Chapter5 - The Discrete-Time Fourier Transform

241-306 Discrete-Time Fourier Transform
1
Chapter 5
The Discrete-Time
Fourier Transform

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

3
Development of the Fourier Transform
Representation of an Aperiodic Signal
1 The Representation of Aperiodic Signals : The
Discrete-Time Fourier Transform
Consider a sequence x[n] that is of finite duration and
x[n] = 0 outside the range -N1
≤ n ≤ N2
. From the
aperiodic signal, we can construct a periodic
sequence when x[n] is one periodx[n]
x[n]= ∑
k=〈N 〉
ak e jk2/ N n

4

5
ak=
1
N
∑
n=〈N 〉
x[n]e− jk 2/N n
ak=
1
N
∑
n=−N1
N 2
x[n]e− jk 2/N n
=
1
N
∑
n=−∞
∞
x[n]e− jk 2/N n
Since for -N1
≤ n ≤ N2
, and also,
x[n] = 0 outside this interval
x[n]=x[n]

6
X e j
= ∑
n=−∞
∞
x[n]e− jn
ak=
1
N
X e
jk 0

x[n]= ∑
k=〈N 〉
1
N
X e
jk 0
e
jk 0 n
Defining the function
we have
we can express

7
x[n]=
1
2
∑
k=〈 N 〉
X e
jk 0
e
jk 0 n
0
since 2π/N = ω0
ω → 0 as N→ ∞, the right-hand side passes to
integral.

8
x[n]=
1
2
∫
2
X e j
e jn
d 
X e
j
= ∑
n=−∞
∞
x[n]e
− jn
Fourier Transform pair
Inverse Fourier Transform(synthesis equation)
Fourier Transform (analysis equation)

9
The discrete-time Fourier transform shares many
similarities with the continuous-time case. The
major differences between the two are the
periodicity (with period of 2π) of the discrete-time
transform X(ejω
) and the finite interval (length 2π)
of integration in the synthesis equation.
Signals at frequencies near even multiple of π are
slowly varying and therefore are all appropriately
thought of as low-frequency signals. Similarly, the
high frequencies in discrete-time are the values of
ω near odd multiples of π.

10
Low frequency signal
High frequency signal

11
Example 5.1
x[n]=a
n
u[n], ∣a∣1
X e j
= ∑
n=−∞
∞
an
u[n]e− jn
=∑
n=0
∞
ae
− j

n
=
1
1−ae
− j 
Find the Fourier Transform for the signal
solution

12
∣X e
j
∣=
1
1−acos
2
a sin
2
∢ X e j 
=−tan−1
 asin
1−a cos
magnitude
phase

13

14

15
Example 5.2
x[n]=a
∣n∣
, ∣a∣1
X e j
= ∑
n=−∞
∞
a∣n∣
e− j n
solution
=∑
n=0
∞
an
e− jn
 ∑
n=−∞
−1
a−n
e− jn

16
=
1
1−ae− j 

ae
j 
1−ae j
X e j
=∑
n=0
∞
ae− j 
n
∑
m=1
∞
ae j
m
replace m=-n
=
1−a2
1−2a cosa
2

17

18
Example 5.3
x[n]=
{1, ∣n∣≤N1
0, ∣n∣N1

19
X e
j
= ∑
n=−N1
N1
e
− j n
solution
=
sin N11/2
sin/2

20
Convergence of Fourier Transforms
∑
n=−∞
∞
∣x[n]∣2
∞
if the sequence x[n] has finite energy
we guaranteed that X(ejω
) is finite
∑
n=−∞
∞
∣x[n]∣∞
The condition on x[n] that guarantee the
convergence of Fourier transform is

21
There are no convergence issues associated with
the synthesis equation, since the integral in this
equation is over a finite interval of integration.
x[n]=
1
2
∫
−W
W
X e j
e jn
d 
x[n]
If we approximate an aperiodic signal x[n] by an
integral
then = x[n] for W = π

22
Example 5.4
Let x[n] be the unit impulse
x[n]=[n]
X e j
=1
Fourier transform of this signal is
x[n]=
1
2
∫
−W
W
e
jn
d =
sinWn
n
We can approximate x[n] by

23

24
2 The Fourier Transform for Periodic Signals
X e
j
= ∑
l=−∞
∞
2−0−2l
Let a signal x[n] is x[n]=e
j
0
n
The Fourier transform of x[n] should have
impulse at ω0
, ω0
±2π, ω0
±4π and so on

25
1
2
∫
2
∑
l=−∞
∞
2−0−2le
jn
d 
In order to check the validity of this expression,
we evaluate its inverse transform
1
2
∫
2
X e
j 
e
jn
d =
If the interval of integration chosen includes the
impulse located at ω0
±2π, then
1
2
∫
2
X e j
e j n
d =e
j02 rn
=e
j 0 n

26
X e j
= ∑
k=−∞
∞
2ak −
2k
N 
x[n]= ∑
k=〈N 〉
ak e jk2/ N n
Consider a periodic sequence x[n] with period N
and with the Fourier series representation
The Fourier transform is

27
Example 5.5
Find Fourier transform of the signal
Solution
By Euler's relation
x[n]=cos0 n
x[n]=
1
2
e
j 0 n

1
2
e
− j 0 n
with 0=
2
5

28
X e j
 = ∑
l=−∞
∞
−
2
5
−2l
− ∑
l=−∞
∞

2
5
−2l
The Fourier transform are
X e
j
 = −
2
5
−
2
5

−≤
That is
for

29
X(ejω
) repeats periodically with a period of 2π

30
Example 5.6
x[n]= ∑
k=−∞
∞
n−kN 
Find The Fourier transform of the impulse train

31
Solution
ak=
1
N
∑
n=〈N 〉
x[n]e
− j k 2/ N n
=
1
N
X e j
=
2
N
∑
k=−∞
∞
−
2k
N 
The Fourier series coefficients(for 0≤n≤N-1) are
given by
The Fourier transform given by

32
(a) periodic impulse train (b) its Fourier transform

33
3 Properties of the Discrete-Time Fourier Transform
Notation x[n] ↔
F
X e
j

X e j
=F {x[n]}
x[n]=F
−1
{X e
j
}

34
Periodicity of the Discrete-Time Fourier Transform
X e
j2
=X e
j

The discrete-time Fourier transform is always
periodic in ω with period 2π

35
Linearity
x1[n] ↔
F
X 1e
j

x2[n] ↔
F
X 2 e
j 

ax1[n]bx2[n] ↔
F
aX 1e
j 
bX 2e
j

If
then

36
Time Shifting and Frequency Shifting
x[n] ↔
F
X e
j

x[n−n0] ↔
F
e
− jn0
X e
j

If
then
e
− j 0 n
x[n] ↔
F
X e
j−0

and

37
Example 5.7
We have depicted the frequency response Hlp
(ejω
)
of a lowpass filter with cutoff frequency ωc

38
Since high frequencies in discrete-time are
concentrated near π, the frequency response
Hhp
(ejω
) :
H hpe
j
=H lpe
j−

that is

39
by the frequency-shifting property
hhp=e j n
hlp[n]
=−1n
hlp[n]

40
Conjugation and Conjugate Symmetry
x[n] ↔
F
X e
j

x
∗
[n] ↔
F
X
∗
e
− j

If
then

41
Conjugate symmetry
If x[n] is real then X(ejω
) has conjugate symmetry
X e
− j 
=X
∗
e
j 
 [x[n] real]
If we express X(ejω
) in rectangular form as
X e
j
=ℜe[ X e
j 
] j ℑm[ X e
j 
]

42
then if x[n] is real
ℜe[ X e
j 
]=ℜe[ X e
− j 
]
ℑm[ X e
j 
]=−ℑm[ X e
− j
]
and
The real part of Fourier transform is an even
function of frequency and the imaginary part is an
odd function of frequency

43
If we express in polar form as
X e
j
=∣X e
j
∣e
j∢ X e j

|X(ejω
)| is an even function of ω and ∢X(ejω
) 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.

44
If x[n] is real then it can always be expressed in
terms of the sum of an even function and an odd
function.
x[n]=xe [n]xo[n]
From the linearity property
F {x[n]}=F {xe[n]}F {xo[n]}
F {xe [n]} is a real function
F {xo [n]} is purely imaginary

45
With x[n] real, we can conclude that
x[n] ↔
F
X e
j

Ev{x[n]} ↔
F
ℜe{X e
j
}
Od {x[n]} ↔
F
j ℑm{X e
j
}

46
Differencing and Accumulation
Let x[n] be a signal with Fourier transform X(ejω
),
the Fourier transform pair for the first-difference
signal x[n]-x[n-1] is given by
x[n]−x[n−1] ↔
F
1−e
− j
 X e
j 


47
∑
m=−∞
n
x[m] ↔
F 1
1−e
− j
X e j 
 X e j0
 ∑
k=−∞
∞
−2k 
Consider the signal
Since y[n] - y[n-1] = x[n] , we might conclude that
the transform of y[n] should be related to the
transform of x[n] by
y[n]= ∑
m=−∞
n
x[m]
The impulse train on the right-hand side reflects
the dc or average value that can result from
summation.

48
Example 5.8
g[n]=[n] ↔
F
Ge j 
=1
x[n]= ∑
m=−∞
n
g[m]
Determine the Fourier transform of x[n] = u[n] by
using the accumulation property and the knowledge
that
solution
We know that

49
X e
j
=
1
1−e
− j 

Ge
j 
Ge
j0
 ∑
k=−∞
∞
−2k 
Taking the Fourier Transform both side
since G(ejω
) =1, we conclude that
X e
j
=
1
1−e
− j 

 ∑
k=−∞
∞
−2k 

50
Time Reversal
x[n] ↔
F
X e
j

Let
that is
x[−n] ↔
F
X e
− j 


51
Time Expansion
xat ↔
F 1
∣a∣
X j 
a 
In continuous-time
Let k be a positive integer, and define the
signal
xk[n]=
{x[n/k], if n isa multipleof k
0, if n is not a multipleof k

52
For example, k = 3, x(k)
[n] is obtained from x[n] by
placing k-1 zeros between successive values of the
original signal.

53
We can think of x(k)
[n] as a slowed-down version
of x[n]. Since x(k)
[n] equals 0 unless n is a
multiple of k, i.e., unless n = rk, we can see that
the Fourier transform of x(k)
[n] is given by
X ke
j 
= ∑
n=−∞
∞
xk[n]e
− jn
= ∑
n=−∞
∞
xk[rk]e
− j rk

54
X ke j
= ∑
r=−∞
∞
xr[n]e− jk r
=X e jk 

Since x(k)
[rk] = x[r]
That is,
xk[n] ↔
F
X e
jk 

Note that as the signal is spread out and
slowed down in time by taking k>1, its Fourier
transform is compressed.

55

56
Example 5.9
Determine the Fourier transform of the sequence
x[n]
x[n]= y2[n]2y2[n−1]
where

57
solution
y2[n]=
{y[n/2], if n iseven
0 , if n isodd

58
Y e
j
=e
− j2 sin5/2
sin/2
By using time shift property to example 5.3
and using time-expansion property,
y2[n]↔
F
e
− j4 sin5
sin

59
2y2[n−1]↔
F
2e
− j5 sin5
sin
and using the linearity and time-shifting
properties,
Combining these two result, we have
X e
j
=e
− j4
12e
− j 

sin5
sin

60
Differentiation in frequency
x[n] ↔
F
X e
j

Let
dX e
j 

d 
= ∑
n=−∞
∞
− j n x[n]e
− jn
Using definition
X e
j
= ∑
n=−∞
∞
x[n]e
− jn

61
The right hand side of this equation is the
Fourier transform of -jnx[n]. Therefore,
multiplying both sides by j, we see that
nx[n] ↔
F
j
d X e
− j 

d 

62
Parseval 's Relation
∑
n=−∞
∞
∣x[n]∣
2
=
1
2
∫
2
∣X e
j 
∣
2
d 
If x[n] and X(ejω
) are Fourier transform pair
Parseval's relation states that this energy can
also be determined by integrating the energy per
unit frequency, |X(ejω
)|2
/2π, over a full 2π interval
of distinct discrete-time frequencies. |X(ejω
)|2
is
referred to as the energy-density spectrum of the
signal x[n].

63
Example 4.14
Determine whether or not , in the time domain, x[n]
whose Fourier transform is depicted below, is
periodic, real, even, and/or of finite energy.

64
solution
We note first that periodicity in the time domain
implies that the Fourier transform is zero,
expect possibly for impulses located at various
integer multiples of the fundamental frequency.
Thus is not true for X(ejω
). We conclude that
x[n] is not periodic.
From the symmetry properties for Fourier
transforms, we know that a real-valued
sequence must have a Fourier transform of
even magnitude and a phase function that is
odd. This is true for the given X(ejω
) and
∢X(ejω
). We conclude that x[n] is real.

65
∑
n=−∞
∞
∣x[n]∣
2
dt =
1
2
∫
2
∣X e
j 
∣
2
d 
If x[n] is an even function, then by the
symmetry properties for real signals, X(ejω
)
must real and even but this X(ejω
) is not a real-
valued function. Consequently, x[n] is not
even.
Finally, for the finite-energy property, we may
use Parseval's relation
It clear that integrating |X(ejω
)|2
from -π to π will
yield a finite quantity. We conclude that x[n]
has finite energy.

66
4 The Convolution Property
h[n] ↔
F
H e
j

y[n] ↔
F
Y e
j 

y[n]=h[n]∗x[n] ↔
F
Y e
j 
=H e
j 
 X e
j

If
then
x[n] ↔
F
X e
j


67
Example 5.11
Consider the LTI system with impulse response
h[n]=n−n0
The frequency response is
H e
j
= ∑
n=−∞
∞
[n−n0]e
− j n
=e
− jn0
For any input x[n] with Fourier transform
H(ejω
), the Fourier transform of the output is
Y ej
=e
−j n0
X e j
 y[n]=xn−n0

68
Example 5.12
Consider the discrete-time ideal lowpass filter
that has the frequency response H(ejω
)
Determine the impulse response of the ideal
lowpass filter

69
h[n]=
1
2
∫
−

H e
j 
e
jn
d 
=
1
2
∫
−c

e
jn
d 
=
sinc n
n
solution

70
Example 5.13
Determine the frequency response of an LTI
system with impulse response
h[n]=
n
u[n], ∣∣1
to the input signal
x[n]=
n
u[n], ∣∣1

71
solution
The Fourier transform of x[n] and h[n] are
X e
j
=
1
1−e
− j H e
j
=
1
1−e− j
Therefore
Y e j
=
1
1−e
− j 
1−e
− j


72
Y e j
=
A
1−e− j

B
1−e− j
A=

−
, B=
−
−
y[n]=

−

n
u[n]−

−

n
u[n]
Assuming that α ≠ β, the partial fraction
expansion for Y(ejω
) takes the form
We find that
Therefore

73
y[n]=
1
−
[n1
u[n]−n1
u[n]]
Y e
j
=
1
1−e
− j 

2
1
1− e− j 
2
=
j

e j  d
d  [ 1
1− e− j ]
or
If α = β
Recognizing this as

74

n
u[n] ↔
F 1
1− e
− j 
n
n
u[n] ↔
F
j
d
d   1
1−e
− j 
We can use the dual of the differentiation
property
For the factor ejω
, we uses time-shifting property
n1
n1
u[n1] ↔
F
j e
j  d
d  1
1−e
− j 

75
Finally, accounting for the factor 1/α, we obtain
y[n]=n1
n
u[n1]
Since (n+1) is zero at n=-1
y[n]=n1
n
u[n]

76
Example 5.14
Consider the system below
The LTI system with frequency response Hlp
(ejω
)
are ideal lowpass filter with cutoff frequency π/4
and unity gain in passband. Find the frequency
response of the overall system.

77
solution
The Fourier transform of the signal w1
[n] can be
obtain by noting that
−1n
=e
jn
so that
w1[n]=e
j n
x[n]
Using the frequency-shifting property
W1 e
j
=X e
j−


78
by convolution property
W 2e
j
=H lpe
j 
 X e
j−

since
w3 [n]=e jn
w2[n]
applying the frequency-shifting property
W 3e j 
=W 2e j−

=H lpe
j−
 X e
j−2


79
W 3e
j 
=Hlp e
j−
 X e
j

since discrete-time Fourier transform are always
periodic with period 2π.
Applying the convolution property to the lower
path, we get
W 4e
j
=H lpe
j
 X e
j 

From the linearity property

80
Y e
j
=W 3e
j
W 4e
j

=[Hlp e
j−
Hlp e
j
] X e
j 

The frequency response of the overall system is
H e j
=[Hlp e j−
H lpe j
]

81
4 The Multiplication Property
y[n]=x1[n] x2[n]↔
F
R j =
1
2
∫
2
X e j
 X e j−
d 
The multiplication in time domain corresponds to
convolution in frequency domain
corresponds to a periodic convolution, and the
integral can be evaluated over any interval of
length 2π

82
Example 5.15
Find the Fourier transform of the signal x[n]
x[n]=x1[n] x2[n]
where
x1[n]=
sin3n/4
n
x2[n]=
sinn/2
n

83
solution
X e j
=
1
2
∫
−

X 1e j
 X 2e j−
d 
X 1e
j
=
{X 1e j
, for−≤
0, otherwise
From multiplication property
We can convert the equation into an ordinary
convolution by defining

84
then replacing X1
(ejθ
)
X e j
=
1
2
∫
−

X 1e j
 X 2e j−
d 
=
1
2
∫−∞
∞
X 1e
j
 X 2e
j−
d 

85

86

87

88
6 Duality
Duality in discrete-time Fourier Series
x[n]= ∑
k=〈 N 〉
ak e
jk 0 n
ak=
1
N
∑
n=〈N 〉
x[n]e
− jk 0 n
x[n] ↔
FS
ak
If we adopt the notation

89
g [n]= ∑
k=〈N 〉
f [k]e
jk 0 n
f [k ]=
1
N
∑
n=〈N 〉
g [n]e
− jk 0 n
x1[n]=g[n] ↔
FS
ak= f [k ]
Suppose f[*] and g[*] are two functions such
that they are a FT pair

90
f [n]=
1
N
∑
k=〈N 〉
g[−k]e
jk 0 n
= ∑
k=〈 N 〉
1
N
g[−k ]e
jk 0 n
then for another time-domain function such that
f [n]= ∑
k=〈N 〉
bk e
jk 0 n
x2[n]= f [n] ↔
FS
bk=
1
N
g[−k ]

91
Example 5.16
Consider the following periodic signal with a
period of N = 9
x[n]=
{
1
9
sin5n/9
sinn/9
,n≠multiple of 9
5
9
, n=multiple of 9

92
From chapter 3, the Fourier series coefficients for
x[n] must be in the form of a rectangular square
wave. Let g[n] be
g [n]=
{1, ∣n∣≤2
0, 2∣n∣≤4
The Fourier series coefficients bk
for g[n] can be
determined from ex. 3.12 as

93
bk=
{
1
9
sin5k /9
sin k /9
,k≠multiple of 9
5
9
, k=multiple of 9
The Fourier series analysis equation for g[n] is
bk=
1
9
∑
n=−2
2
1e
− j 2nk /9

94
x[n]=
1
9
∑
k=−2
2
1e
− j 2n k/9
Let k' = -k
x[n]=
1
9
∑
k '=−2
2
1e
j 2n k ' /9
Interchanging the names of the variables k and n
and noting that x[n] = bn
, we find that

95
Finally, moving the factor 1/9 inside the
summation, we see that the right side of this
equation has the form of the synthesis equation
for x[n]. We conclude that the FS coefficients are
ak=
{1, ∣k∣≤2
0, 2∣k∣≤4

96
x[n]=
1
2
∫
2
X e
j 
e
jn
d 
X e
j
= ∑
n=−∞
∞
x[n]e
jn
xt= ∑
k=−∞
∞
ak e
j k 0 t
ak=
1
T
∫
T
xte
− j k 0t
d t
Duality between the discrete-time Fourier transform
and the continuous-time Fourier series
DTFT
CTFS

97
Example 5.17
Determine the discrete-time Fourier transform
of the sequence
x[n]=
sin n/2
n

98
g t=
{1, ∣t∣≤T1
0, T1∣t∣≤
To use duality, we first must identify a
continuous-time signal g(t) with period T= 2π and
Fourier coefficients ak
= x[k]. From Ex. 3.5, we
know that if g(t) is periodic and with
then
ak=
sink T1
k 
Solution

99
Consequently, if we take T1
= π/2, we will have
ak
=x[k]
sink /2
k 
=
1
2
∫
−

g te
− j k t
dt
=
1
2
∫
−/2
/2
1e
− j k t
dt
Rename k as n and t as ω, we have
sin n/2
n
=
1
2
∫
−/2
/2
1e
− j n
d 

100
Replacing n by -n
sin n/2
n
=
1
2
∫
−/2
/2
1e j n
d 
On the right hand side has the form of the
x[n]=
1
2
∫
2
X e j 
e jn
d 
where
X e
j
=
{1, ∣∣≤/2
0, /2∣∣≤

101

102
7 Systems Characterized by Linear Constant-
Coefficient Difference Equations
A linear constant-coefficient differential equation
with input x[n] and output y[n] is of the form
∑
k=0
N
ak y[n−k ]=∑
k=0
M
bk x[n−k ]

103
Y e
j
=H e
j
 X e
j 

H e
j
=
Y e
j

X e
j

By the convolution property
or
where X(ejω
), Y(ejω
) and H(ejω
) are the Fourier
transforms of the input x[n], output y[n] and
impulse response h[n].

104
F
{∑
k=0
N
ak y[n−k ]
}=F
{∑
k=0
M
bk x[n−k ]
}
∑
k=0
N
ak F {y[n−k ]}=∑
k=0
M
bk F {x[n−k ]}
consider applying the Fourier transform to the
equation in slide before
from linear property

105
Y e j

[∑
k=0
N
ak e− j k 
]=X e j

[∑
k=0
M
bk e− j k 
]
∑
k=0
N
ak e
− j 

k
Y e
j 
=∑
k=0
M
bk e
− j

k
X e
j

and from the differentiation property
or

106
H e
j
=
Y e j

X e
j

=
∑
k=0
M
bk e
− jk 
∑
k=0
N
ak e− jk 
Thus
Observe that H(ejω
) is thus a rational function;
that is, it is a ratio of polynomials in variable e-jω
and the frequency response for the LTI system
can be written directly by inspection

107
Example 5.18
Find the impulse response of the LTI system
y[n]−a y[n−1]=x[n]
with |a| > 1
Solution
Fourier transform of the system is
Y e j
−ae− j 
Y e j 
=X e j 


108
1−ae
− j 
Y e
j 
=X e
j 

H e
j
=
Y e
j

X e
j

=
1
1−ae
− j 
From Example 5.1, the inverse Fourier
Transform of equation above is
h[n]=a
n
u[n]

109
Example 5.19
Find the impulse response of the LTI system
y[n]−
3
4
y[n−1]
1
8
y[n−2]=2x[n]
Solution
The frequency response is
H e
j
=
2
1−
3
4
e
− j 

1
8
e
− j 2

110
We factor the denominator of the right-hand side
By using the partial-fraction expansion
The inverse Fourier transform of each term
h[n]=41
2
n
u[n]−21
4
n
u[n]
H e
j
=
2
1−
1
2
e
− j 
1−
1
4
e
− j 

H e j
=
4
1−
1
2
e− j
−
2
1−
1
4
e− j

111
Example 5.20
Consider the system of Example 5.19, find the
output of the system when the input is
x[n]=1
4
n
u[n]
Solution
Y e
j
=H e
j
 X e
j 

=
[
2
1−
1
2
e
− j 
1−
1
4
e
− j
][
1
1−
1
4
e
− j 
]

112
By using the partial-fraction expansion
Y e
j
=
B11
1−
1
4
e
− j

B12
1−
1
4
e
− j 

2

B21
1−
1
2
e
− j
Y e
j
=
[
2
1−
1
2
e
− j 
1−
1
4
e
− j

2
]
=−
4
1−
1
4
e
− j
−
2
1−
1
4
e
− j 

2

8
1−
1
2
e
− j 

113
The inverse Fourier transform of each term
y[n]=
{−41
4
n
−2n11
4
n
81
2
n
}u[n]

114

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Chapter5 - The Discrete-Time Fourier Transform