Chapter4 - The Continuous-Time Fourier Transform

241-306 The Continuous-Time Fourier Transform
1
Chapter 4
The Continuous-Time
Fourier Transform

2
Outline
1 The Representation of Aperiodic Signals : The
Continuous-Time Fourier Transform
2 The Fourier Transform for Periodic Signals
3 Properties of The Continuous-Time Fourier
Transform
4 The Convolution Property
5 The Multiplication Property
6 Systems Characterized by Linear Constant-
Coefficient Differential Equations

3
Development of the Fourier Transform
Representation of an Aperiodic Signal
1 The Representation of Aperiodic Signals : The
Continuous-Time Fourier Transform
xt=
{1, ∣t∣T1
0, T1∣t∣T /2
We begin with Fourier series representation for the
continuous-time periodic square wave

4
The Fourier series coefficients ak
are
ak=
2sink 0 T1
k 0 T
, k≠0

5
Tak=
2sinT1
 ∣=k 0
An alternative way of interpreting ak
is as
samples of an envelops function :
That is, with ω thought of as a continuous
variable, the function (2sin ωT1
)/ω represents the
envelop of Tak
and the coefficients ak
are simply
equally spaced samples of this envelop.

6
If T increases or
ω0
decreases,
the envelop is
sampled with a
closer ans closer
spacing.

7
As T become large, the periodic square wave
approaches a rectangular pulse. Also, the Fourier
series coefficients, multiplied by T, become more
and more closely spaces samples of envelop, so
that the set of Fourier series coefficients
approaches the envelop function as T → ∞.

8

9
xt= ∑
k=−∞
∞
ak e
jk 0 t
ak=
1
T
∫
−T /2
T /2
xte
− jk 0 t
dt
ak=
1
T
∫
−T /2
T /2
xte
− jk 0 t
dt=
1
T
∫−∞
∞
xte
− jk 0 t
dt
Over the interval -T/2 ≤ t ≤ T/2
where ω0
= 2π/T. Since for |t| < T/2,
and also, since x(t) = 0 outside this interval
xt=xt

10
X  j =∫−∞
∞
xte− j t
dt
ak=
1
T
X  jk 0
xt= ∑
k=−∞
∞
1
T
X  jk 0e
jk 0 t
Therefore, defining the envelop
we have
we can express

11
xt=
1
2
∑
k=−∞
∞
X  jk 0e
jk 0 t
0
since 2π/T = ω0
ω → 0 as T→ ∞, the right-hand side passes to
integral.

12
xt=
1
2
∫−∞
∞
X  j e j t
d 
X  j =∫−∞
∞
xte− j t
dt
Fourier Transform pair
Inverse Fourier Transform
Fourier Transform or Fourier Integral

13
ak=
1
T
∫
s
sT
xte
− jk 0 t
dt=
1
T
∫
s
sT
xte
− jk 0 t
dt
ak=
1
T
∫−∞
∞
xte
− jk 0 t
dt
ak=
1
T
X  j∣=k 0
Let x(t) be a finite-duration signal that is equal to xt
over one period s ≤ t ≤ s+T for some value of s
since x(t) is zero outside the range s≤ t ≤s+T
we conclude that

14
Convergence of Fourier Transforms
xt=
1
2
∫−∞
∞
X  j e
j t
d 
∫−∞
∞
∣xt∣2
dt∞
Let xt
if x(t) has finite energy
we guaranteed that X(jω) is finite

15
∫−∞
∞
∣et∣
2
dt=0
If e(t) denote the error between x(t) and xt
The Dirichlet conditions

16
1 x(t) be absolutely integrable; that is,
∫−∞
∞
∣xt∣dt∞
2 x(t) have finite number of maxima and minima
within any finite interval
3 x(t) have a finite number of discontinuities
within any finite interval. Furthermore, each of
these discontinuities must be finite.

17
Example 4.1
xt=e
−at
ut, a0
X  j =∫
0
∞
e−at
e− jt
dt
=−
1
a j
e
−a jt
∣0
∞
Find the Fourier Transform for the signal
solution

18
X  j =
1
a j
, a0
∣X  j ∣=
1
a
2

2
∢ X  j =−tan
−1

a 
That is
magnitude
phase

19

20
Example 4.2
xt=e
−a∣t∣
, a0
Find the Fourier Transform for the signal
solution
X  j =∫−∞
∞
e
−a∣t∣
e
− j t
dt
=∫−∞
0
e
a t
e
− jt
dt∫
0
∞
e
−a t
e
− j t
dt

21
X  j =
1
a− j

1
a j
=
2a
a
2

2

22

23
Example 4.3
Determine the Fourier Transform of unit
impulse
xt=t
solution
X  j =∫−∞
∞
te
− j t
dt=1
That is, the unit impulse has a Fourier Transform
consisting of equal contributions at all frequencies.

24
Example 4.4
xt=
{1, ∣t∣T1
0, ∣t∣T1
X  j =∫
−T1
T 1
e
− jt
dt=2
sinT1

solution
Determine the Fourier Transform of rectangular
pulse signal

25

26
Example 4.5
X  j =
{1, ∣∣W
0, ∣∣W
xt=
1
2
∫
−W
W
e
jt
d =
sinW t
t
Determine the x(t) whose Fourier Transform is
solution
xt=
1
2
∫−∞
∞
X  j e
j t
d 

27

28
sinc=
sin

The sinc function
Definition
This function arise frequently in Fourier Transform
and in study of LTI systems.

29
2sin T1

=2T1 sincT1
 
sinW t
t
=
W

sincW t
 
The signal in the example 1 and example 2 can be
express in the form of sinc function
example 1
example 2

30
From example 2, 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 Example
4.3.

31

32

33

34
2 The Fourier Transform for Periodic Signals
X  j =2−0
xt=
1
2
∫−∞
∞
2−0e
jt
d =e
j 0 t
Let a signal x(t) with Fourier transform X(jω)
that is a single impulse of area 2π at ω = ω0
Determine the x(t) by using inverse Fourier
transform

35
X  j = ∑
k=−∞
∞
2ak −k 0
xt= ∑
k=−∞
∞
ak e
jk 0 t
If X(jω) is of the form of a linear combination of
impulses equally spaced in frequency
then

36
Example 4.6
ak=
sin k 0 T1
k
Find Fourier transform of the signal
solution
The Fourier series coefficients are

37
X  j = ∑
k=−∞
∞
2sin k 0 T1
k
−k 0
and the Fourier transform of this signal is
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.

38
Plot from
Example 3.5(a)

39
Example 4.7
xt=sin0 t
a1=
1
2j
a−1=−
1
2j
ak=0, k≠1 o r −1
Solution

40
X  j =

j
−0−

j
0
The Fourier transform are

41
Solution
xt=cos0 t
a1=a−1=
1
2
ak=0, k≠1 o r −1

42
X  j =−00
The Fourier transform are

43
Example 4.8
xt= ∑
k=−∞
∞
t−kT 
Find The Fourier transform of the impulse train

44
Solution
ak=
1
T
∫
−T /2
T /2
te
− j k 0 t
dt=
1
T
X  j =
2
T
∑
k=−∞
∞
−
2 k
T 
The Fourier series coefficients are given by
The Fourier transform given by

45
(a) periodic impulse train (b) its Fourier transform

46
3 Properties of the Continuous-Time Fourier
Transform
Notation
F
xt ↔ X  j 
Example
e−at
ut ↔
F 1
a j
1
a j
=F {e−at
ut}
e
−at
ut=F
−1
{ 1
a j }

47
Linearity
xt ↔
F
X  j 
yt ↔
F
Y  j
axtbyt ↔
F
aX  jbY  j
If
then

48
Time Shifting
xt ↔
F
X  j 
xt−t0 ↔
F
e
− jt0
X  j
If
then

49
Example 4.9
Find The Fourier transform of the signal below

50
Solution
xt=
1
2
x1 t−2.5x2t−2.5
x(t) can be expressed as the linear combination

51
By using the result from Example 4.4, we obtain
X 1 j =
2sin/2

X 2 j =
2sin3/2

The Fourier transform of this signal is
X  j =e− j5/2
{sin/22sin3/2
 }

52
Conjugation and Conjugate Symmetry
xt ↔
F
X  j 
x
∗
t ↔
F
X
∗
− j
If
then

53
Conjugate symmetry
If x(t) is real then X(jω) has conjugate symmetry
X − j=X
∗
 j [xt real]
If we express X(jω) in rectangular form as
X  j =ℜe[ X  j] j ℑm[ X  j ]

54
then if x(t) is real
ℜe[ X  j]=ℜe[ X − j ]
ℑm[ X  j]=−ℑm[ X − j ]
and
The real part of Fourier transform is an even
function of frequency and the imaginary part is an
odd function of frequency

55
If we express in polar form as
X  j =∣X  j ∣e
j∢ X  j
|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.

56
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

57
With x(t) real, we can conclude that
xt ↔
F
X  j 
Ev{xt} ↔
F
ℜe{X  j }
Od {xt} ↔
F
j ℑm{X  j}

58
Example 4.10
Find The Fourier transform of the signal
xt=e
−a∣t∣
, a0
by using the symmetry properties
solution
From Example 4.1, we have
e
−at
ut ↔
F 1
a j

59
xt=e
−a∣t∣
=e
−at
ute
at
u−t
=2[e
−at
ute
at
u−t
2 ]=2 Ev {e
−at
ut}
Since e-at
u(t) is real
Ev{e
−at
ut} ↔
F
ℜe{ 1
a j }
It follow that
X  j =2ℜe{ 1
a j}=
2a
a
2

2

60
Differentiation and Integration
Let x(t) be a signal with Fourier transform X(jω)
dxt
dt
=
1
2
∫−∞
∞
j  X  j e
jt
d 
Therefore
dxt
dt
↔
F
j  X  j 

61
∫−∞
t
xd  ↔
F 1
j
X  j  X 0
For integration,
The impulse term on the right-hand side reflects
the DC or average value that can result from
integration.

62
Example 4.11
g t=t ↔
F
G j =1
xt=∫−∞
t
gd 
Determine the Fourier transform of x(t) = u(t) by
using the integration property and the knowledge
that
solution
Noting that

63
X  j =
G j 
j
G0
X  j =
1
j

t=
dut
dt
↔
F
j [ 1
j 
]=1
Taking the Fourier Transform both side
since G(jω) =1, we conclude that
We can recover the impulse by

64
Example 4.12
Find the Fourier transform of the signal x(t)

65
g t=
d
dt
xt
solution
Consider the signal
G j={2sin
 }−e
j
−e
− j 
Fourier transform of g(t) is

66
X  j =
G j 
j
G0
X  j =
2sin 
j
2
−
2cos
j
By using integration property
with G(0) = 0
X(jω) is purely imaginary and odd, which is
consistent with the fact that x(t) is real and odd.

67
Time and Frequency Scaling
xt ↔
F
X  j 
xat ↔
F 1
∣a∣
X j 
a 
Let
then

68
x−t ↔
F
X − j 
If a = -1
That is, reversing a signal in time also reverses
its Fourier transform

69
Duality
By comparing the Fourier transform and the
inverse Fourier transform
xt=
1
2
∫−∞
∞
X  j e j t
d 
X  j =∫−∞
∞
xte− j t
dt
We observe that these equations are similar, but
not quite identical, in form.

70
This symmetry leads to a property of duality. In
example 4.4 and 4.5, we note the relationship
between the Fourier transform pair
x2t=
sin W t
t
↔
F
X 2  j=
{1, ∣∣W
0, ∣∣W
x1t=
{1, ∣t∣T1
0, ∣t∣T1
↔
F
X 1  j=
2sin T1

Example 4.4
Example 4.5

71

72
Example 4.13
g t=
2
1t
2
X  j =
1
1
2
Find the Fourier transform of the signal g(t) by
using duality property
solution
We consider the signal x(t) whose Fourier
transform is

73
From example 4.2 , with a = 1
xt=e
−∣t∣
↔
F
X  j=
2
1
2
e
−∣t∣
=
1
2
∫−∞
∞
 2
12 e
j t
d 
The synthesis equation for this Fourier transform
pair is

74
2e
−∣t∣
=∫−∞
∞
 2
1
2 e
− jt
d 
2e
−∣∣
=∫−∞
∞
 2
1t
2 e
− j t
dt
F
{ 2
1t
2
}=2e
−∣∣
Multiplying this equation by 2π and replacing t by
-t, we obtain :
Interchanging the name of variables t and ω
Right-hand side is the Fourier transform of g(t)

75
Parseval 's Relation
∫−∞
∞
∣xt∣
2
dt =
1
2
∫−∞
∞
∣X  j ∣
2
d 
If x(t) and X(jω) are Fourier transform pair

76
Example 4.14
Evaluate the following expression
by using Fourier transform in the figure below
E=∫−∞
∞
∣xt∣
2
dt D=
d
dt
xt∣t=0

77
Evaluate E infrequency domain
E=
1
2
∫−∞
∞
∣X  j∣
2
d 
which evaluate to 5/8 for figure (a) and to 1 for
figure (b)
Evaluate D infrequency domain
g t=
d
dt
xt ↔
F
j X  j =G j 

78
Noting that
D=g0=
1
2
∫−∞
∞
G j d 
We conclude
D=
1
2
∫−∞
∞
j  X  j d 
which evaluate to zero for figure (a) and to -1/(2√π)
for figure (b)

79
4 The Convolution Property
ht ↔
F
H  j 
yt ↔
F
Y  j
yt=ht∗xt ↔
F
Y  j=H  j X  j
If
then
xt ↔
F
X  j 

80
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.

81
Many of the properties of LTI systems can be
conveniently interpreted in the term of H(jω). For
example,

82
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 integrable; that is ,
∫−∞
∞
∣ht∣dt∞

83
Example 4.15
Consider the LTI system with impulse response
ht=t−t0
The frequency response is
H  j =e
− j
0
For any input x(t) with Fourier transform H(jω),
the Fourier transform of the output is
Y  j =H  j  X  j 
=e
− jt
0
X  j 
yt=xt−t0

84
Example 4.16
Consider the LTI system for which the input x(t)
and the output y(t) are related by
yt=
dxt
dt
From the differentiation property
Y  j= j X  j
The frequency response of the differentiator is
H  j = j

85
Example 4.17
Consider the LTI system for which the input x(t)
and the output y(t) are related by
yt=∫−∞
t
xd 
The impulse response for this system is u(t)
and the frequency response of the system is
H  j=
1
j 


86
By using convolution property
Y  j=H  j X  j 
=
1
j
X  j X  j 
=
1
j
X  j X 0

87
Example 4.18
Consider Ideal low pass filter which have the
frequency response
H  j =
{1, ∣∣c
0, ∣∣c

88
From example 4.5, the impulse response h(t) of
this ideal filter is
ht=
sinc t
t

89
Example 4.19
Determine the response of an LTI system with
impulse response
ht=e
−at
ut, a0
to the input signal
xt=e
−bt
ut, b0

90
solution
Transform the signal into the frequency domain.
From Example 4.1, the Fourier transform of x(t)
and h(t) are
X  j =
1
b j
H  j=
1
a j 
Therefore
Y  j=
1
a j b j 

91
Y  j =
A
a j 

B
b j 
A=
1
b−a
=−B
Y  j =
1
b−a[ 1
a j 
−
1
b j]
Assuming that a ≠ b, the partial fraction
expansion for Y(jω) takes the form
We find that
Therefore

92
yt=
1
b−a
[e
−at
ut−e
−bt
ut]
Y  j=
1
a j 
2
1
a j 2= j
d
d [ 1
a j ]
The output can find by inverse Fourier
transform
If a = b
Recognizing this as

93
e
−at
ut ↔
F 1
a j 
te
−at
ut ↔
F
j
d
d [ 1
a j]
yt=t e
−at
ut
We can use the dual of the differentiation
property
and the consequently

94
Example 4.20
xt=
sini t
t
ht=
sinc t
t
Determine the response of an ideal lowpass filter
to an input
the impulse response of the ideal lowpass filter is

95
Y  j =X  j H  j 
X  j =
{1, ∣∣≤i
0, elsewhere
H  j =
{1, ∣∣≤c
0, elsewhere
solution
The output is the convolution of 2 sinc functions

96
Y  j=
{1, ∣∣≤0
0, elsewhere
yt=
{
sin c t
t
, if c≤i
sin i t
t
, if i≤c
Therefore
The inverse Fourier transform of Y(jω) is

97
4 The Multiplication Property
rt=st pt↔
F
R j =
1
2
∫−∞
∞
S  j P j−d 
The multiplication in time domain corresponds to
convolution in frequency domain
Sometime referred to as the Modulation property

98
Example 4.21
Find the spectrum R(jω) of r(t) = s(t)p(t) when
pt=cos0 t
and

99
solution
P j=−00
R j=
1
2
∫−∞
∞
S  j P j−d 
=
1
2
S j−0
1
2
S  j0

100

101
Example 4.22
Find the spectrum G(jω) when g(t) = r(t)p(t) and
r(t) and p(t) are the signals from Example 4.21
solution
By using linearity property to the spectrum R(jω)

102

103
Example 4.23
xt=
sintsint/2
t
2
xt=sint
t sint/2
t 
Find the Fourier transform of the signal x(t)
solution
The key is to recognize x(t) as the product of
two sinc functions

104
X  j =
1
2
F {sint
t }∗F {sint/2
t }
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)

105

106

107
6 Systems Characterized by Linear Constant-
Coefficient Differential Equations
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
∑
k=0
N
ak
d
k
yt
dt
k
=∑
k=0
M
bk
d
k
xt
dt
k

108
Y  j =H  j  X  j 
H  j =
Y  j
X  j 
By the convolution property
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).

109
F
{∑
k=0
N
ak
dk
yt
dtk }=F
{∑
k=0
M
bk
dk
xt
dtk }
∑
k=0
N
ak F
{dk
yt
dtk }=∑
k=0
M
bk F
{dk
xt
dtk }
consider applying the Fourier transform to the
equation in slide 107
from linear property

110
Y  j
[∑
k=0
N
ak  j
k
]=X  j
[∑
k=0
M
bk  j
k
]
∑
k=0
N
ak  j 
k
Y  j=∑
k=0
M
bk  j
k
X  j
and from the differentiation property
or

111
H  j =
Y  j
X  j 
=
∑
k=0
M
bk  j 
k
∑
k=0
N
ak  j k
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

112
Example 4.24
Find the impulse response of the LTI system
dyt
dt
a yt=xt
with a>0
Solution
Fourier transform of the system is
jY  jaY  j=X  j

113
 jaY  j=X  j
H  j =
Y  j
X  j 
=
1
ja
From Example 4.1, the inverse Fourier
Transform of equation above is
ht=e
−at
ut

114
Example 4.25
Find the impulse response of the LTI system
d2
yt
dt
2
4
dyt
dt
3 yt=
dxt
dt
2xt
Solution
The frequency response is
H  j =
 j 2
 j2
4 j 3

115
We factor the denominator of the right-hand side
H  j =
 j2
 j 1 j3
By using the partial-fraction expansion
H  j =
1/2
j1

1/2
j3
The inverse Fourier transform of each term
ht=
1
2
e
−t
ut
1
2
e
−3t
ut

116
Example 4.26
Consider the system of Example 4.25, find the
output of the system when the input is
xt=e
−t
ut
Solution
Y  j=H  j X  j 
=[  j 2
 j 1 j 3][ 1
j 1]

117
Y  j=
 j 2
 j 1
2
 j3
By using the partial-fraction expansion
Y  j=
A11
j1

A12
 j1
2

A21
j3
Y  j=
1/4
j1

1/2
 j1
2
−
1/4
j3

118
The inverse Fourier transform of each term
yt=
1
4
e
−t
ut
1
2
te
−t
ut−
1
4
e
−3t
ut

119

Chapter4 - The Continuous-Time Fourier Transform

Recomendados

Recomendados

Mais conteúdo relacionado

Mais procurados

Mais procurados (20)

Semelhante a Chapter4 - The Continuous-Time Fourier Transform

Semelhante a Chapter4 - The Continuous-Time Fourier Transform (20)

Mais de Attaporn Ninsuwan

Mais de Attaporn Ninsuwan (20)

Último

Último (20)

Chapter4 - The Continuous-Time Fourier Transform