# The continuous-time Fourier transform

Document Sample

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CHAPTER 4
THE CONTINUOUS-TIME
FOURIER TRANSFORM
Outlines
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Covers O & W pp. 284-309
 Derivation of the CT Fourier Transform pair

 Examples of Fourier Transforms

 Fourier Transforms of Periodic Signals

 Properties of the CT Fourier Transform
Fourier Transform
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 We have shown that Fourier series are useful in analyzing periodic
signals, but many (most) signals are aperiodic. Need a more
general tool –– Fourier transform.
Fourier’s own derivation of the CT Fourier transform
 x(t) – an aperiodic signal

– view it as the limit of a periodic signal as T → ∞
 The harmonic components are spaced ωo = 2π/T apart,        as T
→ ∞, and ωo → 0, then ω = kωo becomes continuous
⇓
Fourier series  Fourier integral
Motivating Examples: Square wave
4
So, on with the derivation of FT …
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Derivation (continued)
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Derivation (continued)
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With CTFT, now the Frequency Response of an LTI
System makes complete sense
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(s = j)
ejt           H(j)              H(j)ejt



CT Frequency response:         H  j    ht e  jt dt


H(j) is the FT of h(t)
For what finds of signals can we do
this?
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It works also even if x(t) is infinite duration, but satisfies:

a) Finite energy
 xt 
2
dt  

In this case, there is zero energy in the error
                             
1
et   x t         X  j e d                 et  dt  0
jt                   2
Then
2                               

b) Dirichlet conditions (including)

(i)
    x t  dt  
2



x t  at points of continuity
1
 X  j e jt d 
1
xt  0  xt  0  

2   
2                         midpoint at discontinuity
(ii) Gibb’s phenomenon
For what finds of signals can we do
this? (Cont.)
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c) By allowing impulses in x(t) or in X(j) , we can represent
even more signals

E.g. It allows us to consider FT for periodic signals
Ex #1
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 Find the Fourier Transform for the signal
(a)    x(t) = (t)

X  j     t e  jt dt  1


1
 t             e jt d - Synthesis equation for  t 
2   

(b)   x(t) = (t – t0)

X  j     t  t0 e  jt dt


 e  jt0          - Linear phase shift in 
Ex #2: Exponential function
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   Find the Fourier Transform for the signal
x(t) = e-atu(t), a > 0
                      
X  j       xt e  jt dt   e  ate  jt dt
                         
e-(a+j)t
1            a  j t         1
 (               )e                 
a  j                       0     a  j
Ex # 3
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   Find the Fourier Transform for the signal

xt   e   a t
,a  0
   Solution

X  j    e
a t
e  jt dt

0                         
  e at e  jt dt   e  ate  jt dt
                         0
Ex #3 (continued)
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1      1
X  j         
a  j a  j
2a
 2
a 2
Ex #4: A square pulse in the time-domain
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   Determine the Fourier Transform of rectangular pulse
signal        1 t  T1

x t   
0
          t  T1

   Fourier transform of this signal is
T1

X  j      x(t )e  jt dt
T1
T1

  (1)e  jt dt
T1

sin T1
2

Ex #5
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   Determine the x(t) whose Fourier Transform is
1   W
X  j   
0   W
   Solution

1
xt         X  j e
jt
d
2   
W
1            sin Wt
xt          e d  t
jt

2   W
The sinc function
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   Definition
sin 
sinc  

   This function arise frequently in Fourier Transform and in
study of LTI systems.
The sinc function (Cont.)
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   The signal in the ex# 4 and ex#5 can be express in the
form of sinc function                 sin 
sinc  

   Ex#4                                  T1 
sin      
X  j   2 sin T1  2T1 T    2T1sinc T1 

     
               1
          

   Ex #5                                 Wt
sin        
sin Wt W                      W      Wt 
xt                                    sinc 
t        Wt
                 

Example #4: (continued)
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Note the inverse relation between the two widths  Uncertainty principle
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   From ex# 5, we see that if W increases, X(jω) becomes
broader, while the main peak of x(t) at t=0 become
higher and the width of the first lobe of this signal
becomes narrower. If W→∞, X(jω) = 1 for all ω. The x(t)
converges to impulse as in ex#1.
Example #5 A Gaussian xt   e                       at2

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
X  j    e    at2  jt
e        dt

 2   j  2   j  2
    a  t  j t    a   
 e        
      a  2a    2a 

dt


  a  t  j     2
2
        
 e     2a 
dt e 4 a
 
                    

 a
2
       
            e       4a
a
CT Fourier Transforms of Periodic
Signals
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Suppose
X(j) = ( - 0)

1                              1 j0t
x t                 0 e d  2 e - Periodic in t with
jt

2                                    frequency 0

That is
e     j0t
 2  - 0 
- All the energy is
concentrated in
More generally, if x(t) = x(t+T), then one frequency 0
                                   
xt     a e     k
jk0t
 X  j      2a    k 
k     0
k                                k  
Example #6:
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                                   
xt     a e     k
jk0t
 X  j      2a    k 
k     0
k                                k  

1 j0t 1  j0t
x t   cos 0t  e  e                               a(1) = ½
2        2                           a(-1) = ½
X  j      0      0 
Example #7
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   Find Fourier transform of the signal

   solution
   The Fourier series coefficients are
sin k0T1
ak 
k

sin k0T1 jk0t
xt                  e
k      k
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sin k0T1 jk0t

xt                  e
k      k
   and the Fourier transform of this signal is
                                       
xt     a e     k
jk0t
 X  j      2a    k 
k     0
k                                    k  

2 sin k0T1

X  j                        k0 
k        k
   For T = 4T1, In comparison with Example 3.5(a), the only
differences are a proportionality factor 2π and the use
of impulses rather than a bar graph.
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   FT

2 sin k0T1
X  j                        k0 
k        k
For T = 4T1, In comparison, the only
differences are a proportionality
factor 2π and the use of impulses
rather than a bar graph.

   FS            
sin k0T1 jk0t
xt                  e
k      k
Example #8
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   Find Fourier transform of the signal
x(t) = sin0t
   Solution
The Fourier series coefficients are
1                  1
a1             a1  
2j                 2j

ak = 0, k ≠ 1 or −1
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   The Fourier transform are
                                     
xt     a e     k
jk0t
 X  j          2a    k 
k        0
k                                k  

                 
X  j                0      0 
j                   j
Example #9 Sampling function
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
x t      t  nT  - Sampling function
n  

T 2
1                   1
x t   ak      2xt e dt  T
 jk0t

T T
                     
1 jk0t
 x t       a e     k
jk0t
  e
k                   k   T
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1  jk0t
xt    e                   xt   X  j 
T k 

2       k 2 
X  j                   
k   T        T 
k0
Properties of the CT Fourier Transform

 1) Linearity
 2) Time Shifting

 3) Conjugate and conjugate Symmetry

 4) Time-Scaling

 5) Differentiation and Integration

 6) Duality
Properties of the CT Fourier Transform
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1) Linearity ax(t) + by(t)                aX(j) + bY(j)
2) Time Shifting     x(t-t0)              e jt0 X  j 
                                 
Proof:            x t  t0 e  jt dt  e  jt  x t e jtdt 
     
0


t
FT magnitude unchanged                                          X(j)

e j0t X  j   X  j 

Linear change in FT phase
                 
 e j0t X  j   X  j   0t
a phase shift
Linear with 
Example #10
33

Calculate Fourier Transform of x(t)

1
x t   x1 t  2.5  x2 t  2.5
2
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sin 2 
12

X 1  j      e
 jt
dt  2
1 2


sin3 2 
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X 2  j      e
 jt
dt  2
3 2


 sin 2  2 sin3 2 
X  j   e    j5t 2
                        
                       
35

12
1
X 1  j    x t  2.5e  jt dt
1 2
2
   Let t = t – 2.5
12                                          12
1                           1  j 5 2
X 1  j    x t e  j t  2.5 
dt  e          2xt e  jtdt 
1 2
2                           2          1

 j 5 2 sin  2 
12
1  j 5 2
2de  e
 jt 
      e
j 2         1

The Properties Keep on Coming…
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3) Conjugate and conjugate Symmetry
if
x(t)  X(j)
then
x*(t)  X*(-j)
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  Conjugate symmetry
If x(t) is real then X(jω) has conjugate symmetry

x(t) real     X(-j) = X*(j)

If we express X(j) in rectangular form as

X(j) = ℜe[X(j)] + ℑm[X(j)]
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then if x(t) is real
ℜe[ X(j)] = ℜe[X(−j)]
and
ℑm[X(j)] = −ℑm[X(−j)]

The real part of Fourier transform is an even function of
frequency and the imaginary part is an odd function of
frequency
39

If we express in polar form as
X(jω) = ∣X(jω)∣ej∢X(jω)
|X(jω)| is an even function of ω and ∢X(jω) is an odd function
of ω

Thus, when computing the Fourier transform of a real signal, the
real and imaginary parts or magnitude and phase of the
transform need only be specified for positive frequencies, as
the values for negative frequencies can be determined
directly from the values for ω > 0 using the relations above.
40

If x(t) is real then it can always be expressed in terms of
the sum of an even function and an odd function.
x(t) = xe(t) + xo(t)

From the linearity property
F{x(t)} = F{xe(t)} + F{xo(t)}

F{xe(t)} is a real function
F{xo(t)} is purely imaginary
41

   With x(t) real, we can conclude that

x(t) ↔ X(j)

Ev{x(t)} ↔ ℜe{X(j)}

Od{x(t)} ↔ jℑm{X(j)}
Example #11
42

 Find The Fourier transform of the signal
x(t) = e−a∣t∣ , a > 0
by using the symmetry properties
solution
From Example #2, we have

1
e u t  
 at    F

a  j
43

x(t) = e−a∣t∣ = e−atu(t) + eatu(−t)
 e  atu t   e atu  t 
 2
2
                 
 2 Ev e atu t 
                            
Since e-atu(t) is real
 1 

Ev e u t  
 at
     e
F

 a  j 
 1          2a
X  j   2e         2
 a  j  a   2
44

4) Time-Scaling x at    1 X  j  
          
a     a
 a = -1
x(-t)         X(-j)

a) x(t) real and even           x(t) = x(-t) = x*(t)
 X(j) = X(-j) = X*(j)         - Real & even
b) x(t) real and odd x(t) = -x(-t) = x*(t)
 X(j) = -X(-j) = -X*(j) – Purely imaginary & odd
45

c) X(j) = Re{X(j)} + Im{X(j)}
          
For real x(t) = Ev{x(t)} + Od{x(t)}
46

5) Differentiation and Integration
x(t)                         X(j)
dx t  1

          jX  j e jt d
dt      2   

dx t  
Differentiation               jX  j 
dt
t
1
x d  j X  j   X 0  
Integration                          


Example #12
47

Determine the Fourier transform X(j) of the unit step
x(t) = u(t)
g t    t  G j   1

t                              
x t      g  d        X  j      x t e  jt dt
                             
Take FT of both sides                                
G  j                             e  jt dt
X  j             G 0  
j                                  0

1
1                                
       G 0                   j
j
Example #13
48

Calculate FT X(j) for the signal x(t)

d
g t   x t 
dt
49

2 sin 
G  j                e j  e j

   Using integration property

G j 
X  j            G0  
j
   with G(0)= 0

2 sin  2 cos 
X  j          
j 2
j
50

   6) Duality

1,
     t T 1                      2 sin T1
x1 t                  X 1  j  


0,
     t T 1                         

sin Wt                1,
           W
x2 t          X 2  j   
t                   0,
           W
51
FT of sinc function
52

                                                jWt            W
sin Wt 1 e              jWt
e              1 jt
x t                                                   e
t    t                    2j             2tj   W
W                         
1              1
x t          e d  2
jt
 H  j e jt d
2   W                        
x(t)
W
X(j)

1



                                         -W            W

W        W
The CT Fourier Transform Pair
53

x(t)                X(j)
                           - FT
X  j         x t e  jt dt   (Analysis Equation)


1                         - Inverse FT
x t            X  j e d (Synthesis Equation)
jt

2   
Example #14
54

   Find the Fourier transform of the signal g(t) by using
duality property
2
g t  
1 t 2

solution
We consider the signal x(t) whose Fourier transform is

1
X  j  
1  2
55

From example #3 , with a = 1
2
xt   e           X  j  
t
 F

1  2
The synthesis equation for this Fourier transform pair is

1       2  jt
 1   2 e d
t
e        
2              
56

Multiplying this equation by 2π and replacing t by -t, we
obtain :              
 2   jt
 
t
2e                  2 
e d

1  
Interchanging the name of variables t and ω

 2   jt
 

2e                    2 
e dt
 
1 t 
Right-hand side is the Fourier transform of g(t)
 2           
F     2
 2e
1  t 
Parseval 's Relation
57

   If x(t) and X(jω) are Fourier transform pair

                   
1
 xt  dt  2    X  j 
2                         2
d
                   
Example #15
58

Evaluate the following expression

d
 xt               D  xt 
2
E                 dt

dt     t 0
by using Fourier transform in the figure below
59

   Evaluate E infrequency domain

1
 X  j 
2
E                       d
2   
which evaluate to 5/8 for figure (a) and to 1 for figure (b)
 Evaluate D infrequency domain
60

   Noting that

   We conclude

   which evaluate to zero for figure (a) and to -1/(2√π) for
figure (b)
The Convolution Property
61

   If

   Then
62

   This property is important property in signals and
system. As expressed in the equation, the Fourier
transform maps the convolution of two signals into the
product of their Fourier transforms. H(jω), is the Fourier
transform of the impulse response, is the frequency
response and capture the change in complex amplitude
of the Fourier transform of the input at each frequency
ω.
   The frequency response H(jω) plays as important a role
in the analysis of LTI systems.
63

   Many of the properties of LTI systems can be
conveniently interpreted in the term of H(jω). For
example,
64

   The convergence of the Fourier transform is guaranteed
only under certain conditions, and consequently, the
frequency response cannot be defined for every LTI
system. If, however, an LTI system is stable, then its
impulse response is absolutely integratble; that is ,
Example #16
65

   Consider the LTI system with impulse response
H(t) = (t−t0)
   The frequency response is
H ( j) = e− jt0
   For any input x(t) with Fourier transform H(jω), the
Fourier transform of the output is
Example #17
66

   Consider the LTI system for which the input x(t) and the
output y(t) are related by

   From the differentiation property

   The frequency response of the differentiator is
Example #18
67

   Consider the LTI system for which the input x(t) and the
output y(t) are related by
t
y t      x d

   The impulse response for this system is u(t) and the
frequency response of the system is
68

   By using convolution property
Example #19
69

   Consider Ideal low pass filter which have the frequency
response
1,     c
H  j   
0,     c
70

   From example #5, the impulse response h(t) of this ideal
filter is
Example #20
71

   Determine the response of an LTI system with impulse
response

to the input signal
72

 solution
Transform the signal into the frequency domain. From
Example #2, the Fourier transform of x(t) and h(t) are

Therefore
73

   Assuming that a ≠ b, the partial fraction expansion for
Y(jω) takes the form

   We find that

   Therefore
74

   The output can find by inverse Fourier transform

   If a = b

   Recognizing this as
75

   We can use the dual of the differentiation property

   and the consequently
Example #21
76

   Determine the response of an ideal lowpass filter to an
input

   the impulse response of the ideal lowpass filter is
77

   solution
   The output is the convolution of 2 sinc functions
78

   Therefore

   The inverse Fourier transform of Y(jω) is
The Multiplication Property
79

   The multiplication in time domain corresponds to
convolution in frequency domain

   Sometime referred to as the Modulation property
Example #22
80

   Find the spectrum R(jω) of r(t) = s(t)p(t) when
p(t) = cosω0t
and
81

   solution

P j      0      0 

1
R  j         S  j P j    d
2   

1                   1
    S  j   0   S  j   0 
2                   2
82
Example #23
83

 Find the spectrum G(jω) when g(t) = r(t)p(t) and r(t) and
p(t) are the signals from Example 22
 solution

By using linearity property to the spectrum R(jω)
84
Example #24
85

   Find the Fourier transform of the signal x(t)

   solution
   The key is to recognize x(t) as the product of two sinc
functions
86

   Applying the multiplication property

   The Fourier transform of each function is a rectangular
pulse, we can proceed to convolution those pulses to
obtain the X(jω)
87
88
Systems Characterized by Linear Constant-
Coefficient Differential quations
89

   A particularly important and useful class of continuous-
time LTI system is those for which the input and output
satisfy a linear constant coefficient differential equation
of the form
90

    By the convolution property
Y(jω) = H(jω)X(jω)

or

where X(jω), Y(jω) and H(jω) are the Fourier transforms of
the input x(t), output y(t) and impulse response h(t).
91

   consider applying the Fourier transform to the equation
in slide 89

   from linear property
92

   and from the differentiation property

   Or
93

   Thus

   Observe that H(jω) is thus a rational function; that is, it is
a ratio of polynomials in (jω) and the frequency
response for the LTI system can be written directly by
inspection
Example #25
94

   Find the impulse response of the LTI system

with a>0
   Solution
Fourier transform of the system is
jY(j) + aY(j) = X(j)
95

   From Example #2, the inverse Fourier Transform of
equation above is
Example #26
96

   Find the impulse response of the LTI system

   Solution
The frequency response is
97

   We factor the denominator of the right-hand side

   By using the partial-fraction expansion

   The inverse Fourier transform of each term
Example #27
98

   Consider the system of Example 26, find the output of
the system when the input is
x(t) = e−tu(t)
   Solution
99

   By using the partial-fraction expansion
100

   The inverse Fourier transform of each term

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Description: Fourier transform, Time Fourier, Fourier series, Signals and Systems, Frequency response, Discrete-time Fourier transform, Periodic Signals, periodic signal, LTI system, Continuous-Time Signals, continuous time, Discrete Fourier Transform, impulse response, Discrete-Time Signals, frequency domain, Digital Signal Processing, sinc function, Discrete Time, discrete-time Fourier, sampling rate, Time domain, Discrete-Time Systems, Laplace Transform, series representation, Fourier transforms, for Real,