Higher Engineering Mathematics, Sixth Edition

650 Higher Engineering Mathematics

thef(x) axis. Thus equation (15) could have been used,

giving:

cn=

2

L

∫ L

2

0

f(x)cos

(

2 πnx

L

)

dx

=

2

4

∫ 2

0

f(x)cos

(

2 πnx

4

)

dx

=

1

2

{∫ 1

0

5cos

(πnx

2

)

dx+

∫ 2

1

0dx

}

=

5

2

⎡

⎢

⎣

sin

(πnx

2

)

πn

2

⎤

⎥

⎦

1

0

=

5

2

(

2

πn

)(

sin

nπ

2

− 0

)

=

5

πn

sin

nπ

2

which is the same answer as in Problem 1; how-

ever, a knowledge of even functions has produced the

coefficient more quickly.

Problem 4. Obtain the Fourier series, in complex

form, for the square wave shown in Figure 71.4.

2

f(x)

0 x

22

2 ^2 ^3 

Figure 71.4

Method A

The square wave shown in Figure 71.4 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

2sin

(

2 πnx

2 π

)

dx

=−j

2

π

∫π

0

sinnxdx=−j

2

π

[

−cosnx

n

]π

0

=−j

2

πn

(

(−cosπn)−(−cos0)

)

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 πLnx

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.