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Chapter 5 The Discrete-Time Fourier Transform
5.0 Introduction • There are many similarities and strong parallels in analyzing continuous-time and discrete-
time signals. • There are also important differences. For example, the Fourier series representation of a
discrete-time periodic signal is finite series, as opposed to the infinite series representation required for continuous-time period signal.
• In this chapter, the analysis will be carried out by taking advantage of the similarities between continuous-time and discrete-time Fourier analysis.
5.1 Representation of Aperiodic Signals: The discrete-Time Fourier Transform
5.1.1 Development of the Discrete-Time Fourier Transform Consider a general sequence that is a finite duration. That is, for some integers 1N and 2N , ][nx equals to zero outside the range 21 NnN ≤≤ , as shown in the figure below.
We can construct a periodic sequence ][~ nx using the aperiodic sequence ][nx as one period. As we choose the period N to be larger, ][~ nx is identical to ][nx over a longer interval, as ∞→N ,
][][~ nxnx = . Based on the Fourier series representation of a periodic signal given in Eqs. (3.80) and (3.81), we have
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∑>=<
=Nk
nNjkk eanx )/2(][~ π , (5.1)
∑=
−=Nk
nNjkk enxa )/2(][~ π . (5.2)
If the interval of summation is selected to include the interval 21 NnN ≤≤ , so ][~ nx can be replaced by ][nx in the summation,
∑∑∞
−∞=
−
=
− ==k
nNjkN
Nk
nNjkk enx
Nenx
Na )/2()/2( ][
1][
1 2
1
ππ , (5.3)
Defining the function
∑∞
−∞=
−=n
njj enxeX ωω ][)( , (5.4)
So ka can be written as
)(1
0ωjkk eX
Na = , (5.5)
Then ][~ nx can be expressed as
0)/2()/2( )(
21
)(1
][~ 00 ωπ
πωπω ∑∑>=<>=<
==Nk
nNjkjk
Nk
nNjkjk eeXeeXN
nx . (5.6)
As ∞→N ][][~ nxnx = , and the above expression passes to an integral,
ωπ π
ωω deeXnx njj∫=2
)(21
][ , (5.7)
The Discrete-time Fourier transform pair:
ωπ π
ωω deeXnx njj∫=2
)(21
][ , (5.8)
∑∞
−∞=
−=n
njj enxeX ωω ][)( . (5.9)
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Eq. (5.8) is referred to as synthesis equation, and Eq. (5.9) is referred to as analysis equation and )( 0ωjkeX is referred to as the spectrum of ][nx .
5.1.2 Examples of Discrete-Time Fourier Transforms Example : Consider ][][ nuanx n= , 1<a . (5.10)
( ) ωωωωω
jn
nj
n
njn
n
njj
aeaeenuaenxeX −
∞
=
−−∞
−∞=
−∞
−∞=
−
−==== ∑∑∑ 1
1][][)(
0
. (5.11)
The magnitude and phase for this example are show in the figure below, where 0>a and 0<a are shown in (a) and (b).
Example : nanx =][ , 1<a . (5.12)
∑∑∑∞
=
−−
−∞=
−−∞
−∞=
− +==0
1
][)(n
njn
n
njn
n
njnj eaeaenuaeX ωωωω
Let nm −= in the first summation, we obtain
2
2
01
cos211
11
1
][)(
aaa
aeaeae
eaeaenuaeX
jj
j
n
njn
m
mjm
n
njnj
+−−=
−+
−=
+==
−
∞
=
−∞
=
∞
−∞=
− ∑∑∑
ωωω
ω
ωωωω
. (5.13)
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Example : Consider the rectangular pulse
>
≤=
,0
,1][
n
nnx , (5.14)
( )( )2/sin
2/1sin)( 1
ωω
ω ω +== ∑
−=
− NejX
n
nj . (5.15)
This function is the discrete counterpart of the sic function, which appears in the Fourier transform of the continuous-time pulse.
The difference between these two functions is that the discrete one is periodic (see figure) with period of π2 , whereas the sinc function is aperiodic.
5.1.3 Convergence
The equation ∑∞
−∞=
−=n
njj enxeX ωω ][)( converges either if ][nx is absolutely summable, that is
∞<∑∞
−∞=n
nx ][ , (5.16)
or if the sequence has finite energy, that is
∞<∑∞
−∞=
2
][n
nx . (5.17)
1N
1N
1N
1N
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And there is no convergence issues associated with the synthesis equation (5.8). If we approximate an aperidic signal ][nx by an integral of complex exponentials with frequencies taken over the interval W≤ω ,
∫−=
W
W
njj deeXnx ωπ
ωω )(21
][ˆ , (5.18)
and ][][ˆ nxnx = for π=W . Therefore, the Gibbs phenomenon does not exist in the discrete-time Fourier transform. Example : the approximation of the impulse response with different values of W . For ππππππ ,8/7,4/3,2/,8/3,4/=W , the approximations are plotted in the figure below. We can see that when π=W , ][][ nxnx =) .
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5.2 Fourier transform of Periodic Signals For a periodic discrete-time signal,
njenx 0][ ω= , (5.19) its Fourier transform of this signal is periodic in ω with period π2 , and is given
∑+∞
−∞=
−−=l
j leX )2(2)(0
πωωπδω . (5.20)
Now consider a periodic sequence ][nx with period N and with the Fourier series representation
nNjk
Nkkeanx )/2(][ π∑
>=<
= . (5.21)
The Fourier transform is
∑+∞
−∞=
−=k
kj
Nk
aeX )2
(2)(π
ωδπω . (5.22)
Example : The Fourier transform of the periodic signal
njnj eennx 00
21
21
cos][ 0ω−ω +=ω= , with
32
0
πω = , (5.23)
is given as
++
−=
32
32
)(π
ωπδπ
ωπδωjeX , πωπ <≤− . (5.24)
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Example : The periodic impulse train
∑+∞
−∞=
−=k
kNnnx ][][ δ . (5.25)
The Fourier series coefficients for this signal can be calculated
∑>=<
−=Nn
nNjkk enxa )/2(][ π . (5.26)
Choosing the interval of summation as 10 −≤≤ Nn , we have
Nak
1= . (5.27)
The Fourier transform is
∑∞
−∞=
−=
k
j
Nk
NeX
πωδ
πω 22)( . (5.28)
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5.3 Properties of the Discrete-Time Fourier Transform Notations to be used
{ }][)( nxFeX j =ω ,
{ })(][ 1 ωjeXFnx −= ,
)(][ ωjF eXnx →← .
5.3.1 Periodicity of the Discrete-Time Fourier Transform The discrete-time Fourier transform is always periodic in ω with period π2 , i.e.,
( ) ( )ωπω jj eXeX =+ )2( . (5.29)
5.3.2 Linearity If )(][ 11
ωjF eXnx →← , and )(][ 22ωjF eXnx →← ,
then
)()(][][ 2121ωω jjF ebXeaXnbxnax +→←+ (5.30)
5.3.3 Time Shifting and Frequency Shifting If )(][ ωjF eXnx →← , then
)(][ 00
ωω−→←− jnjF eXennx (5.31) and
( ))(][ 00 ω−ωω →← jFnj eXnxe (5.32)
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5.3.4 Conjugation and Conjugate Symmetry If )(][ ωjF eXnx →← , then
)(*][* ωjF eXnx −→← (5.33) If ][nx is real valued, its transform )( ωjeX is conjugate symmetric. That is
)(*)( ωω jj eXeX −= (5.34) From this, it follows that { })(Re ωjeX is an even function of ω and { })(Im ωjeX is an odd function of ω . Similarly, the magnitude of )( ωjeX is an even function and the phase angle is an odd function. Furthermore,
{ } { }ωjF eXnxEv (Re][ →← , (5.35) and
{ } { }ωjF eXjnxOd (Im][ →← . (5.36)
5.3.5 Differencing and Accumulation If )(][ ωjF eXnx →← , then
( ) )(1]1[][ ωω jjF eXenxnx −−→←−− . (5.37) For signal
∑−∞=
=n
m
mxny ][][ , (5.38)
its Fourier transform is given as
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∑∑+∞
−∞=−
−∞=
−+−
→←m
jjj
Fn
m
keXeXe
mx )2()()(1
1][ 0 πωδπω
ω . (5.39)
The impulse train on the right-hand side reflects the dc or average value that can result from summation. For example , the Fourier transform of the unit step ][][ nunx = can be obtained by using the accumulation property. We know 1)(][][ =→←= ωδ jF eGnng , so
( ) ( ) ∑∑∑+∞
−∞=−
+∞
−∞=−
−∞=
−+−
=−+−
→←=k
jk
jjj
Fn
m
ke
keGeGe
mgnx )2(1
1)2()()(1
1][][ 0 πωδππωδπ ωω
ω .
(5.40)
5.3.6 Time Reversal If )(][ ωjF eXnx →← , then
)(][ ωjF eXnx −→←− . (5.41)
5.3.7 Time Expansion For continuous-time signal, we have
→←
ajX
aatx F ω1)( . (5.42)
For discrete-time signals, however, a should be an integer. Let us define a signal with k a positive integer,
=kofmultipleanotisnif
kofmultipleaisnifknxnx k ,0
],/[][)( . (5.43)
][)( nx k is obtained from ][nx by placing 1−k zeros between successive values of the original
signal. The Fourier transform of ][)( nx k is given by
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)(][][][)( )()()()(
ωω−+∞
−∞=
ω−+∞
−∞=
ω−+∞
−∞=
ω ==== ∑∑∑ jkrkj
r
rkj
rk
nj
nk
jk eXerxerkxenxeX . (5.44)
That is,
)(][)(ωjkF
k eXnx →← . (5.45) For 1>k , the signal is spread out and slowed down in time, while its Fourier transform is compressed. Example : Consider the sequence ][nx displayed in the figure (a) below. This sequence can be related to the simpler sequence ][ny as shown in (b).
]1[2][][ )2()2( −+= nynynx , where
=oddisnifevenisnifny
ny,0
],2/[][2
The signals ][)2( ny and ]1[2 )2( −ny are depicted in (c) and (d). As can be seen from the figure below, ][ny is a rectangular pulse with 21 =N , its Fourier transform is given by
)2/sin()2/5sin(
)( 2
ωωωω jj eeY −= .
Using the time-expansion property, we then obtain
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)sin()5sin(
][ 4)2( ω
ωωjF eny −→←
)sin()5sin(
2]1[2 5)2( ω
ωωjF eny −→←−
Combining the two, we have
+= −−
)sin()5sin(
)21()( 4
ωωωωω jjj eeeX .
5.3.8 Differentiation in Frequency If )(][ ωjF eXnx →← ,
Differentiate both sides of the analysis equation ∑∞
−∞=
ω−ω =n
njj enxeX ][)(
∑+∞
−∞=
−−=n
njj
enjnxd
edX ωω
ω][
)(. (5.46)
The right-hand side of the Eq. (5.46) is the Fourier transform of ][njnx− . Therefore, multiplying both sides by j , we see that
ω
ω
dedX
jnnxj
F )(][ →← . (5.47)
5.3.9 Parseval’s Relation If )(][ ωjF eXnx →← , then we have
∫∑ =+∞
−∞=π
ω ωπ 2
2)2(
21
][ deXnx j
n (5.48)
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Example : Consider the sequence ][nx whose Fourier transform )( ωjeX is depicted for πωπ ≤≤− in the figure below. Determine whether or not, in the time domain, ][nx is periodic,
real, even, and /or of finite energy.
• The periodicity in time domain implies that the Fourier transform has only impulses located
at various integer multiples of the fundamental frequency. This is not true for )( ωjeX . We conclude that ][nx is not periodic.
• Since real-valued sequence should have a Fourier transform of even magnitude and a phase function that is odd. This is true for )( ωjeX and )( ωjeX∠ . We conclude that ][nx is real.
• If ][nx is real and even, then its Fourier transform should be real and even. However, since ωωω 2)()( jjj eeXeX −= , )( ωjeX is not real, so we conclude that ][nx is not even.
• Based on the Parseval’s relation, integrating 2
)( ωjeX from π− to π will yield a finite
quantity. We conclude that ][nx has finite energy.
5.4 The convolution Property If ][nx , ][nh and ][ny are the input, impulse response, and output, respectively, of an LTI system, so that
][][][ nhnxny ∗= , (5.49) then,
)()()( ωωω jjj eHeXeY = , (5.50)
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where )( ωjeX , )( ωjeH and )( ωjeY are the Fourier transforms of ][nx , ][nh and ][ny , respectively. Example : Consider the discrete-time ideal lowpass filter with a frequency response )( ωjeH illustrated in the figure below. Using πωπ ≤≤− as the interval of integration in the synthesis equation, we have
The frequency response of the discrete-time ideal lowpass filter is shown in the right figure. Example : Consider an LTI system with impulse response
][][ nunh nα= , 1<α , and suppose that the input to the system is
][][ nunx nβ= , 1<β . The Fourier transforms for ][nh and ][nx are
ωω
α jj
eeH −−
=1
1)( ,
and
ωω
β jj
eeX −−
=1
1)( ,
so that
)1)(1(1
)()()( ωωωωω
βα jjjjj
eeeXeHeY −− −−
== .
nnde
deeHnh
cnj
njj
πωω
π
ωπ
ωπ
π
ωπ
π
ω
sin21
)(21
][
==
=
∫
∫
−
−
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If βα ≠ , the partial fraction expansion of )( ωjeY is given by
)1()1()1()1()( ωωωω
ω
ββα
β
αβα
α
βα jjjjj
eeeB
eA
eY −−−− −−
−+
−−=
−+
−= ,
We can obtain the inverse transform by inspection:
( )][][1
][][][ 11 nununununy nnnn ++ −−
=−
−−
= ββαβα
ββα
βα
βαα
.
For βα = ,
2)1(1
)( ωω
α jj
eeY −−
= , which can be expressed as
−
= − ωωω
αωα jjj
edd
ej
eY1
1)( .
Using the frequency differentiation property, we have
−
→← − ωαωα j
Fn
edd
jnun1
1][ ,
To account for the factor ωje , we use the time-shifting property to obtain
−
→←++ −+
ωω
αωα j
jFn
edd
jenun1
1]1[)1( 1 ,
Finally, accounting for the factor α/1 , we have
]1[)1(][ ++= nunny nα . Since the factor 1+n is zero at 1−=n , so ][ny can be expressed as
][)1(][ nunny nα+= . Example : Consider the system shown in the figure below. The LTI systems with frequency response )( ωj
lp eH are ideal lowpass filters with cutoff frequency 4/π and unity gain in the passband.
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• ][][)1(][1 nxenxnw njn π=−= ⇒ )()( )(
1πωω −= jj eXeW .
• )()()( )(
2πωωω −= jj
lpj eXeHeW .
• ][][)1(][ 223 nwenwnw njn π=−= ⇒ )()()()( )2()()(
23πωπωπωω −−− == jj
lpjj eXeHeWeW .
⇒ )()()()( ))(
23ωπωπωω jj
lpjj eXeHeWeW −− == (Discrete-Fourier transforms are always
periodic with period of π2 ). • )()()( )
4ωωω jj
lpj eXeHeW = .
• [ ] )()()()()()( )(
43ωωπωωωω jj
lpj
lpjjj eXeHeHeWeWeY +=+= − .
The overall system has a frequency response
[ ] )()()()( )( ωωπωω jjlp
jlp
jlp eXeHeHeH += − ,
which is shown in figure (b). The filter is referred to as bandstop filter, where the stop band is the region 4/34/ πωπ << . It is important to note that not every discrete-time LTI system has a frequency response. If an LTI system is stable, then its impulse response is absolutely summable; that is,
∞<∑+∞
−∞=n
nh ][ , (5.51)
5.5 The multiplication Property Consider ][ny equal to the product of ][1 nx and ][2 nx , with )( ωjeY , )(1
ωjeX , and )(2ωjeX
denoting the corresponding Fourier transforms. Then
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∫ −→←=π
θωω θπ 2
)(2121 )()(
21
][][][ deXeXnxnxny jjF (5.52)
Eq. (5.52) corresponds to a periodic convolution of )(1
ωjeX and )(2ωjeX , and the integral in
this equation can be evaluated over any interval of length π2 . Example : Consider the Fourier transform of a signal ][nx which the product of two signals; that is
][][][ 21 nxnxnx = where
nn
nxππ )4/3sin(
][1 = , and
nn
nxππ )2/sin(
][2 = .
Based on Eq. (5.52), we may write the Fourier transform of ][nx
∫−
−=π
π
θωωω θπ
deXeXeX jjj )()(21
)( )(21 . (5.53)
Eq. (5.53) resembles aperiodic convolution, except for the fact that the integration is limited to the interval of πθπ <<− . The equation can be converted to ordinary convolution with integration interval ∞<<∞− θ by defining
<<−
=otherwise
foreXeX
jj
0
)()(ˆ 1
1
πωπωω
Then replacing )(1
ωjeX in Eq. (5.53) by )(ˆ1
ωjeX , and using the fact that )(ˆ1
ωjeX is zero for
πωπ <<− , we see that
∫∫∞
∞−
−
−
− == θπ
θπ
θωωπ
π
θωωω deXeXdeXeXeX jjjjj )()(ˆ21
)()(21
)( )(21
)(21 .
Thus, )( ωjeX is π2/1 times the aperiodic convolution of the rectangular pulse )(ˆ
1ωjeX and the
periodic square wave )(2ωjeX . The result of thus convolution is the Fourier transform )( ωjeX ,
as shown in the figure below.
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5.6 Tables of Fourier Transform Properties and Basic Fourier Transform Paris
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5.7 Duality For continuous-time Fourier transform, we observed a symmetry or duality between the analysis and synthesis equations. For discrete-time Fourier transform, such duality does not exist. However, there is a duality in the discrete-time series equations. In addition, there is a duality relationship between the discrete-time Fourier transform and the continuous-time Fourier series.
5.7.1 Duality in the discrete-time Fourier Series Consider the periodic sequences with period N, related through the summation
∑>=<
−=Nr
mNjrergN
mf )/2()(1
][ π . (5.54)
If we let nm = and kr −= , Eq. (5.54) becomes
∑>=<
−=Nk
nNjrergN
nf )/2()(1
][ π . (5.55)
Compare with the two equations below,
∑=
=Nk
nNjkkeanx )/2(][ π
, (3.80)
∑=
−=Nk
nNjkk enx
Na )/2(][
1 π. (3.81)
we fond that )(1
rgN
− corresponds to the sequence of Fourier series coefficients of ][nf . That is
][1
][ kgN
nf FS −→← . (5.56)
This duality implies that every property of the discrete-time Fourier series has a dual. For example,
0)/2(0 ][ nNjk
kFS eannx π−→←− (5.57)
mkFSnNjm ae −→←)/2( π (5.58)
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are dual. Example : Consider the following periodic signal with a period of 9=N .
=
≠=
9,95
9,n/9)sin(
)9/5sin(91
][ofmultiplen
ofmultiplenn
nxππ
(5.59)
We know that a rectangular square wave has Fourier coefficients in a form much as in Eq. (5.59). Duality suggests that the coefficients of ][nx must be in the form of a rectangular square wave. Let ][ng be a rectangular square wave with period 9=N ,
≤<
≤=
42,0
2,1][
n
nng , (5.60)
The Fourier series coefficients kb for ][ng can be given (refer to example on page 27/3)
=
≠=
9,95
9,k/9)sin(
)9/5sin(91
ofmultiplek
ofmultiplekk
bk
ππ
. (5.61)
The Fourier analysis equation for ][ng can be written
∑−=
−=2
2
9/2)1(91
n
nkjk eb π . (5.62)
Interchanging the names of the variable k and n and noting that kbnx =][ , we find that
∑−=
−=2
2
9/2)1(91
][k
nkjenx π .
Let kk −=' in the sum on the right side, we obtain
∑−=
+=2
2
9/'2)1(91
][k
nkjenx π .
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Finally, moving the factor 9/1 inside the summation, we see that the right side of the equation has the form of the synthesis equation for ][nx . Thus, we conclude that the Fourier coefficients for ][nx are given by
≤<
≤=
42,0
2,9/1
k
kak ,
with period of 9=N .
5.8 System Characterization by Linear Constant-Coefficient Difference Equations A general linear constant-coefficient difference equation for an LTI system with input ][nx and output ][nx is of the form
∑∑==
−=−M
kk
N
kk knxbknya
00
][][ , (5.63)
which is usually referred to as Nth-order difference equation. There are two ways to determine )( ωjeH : • The first way is to apply an input njenx ω=][ to the system, and the output must be of the
form njj eeH ωω )( . Substituting these expressions into the Eq. (5.63), and performing some algebra allows us to solve for )( ωjeH .
• The second approach is to use discrete-time Fourier transform properties to solve for )( ωjeH .
Based on the convolution property, Eq. (5.63) can be written as
)()(
)( ω
ωω
j
jj
eXeY
eH = . (5.64)
Applying the Fourier transform to both sides and using the linearity and time-shifting properties, we obtain the expression
∑∑=
−
=
− =M
k
jjkk
N
k
jjkk eXebeYea
00
)()( ωωωω . (5.65)
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or equivalently
∑∑
=−
=−
== N
kjk
k
M
kjk
k
j
jj
ea
eb
eXeY
eH0
0
)()(
)(ω
ω
ω
ωω . (5.66)
Example : Consider the causal LTI system that is characterized by the difference equation,
][]1[][ nxnayny =−− , 1<a . The frequency response of this system is
ωω
ωω
jj
jj
aeeXeY
eH−
==1
1)()(
)( .
The impulse response is given by
][][ nuanh n= . Example : Consider a causal LTI system that is characterized by the difference equation
][2]2[81
]1[43
][ nxnynyny =−+−− .
1. What is the impulse response?
2. If the input to this system is ][41
][ nunxn
= , what is the system response to this input signal?
The frequency response is
)1)(1(2
11
)(41
432
81
21 ωωωω
ωjjjj
j
eeeeeH −−−− −−
=+−
= .
After partial fraction expansion, we have
ωωω
jjj
eeeH −− −
−−
==41
21 1
21
4)( ,
The inverse Fourier transform of each term can be recognized by inspection,
][41
2][21
4][ nununhnn
−
= .
ELG 3120 Signals and Systems Chapter 5
24/5 Yao
Using Eq. (5.64) we have
)1)(1)(1(2
11
)1)(1(2
)()()(
41
41
43
41
41
43
ωωω
ωωωωωω
jjj
jjjjjj
eee
eeeeXeHeY
−−−
−−−
−−−=
−
−−==
.
After partial-fraction expansion, we obtain
( ) ωωωωωω
jjjjjj
eeeeXeHeY −−− −
+−
−−
−==212
41
41 1
8
1
21
4)()()(
The inverse Fourier transform is
][21
841
)1(221
4][2
nunnynn
+
+−
−= .