi t}{T} \mathrm{~d} t= & \int_0^T \frac{a_0}{2} \sin \frac{2 m \pi t}{T} \mathrm{~d} t \\ & +\int_0^T \sum_{n=1}^{\infty}\left(a_n \cos \frac{2 n \pi t}{T}+b_n \sin \frac{2 n \pi t}{T}\right) \sin \frac{2 m \pi t}{T} \mathrm{~d} t \end{aligned}Again assuming that it is legitimate to interchange the order of integration and summation, we obtain
\begin{aligned} \int_0^T f(t) \sin \frac{2 m \pi t}{T} \mathrm{~d} t= & \int_0^T \frac{a_0}{2} \sin \frac{2 m \pi t}{T} \mathrm{~d} t \\ & +\sum_{n=1}^{\infty} \int_0^T\left(a_n \cos \frac{2 n \pi t}{T}+b_n \sin \frac{2 n \pi t}{T}\right) \sin \frac{2 m \pi t}{T} \mathrm{~d} t \end{aligned}The first integral on the r.h.s. is easily shown to be zero. Furthermore, we can use the properties given in Table 23.1 to show that the rest of the integrals on the r.h.s. vanish except for the case when n = m, in which case the r.h.s. reduces to \frac{b_m T}{2}. Hence we find
b_m=\frac{2}{T} \int_0^T f(t) \sin \frac{2 m \pi t}{T} \mathrm{~d} tas required.
Table 23.1
Some useful integral identities.
\begin{aligned}\hline & \int_0^T \sin \frac{2 n \pi t}{T} \mathrm{~d} t=0 \quad \text { for all integers } n \\ & \int_0^T \cos \frac{2 n \pi t}{T} \mathrm{~d} t=0 \quad n=1,2,3, \ldots \\ & \int_0^T \cos \frac{2 n \pi t}{T} \mathrm{~d} t=T \quad n=0 \\ & \int_0^T \cos \frac{2 m \pi t}{T} \cos \frac{2 n \pi t}{T} \mathrm{~d} t= \begin{cases}0 & m \neq n \\ T / 2 & m=n \neq 0\end{cases} \\ & \int_0^T \sin \frac{2 m \pi t}{T} \sin \frac{2 n \pi t}{T} \mathrm{~d} t= \begin{cases}0 & m \neq n \\ T / 2 & m=n \neq 0\end{cases} \\ & \int_0^T \sin \frac{2 m \pi t}{T} \cos \frac{2 n \pi t}{T} \mathrm{~d} t=0 \quad \text { for all integers } m \text { and } n \\\hline & \end{aligned}