646 Higher Engineering Mathematics

Problem 1. Determine the complex Fourier series

for the function defined by:

f(x)=

⎧

⎨

⎩

0 , when− 2 ≤x≤− 1

5 , when− 1 ≤x≤ 1

0 , when 1≤x≤ 2

The function is periodic outside this range of

period 4.

This is the same Problem as Problem 2 on page 631 and

we can use this to demonstrate that the two forms of

Fourier series are equivalent.

The functionf(x) is shown in Figure 71.1, where the

period,L=4.

From equation (11), the complex Fourier series is

given by:

f(x)=

∑∞

n=−∞

cnej

2 πnx

L

wherecnis given by:

cn=

1

L

∫ L

2

−L 2

f(x)e−j

2 πLnx

dx(from equation 12).

With reference to Figure 71.1, whenL=4,

cn=

1

4

{∫− 1

− 2

0dx+

∫ 1

− 1

5e−j

2 π 4 nx

dx+

∫ 2

1

0dx

}

=

1

4

∫ 1

− 1

5e−

jπnx

(^2) dx=^5
4
[
e−
jπnx
2
−jπ 2 n
] 1
− 1

− 5
j 2 πn
[
e−
jπnx
2
] 1
− 1

− 5
j 2 πn
(
e−
jπn
(^2) −e
jπn
2
)

5
πn
(
ej
πn
(^2) −e−j
πn
2
2 j
)

5
πn
sin
πn
2
(from equation (4)).
21 0 1
5
(^2423222345) x
L 54
f(x)
25
Figure 71.1
Hence, from equation (11),the complex form of the
Fourier seriesis given by:
f(x)=
∑∞
n=−∞
cnej
2 πnx
L =
∑∞
n=−∞
5
πn
sin
πn
2
ej
πnx
2
(14)
Let us show how this result is equivalent to the result
involvingsineandcosinetermsdeterminedonpage632.
From equation (13),
c 0 =a 0 =
1
L
∫ L 2
−L 2
f(x)dx=
1
4
∫ 1
− 1
5dx

5
4
[
x
] 1
− 1 =
5
4
[1−(− 1 )]=
5
2
Sincecn=
5
πn
sin
πn
2
,then
c 1 =
5
π
sin
π
2

5
π
c 2 =
5
2 π
sinπ= 0
(in fact, all even terms will be zero since
sinnπ=0)
c 3 =
5
πn
sin
πn
2

5
3 π
sin
3 π
2
=−
5
3 π
By similar substitution,
c 5 =
5
5 π
c 7 =−
5
7 π
,andsoon.
Similarly,
c− 1 =
5
−π
sin
−π
2

5
π
c− 2 =−
5
2 π
sin
− 2 π
2
= 0 =c− 4 =c− 6 ,and so on.
c− 3 =−
5
3 π
sin
− 3 π
2
=−
5
3 π
c− 5 =−
5
5 π
sin
− 5 π
2

5
5 π
,andsoon.