Fourier series over any range 631
Period L 58 ms
0
10
v (t)
(^28244812) t (ms)
Figure 69.1
The square wave is shown in Fig. 69.1. From para. (c),
the Fourier series is of the form:
v(t)=a 0 +
∑∞
n= 1
[
ancos
(
2 πnt
L
)
+bnsin
(
2 πnt
L
)]
a 0 =
1
L
∫ L
2
−L
2
v(t)dt=
1
8
∫ 4
− 4
v(t)dt
1
8
{∫ 0
− 4
0dt+
∫ 4
0
10dt
}
1
8
[10t]^40 = 5
an=
2
L
∫ L
2
−L
2
v(t)cos
(
2 πnt
L
)
dt
2
8
∫ 4
− 4
v(t)cos
(
2 πnt
8
)
dt
1
4
{∫ 0
− 4
0cos
(
πnt
4
)
dt
∫ 4
0
10cos
(
πnt
4
)
dt
}
1
4
⎡
⎢
⎢
⎣
10sin
(
πnt
4
)
(πn
4
)
⎤
⎥
⎥
⎦
4
0
10
πn
[sinπn−sin0]
=0forn= 1 , 2 , 3 ,...
bn=
2
L
∫ L 2
−L
2
v(t)sin
(
2 πnt
L
)
dt
2
8
∫ 4
− 4
v(t)sin
(
2 πnt
8
)
dt
1
4
{∫ 0
− 4
0sin
(
πnt
4
)
dt
∫ 4
0
10sin
(
πnt
4
)
dt
}
1
4
⎡
⎢
⎢
⎣
−10cos
(
πnt
4
)
(πn
4
)
⎤
⎥
⎥
⎦
4
0
− 10
πn
[cosπn−cos0]
Whennis even,bn= 0
Whennis odd,b 1 =
− 10
π
(− 1 − 1 )=
20
π
,
b 3 =
− 10
3 π
(− 1 − 1 )=
20
3 π
,
b 5 =
20
5 π
,andsoon.
Thus the Fourier series for the functionv(t)is given by:
v(t)= 5 +
20
π
[
sin
(
πt
4
)
1
3
sin
(
3 πt
4
)
1
5
sin
(
5 πt
4
)
- ···
]
Problem 2. Obtain the Fourier series for the
function defined by:
f(x)=
⎧
⎪⎨
⎪⎩
0 , when − 2 <x<− 1
5 , when − 1 <x< 1
0 , when 1<x< 2
The function is periodic outside of this range of
period 4.
The functionf(x)is shown in Fig. 69.2 where period,
L=4. Since the functionis symmetrical about thef(x)
axis it is an even functionand theFourier series contains
no sine terms (i.e.bn=0).
L 54
5
f (x)
(^2524232221012345) x
Figure 69.2