Higher Engineering Mathematics, Sixth Edition

Fourier series for periodic functions of period 2π 613

From Section 66.3(i):

a 0 =

1

2 π

∫π

−π

f(x)dx

=

1

2 π

[∫ 0

−π

−kdx+

∫π

0

kdx

]

=

1

2 π

{[−kx]^0 −π+[kx]π 0 }= 0

[a 0 is in fact themean valueof the waveform over a
complete periodof 2πand thiscouldhave been deduced
on sight from Fig. 66.3.]

From Section 66.3(i):

an=

1

π

∫π

−π

f(x)cosnxdx

=

1

π

{∫ 0

−π

−kcosnxdx+

∫π

0

kcosnxdx

}

=

1

π

{[

−ksinnx

n

] 0

−π

+

[

ksinnx

n

]π

0

}

= 0

Hence a 1 , a 2 , a 3 , ...are all zero (since sin0=
sin(−nπ)=sinnπ=0), and therefore no cosine terms
will appear in the Fourier series.

From Section 66.3(i):

bn=

1

π

∫π

−π

f(x)sinnxdx

=

1

π

{∫ 0

−π

−ksinnxdx+

∫π

0

ksinnxdx

}

=

1

π

{[

kcosnx

n

] 0

−π

+

[

−kcosnx

n

]π

0

}

Whennis odd:

bn=

k

π

{[(

1

n

)

−

(

−

1

n

)]

+

[

−

(

−

1

n

)

−

(

−

1

n

)]}

=

k

π

{

2

n

+

2

n

}

=

4 k

nπ

Henceb 1 =

4 k

π

,b 3 =

4 k

3 π

,b 5 =

4 k

5 π

, and so on.

Whennis even:

bn=

k

π

{[

1

n

−

1

n

]

+

[

−

1

n

−

(

−

1

n

)]}

= 0

Hence, from equation (1), the Fourier series for the

function shown in Fig. 66.3 is given by:

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)

= 0 +

∑∞

n= 1

( 0 +bnsinnx)

i.e. f(x)=

4 k

π

sinx+

4 k

3 π

sin3x+

4 k

5 π

sin 5x+···

i.e. f(x)=

4 k

π

(

sinx+

1

3

sin3x+

1

5

sin5x+ ···

)

Problem 2. For the Fourier series of Problem 1

letk=π. Show by plotting the first three partial

sums of this Fourier series that as the series is added

together term by term the result approximates more

and more closely to the function it represents.

Ifk=πin the Fourier series of Problem 1 then:

f(x)= 4 (sinx+^13 sin3x+^15 sin5x+···)

4sinx is termed the first partial sum of the

Fourier series of f(x),(4sinx+^43 sin3x)istermed

the second partial sum of the Fourier series, and

(4sinx+^43 sin3x+^45 sin5x) is termed the third partial

sum, and so on.

Let P 1 =4sinx,

P 2 =

(

4sinx+^43 sin3x

)

and P 3 =

(

4sinx+^43 sin3x+^45 sin5x

)

.

Graphs of P 1 , P 2 and P 3 , obtained by drawing up

tables of values, and adding waveforms, are shown in

Figs. 66.4(a) to (c) and they show that the series is

convergent, i.e. continually approximating towards a

definite limit as more and more partial sums are taken,

and in the limit will have the sumf(x)=π.

Even with just three partial sums, the waveform is start-

ing to approach the rectangular wave the Fourier series

is representing.

Problem 3. If in the Fourier series of Problem 1,

k=1, deduce a series for

π

4

at the pointx=

π

2

Ifk=1 in the Fourier series of Problem 1:

f(x)=

4

π

(

sinx+

1

3

sin3x+

1

5

sin5x+···

)