650 Higher Engineering Mathematics
the f (x) axis. Thus equation (15) could have been used,
giving:
c n =
2
L
L
2
0
f (x) cos
2πnx
L
dx
=
2
4
2
0
f (x) cos
2πnx
4
dx
=
1
2
1
0
5 cos
πnx
2
dx +
2
1
0 dx
=
5
2
⎡
⎢
⎣
sin
πnx
2
πn
2
⎤
⎥
⎦
1
0
=
5
2
2
πn
sin
nπ
2
− 0
=
5
π n
sin
nπ
2
which is the same answer as in Problem 1; however, a knowledge of even functions has produced the
coefficient more quickly.
Problem 4. Obtain the Fourier series, in complex
form, for the square wave shown in Figure 71.4.
2
f (x)
x
0
22
2␲
␲
2␲
3␲
Figure 71.4
Method A
The square wave shown in Figure 71.4 is an odd
function since it is symmetrical about the origin.
The period of the waveform, L =2π.
Thus, using equation (16):
c n = − j
2
L
L
2
0
f (x) sin
2πnx
L
dx
= − j
2
2π
π
0
2 sin
2πnx
2π
dx
= − j
2
π
π
0
sin nx dx = − j
2
π
−cos nx
n
π
0
= − j
2
πn
(−cos πn) − (−cos 0)
i.e. c n = −j
2
πn
[1 − cosπn]
(17)
Method B
If it had not been noted that the function was odd,
equation (12) would have been used, i.e.
c n =
1
L
L
2
−
L
2
f (x) e
− j 2πnx
L dx
=
1
2π
π
−π
f (x) e
− j
2πnx
2π dx
=
1
2π
0
−π
−2 e
− jnx dx +
π
0
2 e
− jnx dx
=
1
2π
−2e − jnx
− j n
0
−π
+
2e − jnx
− j n
π
0
=
1
2π
2
j n
e
− jnx
0
−π
−
e
− jnx
π
0
=
1
2π
2
j n
e
0
− e
+ jnπ
−
e
− jnπ
− e
0
=
1
j πn
1 − e
jnπ
− e
− jnπ
+ 1
=
1
j nπ
2 − 2
e jnπ + e − jnπ
2
by rearranging
=
2
j nπ
1 −
e jnπ + e − jnπ
2
=
2
j nπ
{1 − cos nπ}
from equation (3)
=
− j 2
− j ( j nπ)
{1 − cos nπ}
by multiplying top and bottom by − j
i.e. c n = −j
2
nπ
(1 − cos nπ )
(17)
It is clear that method A is by far the shorter of the two
methods.