[PDF] Chapter 5 - The Discrete Fourier Transform





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Lecture 7 - The Discrete Fourier Transform

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Chapter 5 - The Discrete Fourier Transform

Example. Find the 8-point DFT of the signal x[n] = 6 cos2(? can recover it from xps[n] which in turn can be recovered from the DTFS synthesis equation.

Chapter5

TheDiscreteFourierTransform

Contents

5.1 5.2c

DTFTFT

Sum shifted scaled replicatesSum of shifted replicates DTFS ZDFT

Sinc interpolationRectangular window

Dirichlet interpolationRectangular windowBandlimited:Time-limited:Time-limited:Bandlimited:

SamplingSampling

Sample Unit Circle

Unit Circle

PSfragreplacements

x[n]=xa(nTs) x a(t) X a(F)x[n] X(!)x ps[n] X[k]

X(z)X[k]=X(!)j!=2

Nk

TheFTFamilyRelationships

FT

Xa(F)=R1

1xa(t)e|2Ftdt

xa(t)=R1

1Xa(F)e|2FtdF

DTFT

X(!)=P1

n=1x[n]e|!n=X(z)jz=ej! x[n]=1 2R

X(!)e|!nd!

UniformTime-DomainSampling

x[n]=xa(nTs)

X(!)=1

TsP 1 k=1Xa!=(2)k Ts (sumofshiftedscaledreplicatesofXa())

Xa(F)=Tsrect

F Fs xa(t)=P1 n=1x[n]sinc tnTs Ts DTFS ck=1 NP N1 n=0xps[n]e|2Nkn=1NX(z)jz=e|2Nk xps[n]=PN1 k=0cke|2

Nkn,!k=2Nk

UniformFrequency-DomainSampling

X[k]=X(!)

!=2

Nk;k=0;:::;N1

X[k]=Nck

xps[n]=1 NP N1 k=0X[k]e|2Nkn xps[n]=P1 l=1x[nlN](sumofshiftedreplicatesofx[n]) k=0X[k]P(!2k=N),whereP(!)=1 NP N1 k=0e|!n. c

Overview

Mainpoints

spectrumanalyzerswork.) offrequencies.

Whichfrequencies?

!k=2Nk;k=0;1;:::;N1: canrecoverx[n]fromX2 NkN1 k=0.

X(!)=1X

n=1x[n]e|!n;x[n]=1 2Z

X(!)e|!nd!:

Discrete-timeFourierseries(DTFS)review

RecallthatforaN-periodicsignalx[n],

x[n]=N1X k=0c ke|2

Nknwhereck=1NN1X

n=0x[n]e|2 Nkn: 5.4c

Denition(s)

X[k]4=

PN1 n=0x[n]e|2

Nkn;k=0;:::;N1

??otherwise: Nout fronttomaketheDFTmatchtheDTFS.)

Thisisreasonablemathematicallysince

X[n+N]=N1X

n=0x[n]e|2

N(k+N)n=N1X

n=0x[n]e|(2Nkn+2kn)=N1X n=0x[n]e|2Nkn=X[k]:

2.TreatX[k]asundenedfork=2f0;:::;N1g.

computer.

3.TreatX[k]asbeingzerofork=2f0;:::;N1g.

Thisisavariationonthepreviousoption.

~x[n]= 1 NP N1 k=0X[k]e|2Nkn;n=0;:::;N1 ??;otherwise: x[n]= 1 NP N1 k=0X[k]e|2Nkn;n=0;:::;N1

0;otherwise:

k=0.Theproofofthisis x[n]=1 NN1X k=0X[k]e|2 Nkn:

ThisisindeedaN-periodicexpression.

x ps[n]=1 NN1X k=0X[k]e|2 Nkn; eveninthiscase. c

Examples

X[k]=N1X

n=0x[n]e|2

Nkn=7X

n=0x[n]e|28kn=7X n=0([n]+0:9[n6])ej2kn=8=1+0:9e|28k6:

0123456-0.2

0 0.2 0.4 0.6 0.8 1 n x[n]

DFT Example

012345670

0.5 1 1.5 2 k |X[k]|

01234567-1

-0.5 0 0.5 1 k

Ð X[k]

X[k]=X(z)

z=e|2

Nk=1+0:9e|2

8k:

Example.FindN-pointinverseDFToffX[k]gN1

k=0whereX[k]=1;k=k0

0;otherwise=[kk0];fork02f0;:::;N1g.

Picture.

x[n]=1 NN1X k=0X[k]e|2

Nkn=1Ne|2

Nk0n:

ThuswehavethefollowingimportantDFTpair.

Ifk02f0;:::;N1g,then1Ne|2

Nk0nDFT !N[kk0]:

Example.FindN-pointinverseDFToffX[k]gN1

k=0where

X[k]=8

:e |;k=k0 e|;k=Nk0

0;otherwise;=e|[kk0]+e|[k(Nk0])

fork02f1;:::;N=21;N=2+1;:::;N1g.

Picture.

x[n]=1 NN1X k=0X[k]e|2

Nkn=1Ne|e|2

Nk0n+1Ne|e|2

N(Nk0)n=2Ncos2Nk0n+

5.6c

Expanding:x[n]=3+3cos

2n=3+32e|282n+32e|282n=18h

24+12e|282n+12e|28(82)ni

Sobycoefcientmatching,weseethatX[k]=f24

;0;12;0;0;0;12;0g:

Supposex[n]=e|2

FindtheN-pointDFTofx[n].

X[k]=N1X

n=0x[n]e|2

Nkn=N1X

n=0e |2Nk0ne|2Nkn=N1X n=0e |2N(kk0)n=N;k=k0+lN;l2Z

0;otherwise:

Thus x[n]=e|2Nk0nDFT !NX[k]=N1X l=1[kk0lN]=NN[kk0]; whereN[n]4=P1 l=1[nlN].

Nk0foranyintegerk.

FindtheN-pointDFTofx[n].

X[k]=N1X

n=0e |!0ne|2

Nkn=N1X

n=0 e|(!02Nk)n=1 e |(!02 Nk)N

1e|(!02Nk)=1e|!0N1e|(!02Nk):

ThuswehavethefollowingcuriousDFTpair.

If!0=2Nisnon-integer,thene|!0nDFT !NX[k]=1e|!0N1e|(!02Nk):

Whatisgoingonintheseexamples?

r

N[n]=1;n=0;:::;N1

0;otherwise;

whichhasthefollowingDTFT:

R(!)=N1X

n=0e |!n==e|!(N1)=2Rr(!);whereRr(!)=8 :sin(!N=2) sin(!=2);!6=0

N;!=0Nsinc

N!2

Thenwehave

wheretheaboveisaDiracimpulse.Thus

X[k]=X(!)

!=2

Nk=2R2Nk!0

When!0=2

c

DTFTsamplingpreview

TheDTFTformulaisX(!)=P1

n=0x[n]e|2 Nkn: n=0x[n]e|!n:

X[k]=X(!)

!=2 Nk: synthesisformula:x[n]=1 2R

X(!)e|!nd!:

Doesthisapproachalwayswork?No!

X(!)=8

:3 4 !0 2; !0 1 2 1 2 32
!0 2;1 2< !0 3 2 0;3 2< !0 !0 -2p 0 0.75p p 2p0 0.2 0.4 0.6 0.8

Challenging DTFT H(w)

w H(w) -2p 0 0.75p p 2p0 0.2 0.4 0.6 0.8

Samples, N=16

w

H(2p/N k)

015-0.05

0 0.05 0.1 0.15 0.2 0.25 n x[n]

015-0.05

0 0.05 0.1 0.15 0.2 0.25 n x ps [n] via ifft

Inverse DFT of X[k]

-807 0 0.1 0.2 x[n]

Exact Inverse DTFT

-807 0 0.1 0.2 n x ps [n] via ifft

After "fftshift", for N=16

-807-2 0 2 4

6x 10-4

n Error 5.8c

Themodulofunction

Example

.1mod4=1;7mod4=3;1mod4=3;8mod4=0. x ps[n]4=1X l=1x[nlN]:

Notethatallthevaluesofx[n]affectxps[n].

x((n))N4=x[nmodN]:

Example.x[n]=f4;3;2;1g=4n;0n3

0;otherwise;soL=4.

n 6x[n] -2-1012342 4

N-pointperiodicsuperposition:

n

6xps[n];N=6

-6062 4 n

6xps[n];N=3

0362
3 5

N-pointcircularextension:

n

6x[nmodN];N=6

-6062 4 n

6x[nmodN];N=3

0362
3 4 x[n]!

DTFT!X(!)!sample!X2NkN1

k=0!N-pointinvDFT!xps[n] x[n]!

N-pointDFT!N-pointinvDFT!x[n]orx[nmodN]

c 5.1 X 2 Nk =X(!) !=2 Nk=1X n=1x[n]e|2

Nkn;k=0;1;:::;N1:

Questionsarise:

I.HowisX2

NkN1 k=0.

II.When(ifever)canwerecoverx[n]fromX2

NkN1 k=0orfromDFTfX[k]gN1 k=0?

III.Howcanwerecoverx[n]fromX2

NkN1 k=0orfromDFTfX[k]gN1 k=0? Nk. X 2 Nk =X(!) !=2 Nk=1X n=1x[n]e|2 Nkn 1X l 0=1l

0N+N1X

n

0=l0Nx[n0]e|2

Nkn0(splitupsum)

1X l

0=1N1X

n=0x[n+l0N]e|2

Nk(n+l0N)(n=n0l0N0)

1X l

0=1N1X

n=0x[n+l0N]e|2

Nkn(simplifyexponent,letl=l0)

N1X n=0 1X l=1x[nlN]! e |2

Nkn(exchangesums)

wheren=n0l0Nandl=l0.Thusquotesdbs_dbs17.pdfusesText_23
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