Higher Engineering Mathematics, Sixth Edition

Fourier series for a non-periodic function over range 2π 619

x

f(x) f(x) 5 x

22  2  0



2  3 

Figure 67.3

It is more convenient in this case to take the limits from 0
to2πinstead of from−πto+π.ThevalueoftheFourier
coefficientsareunalteredbythischangeoflimits.Hence

a 0 =

1

2 π

∫ 2 π

0

f(x)dx=

1

2 π

[∫π

0

xdx+

∫ 2 π

π

0dx

]

=

1

2 π

[

x^2

2

]π

0

=

1

2 π

(

π^2

2

)

=

π

4

an=

1

π

∫ 2 π

0

f(x)cosnxdx

=

1

π

[∫π

0

xcosnxdx+

∫ 2 π

π

0dx

]

=

1

π

[

xsinnx

n

+

cosnx

n^2

]π

0

(from Problem 1,by parts)

=

1

π

{[

πsinnπ

n

+

cosnπ

n^2

]

−

[

0 +

cos0

n^2

]}

=

1

πn^2

(cosnπ− 1 )

Whennis even,an=0.

Whennis odd,an=

− 2

πn^2

Hencea 1 =

− 2

π

,a 3 =

− 2

32 π

,a 5 =

− 2

52 π

, and so on

bn=

1

π

∫ 2 π

0

f(x)sinnxdx

=

1

π

[∫π

0

xsinnxdx−

∫ 2 π

π

0dx

]

=

1

π

[

−xcosnx

n

+

sinnx

n^2

]π

0

(from Problem 1,by parts)

=

1

π

{[

−πcosnπ

n

+

sinnπ

n^2

]

−

[

0 +

sin0

n^2

]}

=

1

π

[

−πcosnπ

n

]

=

−cosnπ

n

Henceb 1 =−cosπ=1,b 2 =−

1

2

,b 3 =

1

3

, and so on.

Thus the Fourier series is:

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)

i.e. f(x)=

π

4

−

2

π

cosx−

2

32 π

cos3x

−

2

52 π

cos5x−···+sinx

−

1

2

sin2x+

1

3

sin3x−···

i.e. f(x)

=

π

4

−

2

π

(

cosx+

cos3x

32

+

cos5x

52

+···

)

+

(

sinx−

1

2

sin2x+

1

3

sin3x−···

)

Problem 4. For the Fourier series of Problem 3:

(a) what is the sum of the series at the point of

discontinuity (i.e. atx=π)? (b) what is the

amplitude and phase angle of the third harmonic?

and (c) letx=0, and deduce a series forπ^2 /8.

(a) The sum of the Fourier series at the point of dis-

continuity is given by the arithmetic mean of the

two limiting values off(x)asxapproaches the

point of discontinuity from the two sides.

Hence sum of the series atx=πis

π− 0

2

=

π

2

(b) The third harmonic term of the Fourier series is

(

−

2

32 π

cos3x+

1

3

sin3x

)

This may also bewrittenin theformcsin( 3 x+α),

where amplitude,c=

√√

√

√

[(

− 2

32 π

) 2

+

(

1

3

) 2 ]

= 0. 341