Higher Engineering Mathematics, Sixth Edition

620 Higher Engineering Mathematics

and phase angle,

α=tan−^1

⎛

⎜

⎝

− 2

32 π

1

3

⎞

⎟

⎠

=− 11. 98 ◦ or − 0 .209radians

Hence the third harmonic is given by

0 .341 sin(3x− 0 .209)

(c) Whenx=0,f(x)=0 (see Fig. 67.3).

Hence, from the Fourier series:

0 =

π

4

−

2

π

(

cos0+

1

32

cos0+

1

52

cos0+···

)

+( 0 )

i.e. −

π

4

=−

2

π

(

1 +

1

32

+

1

52

+

1

72

+···

)

Hence

π^2

8

= 1 +

1

32

+

1

52

+

1

72

+···

Problem 5. Deduce the Fourier series for the

functionf(θ )=θ^2 in the range 0 to 2π.

f(θ )=θ^2 is shown in Fig. 67.4 in the range 0 to 2π.

The function is not periodic but is constructed outside

of this range so that it is periodicof period 2π,asshown

by the broken lines.

f() f() 5  2

24  22  0 2  4 

4 ^2



Figure 67.4

For a Fourier series:

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)

a 0 =

1

2 π

∫ 2 π

0

f(θ )dθ=

1

2 π

∫ 2 π

0

θ^2 dθ

=

1

2 π

[

θ^3

3

] 2 π

0

=

1

2 π

[

8 π^3

3

− 0

]

=

4 π^2

3

an=

1

π

∫ 2 π

0

f(θ )cosnθdθ

=

1

π

∫ 2 π

0

θ^2 cosnθdθ

=

1

π

[

θ^2 sinnθ

n

+

2 θcosnθ

n^2

−

2sinnθ

n^3

] 2 π

0

by parts

=

1

π

[(

0 +

4 πcos2πn

n^2

− 0

)

−( 0 )

]

=

4

n^2

cos2πn=

4

n^2

whenn= 1 , 2 , 3 ,...

Hencea 1 =

4

12

,a 2 =

4

22

,a 3 =

4

32

and so on

bn=

1

π

∫ 2 π

0

f(θ )sinnθdθ=

1

π

∫ 2 π

0

θ^2 sinnθdθ

=

1

π

[

−θ^2 cosnθ

n

+

2 θsinnθ

n^2

+

2cosnθ

n^3

] 2 π

0

by parts

=

1

π

[(

− 4 π^2 cos2πn

n

+ 0 +

2cos2πn

n^3

)

−

(

0 + 0 +

2cos0

n^3

)]

=

1

π

[

− 4 π^2

n

+

2

n^3

−

2

n^3

]

=

− 4 π

n

Henceb 1 =

− 4 π

1

,b 2 =

− 4 π

2

,b 3 =

− 4 π

3

, and so on.

Thus f(θ )=θ^2

=

4 π^2

3

+

∑∞

n= 1

(

4

n^2

cosnθ−

4 π

n

sinnθ

)

i.e. θ^2 =

4 π^2

3

+ 4

(

cosθ+

1

22

cos2θ+

1

32

cos3θ+···

)

− 4 π

(

sinθ+

1

2

sin2θ+

1

3

sin3θ+···

)

for values ofθbetween 0 and 2π.

Problem 6. In the Fourier series of Problem 5, let

θ=πand determine a series for

π^2

12