Higher Engineering Mathematics

FOURIER SERIES OVER ANY RANGE 677

L

v(t)

10

− 8 − 4 0 4 8 12 t (ms)

Period L = 8 ms

Figure 72.1

a 0 =

1

L

∫ L

2

−L

2

v(t)dt=

1

8

∫ 4

− 4

v(t)dt

=

1

8

{∫ 0

− 4

0dt+

∫ 4

0

10 dt

}

=

1

8

[10t]^40 = 5

an=

2

L

∫ L

2

−L

2

v(t) cos

(

2 πnt

L

)

dt

=

2

8

∫ 4

− 4

v(t) cos

(

2 πnt

8

)

dt

=

1

4

{∫ 0

− 4

0 cos

(

πnt

4

)

dt

+

∫ 4

0

10 cos

(

πnt

4

)

dt

}

=

1

4

⎡

⎢

⎢

⎣

10 sin

(

πnt

4

)

(πn

4

)

⎤

⎥

⎥

⎦

4

0

=

10

πn

[ sinπn−sin 0]

=0 forn=1, 2, 3,...

bn=

2

L

∫ L

2

−L

2

v(t) sin

(

2 πnt

L

)

dt

=

2

8

∫ 4

− 4

v(t) sin

(

2 πnt

8

)

dt

=

1

4

{∫ 0

− 4

0 sin

(

πnt

4

)

dt

+

∫ 4

0

10 sin

(

πnt

4

)

dt

}

=

1

4

⎡

⎢

⎢

⎣

−10 cos

(

πnt

4

)

(πn

4

)

⎤

⎥

⎥

⎦

4

0

=

− 10

πn

[cosπn−cos 0]

Whennis even,bn= 0

Whennis odd,b 1 =

− 10

π

(− 1 −1)=

20

π

,

b 3 =

− 10

3 π

(− 1 −1)=

20

3 π

,

b 5 =

20

5 π

, and so on.

Thus the Fourier series for the functionv(t)is

given by:

v(t)= 5 +

20

π

[

sin

(

πt

4

)

+

1

3

sin

(

3 πt

4

)

+

1

5

sin

(

5 πt

4

)

+ ···

]

Problem 2. Obtain the Fourier series for the

function defined by:

f(x)=

⎧

⎨

⎩

0, when − 2 <x<− 1

5, when − 1 <x< 1

0, when 1 <x< 2

The function is periodic outside of this range of

period 4.

The functionf(x) is shown in Fig. 72.2 where period,

L=4. Since the function is symmetrical about the

f(x) axis it is an even function and the Fourier series

contains no sine terms (i.e.bn=0).

− 4 − 3 − 2 − 1012345

L = 4

f(x)

5

− 5 x

Figure 72.2

Thus, from para. (c),

f(x)=a 0 +

∑∞

n= 1

ancos

(

2 πnx

L

)

a 0 =

1

L

∫L

2

−L

2

f(x)dx=

1

4

∫ 2

− 2

f(x)dx