Higher Engineering Mathematics

FOURIER SERIES FOR A NON-PERIODIC FUNCTION OVER RANGE 2π 665

L

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 +

cos 0

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 +

sin 0

n^2

]}

=

1

π

[

−πcosnπ

n

]

=

−cosnπ

n

Hence b 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

cos 3x

−

2

52 π

cos 5x−···+sinx

−

1

2

sin 2x+

1

3

sin 3x−···

i.e. f(x)

=

π

4

−

2

π

(

cosx+

cos 3x

32

+

cos 5x

52

+···

)

+

(

sinx−

1

2

sin 2x+

1

3

sin 3x−···

)

Problem 4. For the Fourier series of Prob-

lem 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

discontinuity 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

cos 3x+

1

3

sin 3x

)

This may also be written in the form

csin (3x+α),

where amplitude,c=

√

√

√

√

[(

− 2

32 π

) 2

+

(

1

3

) 2 ]

= 0. 341

and phase angle,

α=tan−^1

⎛

⎜

⎝

− 2

32 π

1

3

⎞

⎟

⎠

=− 11. 98 ◦ or − 0 .209 radians

Hence the third harmonic is given by

0 .341 sin(3x− 0 .209)

(c) Whenx=0,f(x)=0 (see Fig. 70.3).

Hence, from the Fourier series:

0 =

π

4

−

2

π

(

cos 0+

1

32

cos 0+

1

52

cos 0+···

)

+(0)

i.e. −

π

4

=−

2

π

(

1 +

1

32

+

1

52

+

1

72

+···

)