Solutions for Signals Systems And Inference 1st Us Edition by Oppenheim by 6alsm

Signals,Systems&Inference

AlanV.Oppenheim&GeorgeC.Verghese c 2016

Chapter2Solutions

Notefromtheauthors

ThesesolutionsrepresentapreliminaryversionoftheInstructors’SolutionsManual(ISM). Thebookhasatotalof350problems,soitispossibleandevenlikelythatatthispreliminary stageofpreparingtheISMtherearesomeomissionsanderrorsinthedraftsolutions.Itis alsopossiblethatanoccasionalprobleminthebookisnowslightlydiﬀerentfromanearlier versionforwhichthesolutionherewasgenerated.Itisthereforeimportantforaninstructorto carefullyreviewthesolutionstoproblemsofinterest,andtomodifythemasneeded.Wewill, fromtimetotime,updatethesesolutionswithclariﬁcations,elaborations,orcorrections.

Manyofthesesolutionshavebeenpreparedbytheteachingassistantsforthecourseinwhich thismaterialhasbeentaughtatMIT,andtheirassistanceisindividuallyacknowledgedinthe book.Forpreparingsolutionstotheremainingproblemsinrecentmonths,weareparticularly gratefultoAbubakarAbid(whoalsoconstructedthesolutiontemplate),LeightonBarnes, FisherJepsen,TarekLahlou,CatherineMedlock,LucasNissenbaum,EhimwenmaNosakhare, JuanMiguelRodriguez,AndrewSong,andGuolongSu.WewouldalsoliketothankLaura vonBosauforherassistanceincompilingthesolutions.

Solution2.1

(a)Recallthatthephasedelayandgroupdelayofafrequencyresponse H(jω)arerespectively deﬁnedby τp(ω)= ∠H(jω) ω and τg(ω)= d dω ∠AH(jω) where ∠H(jω)maycontaindiscontinuitiesofsize ±π inordertocompensateforthe non-negativityofthemagnitudefunction |H(jω)| whilethefunction ∠AH(jω)doesnot. Forthefrequencyresponse H(jω)=3e j3ω wehavethat A(ω)= |H(jω)| =3forall ω hence ∠AH(jω)= ∠H(jω)= 3ω.Pluggingthisintotheexpressionsaboveyields

τp(ω)= τg(ω)=3forall ω.

UEFALSE

(b)Thisstatementistruebythecombineddeﬁnitionofeigenfunctionandtime-invariance properties.

UEFALSE

(c)Let x(t)bealinearcombinationof K> 1eigenfunctions,i.e. x(t)= K k=1 αkejωk t,and observethat

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whichisingeneralnotequaltoasinglescalartimes x(t

TRUE FA ✗ ✖ ✔ ✕ LSE

Fortheremainingparts,werefertothesystemcharacterizedby:

(d)Thesystemiscausalsince h[n]=0forall n< 0. TR ✗ ✖ ✔ ✕ UEFALSE

(e)ThesystemisBIBOstablesincetheimpulseresponseisabsolutelysummable,i.e.

ThisresultcanalsobeestablishedbythefactthattheROCofthe z-transformof h[n] includestheunitcircle.WemayconcludethissinceweknowthattheFouriertransform H(ejΩ)exists.

✗ ✖ ✔ ✕ UEFALSE

(f)Theinput x[n]=( 1)n forall n isaneigenfunctionofaDTLTIsystem.Thismaybe morevisuallyapparentbyrewriting x[n]as x[n]= ejπn forall n.Therefore,from(b) weareguaranteedthattheoutputispurelya(possiblycomplex)scalartimes x[n]and thereforedoesnottaketheform

since K2 isexpresslyforbiddentobezero.

Solution2.2

(a)Thispartdealswithaninputsignal x[

Theoutputisgivenby(ii) y[n]=2x[n 1].

Ashortderivationrevealsthis:

(b)Thispartdealswithaninputsignal x[

Theoutputisapproximatelygivenby(ii) y

Assume s[n]issuﬃcientlybandlimitedsothatwemaysimplyapply H(ejΩ)tothecomplex exponentials.Wedonotneedtoworryaboutapplyingagroupdelayshiftto s[n]since thephasecurvehasnoslopeatΩ= ± 2π 3

Solution2.3

Wearegiventhesquaredmagnitudefrequencyresponseofaﬁlter H(jω)tobe |H(jω)|2 = ω2 +1 ω2 +100 .

Makingthesubstitution ω2 →−s2 gives H(s)H( s)= s2 +1 s2 +100 = ( s +1)(s +1) ( s +10)(s +10) .

Wenowfactorthetermsaboveinordertoidentify H(s).Sincethesystemisbothcausaland stable,aswellashasacausalandstableinverse,thentheﬁlterspolesandzerosmustlieinthe lefthalfplane.Therefore, H(s)= K s +1 s +10 where K maybeeitherplusorminusone.Substituting s = jω gives H(jω)= jω +1 jω +10

Solution2.4

WearegivenaDTLTIsystemwithimpulseresponse h[n]= δ[n] δ[n 1].

aWewritethefrequencyresponseof h[n]as H ejΩ =1 e jΩ

whereweusetheidentity j = ej π 2 .ThereforeΘ(Ω)= Ω 2 + π 2 for π ≤ Ω <π.The groupdelayofthissystemisthengivenby

g(Ω)= d dΩ Ω 2 + π 2 = 1 2 for π ≤ Ω <π.Thephasedelaymusttakeintoaccountthefactthattheamplitudeis negativefor π< Ω < 0,hence

(b)Wehavethat H ej

Wecannowwrite

where η0 = τp( π 3 )= 1.Moregenerally, η0 = 1+6k foranyinteger k

Looselyspeaking, q[n]isrelatedto p[n]by q[n]= p“[n τg( π 3 )]”wherethequotesindicate furtherclariﬁcationisneeded.Thehalf-sampledelayoftheenvelope p[n]bythegroup delayistobeinterpretedasiftherewasanunderlyingappropriatelybandlimitedCT signal p(t)satisfying p[n]= p(nT )forsomearbitrarysamplinginterval T ,say T =1 withoutlossofgenerality,whichthengotshiftedby T/2= 1 2 andthenresampled,i.e., q[n]= p(n 1 2 ).

Solution2.5

ConsiderthecausalLTIsystemcharacterizedbythediﬀerentialequation dy(t) dt +2y(t)= x(t)

(a)TakingLaplacetransformsofbothsidesgives

Y (s)(s +2)= X(s)

Solvingfor H(s)= Y (s) X(s) andsubstituting s = jω yields

Themagnitudeandphaseresponsesof H(jω)aredepictedwiththeresultof(b)below.

(b)Recallthatthephaseresponseofasystemcorrespondstothecontributionsofthezeros minusthecontributionsfromthepoles.Therefore,from(a)weconcludethat ∠AH(jω)= tan 1 ω 2 .Thegroupdelayisthencomputedby

andisdepictedintheﬁgurebelow.

Solution2.6

(a)First,were-writethepiecewiselinearfunction ∠H(jω)intermsof ω as ∠H(jω)=  

10π 5 2000 ω,ω ∈ [0, 4000π] 10π 5 2000 ω,ω ∈ [ 4000π, 0) 0, elsewhere .

Computingthegroup-delay τg(ω)= d dω ∠H(jω)thengives τg(ω)= 5 2000 ,ω ∈ [ 4000π, 4000π] 0, elsewhere

wherewehavechosenthevalueatDCtomaintaincontinuity.Aplotofthegroupdelay isshownbelow.

(b)Inobtaininganexpressionfor y(t)wemakeuseofthefactthebandwidthof p(t)issmall ascomparedwith w1 and w2 allowingtheapproximationsdiscussedinSection2.2.1to hold.Inparticular,wemakeuseofthefollowingprocessingapproximations: h(t) ∗ anp(t nT )cos(ω1t) ≈ anp(t nT τg(ω1))cos(ω1(t

From(a)wecomputethefollowingvalues:

, and τp(ω2)=0Finally,weobtainanexpressionfortheoutputusingthelinearityofthe channelas:

Solution2.7

y2 [n]

Byexaminingthemagnituderesponseoftheﬁlter,weseethatthepacketwiththelowest frequencyintheinputisattenuated,whiletheintermediatefrequencyisonlyslightlyattenuatedrelativetothehighestfrequency.Thelimitsthechoicesto y2[n]and y4[n].Thegroup delayoftheﬁltershowsthatthepacketwiththehighestfrequencyintheinputisdelayedby approximately75samples,andtheintermediate-frequencypacketisdelayedbyapproximately 40samples.Thesefactspointto y2[n].

Solution2.8

G(z)hasapoleat z =0 7andazeroatinﬁnity.Thissystemdoesnothaveacausaland stableinverse,becausetheinversesystemmusthaveapoleatinﬁnity,whichmeansthatthe systemcannotbebothcausalandstable.

Twodecompositionsof G(z)intoaminimum-phaseandanall-passsystemare:

Solution2.9

(a)Weknow

)

so M (s)musthavetheleft-half-planepolesandzerosof

Thus

. All-pass A(s)= ± s 2 s+2

(b)Similarly,

M (s)= ± s+2 s+1

so N (z)musthavethepolesandzerosofthisthatareinsidetheunitcircle.Thezeros areat 1 6 and6,whilethepolesareat0and ∞.Hence N (z)= K z 1 6 z = K(1 1 6 z 1) , andcomparingwiththeprecedingexpressionshows K = ±6.Then B(z)= H(z) N (z) = ∓z.

All-pass B(z)= ∓z,

Solution2.10

Ifwerepresentthefrequencyresponsesusingamagnitudeandphasedecomposition,

thenthecascadeofthetwosystemshasfrequencyresponse

Therefore,thegroupdelayofthecascadedsystemwillbe

Solution2.11

y2[n] isthemostlikelyoutputsignalforthegivensystem.

Theﬁlterisalow-pass,sothehighfrequencycomponentsareremoved,soitcannotbe either y1[n]or y3[n]sincetheystillcontainthehighfrequencypulse.

Thereshouldbeabouta40-sampledelayofthelow-frequencypulseandan80-sampledelay ofthemid-frequencycomponent. y4[n]hasnodelayforthemid-frequencycomponent,but y2[n]hasbothofthesedelays.

Solution2.12

ConsidertheDTcausalLTIsystemwithfrequencyresponse

(a)Themagnitudeofthecascadeoftwosystemsistheproductofthemagnitudesandso

since e j4Ω isalinearphasefactorandthusisanall-passsystemwithunitygain.ConsistentwithEq.(2.28),theremainingtermisanall-passtermsinceitcorrespondstoa systemfunction H(z)whosepoleandzeroareinaconjugatereciprocallocation.This canbeseenbyreplacing ejΩ by z andobservingthepoleisat z = 1 2 andthezeroisat z =2.Sincethefrequencyresponsemagnitudeisconstant,evaluatingitforanyvalueof ΩprovidesitsvalueforallΩ.SubstitutingΩ=0abovegives

(b)Wearegiventhatthegroupdelayofthesystem

Notethat F (ejΩ)onlydiﬀersfrom H ejΩ byalinearphasefactor.Moreover,thephases (andgroupdelays)ofacascadeofsystemsarethesumofthephases(andgroupdelays), thereforeitfollowsthatthegroupdelayof H ejΩ isthegroupdelayof F (ejΩ)raisedby 3forallΩ,i.e.

Thegroupdelaycurve τg,H (Ω)isplottedbelowforΩ ∈ [0,π].

(c)Weknowthemagnitudeoftheinputsignal x[n]=cos(0.1n)cos π 3 n ispreservedsince thesystemisanall-passwithunitygain.Wemakeuseofthefactthatthefrequency contentofcos(0.1n)isconcentrated(impulsesinfact)andthatΩ=0.1issuﬃciently smallascomparedtoΩ= π 3 sothattheapproximation

isreasonable.Usingtheexpressionobtainedin(b)wehavethat

g,H

3 =4andthat τp,H π 3 =1therefore

Solution2.13

Theinputsignalis x(t)= m(t)cos(ω0t)where m(t)= sin(πt/T ) πt/T

(a)Sketch X(jω).RecalltheFouriertransformpairs cos(ω0t) FT ←→

[

(

)+ δ(ω + ω0)] sin(πt/T ) πt/T FT ←→ T,ω ∈ [ π T , π T ] 0elsewhere

Utilizingthefactthatmultiplicationoftwosignalsinthetimedomaincorrespondstothe convolutionoftheirFouriertransformsinthefrequencydomain,weobtainthefollowing plotfor X(jω).

(b)Express y(t)if H (jω)= e jω(4 10 6)

Wehavethatthecorrespondingimpulseresponseis h(t)= δ(t 4 10 6).Recallfrom thesiftingpropertythat x(t) ∗ δ(t T )= x(t T ).Therefore, y(t)correspondstoa time-delayoftheinput,i.e. y(t)= x(t 4 10 6) =

ThiscanequivalentlybeseenbyusingtheapproximationfoundinSection2.2.1.In particularwecanexpresstheoutputas y(t) ≈ m (t τg (ω0))cos(ω0 (t τp (ω0))) .

Foranimpulseresponse h(t)correspondingtoasimpledelaythephaseandgroupdelays arethesame.(SeeProblem2.1(a)foranexample)

(c)Weshallmakeuseofthefactthatthebandwidthofthesincfunction m(t)israthersmall ascomparedtothefrequencyofthecosine.Specifically,wearegiventhat ω0 =2600π while 1 T =75kHz.ThisvalidatestheapproximationmodeldiscussedinSection2.2.1 wheretheeffectofthefilterontheenvelopeisshowntobeapproximatelyequaltoa timedelaybythegroupdelayofthefilterat ω0 andtheeffectonthefilteronthecarrier sinusoidisadelaybythephasedelayofthesystem.Thisissummarizedas

FromFig.P2.13-2weﬁndthat

1 650 seconds.Notethatit isalsosuﬃcienttotake τp(ω)=0sincewhenmultipliedby ω0 thequantity 1 650 becomes amultipleof2π.Therefore,ourﬁnalapproximationisgivenby

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Notethatifweweretousethesameapproximationinpart(b)thesolutionto(b)would notchange.Thisfollowsfromthefactthatthephasedelayandgroupdelayforasystem withanimpulseresponse δ(t ∆)areequal.

Solution2.14

(a)Calculatethegroupdelayoftheﬁlter H(ejΩ)andsketch w[n]and |W (ejΩ)|

Wearetoldthattheﬁrstﬁlter h[n]anditsassociatedfrequencyresponse H(ejΩ)are purelyreal.Fromthisinformationwemayconcludethattheimpulseresponse h[n]is evensymmetricabout0,i.e. h[ n]= h[n]forall n,andconsequentlythatthephase ∠H(ejΩ)iszeroforallΩ ∈ [ π,π).Therefore, τg(Ω)=0, π ≤ Ω <π.

Themagnitudeof H(ejΩ)issuchthatthehighfrequencypulseassociatedwith x2[n 150] willbeattenuatedby 60dBormorewhilethelowfrequencypulseassociatedwith x1[n] ispassedwithnomagnitudechange.Thehighfrequencypulsehasastrengthofabout 20dBandthuswillhaveamagnitudeofapproximately 40dBintheoutput.Therefore, theoutput w[n]isapproximatelygivenby w[n] ≈ x1[n]+0 001x2[n 150].Arough approximationtothisisdepictedintheﬁgurebelow.

(b)Sketchwhattheoutput y[n]willlooklike.

Oursketchof y[n]willbefocusedontheeﬀectoftheall-passﬁlterontheenvelopeofthe waveform x1[n]andnotthephaseshiftofthehigherfrequencycarrier.Tothisend,we approximatethegroupdelayatΩ=0.25π fromthephasecurveinFigureP2.14-5as:

=40samples.

Therefore,weconcludethat y[n] ≈ x1[n 40]asisdepictedintheﬁgurebelow.The signal x1[n]isalsodepictedforcomparison. n (samples)

Solution2.15

TheoutputofﬁlterAforthedepictedinput x[n]is y6[n].

ThesharppeaksintheFouriertransformmagnitudeoftheinputsignal x[n]corresponding tohighfrequencycontent,i.e.thepeaksclosesttoΩ= π,areattenuatedbyapproximately 100dBormoreandthustheoutput y[n]oftheﬁlterwillessentiallynotcontainsignalcontent atthesefrequencies.Thefrequencyresponsemagnitudeoftheﬁlterisalsosuchthatthefrequencycontentassociatedwiththeslowandmediumvaryingpulsespassessentiallyuntouched inmagnitude.Fromthisanalysiswecaneliminatethepossibleoutputs yi[n]for i =1, 3, 4, 5, and7sincetheyeachcontainthehighfrequencypulse.

Thegroupdelayexperiencedbythelowfrequencypulseisapproximately220samples.If weapproximatethecenterofthispulseat n =100thenthepulseshouldappearintheoutput centeredat n =320.Thegroupdelayexperiencedbythemediumfrequencypulseisapproximately60samples.Ifweapproximatethecenterofthispulseat n =170thenthepulseshould appearintheoutputcenteredat n =230.Theonlyﬁlteroutput yi[n]ofthoseremaining consistentwiththeseobservationsis y6[n].

Solution2.16

Theimpulseresponses h1[n]and h2[h]arebothzerobeforetime n =20whichmanifestsitself inthegroupdelaycurvesasanadditiveoﬀsetof20samples.Weconcludefromthisobservation thatwemustselectbetween A and C.Ascomparedwith h2[n],theimpulseresponse h1[n] appearstocontainlowerfrequencycontentthatemergeslater,atatimethatcorrespondsto theassociatedgroupdelayof40inplot C.Ontheotherhand, h2[n]hashigher-frequency contentthatdominatesaroundtheassociatedgroupdelayof60inplot A.

(1)Impulseresponse h1[n]correspondstogroupdelayplot C

(2)Impulseresponse h2[n]correspondstogroupdelayplot A.

Theremainingimpulseresponses h3[n]and h4[h]mustcorrespondtochoices B and D.The impulseresponse h3[n]endswithalow-frequencycomponentthatdominatesaroundtheassociatedgroupdelayof10inplot D,whereas h4[n]hadahigher-frequencycomponentdominating ataroundthecorrespondinggroupdelayof10inplot B

(3)Impulseresponse h3[n]correspondstogroupdelayplot D.

(4)Impulseresponse h4[n]correspondstogroupdelayplot B.

Solution2.17

(i)Theimpulseresponse h3(t)correspondstogroupdelayplot B

Afeatureoftheimpulseresponse h3(t)thatclearlydistinguishesitfromtheothersis thatitiszeroforall t approximatelylessthan0 01,whichmanifestsitselfinthegroup delaycurveasanadditiveoﬀsetof0 01seconds.

(ii)Theimpulseresponse h4(t)correspondstogroupdelayplot A.

Theimpulseresponse h4(t)exhibitsstrongerhigh-frequencyoscillationsthatpersistto timesaround0.06,andtheseareunaccompaniedbylow-frequencyoscillations.Contrast thiswith h1(t),forwhichthelow-frequencyandhigh-frequencyoscillationshaveequal groupdelaysandarebothvisibleouttotimearound0.07.

(iii)Theimpulseresponse h2(t)correspondstogroupdelayplot C.

Notethat h2(t),incontrastto h4(t),exhibitsstrongerlow-frequencyoscillationsthat persisttotimesaround0.08,andtheseareunaccompaniedbyhigh-frequencyoscillations.

(iv)Theimpulseresponse h1(t)correspondstogroupdelayplot D.

Theimpulseresponse h1(t)appearstohaveanequalmixoflowandhighfrequenciesin itsimpulseresponse,andtheybothsettlebytimesaround0.02.

Solution2.18

(a)Express Sc(jω)intermsof Sd(ejΩ)andgiveafullylabeledsketchof Sc(jω)intheinterval |ω| < 2π × 2000rad/sec.

RecallfromthediscussionofperiodicsamplingandreconstructioninChapter1that Sc(jω)= TSd(ejΩ)|Ω=ωT |ω|≤ π T 0, elsewhere

Usingthisequationweassemblethesketchof Sc(jω)depictedbelow.Noticetheperiodic replicationsdonotpersistinthe ω domainduetotheinterpolatingLPFinthereconstructionprocess.Thisisrelevantsinceweareaskedtoproducetheplotfor |ω| < 2π × 2000 and π T =2π × 1000.

(b)Drawadetailedsketchof Xc(jω)for |ω| < 2π × 2000rad/sec.

Theeﬀectofthesinusoidtranslatestwoduplicatesofthespectrumof Sc(jω)scaledby π totheCTequivalentof ± 4π 5 .Thisisdepictedintheﬁgurebelow.

(c)Giveatime-domainexpressionfor xc(t)intermsof sc(t).

Wearegiven x[n]= sd[n]cos 4π 5 n .Goingthroughthe D/C converterwith T =0 5ms translatestheDTfrequencyΩ= 4π 5 totheCTfrequency ω =2π × 800.Theoutputof theD/Cdueto sd[n]willbedenoted sc(t).Therefore, xc(t)= sc(t)cos(1600πt).

(d)Determineanapproximatetime-domainexpressionfor yc(t)and yd[n]intermsof sc(t).

Indetermininganexpressionfor yc(t)wemaketheapproximationsdiscussedinSection2.2.1.Inparticular,weapproximatetheoutputoftheﬁlterby

c(t) ≈ sc (t τg (2π × 800))cos(2π × 800(t τp (2π × 800))) .

Wejustifytheseapproximationsfromthesketchinpart(b).Inparticular,thefigure illustratesthatthespectrumof sc(t)issufficientlyconcentratedandsmallascompared withthecarriersfrequencyof2π × 800.FromthephaseandgroupdelaycurvesinFigure P2.18-3wefindthat τg(2π × 800)=0.25msand τp(2π × 800)= 1 2π×800 .Finally,weobtain theexpression

Usingtherelation yd[n]= yc(nT )for T =0 5msweobtaintheexpression

[n]=

(e)Themappingfrom xd[n]to yd[n]doesindeedcorrespondtoanLTIsystemsinceno aliasingisintroducedduringeitherofthedomainconversionsystems.Subsequently,the frequencyresponse H(ejΩ)isrelatedto Hc(jω)via

Thegroupdelayof H(ejΩ)atΩ= 4π 5 is 1 2 sample,asisillustratedintheexpression for yd[n]inpart(d).Wecandevelopaniceinterpretationforwhatthismeansusing theproceduredevelopedinthisproblem.Namely,theactionofadiscretetimesystem correspondingtoahalf-sampledelayisequivalentlythoughtofasperiodicallysampling acontinuoustimesignalevery T secondswherethecontinuoustimesignalisa T 2 delayed versionofthebandlimitedinterpolation(usingparameter T )oftheoriginaldiscretetime signal.Noticethatthisinterpretationdoesnotdependonthevaluetakenby T solong astheD/CandC/Dbothusethesamevalueandthatthetime-shiftisby T 2

Solution2.19

ThisproblempertainstoaCTall-passsystemoftheform

(a)Drawthepole-zeroplotfor HAP (s)with M =1and bk real.

Thepole-zeroplotfor HAP (s)isdepictedbelow.ConsistentwiththeconventionestablishedbyEq.(1),thepoleisfoundinthe s-planeatthevalue s = bk whilethezerois foundat s = b∗ k.Inourillustrationwehaveselected b< 0hencethepoleiscontained withinthestrictlefthalfplane.Further,since b ischosentoberealwehavethat b∗ = b

(b)Inthispartwelet bk becomplexvalued.However,inordertoensurethesystem HAP (s) isbothcausalandstable,werequireRe(b) < 0,asisdiscussedin(a).Wenowarguethe factthatthegroupdelayofsuchasystemisalwayspositiveateveryfrequency.

Toseethisgeometrically,considertheexamplepole-zeroplotintheﬁgurebelowwith M =1and b = 1+ j.Recallthatthefrequencyresponsephaseatanyvalue ω isthe angleofthevectorfromthezeroto s = jω (shownwithadottedline)minustheangle ofthevectorfromthepoleto s = jω (alsoshownwithadottedline).Thegroupdelay correspondstothenegativerateofchangeofthesevectorsaswemove ω.Sincethepole isalwaysinthelefthalfplaneandthezeroisintherighthalfplaneitisstraightforward toseethatthecontributionofeachispositive,hencetheoverallgroupdelayispositive atallfrequencies.

(c)Forageneralstable,causalCTall-passsystemoftheformofEq.(1),wecanconclude thatthegroupdelayisalwayspositiveforeachfrequency.From(b)weconcludethat for M =1thisistrue.Toextendourreasoningto M> 1werelyuponthelinearityof thephaseresponseandthederivativerequiredtocomputethegroupdelay.Recallthat thephaseresponsefor M> 0isthesumofthephaseresponsesofeachfactordescribed in(b).Sincethederivativeisalsoalinearoperator,thegroupdelayisthesumofthe groupdelayseachfactorinEq.(1),eachsatisfying(b).Finally,sincethesumofpositive numbersisnecessarilyapositivenumberweconcludethatthegroupdelayofacausal, stable,all-passCTsystemispositiveeverywhere.

Solution2.20

Wearegivenaninputsignal x(t)oftheform x(t)= s(t)cos(ω0t)where s(t)= e t2 T 2 .The ﬁgurebelowsketchestheFouriertransform X(jω)ofthesignal x(t).Thewidthof S(jω)is overexaggeratedforclarity;thefrequency ω0 =4π × 1014 isabout108 timesgreaterthan

π T = π × 106.Weremarkthatthesignal s(t)isaGaussianfunctionwhichhasaFourier transformwhichisalsoaGaussianfunction.

Wenowmakeuseofthefactthattheresponseoftheopticalfibertotheinputsignalcos(ωt) isgivenby10 α(ω)L cos(ωt β(ω)L).Referringtotheinputsignal x(t),thebandwidthof envelopesignal s(t)issufficientlysmallascomparedto ω0 sothattheoutputoftheoptical fiberisapproximatedby

y(t)= s(t τg(ω0))10 α(ω0)L cos(ω0

where τg(ω0)isthegroupdelayoftheﬁberat ω0.Wenowjustifythisformula.Thestandard approximationformulawhenthefrequencyresponseisall-pass(Eq.(2.23a)orEq.(2.23b)),is

y(t)= s(t τg(ω0))cos(ω0 (t τp(ω0))) .

Comparingthesetwoequationsweseethat β(ω0)L playstheroleof ω0τp(ω0).Usingthefact that τp(ω0)= ∠HL(jω0) ω0 weconcludethat

∠HL(jω0)= β(ω0)L

Moreover,thefunction β(ω0)isessentiallylinearoverthefrequencybandofinterest,asis depictedbythecurve dβ(ω) dω inFigureP2.20-2.So,fromtheperspectiveofourinputsignal theopticalﬁberphaseresponseislinearhencetheﬁberactsasasimpledelaywithmagnitude 10 α(ω)L.Withthisapproximationwecomputethegroupdelayas τg(ω0)= τp(ω0)sincelinear phasesystemshaveequalphaseandgroupdelays.(Seeproblem2.1(a)foranexampleofthis). Thereforewehavethat τg(ω0)= β(ω0)L ω0 .Wealsoobtainthevalues α(ω0)=0 1and β(ω0)=109 fromFigureP2.20-2andaregiventhat L =10and T =10 6.Finally,wewritetheoutputas:

Notethattheoutputisequivalentlyexpressedas

MassachusettsInstituteofTechnology DepartmentofElectricalEngineeringandComputerScience

6.011: Communication,Control andSignalProcessing

Spring2015

ProblemSet2Solutions StateSpaceModels

Issued: Thursday,26February2015

Problem2.1

(i)(a)Statevariablechoice:

Comments: lbarnes@mit.eduandverghese@mit.edu

Anaturalchoiceis q1 (t)=position y(t),and q2 (t)=velocity˙ y(t). State-spacemodel(stateevolutionequationsandinstantaneousoutputequation):

(b)Equilibriumvaluesofstatevariableswhen x(t) ≡ 8: Set q1 (t) ≡

and

,andaccordinglyset˙

precedingstate-spacemodel.Thisimmediatelyyields

(c)Linearizedstate-spacemodelattheequilibriumabove: Thekeystephereistolinearizethenonlinearterm y3 (t)or q3 1 (t).Notethat d(y3 )= 3y2 dy,sowith

(andsimilarnotationforalltheotherperturbationsfromequilibrium)weget,for smallperturbations,

Thelinearizedmodelisthen

(ii)Nowwearesupposingthatthegivensystemisdescribedby

) .

(a)Theobviousstatevariableschosenin(i)willnotworkasthereisnowaytoaccount forthe dx(t) dt term.Notethatwecannothavedifferentialsinthefunctions fi because ˙ q(t)= f (q(t),x(t))mustfitthestandardformforasystemoffirstorderODE’s.

Thechoice q1 (t)= y(t)and q2 (t)=˙ y(t) x(t)gives

(b)Setting˙ q1 (t)=˙ q2 (t)=0and x(t) ≡ 8weget q1 =2and q2 = 8 (c)Again,theonlynonlineartermis q3 1 (t)and q1 hasnotchanged.Thuswehave

Problem2.2

Usingthegiveninstantaneousoutputequation,wecanwrite

whichwillbecometheinstantaneousoutputequationfortheinversesystem.Substitutingthe aboveequationinthegivenstateevolutionequation,weobtain

, whichisnowastateevolutionequationdrivenby y(t)ratherthan x(t),butwiththesamestate vector q(t)asbefore.Thus,

Problem2.3

(a)Assumingtemperatureisheldconstant,thereactionrates kf , kr ,and kc arealsoconstant. Inthiscasethemodelistime-invariant.However,themodelisnonlinearbecausethe q1 (t)q2 (t)termsensurethatsomeofthe˙ qi(t)cannotbewrittenasalinearcombination ofthestatevariablesandtheinput.

(b)Firstwelookattherateofchange

Thismeansthat q2 (t)+ q3 (t)willstayconstantatitsinitialvalue q2 (0)+ q3 (0).

(c)Nowweareassumingthat x(t) ≡ 0andwewouldliketoﬁndtheequilibriumvalues¯ qi in termsof¯ q2 +¯ q3 = E0 > 0.Setting q(t)=0thestateevolutionequationsbecome 0= kf q1 q2 +

r q3 0= kf q1 q2 +(kr + kc)¯ q3 0= kf q1 q2 (kr + k

Immediatelywegetthat

andthus

and

Finallynotethattheequilibriumvalueof q4 (t)willdependonthedynamicsofthereaction asitapproachesequilibrium.Namely,

Thelinearizedmodelaroundthisequilibriumwillhave

(d)Supposenowwearelookingforanequilibriumwhen x(t) ≡ x> 0.Thistimethestate evolutionequationsgive

Again¯ q3 =0and¯ q2 = E0 .Thistime,however,(2)and(3)requirethat¯ q1 =0while(1) requires¯ q1 =¯x/(E0 kf ) =0.Thismeansthatnofullequilibriumispossible! Ifweonlyrequirethattheﬁrstthreestatevariablesareinequilibriumweinsteadget

Solvingthissystemgives

Therateofchangeof

3 (t)whichinthispartialequilibirumissimply kcq3 =¯ x.

Problem2.4

(a)ThefollowingﬁgureshowstheresponseoftheSIRmodelforavarietyofinitialconditions. Inallcasestheparametersusedwere P =10000, β = 01, γ = 2, ρ = 1,and x[n]=¯ x = 2.Thebluecurvesshow s[n]andtheredcurvesshowthecorresponding i[n].

(b)Inthesimulationsfrom(a)weseethat i[n]settlestoapproximately227.3forlargevalues of n.Thisisconsistentwiththecomputedequilibriumvaluei.e.

Thematrix AEE haseigenvalues

withassociatedeigenvectors

Notethatsincetheeigenvalueshavemagnitudelessthanone,thelinearizedmodelisa stablesystem.

(d)Ifweusethefeedbackrule x[n]= gi[n]/P ,thentheendemicequilibriumvalueof i changes alongwiththegain g.See,forexample,theplotwith g =1andvariousinitialconditions below.

Wecanﬁndthenewendemicequilibriumvaluesbyplugging x[n]= gi[n]/P intothe stateevolutionequations:

Theendemic(i.e.nonzero)solutionfor i is

whichforourvaluesgives i =375asintheplot.Notethatas g increases,thevalueof i willdecrease.However,thereisnopointatwhichahighenoughgainwillgetridofthe endemicequilibriumentirely.