Fourier series for periodic functions of period 2π 615
=
5
2π
−cos
2π
1
2 + n
1
2 + n
−
cos
2π
1
2 − n
1
2 − n
−
−cos 0
1
2 + n
−
cos 0
1
2 − n
When n is both odd and even,
a n =
5
2π
1
1
2 + n
+
1
1
2 − n
−
−1
1
2 + n
−
1
1
2 − n
=
5
2π
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
−
1
2
=
5
π
2
3
−
2
1
=
−20
3π
a 2 =
5
π
1
5
2
+
1
−
3
2
=
5
π
2
5
−
2
3
=
−20
(3)(5)π
a 3 =
5
π
1
7
2
+
1
−
5
2
=
5
π
2
7
−
2
5
=
−20
(5)(7)π
and so on
b n =
1
π
2π
0
5 sin
θ
2
sin nθ dθ
=
5
π
2π
0
−
1
2
cos
θ
1
2
+ n
− cos
θ
1
2
− n
dθ
from Chapter 40
=
5
2π
sin
θ
1
2 − n
1
2 − n
−
sin
θ
1
2 + n
1
2 + n
2π
0
=
5
2π
sin 2π
1
2 − n
1
2 − n
−
sin 2π
1
2 + n
1
2 + n
−
sin 0
1
2 − n
−
sin 0
1
2 + n
When n is both odd and even, b n = 0 since sin(−π),
sin 0, sin π, sin3π, ... are all zero. Hence the Fourier
series for the rectified sine wave,
i = 5 sin
θ
2
is given by:
f (θ) = a 0 +
∞
n=1
(a n cos nθ + b n sin nθ)
i.e. i = f (θ) =
10
π
−
20
3π
cos θ −
20
(3)(5)π
cos 2θ
−
20
(5)(7)π
cos 3θ − · · ·
i.e. i =
20
π
1
2
−
cos θ
(3)
−
cos 2θ
(3)(5)
−
cos 3θ
(5)(7)
− · · ·
Now try the following exercise
Exercises 228 Further problems on Fourier
series of periodic functions of period 2π
1. 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
sin x +
1
3
sin 3x
+
1
5
sin 5x + · · ·
⎤
⎥
⎥
⎦
2. For the Fourier series in Problem 1, deduce a
series for
π
4
at the point where x =
π
2
π
4
= 1 −
1
3
+
1
5
−
1
7
+ · · ·