chapter 5 the discrete-time fourier transform

ELG 3120 Signals and Systems Chapter 5

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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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Pencil

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Pencil


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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

−

= .


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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

+

+−

−= .

chapter 5 the discrete-time fourier transform

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