Higher Engineering Mathematics, Sixth Edition

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