Fourier series for periodic functions of period 2π 615
=
5
2 π
{[
−cos
[
2 π
( 1
2 +n
)]
( 1
2 +n
)
−
cos
[
2 π
( 1
2 −n
)]
( 1
2 −n
)
]
−
[
−cos0
( 1
2 +n
)−
cos0
( 1
2 −n
)
]}
Whennis both odd and even,
an=
5
2 π
{[
1
( 1
2 +n
)+
1
( 1
2 −n
)
]
−
[
− 1
( 1
2 +n
)−
1
( 1
2 −n
)
]}
=
5
2 π
{
2
( 1
2 +n
)+
2
( 1
2 −n
)
}
=
5
π
{
1
( 1
2 +n
)+
1
( 1
2 −n
)
}
Hence
a 1 =
5
π
[
1
3
2
+
1
−^12
]
=
5
π
[
2
3
−
2
1
]
=
− 20
3 π
a 2 =
5
π
[
1
5
2
+
1
−^32
]
=
5
π
[
2
5
−
2
3
]
=
− 20
( 3 )( 5 )π
a 3 =
5
π
[
1
7
2
+
1
−^52
]
=
5
π
[
2
7
−
2
5
]
=
− 20
( 5 )( 7 )π
andsoon
bn=
1
π
∫ 2 π
0
5sin
θ
2
sinnθdθ
=
5
π
∫ 2 π
0
−
1
2
{
cos
[
θ
(
1
2
+n
)]
−cos
[
θ
(
1
2
−n
)]}
dθ
from Chapter 40
=
5
2 π
[
sin
[
θ
( 1
2 −n
)]
( 1
2 −n
) −
sin
[
θ
( 1
2 +n
)]
( 1
2 +n
)
] 2 π
0
=
5
2 π
{[
sin2π
( 1
2 −n
)
( 1
2 −n
) −
sin2π
( 1
2 +n
)
( 1
2 +n
)
]
−
[
sin0
( 1
2 −n
)−
sin0
( 1
2 +n
)
]}
Whennis both odd and even,bn=0sincesin(−π),
sin0, sinπ,sin3π,...are all zero. Hence the Fourier
series for the rectified sine wave,
i=5sin
θ
2
is given by:
f(θ )=a 0 +
∑∞
n= 1
(ancosnθ+bnsinnθ)
i.e. i=f(θ )=
10
π
−
20
3 π
cosθ−
20
( 3 )( 5 )π
cos2θ
−
20
( 5 )( 7 )π
cos3θ−···
i.e. i=
20
π
(
1
2
−
cosθ
( 3 )
−
cos2θ
( 3 )( 5 )
−
cos3θ
( 5 )( 7 )
−···
)
Now try the following exercise
Exercises228 Further problemson Fourier
series of periodic functions of period 2π
- Determine the Fourier series for the periodic
function:
f(x)=
{
− 2 , when −π<x< 0
+ 2 , when 0<x<π
which is periodic outside this range of
period 2π.
⎡
⎢
⎢
⎣
f(x)=
8
π
(
sinx+
1
3
sin3x
+
1
5
sin5x+···
)
⎤
⎥
⎥
⎦
- For the Fourier series in Problem 1, deduce a
series for
π
4
at the point wherex=
π
2
[
π
4
= 1 −
1
3
+
1
5
−
1
7
+···
]