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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

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