624 Higher Engineering Mathematics
Problem 1. Determine the Fourier series for the
periodic function defined by:
f(x)=
⎧
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎩
− 2 , when−π<x<−
π
2
2 , when−
π
2
<x<
π
2
− 2 , when
π
2
<x<π.
and has a period of 2π.
The square wave shown in Fig. 68.1 is an even function
since it is symmetrical about thef(x)axis.
Hence from para. (a) the Fourier series is given by:
f(x)=a 0 +
∑∞
n= 1
ancosnx
(i.e. the series contains no sine terms.)
2
f(x)
2
3 /2 /2 0 /2 3 /2 2 x
Figure 68.1
From para. (a),
a 0 =
1
π
∫π
0
f(x)dx
=
1
π
{∫π/ 2
0
2dx+
∫π
π/ 2
−2dx
}
=
1
π
{
[2x]π/ 02 +[− 2 x]ππ/ 2
}
=
1
π
[(π )+[(− 2 π)−(−π)]= 0
an=
2
π
∫π
0
f(x)cosnxdx
=
2
π
{∫π/ 2
0
2cosnxdx+
∫π
π/ 2
−2cosnxdx
}
=
4
π
{[
sinnx
n
]π/ 2
0
+
[
−sinnx
n
]π
π/ 2
}
=
4
π
{(
sin(π/ 2 )n
n
− 0
)
+
(
0 −
−sin(π/ 2 )n
n
)}
=
4
π
(
2sin(π/ 2 )n
n
)
=
8
πn
(
sin
nπ
2
)
Whennis even,an= 0
Whennis odd, an=
8
πn
forn= 1 , 5 , 9 ,...
and an=
− 8
πn
forn= 3 , 7 , 11 ,...
Hencea 1 =
8
π
,a 3 =
− 8
3 π
,a 5 =
8
5 π
, and so on.
Hence the Fourier series for the waveform of Fig. 68.1
is given by:
f(x)=
8
π
(
cosx−
1
3
cos3x+
1
5
cos 5x
−
1
7
cos 7x+···
)
Problem 2. In the Fourier series of Problem 1 let
x=0 and deduce a series forπ/4.
Whenx=0,f(x)=2 (from Fig. 68.1).
Thus, from the Fourier series,
2 =
8
π
(
cos0−
1
3
cos0+
1
5
cos0
−
1
7
cos0+···
)
Hence
2 π
8
= 1 −
1
3
+
1
5
−
1
7
+···
i.e.
π
4
= 1 −
1
3
+
1
5
−
1
7
+···
Problem 3. Obtain the Fourier series for the
square wave shown in Fig. 68.2.