Higher Engineering Mathematics

678 FOURIER SERIES

=

1

4

{∫− 1

− 2

0dx+

∫ 1

− 1

5dx+

∫ 2

1

0dx

}

=

1

4

[5x]^1 − 1 =

1

4

[(5)−(−5)]=

10

4

=

5

2

an=

2

L

∫L

2

−L

2

f(x) cos

(

2 πnx

L

)

dx

=

2

4

∫ 2

− 2

f(x) cos

(

2 πnx

4

)

dx

=

1

2

{∫− 1

− 2

0 cos

(πnx

2

)

dx

+

∫ 1

− 1

5 cos

(πnx

2

)

dx

+

∫ 2

1

0 cos

(πnx

2

)

dx

}

=

5

2

⎡

⎢

⎣

sin

πnx

2

πn

2

⎤

⎥

⎦

1

− 1

=

5

πn

[

sin

(πn

2

)

−sin

(

−πn

2

)]

Whennis even,an= 0
Whennis odd,

a 1 =

5

π

(1−(−1))=

10

π

a 3 =

5

3 π

(− 1 −1)=

− 10

3 π

a 5 =

5

5 π

(1−(−1))=

10

5 π

and so on.

Hence the Fourier series for the functionf(x)is

given by:

f(x)=

5

2

+

10

π

[

cos

(πx

2

)

−

1

3

cos

(

3 πx

2

)

+

1

5

cos

(

5 πx

2

)

−

1

7

cos

(

7 πx

2

)

+ ···

]

Problem 3. Determine the Fourier series for

the functionf(t)=tin the ranget=0tot=3.

The functionf(t)=tin the interval 0 to 3 is shown
in Fig. 72.3. Although the function is not periodic
it may be constructed outside of this range so that

Period L = 3

− 3 0 3 6 t

f(t)

f(t) = t

Figure 72.3

it is periodic of period 3, as shown by the broken

lines in Fig. 72.3. From para. (c), the Fourier series

is given by:

f(t)=a 0 +

∑∞

n= 1

[

ancos

(

2 πnt

L

)

+bnsin

(

2 πnt

L

)]

a 0 =

1

L

∫L 2

−L

2

f(t)dx=

1

L

∫L

0

f(t)dx

=

1

3

∫ 3

0

tdt=

1

3

[

t^2

2

] 3

0

=

3

2

an =

2

L

∫L

2

−L

2

f(t) cos

(

2 πnt

L

)

dt

=

2

L

∫L

0

tcos

(

2 πnt

L

)

dt

=

2

3

∫ 3

0

tcos

(

2 πnt

3

)

dt

=

2

3

⎡

⎢

⎢

⎢

⎣

tsin

(

2 πnt

3

)

(

2 πn

3

) +

cos

(

2 πnt

3

)

(

2 πn

3

) 2

⎤

⎥

⎥

⎥

⎦

3

0

by parts

=

2

3

⎡

⎢

⎢

⎢

⎣

⎧

⎪⎪

⎪⎨

⎪⎪

⎪⎩

3 sin 2πn

(

2 πn

3

) +

cos 2πn

(

2 πn

3

) 2

⎫

⎪⎪

⎪⎬

⎪⎪

⎪⎭

−

⎧

⎪⎪

⎪⎨

⎪⎪

⎪⎩

0 +

cos 0

(

2 πn

3

) 2

⎫

⎪⎪

⎪⎬

⎪⎪

⎪⎭

⎤

⎥

⎥

⎥

⎦

= 0