113D 1 SayedText Ch12 DiscreteTimeFourierXform

8/12/2019 113D 1 SayedText Ch12 DiscreteTimeFourierXform

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Chapter 12

DISCRETE-TIME FOURIERTRANSFORM

Copyright c 1996 by Ali H. Sayed. All rights reserved. These notes are distributed only to thestudents attending the undergraduate DSP course EE113 in the Electrical Engineering Departmentat UCLA. The notes cannot be reproduced without written consent from the instructor: Prof. A.H. Sayed, Electrical Engineering Department, UCLA, CA 90095, [email protected].

From this point onwards in the book we proceed to examine how the transform domainrepresentation of discrete-time signals and systems provides a very useful characterizationof their behavior in the so-called frequency domain as opposed to the time domain. Therst step in our exposition is to introduce the Discrete-Time Fourier Transform (DTFT) of a sequence and to study some of its properties.

The Discrete-Time Fourier Transform (DTFT)

Assume X (z) is the z transform of a sequence x(n) and that its ROC includes the unitcircle |z| = 1. That is,

n = |x(n)z n

| < for any |z| = 1 .

This condition is equivalent to saying that the sequence x(n) is absolutely summable :

n = |x(n)| < .

For such sequences we can evaluate X (z) at any point on the unit circle, i.e., at any pointz of the form z = ej for any [ , ], and get

X (ej ) =

n = x(n)e jn

The quantity X (ej ) so-dened is referred to as the discrete-time Fourier transform (DTFT)of the sequence x(n).

The DTFT of an absolutely summable sequence x(n) can therefore be computed in oneof two ways:

1. Either nd the z transform X (z) and replace the variable z by e j or

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2. Evaluate the series

X (ej ) =

n = x(n)e jn

directly.

Examples of DTFTs

As a rst example, the DTFT of the sequence x(n) = {0.5, 1 , 0.5} is, by denition,

X (ej ) = 1 + 0 .5ej + 0 .5e j = 1 + cos .

Some common DTFTs arise from the following additional examples:

1. Unit-sample sequence . The DTFT of (n) is X (ej ) = 1. Likewise, The DTFT of (n no ) is X (ej ) = e jn o for any integer no .

2. Exponential sequence. The DTFT of the exponential x(n) = n u(n) is, by denition,

X (ej ) =

n =0

(e j )n .

We therefore have the sum of the terms of a geometric series with ratio e j and,consequently,

X (ej ) = 11 e j if | | < 1

Alternatively, we already know that the z transform of x(n) is z/ (z ) for |z | > | |.Replacing z by ej we obtain the above DTFT. However, for this substitution to bevalid we need to guarantee that the ROC of X (z) includes the unit circle, which inthis case requires that we impose the condition | | < 1.Thus note that the DTFT of the exponential sequence n u(n) is dened only for| | < 1. In contrast, the z transform is dened for any over a ROC of the form|z | > | |.

3. Rectangular pulse. Consider now the rectangular pulse sequence

x(n) =1 0 n L 1

0 otherwise

Its DTFT is given by

X (ej ) =L 1

n =0e jn

= 1 e jL

1 e j

=e j

L2 ej

L2 e j

L2

e j12 ej

12 e j

12

= e j( L 1)

2sin L2sin 2


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118 Discrete-Time Fourier Transform Chapter 12

Therefore, we have the important DTFT pair

x(n) =1 0 n L 1

0 otherwise

DTFT X (ej ) =

L = 0

e j( L 1)

2 .sin ( L2 )sin ( 2 )

otherwise

The value at = 0 of X (ej ) is obtained by applying LHospitals rule to the ratio

sin L2 / sin 2 . Observe that in evaluating X (ej ) we used the following generalresult for adding a nite number of terms of a geometric series:

n 2

k = n 1

k = n 1 n 2 +1

1 , for n2 n1 .

Uniform Convergence

The DTFT of a sequence x(n) as dened so far requires x(n) to be absolutely summableor, equivalently, that the ROC of X (z) include the unit circle. This condition guarantees thatthe series dening the DTFT converges in a desirable sense known as uniform convergence .This means the following. If, for any N > 0, we consider the partial sum

X N (ej ) =N

n = N x(n)e jn .

Then the uniform convergence of the series dening X (ej ) means that for large enoughN we can use X N (ej ) as a good approximation for X (ej ). More explicitly, the followingthree conclusions hold:

1. Uniform convergence means that no matter what we pick, we can always nd asmall enough number and a large enough N o such that

|X N (ej ) X (ej )| for every N > N o

The word uniform is used to highlight the fact that is the same for all . Hence,at all , the approximation X N (ej ) will be close to X (ej ).

2. Uniform convergence also implies point-wise convergence at each . That is,

limN

X N (ej ) = X (ej ).

3. The limiting function X (ej ) is necessarily continuous in . This is because it is thelimit of a uniformly convergent sequence {X N (ej )} of continuous functions X N (ej ).Therefore, the DTFT of an absolutely summable sequence is always a continuousfunction of .

Mean-Square Convergence


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There are several important sequences x(n) that are not absolutely summable so thatthe DTFT of such sequences cannot be dened as the power series

X (ej ) =

n = x(n)e jn

since, strictly speaking, the series will not converge in this case.However, there are other (weaker) notions of convergence that have been widely studied

in the (mathematical) literature. By appealing to these alternative notions of convergence,as opposed to solely uniform convergence, we are able to extend the denition of the DTFTto a larger class of sequences.

Thus consider a sequence x(n) that has nite energy (rather than being absolutelysummable), i.e.,

n = |x(n)|2 <

One notion of convergence that is useful for dening the DTFT of such sequences x(n) ismean-square convergence. For such sequences, the series

X (ej ) =

n =

x(n)e jn

is said to converge in the mean-square sense. What this means is the following. The partialsums X N (ej ) now satisfy

limN

|X N (ej ) X (ej ) |2 d = 0 .

where the difference |X N (ej ) X (ej )|2 is the square error between X N (ej ) and X (ej ).The integral expression measures the area under the square-error curve over the interval[ , ]. The reason for restricting ourselves to this interval is that X N (ej ) is periodic of period 2. Convergence in the mean-square sense therefore means that this area tends tozero as N . In particular, this means that there exists a large enough N o and a small

enough such that

|X N (ej ) X (ej ) |2 d < for every N > N o

Contrary to uniform convergence, this does not mean that at every the function X N (ej )will be close to X (ej ); the function X N (ej ) can be closer to X (ej ) at some s andfurther away at others. But the overall effect will be that the square-error curve has arealess than .

This is a weaker notion of convergence than uniform convergence (in other words, uniformconvergence implies mean-square convergence). Therefore, for a square-summable (nite-energy) sequence, while the series

n = x(n)e jn

does not converge uniformly, we still say that it converges close enough (in the mean-squaresense) to a function X (ej ) that we therefore take to represent the desired value for theseries.


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As an example, consider the sequence

x(n) =

w c n = 0

sin ( c n )n n = 0

where c < . The DTFT of x(n) is by denition

X (ej ) = wc

+

n = n = 0

sin(c n)n

e jn .

The series on the right-hand side does not converge uniformly since x(n) is not absolutelysummable. However, x(n) is square-summable since

E x =

n = |x(n)|2

= w2c2

+ 2

n =1

sin2(c n)2n2

w2c2

+ 2

n =1

12n2

<

where we used the wellknown fact that

n =1

1n2

< .

Therefore, the DTFT of x(n) is only guaranteed to exist in the mean-square sense. Buthow do we nd what X (ej ) is in this case? We shall address this issue in the next chapter

when we study the inverse DTFT transform. For now we simply state that X (ej

) will turnout to be

X (ej ) =1 |w| < w c

0 wc | w|

That is, it is unity over [ c , c ] and zero otherwise. Notice in particular, that X (ej ) isnot a continuous function of in this case.

Notation . From now on, whenever we refer to the DTFT of a sequence we mean that it isa function that results from the convergence of the series

n = x(n)e jn

either uniformly or in the mean-square sense. In either case, we shall write

X (ej ) =

n = x(n)e jn


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where the equality sign should be understood to mean convergence in one sense or another.

Gibbs Phenomenon

The previous example of a square-summable sequence with a mean-square convergentDTFT serves as a good illustration of the so-called Gibbs phenomenon. Thus consider thepartial sum

X N (ej ) = wc +

N

n = N n = 0

sin (c n)n

e jn

which corresponds to a truncation to the interval n [ N, N ] of the series dening theDTFT of the sequence in the previous example. Observe that we can write

X N (ej ) = wc

+

N

n =1

sin (c n)n

[e jn + ejn ]

= wc

+ 2

N

n =1

sin (c n)n

cos(n)

Hence, X N (ej ) will be a real-valued function of . It is also an even function in sinceX N (ej ) = X N (e j ).

Figure 12.1 shows plots of X N (ej ) for different values of N superimposed on the originalrectangular pulse describing X (ej ), with c = / 2. For increasing values of N , we expectX N (ej ) to provide increasingly better ts for X (ej ). The gure shows that this is indeedthe case and that X N (ej ) is tending to X (ej ) as N . However, since the convergenceis not uniform, there are some annoying discrepancies that persist in the form of ripplesthat occur around the discontinuities at c regardless of the values of N . Also, their peakvalue does not seem to decrease with increasing N . This is known as Gibbs phenomenon(so-called since this phenomenon was rst mathematically explained by J. Gibbs around1899). Gibbs showed that the size of the peak is actually independent of N (the maximumvalue is 1.09, which corresponds to a 9% overshoot). Gibbs also showed that, for every N ,

the value of X N (ej

) at each point of discontinuity is the average value of X (ej

) at thepoint. In this example, we have two such points at c , with average value at each equalto 1/ 2.

The practical implication of these facts is that any truncated approximation of a dis-continuous pulse like X (ej ) will exhibit high frequency ripples near the discontinuities.This suggests that a sufficiently large N should be used in order to guarantee that the totalenergy of the ripples is sufficiently small.

Convergence in a Distributional Sense

Consider now the exponential sequence

x(n) = ej o n .

As we saw in Chapter 3, it may or may not be periodic depending on the value of o . Inaddition, this sequence is neither absolutely summable nor square-summable! Hence, itsDTFT cannot be obtained as the convergent value of the series

n = ej o n e jn =

n = e j ( o )n


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4 2 0 2 40.5

0

0.5

1

1.5N=1

4 2 0 2 40.5

0

0.5

1

1.5N=3

4 2 0 2 40.5

0

0.5

1

1.5N=5

4 2 0 2 40.5

0

0.5

1

1.5N=10

4 2 0 2 40.5

0

0.5

1

1.5N=20

4 2 0 2 40.5

0

0.5

1

1.5N=50

Figure 12.1. Plots of X N (e j ) for several values of N : Gibbs Phenomenon.The rectangular pulse occurs in the interval [ 2 ,

2 ]. The values of X N (e

j )are shown in the interval [ , ].

neither in the uniform sense nor in the mean-square sense. In this case we need a weakernotion of convergence (in a so-called distributional sense). We shall not discuss it hereexcept to say that the series

n = e

j (

o )n

can be shown to converge in this new sense to another series, viz.,

2

n = (w o 2n )

where the () are impulse functions in the continuous variable . [Recall that in continuoustime, an impulse function (t) is dened by the properties

(t)dt = 1 ,

f (t) (t)dt = f (0)

where the second property is known as the sifting property.]We therefore say that the DTFT of ej o n consists of a periodic train of impulses. In

the interval [ , ], we have a single impulse at o with amplitude 2 :

ej o n DTFT 2 n = (w 0 2n )


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Sequence x(n) DTFT X (ej ) over [ , ]

x(n) = (n) X (ej ) = 1delta sequence

x(n) =1 0 n L 1

0 otherwiseX (ej ) =

L = 0

e j ( L 1)2 . sin ( L2 )sin ( 2 )

otherwise

rectangular pulse

x(n) = n u(n), | | < 1 X (ej ) = 11 e jexponential sequence

x(n) =

w c n = 0

sin( c n )n n = 0

X (ej ) =1 |w| < w c

0 wc | w|

sinc sequence

x(n) = ej o n X (ej ) = 2 (w o )unit-norm exponential

x(n) = cos( o n) X (ej ) = [ (w 0) + (w + 0)]sinusoidal sequence

Table 12.1. Some useful DTFT pairs.

In a similar vein, consider the sinusoidal sequence x(n) = cos 0n. This again may ormay not be periodic. Still, using x(n) = 12 [e

j 0 n + e j 0 n ] and by linearity we obtain

cos 0n DTFT n = [ (w 0 2n ) + (w + 0 2n )]

Hence, we obtain two impulses in the domain [ , ], one at 0 and the other at 0 .

Some useful DTFTs .

The designation sinc for the fourth sequence in Table 12.1 will be explained later inthis chapter. In particular, we shall write sinc( a) to denote

sinc(a) = sin aa

with value 1 when a = 0.


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By comparing coefficients of powers of e j on both sides of the above equality we nd thatA = 1 and B = 2. Therefore,

3 43 e j

1 56 e j + 16 e

2j = 1

1 12 e j +

21 13 e

j

By inverse transforming we obtain

x(n) =12

n

+ 213

n

u(n) .

Alternatively, we can replace ej by z and write rst

X (z) = 3 43 z

1

1 56 z 1 + 16 z

2 .

We then choose a ROC for this z-transform that includes |z| = 1. In this case, the ap-propriate ROC should be |z| > 12 . We now invert X (z) by partial fractions by notingthat

3 43 z 1

1 56 z 1 + 16 z

2 = zz 12

+ 2zz 13

which leads to the same sequence as above.

The Inverse DTFT

There are cases for which the DTFT is not a rational function of e j so that inversionby partial fractions is not possible. Still, we can determine the corresponding sequence x(n)by using the following inversion formula

x(n) = 12 2 X (ej )ejn dwhere the integration is over a 2 -long interval, say [0 , 2] or [ , ].Proof: Assume rst that the sequence x(n ) that we are seeking is one for which the DTFT seriesconverges uniformly (i.e., that x(n ) is absolutely summable), so that x(n ) and X (e j ) are relatedvia the dening relation

X (e j ) =

k = x (k )e jk .

In order to establish the validity of the above integral inversion formula, we note that

X (e j )e jn =

k = x (k )e jk e jn .

If we now integrate over any interval of length 2 and use the very useful identify

2 ej ( n k )

d = 0 k = n2 k = n

we obtain the desired formula. Although not proved here, we shall assume the validity of theinversion formula for general sequences x(n ) that may not be absolutely summable (e.g., for caseswhen X (e j ) is dened in a mean-square sense or a distributional sense).


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As an example, let us determine the inverse DTFT of

X (ej ) =1 |w| < w c

0 wc | w|

That is, the DTFT is unity over [ c , c ] and zero otherwise. Using the inversion formulawe see that

x(n) = 12 w

c

w cejn d =

w c

n = 0

w c

sin ( c n ) c n n = 0

This is the same sequence we considered earlier in the chapter and for which we stated with-out justication that its DTFT is given by the above X (ej ). We now have a justicationfor the result.

The Sinc Function

It is common to introduce the special notation

sinc() = sin

This is a function of the continuous variable . The function sinc( ) is such that it attains itsmaximum value of unity at = 0 (i.e., when its argument is zero). It is also zero whenever is an integer multiple of , sinc(k ) = 0 for any integer k = 0. The gure shows a plot of sinc() for values of in the interval [ 20, 20].

We can use the sinc notation to dene sequences. For example, the notation sinc( 4 n)refers to the sequence

sinc4

n = sin 4 n

4 n

.

This sequence is zero at all integer multiples of 4, as shown in the gure for values of the

integer n in the interval [ 20, 20].

Properties of the DTFT

The DTFT shares several properties with the z-transform. They are listed in the tablebelow. Some of the properties are straightforward to verify.

1. Shifting in one domain corresponds to phase change in the other domain . Consider,for example, the statement

x(n n o ) e jn o X (ej )

It shows that shifting in time corresponds only to phase change in the frequencydomain; no magnitude change is introduced!

We can establish this fact by simply invoking the denition of the DTFT. Al-ternatively, if we assume the ROC of x(n) includes the unit circle and let X (z)denote the z-transform of x(n). Then using the the z transform property thatx(n n o ) z n o X (z), we can replace z by ej to obtain the desired result.


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20 15 10 5 0 5 10 15 200.4

0.2

0

0.2

0.4

0.6

0.8

1

s i n

c

20 15 10 5 0 5 10 15 200.4

0.2

0

0.2

0.4

0.6

0.8

1

n

s i n

c (

n

/ 4 )

Figure 12.3. Plots of the functions sinc( ) and sinc 4 n .

Likewise, consider the statement

ej o n x(n) X (ej ( o ) )

It shows that phase change in the time domain corresponds to shifting in the frequencydomain. This is the dual property to the above one.

This fact can also be derived from the denition of the DTFT. Alternatively, if we assume that the ROC of x(n) includes the unit circle and let X (z) denote the z-transform of x(n). Then using the the z transform property that an x(n) X (z/a )and by choosing z = ej and a = ej o , we obtain the desired result.

2. Modulation property . Consider now the the property

cos(o n)x(n) 12 X (ej ( o ) ) + 12 X (e

j ( + o ) )

By writing cos( o n) = 12 [ej o n + e j o n ] and by using linearity, we can establish the

result in a straightforward manner. This is an important conclusion that is known asthe modulation property. It states that multiplying a sequence by cos o n shifts itsDTFT both to the left and to right by o and scales it by 1 / 2.

3. Parsevals Theorem . Another important result concerning the DTFT is Parsevals

relation. It states that for any two sequences {x(n), y(n)} with given DTFTs, it holdsthat

n = x(n)y (n) =

12 2 X (ej )Y (ej )d .

Proof: We establish the result for the case in which the sequences are absolutely summable


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Sequence DTFTx(n) X (ej )

y(n) Y (ej )

ax (n) + by(n) aX (ej ) + bY (ej )

x(n n0) e jn 0 X (ej )

ej o n x(n) X (ej ( o ) )

cos(o n)x(n) 12 X (ej ( o ) ) + 12 X (e

j ( + o ) )

x( n) X (e j )

nx (n) j dX (ej )

dw

x(n) y(n) X (ej )Y (ej )

x(n)y(n) 12

2 X (e

j )Y (ej ( ) )d

Table 12.2. Properties of the DTFT.

so that x(n ) and X (e j ) are related via the dening relation

X (e j ) =

k = x (k )e jk .

Similarly for Y (e j ). We now use the identities:

12 2 X (e j )Y (e j )d = 12 2

n = x (n )e jn Y (e j )d

=

n = x (n ) 1

2 2 Y (e j )e jn d=

n = x (n )y (n )

A special case of Parsevals relation is the conclusion that the energy of a sequencecan be evaluated from its DTFT as follows:

n = |x(n)|2 = 12 2 |X (ej )|2d

That is, we simply evaluate the area underneath its spectrum over an interval of length2 and scale the result by 2 . The squared DTFT of a sequence is dened as its


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113D 1 SayedText Ch12 DiscreteTimeFourierXform

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