The complex or exponential form of a Fourier series 651
From equation (11), the complex Fourier series is
given by:
f (x) =
∞
n=−∞
c n e
j
2πnx
L
=
∞
n=−∞
−j
2
nπ
(1 − cos nπ )e
jnx
(18)
Problem 5. Show that the complex Fourier series
obtained in problem 4 above is equivalent to
f (x) =
8
π
sin x +
1
3
sin 3x +
1
5
sin 5x
+
1
7
sin 7x + · · ·
(which was the Fourier series obtained in terms of
sines and cosines in Problem 3 on page 625).
From equation (17) above, c n = − j
2
nπ
(1 − cos nπ)
When n =1,
c 1 = − j
2
(1)π
(1 − cos π)
= − j
2
π
1 − (−1)
= −
j 4
π
When n =2,
c 2 = − j
2
2π
(1 − cos 2π) = 0;
in fact, all even values of c n will be zero.
When n =3,
c 3 = − j
2
3π
(1 − cos 3π)
= − j
2
3π
(1 − (−1)) = −
j 4
3π
By similar reasoning,
c 5 =−
j 4
5π
, c 7 =−
j 4
7π
, and so on.
When n =−1,
c −1 = − j
2
(−1)π
(1 − cos(−π))
= + j
2
π
(1 − (−1)) = +
j 4
π
When n =−3,
c −3 = − j
2
(−3)π
(1 − cos(−3π))
= + j
2
3π
(1 − (−1)) = +
j 4
3π
By similar reasoning,
c −5 = +
j 4
5π
, c −7 = +
j 4
7π
, and so on.
Since the waveform is odd, c 0 = a 0 = 0.
From equation (18) above,
f (x) =
∞
n=−∞
− j
2
nπ
(1 − cos nπ)e
jnx
Hence,
f (x) = −
j 4
π
e
j x
−
j 4
3π
e
j 3x
−
j 4
5π
e
j 5x
−
j 4
7π
e
j 7x
− · · · +
j 4
π
e
− j x
+
j 4
3π
e
− j 3x
+
j 4
5π
e
− j 5x
+
j 4
7π
e
− j 7x
+ · · ·
=
−
j 4
π
e
j x
+
j 4
π
e
− j x
+
−
j 4
3π
e
3x
+
j 4
3π
e
−3x
+
−
j 4
5π
e
5x
+
j 4
5π
e
−5x
+ · · ·
= −
j 4
π
e
j x
− e
− j x
−
j 4
3π
e
3x
− e
−3x
−
j 4
5π
e
5x
− e
−5x
+ · · ·
=
4
j π
e
j x
− e
− j x
+
4
j 3π
e
3x
− e
−3x
+
4
j 5π
e
5x
− e
−5x
+ · · ·
by multiplying top and bottom by j
=
8
π
e j x − e − j x
2 j
+
8
3π
e j 3x − e − j 3
2 j
+
8
5π
e j 5x − e − j 5x
2 j
+ · · ·
by rearranging
=
8
π
sin x +
8
3π
sin 3x +
8
3x
sin 5x + · · ·
from equation (4), page 644