Orthonormal Basis Functions

A set of signals {0 (t ),, n (t )} are called orthogonal on an
interval (a, b) if any two signals m (t ) and k (t ) in the set satisfy
0,
a m (t ) k (t )dt   ,
b
*
mk
mk

If the magnitude of each signal i (t ) is set to one, it is called
normalized.

A set of normalized orthogonal functions can form an
orthonormal basis set。

The complex exponential signals {e j 2nf t , n  0,  1,  2,  3, }
are orthogonal on any interval over a period T0  1/ f0 .
教育部網路通訊人才培育先導型計畫
0
Principles of Communications 2.3
1
Generalized Transformation of Signal

A signal x(t) on any interval over a period T0, i.e., (t0 , t0+T0 ), can
be expressed as following in terms of {0 (t ),, n (t ),}

x(t )   cnn (t ), t0  t  t0  T0
n 0
1
cn 
T0


t0 T0
t0
x(t ) n* (t )dt
Parseval theorem：(Will be discussed later)
1
P
T0

T0

x(t ) dt   |cn |2 ,
2
教育部網路通訊人才培育先導型計畫
n 0

T0
: Integrate over a period.
Principles of Communications 2.3
2
Fourier theory
 Jean Baptiste Joseph Fourier (1768-1830) proposed that a periodic signal can
be represented by a summation of a (possibly infinite) number of sinusoids each
with a particular amplitude and phase.
 Assume that x(t) is a periodic signal with fundamental period T0, then x(t) can
be represented as

x(t )  a0  [an cosn0t  bn sin n0t ],
n 1
or

x(t )  a0   [an cos 2nf0t  bn sin 2nf0t ],
n 1
0  2f 0 
2
T0
 The series is called Trigonometric Fourier series because it is represented in
terms of sinusoids.
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
3
Coefficients of trigonometric Fourier series (1/7)
 Given a periodic signal with fundamental period T0 that can be represented by
the trigonometric Fourier series

x(t )  a0   [an cos n0t  bn sin n0t ]
n 1
 The problem is how to find the coefficients a0, an and bn.
 We begin with a0
 Integrating the series term by term over one period T0, we obtain

T0
x(t )dt  
T0
0
0



 

a0 dt   a1 cos0tdt   a2 cos 20tdt  
T0
T0
  b1 sin 0tdt   b2 sin 20tdt  
T0
T0



 


0
0
Note：Integrating the sinusoidal over one or integer number of periods is 0.
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
4
Coefficients of trigonometric Fourier series (2/7)


T0
x(t )dt  a0T0
1
 a0 
T0

T0
x(t )dt
a0 is the average value of
the waveform (signal).
 Find coefficients an
 Multiplying both sides of the series equation by cos k0t and integrating
over one period T0, we obtain
 

a0 cos k0tdt     an cos n0t  cos k0tdt
T0 x(t ) cosk0tdt  T0 

 T0  n 1
0
 

    bn sin n0t  cos k0tdt
T0
 n 1

教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
5
Coefficients of trigonometric Fourier series (3/7)

  x(t ) cos k0tdt   [an  cos n0t cos k0tdt  bn  sin n0t cos k0t dt]
T0
T0
T0
n 1





 I1  0

 T0 n  k
 I3  0
nk

2
The terms I1 and I3 will be
discussed in details later
T0
  x(t ) cos k0tdt  ak
T0
2
2
2
 ak   x(t ) cos k0tdt  an 
T0 T0
T0
教育部網路通訊人才培育先導型計畫

T0
x(t ) cos n0tdt
Signals and Systems 4.3
6
Coefficients of trigonometric Fourier series (4/7)
 Find coefficients bn
 Similarly we can obtain
 

  x(t ) sin k0tdt   a0 sin k0tdt     an cos n0t  sin k0tdt
T0
T0



 T0  n 1
0
 

    bn sin n0t  sin k0tdt
T0
 n 1


  x(t ) sin k0tdt  [an  cosn0t sin k0tdt  bn  sin n0t sin k0t dt]
T0
T0
T0
n 1






 I1 0
  x(t ) sin k0tdt  bk
T0
2
 bk 
T0

T0
T0
2
x(t ) sin k0tdt
教育部網路通訊人才培育先導型計畫

 0 nk
 I 2   T0
nk

2
2
 bn 
T0
I2 will be discussed
in details later

T0
x(t ) sin n0tdt
Signals and Systems 4.3
7
Coefficients of trigonometric Fourier series (5/7)
 In the procedure of finding coefficients an and bn , we used the properties of
integrals involving products of sines and cosines.
 Consider
I1   cosn0t sin k0tdt
T0
0
0

 

1
n

k
,
I

[  sin( n  k )0tdt   sin( n  k )0tdt ]  0
If
1
T0
2 T0
1
If n  k , I1   cos k0t sin k0tdt   sin 2k0tdt  0
T0
2 T0
 I1   cosn0t sin k0tdt  0
T0
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
8
Coefficients of trigonometric Fourier series (6/7)
 Consider
I 2   sin n0t sin k0tdt
T0
If
n  k,
0
0



 



1
I 2  [  cos( n  k )0tdt   cos( n  k )0tdt ]  0
T0
2 T0
If
n  k,
I 2   sin k0t sin k0tdt   sin 2 k0tdt
T0
T0
0


1  cos 2k0t
T0
1
1

dt   dt   cos 2k0tdt 
T0
T0 2
T0 2
2
2

 0, n  k
I 2   T0
2 , nk
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
9
Coefficients of trigonometric Fourier series (7/7 )
 Consider
I 3   cos n0t cosk0tdt
T0
If
n  k,
If
n  k,
0
0



 



1
I 3  [  cos( n  k )0tdt   cos( n  k )0tdt ]  0
T0
2 T0
1  cos 2k0t
I 3   cos k0tdt  
dt
T0
T0
2
0


T0
1
1
  dt   cos 2k0tdt 
T0 2
T0 2
2
2
 0, n  k
 I 3   T0
2 , nk
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
10
Trigonometric Fourier series
 Given a periodic signal with period T0 that can be represented by the
trigonometric Fourier series,

x(t )  a0   [an cos n0t  bn sin n0t ]
n 1
where
1
a0 
T0

x(t )dt
T0
the average value of the signal
2
an 
T0

x(t ) cos n0tdt
2
bn 
T0

x(t ) sin n0tdt
教育部網路通訊人才培育先導型計畫
T0
T0
n0
n0
Signals and Systems 4.3
11
Harmonic form Fourier series
 By using the Trigonometric equality
an cosn0t  bn sin n0t  cn cos(n0t  n )
where
we have
cn  an  bn ;
2
2
 n  tan1 (
 bn
)
an

x(t )  a0   (an cos n0t  bn sin n0t )
n 1

 a0   cn cos(n0t   n )
n 1
 Let c0 = a0 and we have the harmonic form Fourier series of x(t)

x(t )  c0   cn cos(n0t   n )
n 1
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
12
Limits of Fourier series at the discontinuities
 The Fourier series of waveform converges to the mean of the right- and lefthand limits at the discontinue point.

x(t )  a0   (an cos n0t1  bn sin n0t1 )
n 1
x(t )
limits of Fourier series
b
x(t1 )  a or b
a
Discontinue at t = t1+ nT0
t1  T0
t1
t1  2T0
t1  3T0
t
 Limits of Fourier series
ab
a0   (an cos n0t1  bn sin n0t1 ) 
2
n 1

教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
13
Dirichlet conditions for Fourier series
 A periodic signal x(t) can be represented as a Fourier series only if it
satisfy the following Dirichlet conditions:
 x(t) is absolutely integrable over any period, that is

t1 T0
t1
x(t ) dt  , for any t1
 x(t) has a finite number of maxima and minima within any finite interval of t.
 x(t) has a finite number of discontinuities within any finite interval of t, and
each of these discontinuities is finite.
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
14
Example 4-9 (1/6)
 Please find the trigonometric Fourier series for the periodic
rectangular pulse train signal with period 2.
x(t)
1
3 
2 
教育部網路通訊人才培育先導型計畫
 /2 /2 
2
3
Signals and Systems 4.3
t
15
Example 4-9 (2/6)
【Sol.】
1) The trigonometric Fourier series with 0 = 2 /T0= 2 /2 = 1

x(t )  a0  [an cos nt  bn sin nt]
n 1
2) Find a0
1
a0 
T0
教育部網路通訊人才培育先導型計畫
1
T0 x(t )dt  2



2
2
1
1dt 
2
Signals and Systems 4.3
16
Example 4-9 (3/6)
3) Find an
2
an 
T0

T0
x(t ) cos ntdt
2
n
   cos ntdt 
sin( )
  2
n
2
1

2

 0,

 2
 ,
 n
 2

,

 n
教育部網路通訊人才培育先導型計畫
n is even
n  1,5,9,13,...
n  3,7,11,15,...
Signals and Systems 4.3
17
Example 4-9 (4/6)
4) Find bn
2
bn 
T0

T0
x(t ) sin ntdt 
5) The signal x(t) is written by
x(t ) 

1

 

2
sin ntdt  0
2
 cos x  cos(x   )
1 2
1
1
1
 (cos t  cos 3t  cos 5t  cos 7t  ...)
2 
3
5
7
1 2
1
1
1
 [cost  cos(3t   )  cos5t  cos(7t   )  ...]
2 
3
5
7

 c0   cn cos(nt   n )
n 1
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
18
Example 4-9 (5/6)
6) The trigonometric Fourier series for the periodic rectangular pulse train
signal with period 2 is obtained.

x(t )  c0   cn cos(nt   n )
n 1
1
2
 0,

cn   2
,

 n
  ,
n  
 0,
c0 
教育部網路通訊人才培育先導型計畫
n is even
n is even
n  3,7,11,15,...
else
Signals and Systems 4.3
19
Example 4-9 (6/6)
 Plot the series in frequency domain.
1 2
1
1
1
x(t )   [cos t  cos( 3t   )  cos 5t  cos( 7t   )  ...]
2 
3
5
7
Note: The phases are either 0 or  ,
cne jn
教育部網路通訊人才培育先導型計畫
the amplitude and phase are
combined in this special case.
Signals and Systems 4.3
20
Example 4-10 (1/6)
 Please find the trigonometric Fourier series for the periodic
triangle pulse train signal with period 2.
x(t )
A
0
t
A
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
21
Example 4-10 (2/6)
【Sol.】
1) The trigonometric Fourier series with 0 = 2 /T0= 2 /2 =  .

x(t )  a0  [an cos nt  bn sin nt]
n 1
2) Set the period as [1/2, 3/2], then x(t) in this period is written as

2 At,


x(t )  
2 A(1  t ),


1
t 
2
1
3
t 
2
2
3) Obtain a0 = 0 . (the average of x(t) is 0)
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
22
Example 4-10 (3/6)
4) Find an
2
an 
T0

T0
x(t ) cos ntdt
2 3/ 2
  x(t ) cos ntdt
2 1/ 2
1/ 2

1 / 2
3/ 2
2 At cos ntdt  1/ 2 2 A(1  t ) cos ntdt
0
Note: use the odd/even properties
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
23
Example 4-10 (4/6)
4) Find bn
bn 
2
T0

T0
x(t ) sin(nt )dt
2 3/ 2
  x(t ) sin(nt )dt
2 1/ 2
1/ 2

1 / 2
2 At sin(nt )dt  
3/ 2
1/ 2
2 A(1  t ) sin(nt )dt

 0,

8A
n
 8A
 2 2 sin( )   2 2 ,
n
2
n 
 8A
 2 2,

 n
教育部網路通訊人才培育先導型計畫
n is even
n  1, 5, 9, 13....
n  3, 7, 11, 15....
Signals and Systems 4.3
24
Example 4-10 (5/6)
5) The signal x(t) is written by
8A
1
1
1
x(t )  2 [sin t  sin 3t 
sin 5t 
sin 7t  ...)

9
25
49
Using  sin kt  cos(kT  900 ) , we rewrite x(t) as
8A
1
x(t )  2 [cos(t  90 )  cos(3t  900 )

9
1
1

cos(5t  900 ) 
cos(7t  900 )  ...]
25
49
教育部網路通訊人才培育先導型計畫
0
Signals and Systems 4.3
25
Example 4-10 (6/6)
 The signal is expressed as harmonic form Fourier series and plotted in
frequency domain.

x(t )  c0   cn cos(nt   n )
n 1
Only odd order harmonic (n
times of fundamental
frequency) components exist.
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.3
26
Exponential Fourier series
 Assume that x(t) is a periodic signal with fundamental period T0,
then x(t) can be represented as the exponential Fourier series,
x(t )    Χ  2 e  j 20t  Χ 1e  j0t  Χ 0  Χ 1e j0t  Χ 2 e j 20t  


jn0t
Χ
e
 n
n  


Χ
n  
n
e
j 2nf 0t
教育部網路通訊人才培育先導型計畫
,
2
0  2f 0 
T0
Signals and Systems 4.4
27
Coefficients of exponential Fourier series (1/6)
 Review the trigonometric Fourier series

x(t )  a0   (an cosn0t  bn sin n0t )
n 1
 Review the Euler’s equality
e jn0t  e  jn0t
sin n0t 
2j
e jn0t  e  jn0t
cos n0t 
2
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
28
Coefficients of exponential Fourier series (2/6)
 By using Euler’s equality, we rewrite the trigonometric Fourier
series and obtain
e jn0t  e  jn0t
e jn0t  e  jn0t
x(t )  a0   (an
 bn
)
2
2j
n 1



1
1
 a0   (an  jbn )e jn0t   (an  jbn )e  jn0t
n 1 2
n 1 2


1
1
 jn0t
  (an  jbn )e
 a0   (an  jbn )e jn0t
n 1 2
n 1 2
 x(t )  ...  Χ  2 e  j 20t  Χ 1e  j0t  Χ 0  Χ 1e j0t  Χ 2 e j 20t  ...


jn0t
Χ
e
 n
n  
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
29
Coefficients of exponential Fourier series (3/6)
 The relationship between the coefficients of exponential Fourier
series and those of trigonometric Fourier series
1
 2 ( a n  jb n ),

Χ n   a0 ,
1
 ( an  jbn ),
2
教育部網路通訊人才培育先導型計畫
n0
n0
n0
Signals and Systems 4.4
30
Coefficients of exponential Fourier series (4/6)
 Review the harmonic form Fourier series

x(t )  c0   cn cos(n0t   n )
n 1
 By using Euler’s equality, we rewrite the trigonometric Fourier
series and obtain
cn j ( n0t  n )  j ( n0t  n )
[e
e
]
2
c
c
 ( n e j n )e jn0t  ( n e  j n )e  jn0t
2
2

cn j n jn0t  cn  j n  jn0t
 x(t )  c0   ( e )e
  ( e )e
n 1 2
n 1 2
cn cos(n0t   n ) 


 e
n  
教育部網路通訊人才培育先導型計畫
jn0t
n
Signals and Systems 4.4
31
Coefficients of exponential Fourier series (5/6)
 The relationship between the coefficients of exponential Fourier
series and those of trigonometric Fourier series:
 c  n  j  n
,
 2 e

Χ n  c0 ,
c
 n e j n ,
2
n0
n0
n0
 The exponential Fourier series and trigonometric Fourier series are
equivalent.
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
32
Coefficients of exponential Fourier series (6/6)
 An alternative way to find the coefficients of exponential Fourier series.
 Multiplying both sides of the series equation by e - jk 0t and integrating
over one period T0, we obtain

T0
x(t )e
- jk0t
dt 



T0
(  Χ n e jn0t )e - jk0t dt
n  


n  
Χ n  e j ( n  k )0t dt
T0
 Χ k T0
1
 Χk 
T0

x(t )e  jk0t dt
1
 Χn 
T0

x(t )e  jn0t dt
T0
T0
教育部網路通訊人才培育先導型計畫

T0
e
j ( n  k ) 0 t
 0, n  k
dt  
T0 , n  k
Signals and Systems 4.4
33
Example 4-11 (1/6)
 Please find the exponential Fourier series for the periodic
rectangular pulse train signal with period T0.
.
T0
教育部網路通訊人才培育先導型計畫
T0
Signals and Systems 4.4
34
Example 4-11 (2/6)
【Sol.】
1) The exponential Fourier series with 0 = 2 /T0= 2 f0.
x(t ) 

 Χ ne
jn0t
n  


 Χ ne j 2nf 0t ,
0  2f 0 
n  
2
T0
2) Find X0
1
Χ0 
T0
教育部網路通訊人才培育先導型計畫
1
T0 / 2 x(t )dt  T0
T0 / 2
1
T0 / 41dt  2
T0 / 4
Signals and Systems 4.4
35
Example 4-11 (4/6)
3) Find Xn
1
Χn 
T0

T0 / 2
-T0 / 2
x(t )e
 jn0t
1
dt 
T0

T0 / 4
- T0 / 4
e  jn0t dt
jn / 2
 jn / 2
1
1
e

e

[e  jn / 2  e jn / 2 ] 
[
]
 j 2n
n
j2
0,

sin(n / 2) 
1


k
(1)
,
n

(2k  1)
教育部網路通訊人才培育先導型計畫
n  2k  0
n  2k  1
Signals and Systems 4.4
36
Example 4-11 (5/6)
4) Rewrite the coefficients
1
Χ0  ;
2
Χ 2 k  0, k  0;
Χ 2 k 1  (1) k
1
(2k  1)
5) The exponential Fourier series
x(t ) 

Χe
n  
jn0t
n
1  (1) k
 
e j ( 2 k 1)0t
2 k  (2k  1)
or the trigonometric Fourier series
1 
2
x(t )    (1) k
cos[2 (2k  1) f 0 t ]
2 k 0
(2k  1)
1 2
1
1
1
  [cos(2f 0 t )  cos(6f 0 t )  cos(10f 0 t )  cos(14f 0 t )  ]
2 
3
5
7
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
37
Example 4-11 (6/6)
 Double-sided spectrums of the periodic rectangular pulse train signal.
Double-sided amplitude spectrum
Double-sided phase spectrum
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
38
Example 4-12 (1/2)
 Please find the exponential Fourier series for the signal
xc (t )  3e j ( 2000t  / 6)  4e j ( 4000t  / 3)  e j (6000t  / 6)
【Sol.】
1) The signal consists of 3 complex exponential components with
frequencies 1000, 2000 and 3000, respectively. The fundamental
frequency of xc(t) is given by GCD(1000, 2000, 3000) = 1000.
2) The expression of xc(t) is in the form of exponential Fourier series.
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
39
Example 4-12 (2/2)
3) Rewrite xc(t)
xc (t )  3e j ( 2000t  / 6)  4e j ( 4000t  / 3)  e  j ( 6000t  / 6)
 3e j / 6e j 2000t  4e j / 3e j 4000t  e  j / 6e  j 6000t
 3e j / 6e j 2f 0t  4e j / 3e j 2 2 f 0t  e  j / 6e  j 2 3 f 0t ,
f 0  1000
 3e j / 6e j0t  4e j / 3e j 20t  e  j / 6e  j 30t , 0  2000
x(t ) 

 e
n  
n
jn0t


 e
n  
n
j 2nf 0t
,
2
0  2f 0 
T0
Χ1  3e j / 6 ; Χ 2  4e j / 3 ; Χ 3  e j / 6 ; Χ n  0, n  1, 2,  3
Note that xc(t) is a complex signal and thus the symmetry property of
Fourier series discussed later is not held.
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
40
Example 4-13 (1/5)
 Please find the exponential Fourier series for the periodic signal
with period T0.
 A sin 0t ,
x(t )  
0,
0  t  T0 / 2
T0 / 2  t  T0
A
T0 / 2
教育部網路通訊人才培育先導型計畫
T0
Signals and Systems 4.4
41
Example 4-13 (2/5)
【Sol.】
1) The exponential Fourier series with 0 = 2 /T0.
x(t ) 

Χ
n  
2) Find Xn
n
e
jn0t
1
Χn 
T0

,
T0 / 2
0
A

2 jT0
2
0 
T0

A sin 0te  jn0t dt
T0 2
0
(e j0t  e  j0t )e  jn0t dt
T0 2
T0 2
A
j0 (1 n ) t

[ e
dt   e  j0 (1 n )t dt]
0
2 jT0 0
A e j (1 n )  1 e  j (1 n )  1

[

] , n  1
4
1 n
1 n
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
42
Example 4-13 (3/5)
【Sol.】
1) The trigonometric Fourier series with 0 = 2 /T0.
x(t ) 

Χ
n  
2) Find Xn
n
e
jn0t
1
n 
T0

,
T0 / 2
0
A

2 jT0
2
0 
T0

A sin 0te  jn0t dt
T0 2
0
(e j0t  e  j0t )e  jn0t dt
T0 2
T0 2
A
j0 (1 n ) t

[ e
dt   e  j0 (1 n )t dt]
0
2 jT0 0
A e j (1 n )  1 e  j (1 n )  1

[

] , n  1
4
1 n
1 n
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
43
Example 4-13 (4/5)
Using
e j (1 n )  cos(1  n)  j sin(1  n)
 (1)n
we obtain the coefficients
0,
A 1
,

2
 1 n
Χn   A
j ,
 4

A
 j ,
4

教育部網路通訊人才培育先導型計畫
n is odd and n  1
n is even
n  1
 see next page
n 1
Signals and Systems 4.4
44
Example 4-13 (5/5)
A
Χ1 
2 jT0
A

2 jT0

T0 2
0

T0 2
0
(e j0t  e  j0t )e  j0t dt
(1  e  j 20t )dt
A
A

j
4j
4
Similarly
Χ 1
A
A

 j
4j
4
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
45
Example 4-14 (1/4)
 Please find the exponential Fourier series for the periodic impulse
train T (t ) with period T0.
0
 T (t ) 
0

  (t  m T )
0
m  
T (t )
0
1
…
3T0
2T0
教育部網路通訊人才培育先導型計畫
T0
0
…
T0
2T0
3T0
Signals and Systems 4.4
t
46
Example 4-14 (2/4)
【Sol.】
1) The exponential Fourier series with 0 = 2 /T0.
x(t ) 

 Χ ne jn0t ,
n  
0 
2
T0
2) Find Xn
1
Xn 
T0
1

T0


T0
 T (t )e  jn t dt
T0 2

T0 2
0
0
 (t )e  jn t dt
 T (t )   (t )
0
in t herange [
0
T0 T0
,
]
2
2
1
T0
1
  T0 (t ) 
T0
教育部網路通訊人才培育先導型計畫

jn0t
e

n  
Signals and Systems 4.4
47
Example 4-14 (3/4)
 Double-sided amplitude spectrum of the periodic pulse train T (t ) .
0
Phase is 0, thus the
phase spectrum is not
shown
| X nX|n



1/ T0

教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
48
Example 4-14 (4/4)
e jn0t  e jn0t  2 cosn0t
 Using
we obtain
1
 T0 (t ) 
T0

 e jn0t 
n  
1
[1  2(cos0t  cos 20t  ...)]
T0

1
 [1   2 cos(n0t )]
T0
n 1
 Single-sided amplitude spectrum of the periodic pulse train T (t ) .
0
教育部網路通訊人才培育先導型計畫
Signals and Systems 4.4
49
Exponential Fourier series of common used signals
Name
Waveform
X0
Square
wave
0
Sawtooth
wave
A
2
A
2
2A

Triangular
wave
Fullrectified sine
wave
Half-rectified
sine wave
Rectangular
pulse train
Pulse
train
教育部網路通訊人才培育先導型計畫
A

TA
T0
A
T0
Xn, n  0
Note
2A
n
A
j
2n
X n  0, n is even
j
 2A
(n) 2
X n  0, n is even
 2A
 (4n 2  1)
A
 (n 2  1)
Tn0
TA
sinc
T0
2
A
T0
X n  0, n is odd except
X1   j
A
A
and X 1  j
4
4
Tn0 Tn

2
T0
Signals and Systems 4.4
50
From Fourier series to Fourier transform (1/5)
 Rewrite the exponential Fourier series of xE(t)
xE (t ) 

j 2nf 0t
Χ
e
 n
n  
where
1
Xn 
T0
1

T0


 x(t ),
xE (t )  
 0,
教育部網路通訊人才培育先導型計畫
T0 / 2
T0 / 2


xE (t )e  j 2nf 0t dt
x(t )e  j 2nf 0t dt
| t |  T0 / 2
else
Signals and Systems 5.1
51
From Fourier series to Fourier transform (2/5)
 Let’s define a function

X ( f )   x(t )e j 2ft dt

 Then the coefficients Xn can be expressed as
1
X n X (nf0 )
T0
 The exponential Fourier series of xE(t) can be written as


1
j 2nf 0t
xE (t )   Χ (nf0 )e
  Χ (nf0 )e j 2nf 0t f 0
n   T0
n  
教育部網路通訊人才培育先導型計畫
 f0 
Signals and Systems 5.1
1
T0
52
From Fourier series to Fourier transform (3/5)
 As T0  , f0 = 1/T0 become infinitesimal ( f0  0), Thus let f0 =  f . Then we
have
x(t )  lim xE (t )  lim
T0 
f 0

j 2nft
Χ
(
n

f
)
e
f

n  
 The sum on the right-hand side of the above equation can be viewed as the area
under the function Χ (nf )e j 2ft f
as shown in the figure.
Χ ( f )e j 2ft
f
j 2ft
f
Area = Χ (nf )e
Χ (nf )e j 2nft
f
nf
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.1
53
From Fourier series to Fourier transform (4/5)
 Therefore, we have the Fourier representation (Fourier integral) of a
nonperiodic signal x(t).
x(t )  


X ( f )e j 2ft df
 Similarly, we rewrite the coefficients Xn multiplied by T0
1
Xn 
T0
In the limit case
f = nf0 = n f



x(t )e
 j 2nf 0t
dt

lim X nT0   x(t )e
T0 0
教育部網路通訊人才培育先導型計畫


X nT0   x(t )e  j 2nf 0t dt

 j 2nft

dt   x(t )e j 2ft dt  X ( f )

Signals and Systems 5.1
54
From Fourier series to Fourier transform (5/5)
 The function X(f ) is called the Fourier transform of x(t), and the Fourier
representation defines the inverse Fourier transform of X(f ).
 Symbolically, they are denoted by

X ( f )  F [ x(t )]   x(t )e j 2ft dt
-

x(t )  F [ X ( f )]   X ( f )e j 2ft df
1
Keep in mind that the Fourier
integral is the nature of a
Fourier series with
fundamental frequency f0  0.
-
 x(t) and X(f ) are a Fourier transform pair expressed as
F
x(t )
X( f )
F-1
教育部網路通訊人才培育先導型計畫
or
x(t )
X( f )
Signals and Systems 5.1
55
Fourier transform pair (use )
 If the angular frequency  is used, the Fourier transform pair is
expressed as
x(t )  Χ  ()
 Fourier transform:

Χ  ( )   x(t )e

 j t

dt  Χ ( )
2
 Inverse Fourier transform:

x(t )   Χ ( f )e j 2ft df


  Χ(

1

2
教育部網路通訊人才培育先導型計畫



 jt 
)e d ( )
2
2
Χ  ( )e jt d
Signals and Systems 5.1
56
Existence conditions of the Fourier transform
 Not all signals are Fourier transformable. The existence conditions for the
Fourier transform of x(t) are
 The signal x(t) is absolutely integrable, i.e.,



| x(t ) | dt  .
 The signal x(t) has a finite number of maxima and minima within any finite interval.
 The signal x(t) has a finite number of discontinuities within any finite interval, and
each of these discontinuities is finite.
 Though the above conditions guarantee the existence of the Fourier transform
for signal. If the impulse function is allowed in the Fourier transform, some
signals ( e.g. impulse, unit step, sinusoidal, complex exponential signals) can
have Fourier transforms (called generalized Fourier transform that will be
discussed later).
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.1
57
Example 5-1 (1/2)
 Determine the Fourier transform of y(t) = et u( t),  > 0 shown in the figure,
and plot its spectra.
【Sol.】
1) Compute the Fourier transform

Υ ( f )   y (t )e
 j 2ft


dt   e t u (t )e  j 2ft dt
  e (  j 2f )t dt 
0


1
e (  j 2f )t
  j 2f

0
1

  j 2f
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.3
58
Example 5-1 (2/2)
2) Compute the spectra and plot them shown in the figures.
Υ( f ) 
1
1

  j 2f
 2  4 2 f 2
Υ ( f )  (
1
)  1  (  j 2f )  0  t an1 (2f /  )   t an1 (2f /  )
  j 2f
1/ 
 /2
  / 2
 / 2
 / 2
  / 2
 / 2
Double-sided amplitude spectrum
教育部網路通訊人才培育先導型計畫
Double-sided phase spectrum
Signals and Systems 5.3
59
Example 5-2
 Determine the Fourier transform of x(t) = ea|t|, a > 0 shown in the figure, and
plot its spectra.
【Sol.】Compute the Fourier transform

0



0
Χ ( f )  x(t )   x(t )e  j 2ft dt   e at e  j 2ft dt   e at e  j 2ft dt

1
1
2a

 2
a  j 2f a  j 2f a  (2f ) 2
a2
1.5
1.5
x(t) =
Χ( f ) 
ea|t|
1
Χ( f )
X2(f)
x2(t)
1
x(t )
0.5
0.5
0
0
-0.5
-5
2a
a  (2f ) 2
2
-4
-3
-2
-1
0
t
t
1
2
教育部網路通訊人才培育先導型計畫
3
4
5
-0.5
-5
-4
-3
-2
-1
0
f
1
2
3
4
5
f
Signals and Systems 5.3
60
Example 5-3 (1/5)
t
 Determine the Fourier transform of the rectangular pulse signal x (t )  rect ( )

shown in the figure, and plot its spectra.
The rectangular pulse signal
| t |  / 2
1,
x(t )  rect( )  
 0,
t
| t |  / 2
t
x (t )  rect ( )

 / 2
教育部網路通訊人才培育先導型計畫
1
0  /2
t
Signals and Systems 5.3
61
Example 5-3 (2/5)
【Sol.】
1) Compute the Fourier transform ( use f )

Χ ( f )  x(t )   x(t )e  j 2ft dt


t


  rect( )e

 j 2ft
1
e  j 2ft
 j 2f
 /2
dt   1e  j 2ft dt
 / 2
 /2

 / 2
1
(e jf  e  jf )
 j 2f
sin(f )

  sinc(f )
f
sin(x)
sinc( x) 
x
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.3
62
Example 5-3 (3/5)
2) Compute the Fourier transform ( use  )

Χ  ( )   x(t )e  jt dt


t


  rect( )e
 j t
 /2
dt   1e  jt dt
 / 2
1
(e  j / 2  e j / 2 )  

 j
 Sa(

教育部網路通訊人才培育先導型計畫
2

sin(
2
)

2
)
sin( y )
Sa( y ) 
y
Signals and Systems 5.3
63
Example 5-3 (4/5)
 Define and plot sinc(.) and Sa( .)
sin(x)
sinc(x) 
x
sin( y )
Sa( y ) 
y
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.3
64
Example 5-3 (5/5)
Double-sided
amplitude spectrum
Double-sided
phase
spectrum
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.3
65
Example 5-4 (1/2)
 Determine the Fourier transform of the signal xb (t )  rect (t 
1
1
)  rect (t  )
2
2
【Sol.】 Compute the Fourier transform
0
Χb( f )   e
1
 j 2ft
1
dt   e  j 2ft dt
0
1
 j 2ft 0
 j 2ft 1

[e
e
]

1
0
j 2f
1

(1  e j 2f  e  j 2f  1)
j 2f
1

[2  2 cos(2f )]
j 2f
j

(1  cos 2f )
f
1  cos 2f
sin 2 f
 j2
f
e j 2f  e  j 2f
 2 cos 2f
xb (t )
t
 2 sin (f )
2
 j 2f sinc2 ( f )
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.3
66
Example 5-4 (1/2)
Double-sided
amplitude spectrum
Hz
Double-sided
phase spectrum
Hz
 b ( f )  j 2f sinc2 ( f )
 

sin(
)

sin f 2
2 
2
)  j 

j

Sa
( )
If  is used   b ( )  j (


f
2
 ( ) 
 2 
2
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.3
67
Example 5-5
 Determine the Fourier transform of the signal x(t )  1,
  t  .
【Sol.】 Compute the Fourier transform

Χ ( f )   1e  j 2ft dt

 lim 
T /2
1
e  j 2ft
T   j 2f
T /2
1e  j 2ft dt  lim
T  T / 2
T / 2
1
1
 jfT
jfT
 lim
[e
e ]
lim sin(fT )
T   j 2f
T
f 
The Fourier transform of x(t) = 1 does not exist.
(The signal x(t) is not absolutely integrable)
A generalized Fourier transform is introduced next to address this situation.
教育部網路通訊人才培育先導型計畫
Signals and Systems 5.3
68