The complex or exponential form of a Fourier series 651
From equation (11), the complex Fourier series is
given by:
f(x)=
∑∞
n=−∞
cnej
2 πnx
L
=
∑∞
n=−∞
−j
2
nπ
(1−cosnπ)ejnx (18)
Problem 5. Show that the complex Fourier series
obtained in problem 4 above is equivalent to
f(x)=
8
π
(
sinx+
1
3
sin3x+
1
5
sin5x
+
1
7
sin7x+···
)
(which was the Fourier series obtained in terms of
sines and cosines in Problem 3 on page 625).
From equation (17) above,cn=−j
2
nπ
( 1 −cosnπ)
Whenn=1,
c 1 =−j
2
( 1 )π
( 1 −cosπ)
=−j
2
π
(
1 −(− 1 )
)
=−
j 4
π
Whenn=2,
c 2 =−j
2
2 π
( 1 −cos2π)= 0 ;
in fact, all even values ofcnwill be zero.
Whenn=3,
c 3 =−j
2
3 π
( 1 −cos3π)
=−j
2
3 π
( 1 −(− 1 ))=−
j 4
3 π
By similar reasoning,
c 5 =−
j 4
5 π
,c 7 =−
j 4
7 π
,andsoon.
Whenn=−1,
c− 1 =−j
2
(− 1 )π
( 1 −cos(−π))
=+j
2
π
( 1 −(− 1 ))=+
j 4
π
Whenn=−3,
c− 3 =−j
2
(− 3 )π
( 1 −cos(− 3 π))
=+j
2
3 π
( 1 −(− 1 ))=+
j 4
3 π
By similar reasoning,
c− 5 =+
j 4
5 π
,c− 7 =+
j 4
7 π
,andsoon.
Since the waveform is odd,c 0 =a 0 = 0.
From equation (18) above,
f(x)=
∑∞
n=−∞
−j
2
nπ
( 1 −cosnπ)ejnx
Hence,
f(x)=−
j 4
π
ejx−
j 4
3 π
ej^3 x−
j 4
5 π
ej^5 x
−
j 4
7 π
ej^7 x−···+
j 4
π
e−jx+
j 4
3 π
e−j^3 x
+
j 4
5 π
e−j^5 x+
j 4
7 π
e−j^7 x+···
=
(
−
j 4
π
ejx+
j 4
π
e−jx
)
+
(
−
j 4
3 π
e^3 x+
j 4
3 π
e−^3 x
)
+
(
−
j 4
5 π
e^5 x+
j 4
5 π
e−^5 x
)
+···
=−
j 4
π
(
ejx−e−jx
)
−
j 4
3 π
(
e^3 x−e−^3 x
)
−
j 4
5 π
(
e^5 x−e−^5 x
)
+···
=
4
jπ
(
ejx−e−jx
)
+
4
j 3 π
(
e^3 x−e−^3 x
)
+
4
j 5 π
(
e^5 x−e−^5 x
)
+···
by multiplying top and bottom byj
=
8
π
(
ejx−e−jx
2 j
)
+
8
3 π
(
ej^3 x−e−j^3
2 j
)
+
8
5 π
(
ej^5 x−e−j^5 x
2 j
)
+···
by rearranging
=
8
π
sinx+
8
3 π
sin3x+
8
3 x
sin5x+···
from equation (4), page 644