692 FOURIER SERIES

f(x)

− 5 − 4 − 3 − 2 −10 1

5

2345

L = 4

x

Figure 74.1

wherecnis given by:

cn=

1

L

∫L

2

−L 2

f(x)e−j

2 πnx

L dx (from equation 12).

With reference to Figure 74.1, whenL=4,

cn=

1

4

{∫− 1

− 2

0dx+

∫ 1

− 1

5e−j

2 πnx

(^4) 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)).
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 involving sine and cosine terms determined
on page 678.
From equation (13),
c 0 =a 0 =
1
L
∫L
2
−L 2
f(x)dx=
1
4
∫ 1
− 1
5dx

5
4
[x]−^11 =
5
4
[1−(−1)]=
5
2
Since cn=
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
, and so on.
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
, and so on.
Hence, the extended complex form of the Fourier
series shown in equation (14) becomes:
f(x)=
5
2

• 
  

  5
  π
  ej
  πx
  (^2) −^5
  3 π
  ej
  3 πx
  (^2) +^5
  5 π
  ej
  5 πx
  2
  −
  5
  7 π
  ej
  7 πx
  (^2) +···+
  5
  π
  e−j
  πx
  2
  −
  5
  3 π
  e−j
  3 πx
  (^2) +
  5
  5 π
  e−j
  5 πx
  2
  −
  5
  7 π
  e−j
  7 πx
  (^2) + ···

  
  

  5
  2

  
  


• 
  

  5
  π
  (
  ej
  πx
  (^2) +e−j
  πx
  2
  )
  −
  5
  3 π
  (
  ej
  3 πx
  (^2) +e−j
  3 πx
  2
  )

  
  


• 
  

  5
  5 π
  (
  e
  5 πx
  (^2) +e−j
  5 πx
  2
  )
  − ···