Higher Engineering Mathematics

FOURIER SERIES OVER ANY RANGE 679

L

bn=

2

L

∫ L

2

−L

2

f(t) sin

(

2 πnt

L

)

dt

=

2

L

∫L

0

tsin

(

2 πnt

L

)

dt

=

2

3

∫ 3

0

tsin

(

2 πnt

3

)

dt

=

2

3

⎡

⎢

⎢

⎢

⎣

−tcos

(

2 πnt

3

)

(

2 πn

3

) +

sin

(

2 πnt

3

)

(

2 πn

3

) 2

⎤

⎥

⎥

⎥

⎦

3

0

by parts

=

2

3

⎡

⎢

⎢

⎢

⎣

⎧

⎪⎪

⎪⎨

⎪⎪

⎪⎩

−3 cos 2πn

(

2 πn

3

) +

sin 2πn

(

2 πn

3

) 2

⎫

⎪⎪

⎪⎬

⎪⎪

⎪⎭

−

⎧

⎪⎪

⎪⎨

⎪⎪

⎪⎩

0 +

sin 0

(

2 πn

3

) 2

⎫

⎪⎪

⎪⎬

⎪⎪

⎪⎭

⎤ ⎥ ⎥ ⎥ ⎦ =

2

3

⎡

⎢

⎢

⎣

−3 cos 2πn

(

2 πn

3

)

⎤

⎥

⎥

⎦=

− 3

πn

cos 2πn=

− 3

πn

Henceb 1 =

− 3

π

,b 2 =

− 3

2 π

,b 3 =

− 3

3 π

and so on.

Thus the Fourier series for the functionf(t)inthe
range 0 to 3 is given by:

f(t)=

3

2

−

3

π

[

sin

(

2 πt

3

)

+

1

2

sin

(

4 πt

3

)

+

1

3

sin

(

6 πt

3

)

+ ···

]

Now try the following exercise.

Exercise 244 Further problems on Fourier

series over any rangeL

1. The voltage from a square wave generator is
   of the form:

v(t)=

{

0, − 10 <t< 0

5, 0 <t< 10

and is periodic of period 20. Show that the

Fourier series for the function is given by:

v(t)=

5

2

+

10

π

[

sin

(

πt

10

)

+

1

3

sin

(

3 πt

10

)

+

1

5

sin

(

5 πt

10

)

+···

]

2. Find the Fourier series forf(x)=xin the
   rangex=0tox=5.
   ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
   f(x) =

5

2

−

5

π

[

sin

(

2 πx

5

)

+

1

2

sin

(

4 πx

5

)

+

1

3

sin

(

6 πx

5

)

+···

]

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

3. A periodic function of period 4 is defined by:

f(x)=

{

−3, − 2 <x< 0

+3, 0 <x< 2

Sketch the function and obtain the Fourier

series for the function.

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

f(x) =

12

π

(

sin

(πx

2

)

+

1

3

sin

(

3 πx

2

)

+

1

5

sin

(

5 πx

2

)

+···

)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

4. Determine the Fourier series for the half
   wave rectified sinusoidal voltage Vsinωt
   defined by:

f(t)=

⎧

⎪⎨

⎪⎩

Vsinωt,0<t<

π

ω

0,

π

ω

<t<

2 π

ω

which is periodic of period

2 π

ω ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

f(t) =

V

π

+

V

2

sinωt

−

2 V

π

(

cos 2ωt

(1)(3)

+

cos 4ωt

(3)(5)

+

cos 6ωt

(5)(7)

+···

)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦