midexam_soln

7/28/2019 midexam_soln

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1MASSACHUSETTSINSTITUTEOFTECHNOLOGY

DepartmentofElectricalEngineeringandComputerScience

6.341Discrete-TimeSignalProcessingFall2005

MIDTERMEXAMTuesday,November8,2005

SOLUTIONSDisclaimer: These are not meant to be full solutions, but rather an indication of how toapproacheachproblem.

Problem Grade Points Grader1 /15

2(a) /122(b) /62(c) /123(a) /103(b) /10

4 /155(a) /105(b) /10Total /100


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2Problem1 (15%)A stable system with system function H(z) has the pole-zero diagram shown in Figure 1-1.Itcanberepresentedasthecascadeofastableminimum-phasesystem Hmin(z)andastableall-passsystemHap(z).

Re(z)

Im(z)

12 3 4

Figure1-1: Pole-zerodiagramforH(z).DetermineachoiceforHmin(z)andHap(z)(uptoascalefactor)anddrawtheircorrespondingpole-zeroplots. Indicatewhetheryourdecompositionisuniqueuptoascalefactor.Wefindaminimum-phasesystemHmin(z)thathasthesamefrequencyresponsemagnitudeasH(z)uptoascalefactor. Polesandzerosthatwereoutsidetheunitcirclearemovedtotheirconjugatereciprocal locations(3to 1 4,to0).3, 4 to 1

(11z1)4Hmin(z) = K1(11z1)(11z1)2 3

There isnoneedto includeanexplicitz termtoaccountforthezeroattheorigin.Wenowincludeall-passtermsinHap(z)tomovepolesandzerosbacktotheiroriginallocations

1z 3inH(z). Theterm 13z1 movesthepoleat 1 backto3,thetermz1 movesthezerofrom03to,andsoon:

Hap(z) = z1 z13 z11413z1 11z14

Thedecomposition isuniqueuptoascalefactor. Wecannot includeadditionalall-passtermsin Hap(z), since it is not possible for Hmin(z) to cancel the resulting poles and zeros outsidetheunitcircle.


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3Name:

Re(z)

Im(z)

12

13

14

Figure1-2: Pole-zerodiagramforHmin(z).

Re(z)

Im(z)

13

14 3 4

zeroat

Figure1-3: Pole-zerodiagramforHap(z).


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4Problem2 (30%)

xc(t) xL[n]C/D

x[n] 2

y

H(ej

)yM[n[n]

2 yc(t)]

D/C

T TFigure2-1:

Forparts(a)and(b)only,Xc(j)=0for|| > 2 103 andH(ej )isasshowninFigure2-2(andofcourseperiodicallyrepeats).

j )H(e

A

4 4

Figure2-2:(12%)(a) Determine the most general condition on T, if any, so that the overall continuous-time

systemfromxc(t) to yc(t) is LTI. We can simplify Figure 2-1 by replacing the expander, H(ej ) and compressor with asingleDTLTIsystem. The impulse responseoftheequivalentDTLTIsystem ish[2n],where h[n] is the inverse DTFT of H(ej ). We saw this result in Problem Set 6 wherewederived itbypolyphasedecompositionandsubsequentsimplification. The frequencyresponseoftheequivalentDTLTIsystem is 12H(ej 2 j() + 1H(e2 2)).Alternatively,wecanreachthesameconclusionbyrelatingYM(ej ) to XL(ej ):

1 j( 2))XL(ej2(

2))j2j ) = jYM(e H(e )XL(e ) + H(e2 2

2BothXL()termsreducetoXL(ej ),thelatterbecauseof2 periodicity,leavingbehindthesameequivalentfrequencyresponse.TheoverallCTsystemwillbeLTI ifthere isnoaliasingattheC/Dconverter,or iftheDTfilterrejectsanyfrequencycontentcontaminatedbyaliasing. SincetheequivalentDTfrequencyresponse is0for < || ,thecopyofXc(j)centredat =0canextend2

3to = 2 = without passing any aliased components. Applying the = T2 2mapping:

max = maxT = (2 103)T < 32

andsowerequirethatT < 3 103.4


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5Name:(6%)(b) Sketchandclearlylabeltheoverallequivalentcontinuous-timefrequencyresponseHeff(j)

thatresultswhentheconditiondetermined in(a)holds.

H (j)

2T 2T

A/2ef f

(12%)(c) For thispartonlyassumethatXc(j) inFigure2-1 isbandlimitedtoavoidaliasing,i.e.

Xc(j)

=

0for

||

.

Forageneral

sampling

period

T

,we

would

like

to

choose

TthesystemH(ej )inFigure2-1sothattheoverallcontinuous-timesystemfromxc(t) to

yc(t) isLTIforany inputxc(t)bandlimitedasabove.DeterminethemostgeneralconditionsonH(ej ), ifany,sothattheoverallCTsystemis LTI.Assuming that theseconditions hold, also specify interms ofH(ej ) the overallequivalentcontinuous-timefrequencyresponseHeff(j).Thesimplification inpart(a)yieldsanequivalentDTLTIsystembetweentheC/DandD/Cconverters. ThustheoverallsystemwillbeLTIifxc(t)isappropriatelybandlimitedtoavoidaliasing,asassumed inthispart. TherearenoconditionsonH(ej ).


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6Problem3 (20%)Forthisproblemyoumayfindthe informationonpage12useful.ConsidertheLTIsystemrepresentedbytheFIRlatticestructure inFigure3-1.

x[n] y[n]

v[n]

12

12

3

3

2

2

1 1 1z z zFigure3-1:

(10%)(a) Determinethesystemfunctionfromthe inputx[n]totheoutputv[n](NOTy[n]).Theoutputv[n]istakenaftertwostages,soweperformthelatticerecursionuptoorderp = 2.

1k1 =2, k2 = 3, k3 = 2

a(1)1 =k1 =12

a(2)1

a(2) = a

(1)10 k2 a

(1)1

1 =

13

2V(z)

2= 1 z13zX(z)

Notethechangeofsigns ingoingfroma(2) tothesystemfunction.k


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7Name:(10%)(b) LetH(z)bethesystem function fromthe inputx[n]totheoutputy[n],and letg[n] be

theresultofexpandingtheassociated impulseresponseh[n] by 2: h[n] g[n] 2

The impulseresponseg[n]definesanewsystemwithsystemfunctionG(z).Wewould liketo implementG(z)usinganFIR latticestructureasdefinedbythefigureonpage12. Determinethek-parametersnecessary foranFIR lattice implementationofG(z).Note: Youshouldthinkcarefullybeforediving intoa longcalculation.Since g[n] is h[n] expanded by 2, G(z) = H(z2). We replace z by z2 in Figure 3-1. Wethenexpandeachofthethreesectionsasshownbelow:

kp

kp

kp

kp

0

0

2 1 1z z zThe resulting flowgraph for G(z) is in the form of a 6th-order FIR lattice. We read offthesixk-parametersas:

1k2 = 2k4 = 3 k6 = 2 kp = 0, p = 1, 3, 5


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8Problem4 (15%)

Filter

response

Wefind inatreasurechestaneven-symmetricFIRfilterh[n]of length2L + 1, i.e.h[n]=0for |n|> L,

h[n] = h[n].H(ej ),theDTFTofh[n], isplottedover inFigure4-1.

1.2

1

0.8

0.6

0.4

0.2

0

0.21 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1

Normalizedfrequency(radians/sample)

Figure4-1: PlotofH(ej

) over .


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9Name:WhatcanbeinferredfromFigure4-1aboutthepossiblerangeofvaluesofL? Clearlyexplainthereason(s)foryouranswer. Donotmakeanyassumptionsaboutthedesignprocedurethatmighthavebeenusedtoobtainh[n].As discussed in Section 7.4 of OSB, H(ej ) can be represented as an Lth order polynomialincos sinceh[n] iseven-symmetricandfinite-length. AnLthorder polynomial incos canhavenomorethanL +1extremaon [0, ]. (AnLthorderpolynomial inxcanhavenomorethanL 1extremainanopeninterval,butanLthorderpolynomialincoswillalwayshaveadditionalextremaat=0and= foratotalofL +1extrema.)H(ej )has6extremaon [0, ],so6L + 1 or L5.


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10Problem5 (20%)InFigure5-1,x[n]isafinitesequenceoflength1024. ThesequenceR[k]isobtainedbytakingthe1024-pointDFTofx[n]andcompressingtheresultby2.

DFT1024-pointX[k]

2R[k]

2Y[k]x[n]

IDFT1024-pointy[n]

IDFT512-point

r[n]

Figure5-1:

(10%)(a) Choosethemostaccuratestatementforr[n],the512-point inverseDFTofR[k]. Justifyyourchoiceinafewconcisesentences.A. r[n] = x[n], 0n511B. r[n] = x[2n], 0n511C. r[n] = x[n] + x[n+512], 0n511D.

r[n] =

x[n] +

x[n

+

512],

0

n

511

E. r[n] = x[n] + x[1023n], 0n511

Inallcasesr[n]=0outside0n511.Answer: CCompressingthe1024-pointDFTX[k]by2undersamplestheDTFTX(ej ). Undersam-plinginthefrequencydomaincorrespondstoaliasinginthetimedomain. Inthisspecificcase,thesecondhalfofx[n]isfoldedontothefirsthalf,asdescribedbystatementC.


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Name: 11(10%)(b) ThesequenceY[k] isobtainedbyexpandingR[k]by2. Choosethemostaccuratestate-

ment for y[n], the 1024-point inverse DFT ofY[k]. Justify your choice in a few concisesentences.

A.

y[n] = 12

12(x[n] + x[n+512]), 0n511(x[n] + x[n512]), 512n1023

x[n], 0n511B. y[n] =

x[n512], 512n1023x[n], n even

C. y[n] = 0, noddx[2n], 0n511

D. y[n] = x[2(n512)], 512n1023

E. y[n] = 12 (x[n] + x[1023n]), 0n1023Inallcasesy[n]=0outside0n1023.Answer: AWe can first think about how expanding by 2 in the time domain affects the DFT. Ex-pandingatimesequencex[n]by2compressestheDTFTX(ej )by2infrequency. Asaresult,the2N-pointDFToftheexpandedsequencesamples two periodsofX(ej )andequalstwocopiesoftheN-pointDFTX[k].By duality, expanding the DFT R[k] by 2 corresponds to repeating r[n] back-to-back,with

an

additional

scaling

by

12 .

Thus

statement

A

is

correct.

Alternatively,Y[k] = 12 1 + (1)k X[k]. ModulatingtheDFTby(1)k correspondstoacirculartimeshiftofN/2=512.

ENDOFEXAM


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12YOUCANUSETHEBLANKPARTOFTHISPAGEASSCRATCHPAPERBUT

NOTHINGONTHISPAGEWILLBECONSIDEREDDURINGGRADING

Ap+1(z) = Ap(z) kp+1Bp(z) (1a)(p+1)Ap(1/z)Bp(z) = z (3a)

A0(z)=1 and B0(z) = z1 (2)(p+1) (p)

ak = ak kp+1ap(p)k+1 , k= 1, . . . , p (8a)(p+1)

(8b)ap+1 =kp+1Define: T

p = a(1p) (p) (p)a2 ap T(p)

= a(p) ap1 ap p (p) 1Thenequations(8)become: p p

p+1 = kp+1 (9)0 1

Reverserecursion:(M) (M)

a(M1)

= ak +kMaMk, k= 1, . . . , M 1 (11)k 1 k2M(M) (M1)

withkM =aM , kM1 =aM1 .

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