082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

23.5 Fourier series 739
Example23.16 Obtain the expressions for the Fourier coefficients a 0
, a n
and b n
in Equations (23.3),
(23.4) and (23.5).
Solution Assumethat f (t)can be expressed inthe form
f(t)= a ∞
0
2 + ∑
(
a n
cos 2nπt +b
T n
sin 2nπt )
T
n=1
(23.6)
Multiplying Equation (23.6) through by cos 2mπt and integrating from 0 toT wefind
T
∫ T
0
f(t)cos 2mπt
T
dt =
∫ T
a 0
2
0
∫ T
+
0
2mπt
cos dt
T
∞∑
(
a n
cos 2nπt
T
n=1
+b n
sin 2nπt )
cos 2mπt dt
T T
If we now assume that it is legitimate to interchange the order of integration and
summation weobtain
∫ T
0
f(t)cos 2mπt
T
dt =
∫ T
a 0
0
+
2
∞∑
n=1
2mπt
cos dt
T
∫ T
0
(
a n
cos 2nπt +b
T n
sin 2nπt
T
)
cos 2mπt dt
T
The first integral on the r.h.s.iseasily shown tobe zero unlessm = 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 whenn = m, in which case the
r.h.s.reduces to a m T
2 . Consequently,
a m
= 2 T
∫ T
0
f(t)cos 2mπt
T
dt
forall positive integersm
as required. Whenm = 0 all terms on the r.h.s.except the first vanish and weobtain
so that
∫ T
0
a 0
= 2 T
f(t)dt =
∫ T
0
∫ T
0
= a 0 T
2
f(t)dt
a 0
2 dt
To obtain the formula for the b n
multiply Equation (23.6) through by sin 2mπt and
T
integrate from 0 toT.
∫ T
0
f(t)sin 2mπt
T
dt =
∫ T
a 0
2
0
∫ T
+
0
2mπt
sin dt
T
∞∑
(
a n
cos 2nπt
T
n=1
+b n
sin 2nπt )
sin 2mπt dt
T T