Higher Engineering Mathematics, Sixth Edition

624 Higher Engineering Mathematics

Problem 1. Determine the Fourier series for the

periodic function defined by:

f(x)=

⎧

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎩

− 2 , when−π<x<−

π

2

2 , when−

π

2

<x<

π

2

− 2 , when

π

2

<x<π.

and has a period of 2π.

The square wave shown in Fig. 68.1 is an even function

since it is symmetrical about thef(x)axis.

Hence from para. (a) the Fourier series is given by:

f(x)=a 0 +

∑∞

n= 1

ancosnx

(i.e. the series contains no sine terms.)

2

f(x)

 2

 3 /2 /2 0 /2  3 /2 2 x

Figure 68.1

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/ 02 +[− 2 x]ππ/ 2

}

=

1

π

[(π )+[(− 2 π)−(−π)]= 0

an=

2

π

∫π

0

f(x)cosnxdx

=

2

π

{∫π/ 2

0

2cosnxdx+

∫π

π/ 2

−2cosnxdx

}

=

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

(

2sin(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. 68.1

is given by:

f(x)=

8

π

(

cosx−

1

3

cos3x+

1

5

cos 5x

−

1

7

cos 7x+···

)

Problem 2. In the Fourier series of Problem 1 let

x=0 and deduce a series forπ/4.

Whenx=0,f(x)=2 (from Fig. 68.1).

Thus, from the Fourier series,

2 =

8

π

(

cos0−

1

3

cos0+

1

5

cos0

−

1

7

cos0+···

)

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. 68.2.