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