The complex or exponential form of a Fourier series 647

Hence, the extended complex form of the Fourier series

shown in equation (14) becomes:

f(x)=

5

2

+

5

π

ej

πx

(^2) −
5
3 π
ej
3 πx
(^2) +
5
5 π
ej
5 πx
2
−
5
7 π
ej
7 πx
(^2) +···+
5
π
e−j
πx
2
−
5
3 π
e−j
3 πx
(^2) +
5
5 π
e−j
5 πx
2
−
5
7 π
e−j
7 πx
(^2) + ···

5
2

• 5
  π
  (
  ej
  πx
  (^2) +e−j
  πx
  2
  )
  −
  5
  3 π
  (
  ej
  3 πx
  (^2) +e−j
  3 πx
  2
  )


• 
  

  5
  5 π
  (
  e
  5 πx
  (^2) +e−j
  5 πx
  2
  )
  − ···

  
  

  5
  2

  
  


• 
  

  5
  π
  ( 2 )
  (
  ej
  πx
  (^2) +e−j
  πx
  2
  2
  )
  −
  5
  3 π
  ( 2 )
  (
  ej
  3 πx
  (^2) +e−j
  3 πx
  2
  2
  )

  
  


• 
  

  5
  5 π
  ( 2 )
  (
  ej
  5 π 2 x
  +e−j
  5 π 2 x
  2
  )
  − ···

  
  

  5
  2

  
  


• 
  

  10
  π
  cos
  (πx
  2
  )
  −
  10
  3 π
  cos
  (
  3 πx
  2
  )

  
  


• 
  

  10
  5 π
  cos
  (
  5 πx
  2
  )
  − ···
  (from equation (3))
  i.e.f(x)=
  5
  2

  
  


• 
  

  10
  π
  [
  cos
  (πx
  2
  )
  −
  1
  3
  cos
  (
  3 πx
  2
  )

  
  


• 
  

  1
  5
  cos
  (
  5 πx
  2
  )
  −···
  ]
  which is the same as obtained on page 632.
  Hence,
  ∑∞
  n=−∞
  5
  πn
  sin
  nπ
  2
  ej
  π 2 nx
  is equivalent to
  5
  2

  
  


• 
  

  10
  π
  [
  cos
  (πx
  2
  )
  −
  1
  3
  cos
  (
  3 πx
  2
  )

  
  


• 
  

  1
  5
  cos
  (
  5 πx
  2
  )
  −···
  ]
  Problem 2. Show that the complex Fourier series
  for the functionf(t)=tin the ranget=0tot=1,
  and of period 1, may be expressed as:
  f(t)=
  1
  2

  
  


• 
  

  j
  2 π
  ∑∞
  n=−∞
  ej^2 πnt
  n
  The saw tooth waveform is shown in Figure 71.2.
  Period L 51
  f(t)
  f(t) 5 t
  2 10 12t
  Figure 71.2
  From equation (11), the complex Fourier series is
  given by:
  f(t)=
  ∑∞
  n=−∞
  cnej
  2 πnt
  L
  and when the period,L=1, then:
  f(t)=
  ∑∞
  n=−∞
  cnej^2 πnt
  where, from equation (12),
  cn=
  1
  L
  ∫ L
  2
  −L 2
  f(t)e−j
  2 πnt
  L dt=^1
  L
  ∫ L
  0
  f(t)e−j
  2 πnt
  L dt
  and whenL=1andf(t)=t, then:
  cn=
  1
  1
  ∫ 1
  0
  te−j
  2 π 1 nt
  dt=
  ∫ 1
  0
  te−j^2 πntdt
  Using integration by parts (see Chapter 43), letu=t,
  from which,
  du
  dt
  = 1 ,and dt=du,and
  let dv=e−j^2 πnt, from which,
  v=
  ∫
  e−j^2 πntdt=
  e−j^2 πnt
  −j 2 πn