Higher Engineering Mathematics

(Greg DeLong) #1
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
    )
    − ···



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