Higher Engineering Mathematics

696 FOURIER SERIES

Problem 4. Obtain the Fourier series, in com-

plex form, for the square wave shown in

Figure 74.4.

f(x)

x

2

0

− 2

−π π 2 π 3 π

Figure 74.4

Method A

The square wave shown in Figure 74.4 is anodd
functionsince it is symmetrical about the origin.
The period of the waveform,L= 2 π.

Thus, using equation (16):

cn=−j

2

L

∫ L

2

0

f(x) sin

(

2 πnx

L

)

dx

=−j

2

2 π

∫π

0

2 sin

(

2 πnx

2 π

)

dx

=−j

2

π

∫π

0

sinnxdx=−j

2

π

[

−cosnx

n

]π

0

=−j

2

πn

(

(−cosπn)−(−cos 0)

)

i.e.cn=−j

2

πn

[1−cosπn] (17)

Method B

If it hadnotbeen noted that the function was odd,

equation (12) would have been used, i.e.

cn=

1

L

∫L

2

−L 2

f(x)e−j

2 πnx

L dx

=

1

2 π

∫π

−π

f(x)e−j

2 πnx

2 π dx

=

1

2 π

{∫ 0

−π

−2e−jnxdx+

∫π

0

2e−jnxdx

}

=

1

2 π

{[

−2e−jnx

−jn

] 0

−π

+

[

2e−jnx

−jn

]π

0

}

=

1

2 π

(

2

jn

){

[

e−jnx

] 0

−π−

[

e−jnx

]π

0

}

=

1

2 π

(

2

jn

){

[

e^0 −e+jnπ

]

−

[

e−jnπ−e^0

]}

=

1

jπn

{

1 −ejnπ−e−jnπ+ 1

}

=

1

jnπ

{

2 − 2

(

ejnπ+e−jnπ

2

)}

by rearranging

=

2

jnπ

{

1 −

(

ejnπ+e−jnπ

2

)}

=

2

jnπ

{ 1 −cosnπ} from equation (3)

=

−j 2

−j(jnπ)

{ 1 −cosnπ}

by multiplying top and bottom by−j

i.e.cn=−j

2

nπ

(1−cosnπ) (17)

It is clear that method A is by far the shorter of the

two methods.

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 equiva-

lent to

f(x)=

8

π

(

sinx+

1

3

sin 3x+

1

5

sin 5x

+

1

7

sin 7x+···

)

(which was the Fourier series obtained in terms

of sines and cosines in Problem 3 on page 671).