Even and odd functions and half-range Fourier series 625
2
f(x)
(^0) x
22
2 2 3
Figure 68.2
The square wave shown in Fig. 68.2 is an odd function
since it is symmetrical about the origin.
Hence, from para. (b), the Fourier series is given by:
f(x)=
∑∞
n= 1
bnsinnx
The function is defined by:
f(x)=
{
− 2 , when−π<x< 0
2 , when 0<x<π
From para. (b),bn=
2
π
∫π
0
f(x)sinnxdx
2
π
∫π
0
2sinnxdx
4
π
[
−cosnx
n
]π
0
4
π
[(
−cosnπ
n
)
−
(
−
1
n
)]
4
πn
( 1 −cosnπ)
Whennis even, bn=0.
Whennis odd, bn=
4
πn
( 1 −(− 1 ))=
8
πn
Hence b 1 =
8
π
,b 3 =
8
3 π
,b 5 =
8
5 π
,
and so on
Hence the Fourier series is:
f(x)=
8
π
(
sinx+
1
3
sin3x+
1
5
sin5x
1
7
sin7x+···
)
Problem 4. Determine the Fourier series for the
functionf(θ )=θ^2 in the range−π<θ<π.The
function has a period of 2π.
A graph off(θ )=θ^2 isshowninFig.68.3intherange
−πtoπwith period 2π. The function is symmetrical
about thef(θ )axis and is thus an even function. Thus
a Fourier cosine series will result of the form:
f(θ )=a 0 +
∑∞
n= 1
ancosnθ
22 ^0
f()
f() 5 ^2
^2
2 2
Figure 68.3
From para. (a),
a 0 =
1
π
∫π
0
f(θ )dθ=
1
π
∫π
0
θ^2 dθ
1
π
[
θ^3
3
]π
0
π^2
3
and an=
2
π
∫π
0
f(θ )cosnθdθ
2
π
∫π
0
θ^2 cosnθdθ
2
π
[
θ^2 sinnθ
n
2 θcosnθ
n^2
−
2sinnθ
n^3
]π
0
by parts
2
π
[(
0 +
2 πcosnπ
n^2
− 0
)
−( 0 )
]
4
n^2
cosnπ
When n is odd, an=
− 4
n^2
. Hence a 1 =
− 4
12
,
a 3 =
− 4
32
,a 5 =
− 4
52
, and so on.
Whennis even,an=
4
n^2
. Hencea 2 =
4
22
,a 4 =
4
42
,and
so on.