Higher Engineering Mathematics

670 FOURIER SERIES

f(x)

2

− 3 π/2 −π −π/2 0 π/2 π 3 π/2 2 π x

− 2

Figure 71.1

From para. (a),

a 0 =

1

π

∫π

0

f(x)dx

=

1

π

{∫π/ 2

0

2dx+

∫π

π/ 2

−2dx

}

=

1

π

{

[2x]

π/ 2

0 +[−^2 x]

π

π/ 2

}

=

1

π

[(π)+[(− 2 π)−(−π)]= 0

an=

2

π

∫π

0

f(x) cosnxdx

=

2

π

{∫π/ 2

0

2 cosnxdx+

∫π

π/ 2

−2 cosnxdx

}

=

4

π

{[

sinnx

n

]π/ 2

0

+

[

−sinnx

n

]π

π/ 2

}

=

4

π

{(

sin (π/2)n

n

− 0

)

+

(

0 −

−sin (π/2)n

n

)}

=

4

π

(

2 sin (π/2)n

n

)

=

8

πn

(

sin

nπ

2

)

Whennis even,an= 0

Whennis odd,an=

8

πn

forn=1, 5, 9,...

and an=

− 8

πn

forn=3, 7, 11,...

Hencea 1 =

8

π

,a 3 =

− 8

3 π

,a 5 =

8

5 π

, and so on.

Hence the Fourier series for the waveform of

Fig. 71.1 is given by:

f(x)=

8

π

(

cosx−

1

3

cos 3x+

1

5

cos 5x

−

1

7

cos 7x+···

)

Problem 2. In the Fourier series of Problem 1

letx=0 and deduce a series forπ/4.

Whenx=0,f(x)=2 (from Fig. 71.1).

Thus, from the Fourier series,

2 =

8

π

(

cos 0−

1

3

cos 0+

1

5

cos 0

−

1

7

cos 0+···

)

Hence

2 π

8

= 1 −

1

3

+

1

5

−

1

7

+···

i.e.

π

4

= 1 −

1

3

+

1

5

−

1

7

+···

Problem 3. Obtain the Fourier series for the

square wave shown in Fig. 71.2.

2

0

− 2

−π π 2 π 3 π x

f(x)

Figure 71.2

The square wave shown in Fig. 71.2 is an odd

function since it is symmetrical about the origin.

Hence, from para. (b), the Fourier series is

given by:

f(x)=

∑∞

n= 1

bnsinnx