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

740 Chapter 23 Fourier series
Again assuming that it is legitimate to interchange the order of integration and summation,
weobtain
∫ T
0
f(t)sin 2mπt
T
dt =
∫ T
a 0
0
+
2
∞∑
n=1
2mπt
sin dt
T
∫ T
0
(
a n
cos 2nπt +b
T n
sin 2nπt
T
)
sin 2mπt dt
T
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
exceptforthecasewhenn =m,inwhichcasether.h.s.reducesto b m T
2 .Hencewefind
b m
= 2 T
as required.
∫ T
0
f(t)sin 2mπt
T
dt
23.5.1 Fourierseriesofoddandevenfunctions
LetusnowconsiderwhathappenswhenwedetermineFourierseriesoffunctionswhich
are either oddoreven.
Example23.17 Findthe Fourier series forthe functionwith period2π definedby
⎧
⎪⎨ 0 −π<t<−π/2
f(t)= 4 −π/2tπ/2
⎪⎩
0 π/2<t<π
Solution As usual we sketch the function first (Figure 23.17). Inspection of Figure 23.17 shows
that the Dirichlet conditions (page 735) are satisfied. We note from the graph that the
function is symmetrical about the vertical axis; that is, it is an even function. We shall
see shortly that this fact has important implications for the Fourier series representation.
For convenience we consider the period of integration to be
Equation (23.4) becomes
[
− T 2 ,T 2
]
. Hence
a n
= 2 T
∫ T/2
−T/2
f(t)cos 2nπt
T
dt
for all positive integersn
Inthis example the periodT equals 2π. The formulafora n
then simplifies to
a n
= 1 π
∫ π
−π
f(t)cosntdt
The interval of integration is fromt = −π tot = π. However, a glance at the graph
shows that the function is zero outside the interval − π 2 t π , and takes the value 4
2