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Discrete-timeSignalsandSystems
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ii
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Discrete-timeSignalsandSystemsAnOperatorApproach
SanjoyMahajanandDennisFreemanMassachusettsInstituteofTechnology
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TypesetinPalatinoandEulerbytheauthorsusingConTEXtandPDFTEX
Copyright2009SanjoyMahajanandDennisFreemanSourcerevision:66261db0f9ed+ (2009-10-18 13:33:48 UTC)Discrete-timeSignalsandSystemsbySanjoyMahajanandDennisFreeman(authors)and??(publisher)islicensedunderthe. . . license.
C
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Briefcontents
Preface ix1 Differenceequations 12 Differenceequationsandmodularity 173 Blockdiagramsandoperators:Twonewrepresentations 334 Modes 515 Repeatedroots 636 Theperfect(sine)wave 717 Control 838 Proportionalandderivativecontrol 95
Bibliography 105Index 107
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vi
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Contents
Preface ix1 Differenceequations 1
1.1 Rabbits 21.2 Leakytank 71.3 Fallofafogdroplet 111.4 Springs 14
2 Differenceequationsandmodularity 172.1 Modularity:Makingtheinputliketheoutput 172.2 Endowmentgift 212.3 Rabbits 25
3 Blockdiagramsandoperators:Twonewrepresentations 333.1
Disadvantages
of
difference
equations
34
3.2 Blockdiagramstotherescue 353.3 Thepowerofabstraction 403.4 Operationsonwholesignals 413.5 Feedbackconnections 453.6 Summary 49
4 Modes 514.1 GrowthoftheFibonacciseries 524.2 TakingoutthebigpartfromFibonacci 554.3 Operatorinterpretation 574.4 Generalmethod:Partialfractions 59
5 Repeatedroots 635.1 Leaky-tankbackground 645.2 Numericalcomputation 655.3 Analyzingtheoutputsignal 67
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viii
5.4 Deformingthesystem:Thecontinuityargument 685.5 Higher-ordercascades 70
6 Theperfect(sine)wave 716.1 ForwardEuler 726.2 BackwardEuler 766.3 Leapfrog 796.4 Summary 82
7 Control 837.1 Motormodelwithfeedforwardcontrol 837.2 Simplefeedbackcontrol 857.3 Sensordelays 877.4 Inertia 90
8 Proportionalandderivativecontrol 958.1 Whyderivativecontrol 958.2 Mixingthetwomethodsofcontrol 968.3 Optimizingthecombination 988.4 Handlinginertia 998.5 Summary 103Bibliography 105Index 107
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Preface
Thisbookaimstointroduceyoutoapowerfultoolforanalyzinganddesigningsystemswhetherelectronic,mechanical,orthermal.ThisbookgrewoutoftheSignalsandSystemscourse(numbered6.003)thatwehavetaughtonandofftoMITsElectricalEngineeringandComputerSciencestudents.The traditional signals-and-systems course for example [17] emphasizestheanalysisofcontinuous-timesystems,inparticularanalogcircuits.However,mostengineerswillnotspecializeinanalogcircuits.Rather,digitaltechnologyofferssuchvastcomputingpowerthatanalogycircuitsareoftendesignedthroughdigitalsimulation.Digitalsimulationisaninherentlydiscrete-timeoperation. Furthermore,almostallfundamentalideasofsignalsandsystemscanbetaughtusingdiscrete-timesystems. Modularityandmultiplerepresentations , forexample,aidthedesignofdiscrete-time(orcontinuous-time)systems.Similarly,theideasformodes,poles,control,andfeedback.Furthermore,byteachingthematerialinacontextnotlimitedtocircuits,weemphasizethegeneralityofthesetools.Feedbackandsimulationaboundinthenaturalandengineeredworld,andwewouldlikeourstudentstobeflexibleandcreativeinunderstandinganddesigningthesesystems.Therefore,webeginour SignalsandSystemscoursewithdiscrete-timesystems,andgiveourstudentsthisbook. Afundamentaldifferencefrommostdiscussionsofdiscrete-timesystemsistheapproachusingoperators.Operatorsmakeitpossibletoavoidtheconfusingnotionoftransform.Instead,theoperatorexpressionforadiscrete-timesystem,andthesystemsimpulse response are two representations for the same system; theyarethecoordinatesofapointasseenfromtwodifferentcoordinatesystems.Then a transformationofa system hasanactive meaning: for example,composingtwosystemstobuildanewsystem.
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x Preface
HowtousethisbookAristotlewastutortotheyoungAlexanderofMacedon(later,theGreat).As ancient royalty knew, a skilled and knowledgeable tutor is the mosteffectiveteacher[3]. Askilledtutormakesfewstatementsandasksmanyquestions,forsheknowsthatquestioning,wondering,anddiscussingpromote long-lasting learning. Therefore, questions of two types are interspersedthroughthebook:questionsmarkedwitha inthemargin: Thesequestionsarewhatatutormightaskyouduringatutorial,andaskyoutoworkoutthenextstepsinananalysis. Theyareansweredinthesubsequenttext,whereyoucancheckyoursolutionsandmyanalysis.numbered
questions:
These
problems,
marked
with
a
shaded
background,
arewhata tutormightaskyou totakehomeaftera tutorial. Theygivefurtherpracticewiththetooloraskyoutoextendanexample,useseveraltoolstogether,orresolveparadoxes.Trylotsofquestionsofbothtypes!
CopyrightlicenseThisbookislicensedunderthe. . . license.AcknowledgmentsWegratefullythankthefollowingindividualsandorganizations:Forsuggestionsanddiscussions:. . . For thefree softwarefor typesetting: Donald Knuth (TEX); Han The Thanh(PDFTEX);HansHagenandTacoHoekwater(ConTEXt);JohnHobby(Meta-Post); AndyHammerlindl,JohnBowman,andTomPrince(asymptote);Richard
Stallman
(emacs);
the
Linux
and
Debian
projects.
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1Differenceequations
1.1 Rabbits 21.2 Leakytank 71.3
Fall
of
afog
droplet
11
1.4 Springs 14
Theworld is toorichand complex for ourminds to grasp itwhole, forourmindsarebutasmallpartoftherichnessoftheworld. Tocopewiththecomplexity,wereasonhierarchically. Wedividetheworldintosmall,comprehensiblepieces:systems.Systemsareubiquitous:aCPU,amemorychips,amotor,awebserver,ajumbojet,thesolarsystem,thetelephonesystem,oracirculatorysystem. Systemsareausefulabstraction,chosen
becausetheirexternalinteractionsareweakerthantheirinternalinteractions.Thatpropertiesmakesindependentanalysismeaningful.Systemsinteractwithothersystemsviaforces,messages,oringeneralviainformationorsignals. Signalsandsystemsisthestudyofsystemsandtheirinteraction.Thisbook studies only discrete-time systems, where timejumps ratherthanchangescontinuously. Thisrestriction isnotassevereas itsseems.First,digitalcomputersare,bydesign,discrete-timedevices,sodiscrete-time
signals
and
systems
includes
digital
computers.
Second,
almost
all
theimportantideasindiscrete-timesystemsapplyequallytocontinuous-timesystems.Alas,evendiscrete-timesystemsaretoodiverseforonemethodofanalysis.Thereforeeventheabstractionofsystemsneedssubdivision.Theparticularclassofso-calledlinearandtime-invariantsystemsadmitspowerfultoolsofanalysisanddesign.Thebenefitofrestrictingourselvestosuch
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2 1.1 Rabbits
systems,andthemeaningoftherestrictions,willbecomeclearinsubsequentchapters.
1.1 RabbitsHere isFibonaccisproblem [6, 10], a famousdiscrete-time, linear, time-invariantsystemandsignal:
Acertainmanputapairofrabbitsinaplacesurroundedonallsidesbyawall.How many pairs of rabbits canbe produced from that pair in a year if it issupposedthateverymontheachpairbegetsanewpairwhichfromthesecondmonthonbecomesproductive?
1.1.1 MathematicalrepresentationThissystemconsistsoftherabbitpairsandtherulesofrabbitreproduction.Thesignal is thesequence fwhere f[n] is the number of rabbit pairs atmonthn(theproblemasksaboutn= 12).Whatisfinthefirstfewmonths?Inmonth0, onerabbitpair immigrates into thesystem: f[0] = 1. Letsassumethattheimmigrantsarechildren.Thentheycannothavetheirownchildren
in
month
1
they
are
too
young
so
f[1] = 1.But
this
pair
is
an
adultpair,soinmonth2thepairhaschildren,makingf[2] = 2.Findingf[3] requiresconsideringtheadultandchildpairsseparately(hierarchicalreasoning),becauseeachtypebehavesaccordingtoitsownreproductionrule.Thechildpairfrommonth2growsintoadulthoodinmonth3,andtheadultpairfrommonth2begetsachildpair. Soinf[3] = 3: twoadultandonechildpair.The two adultpairs contribute two child pairs in month4, and the onechildpairgrowsup,contributinganadultpair.Somonth4hasfivepairs:two
child
and
three
adult
pairs.
To
formalize
this
reasoning
process,
define
twointermediatesignalscanda:
c[n] = numberofchildpairsatmonthn;a[n] = numberofadultpairsatmonthn.
Thetotalnumberofpairsatmonthnisf[n] = c[n] + a[n].Hereisatableshowingthethreesignalsc,a,andf:
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31 Differenceequations
n 0 1 2 3c 1 0 1 1a 0 1 1 2f 1 1 2 3
Thearrowsinthetableshowhownewentriesareconstructed.Anupwarddiagonalarrowrepresentstheonlymeanstomakenewchildren,namelyfromlastmonthsadults:
a[n 1] c[n] or c[n] = a[n 1].Ahorizontalarrowrepresentsonecontributiontothismonthsadults,thatadultslastmonthremainadults:a[n 1]
a[n]. Adownwarddiagonal
arrow represents the other contribution to this months adults, that lastmonthschildrengrowupintoadults:c[n 1] a[n].Thesumofthetwocontributionsis
a[n] = a[n 1] + c[n 1].Whatisthedifferenceequationforfitself?Tofindtheequationforf,onehasatleasttwomethods:logicaldeduction(Problem1.1)ortrialanderror. Trialanderrorisbettersuitedforfindingresults,andlogicaldeductionisbettersuitedforverifyingthem.Therefore,usingtrialanderror,lookforapatternamongadditionsamplesoff:
n 0 1 2 3 4 5 6c 1 0 1 1 2 3 5a 0 1 1 2 3 5 8f 1 1 2 3 5 8 13
Whatusefulpatternsliveinthesedata?Oneprominentpatternisthatthesignalsc,a,andflooklikeshiftedversionsofeachother:
a[n] = f[n 1];c[n] = a[n 1] = f[n 2].
Sincef[n] = a[n] + c[n],
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4 1.1 Rabbits
f[n] = f[n 1] + f[n 2].withinitialconditionsf[0] = f[1] = 1.Thismathematicaldescription, orrepresentation, clarifiesapoint that isnotobvious intheverbaldescription: thatthenumberofrabbitpairs inany month depends on the number in the two preceding months. Thisdifferenceequationissaidtobeasecond-orderdifferenceequation. Sinceits coefficientsare all unity, and thesigns arepositive, it is thesimplestsecond-orderdifferenceequation.Yetitsbehaviorisrichandcomplex.
Problem 1.1 VerifyingtheconjectureUsethetwointermediateequations
c[n] = a[n 1],a[n] = a[n 1] + c[n 1];
andthedefinitionf[n] = a[n] + c[n] toconfirmtheconjecturef[n] = f[n 1] + f[n 2].
1.1.2 Closed-formsolutionTherabbitdifferenceequationisanimplicitrecipethatcomputesnewvaluesfromoldvalues. Butdoesithaveaclosedform: anexplicitformulaforf[n] thatdependsonnbutnotonprecedingsamples? Asasteptoward finding a closed form, lets investigate how f[n] behaves as nbecomeslarge.Doesf[n] growlikeapolynomialinn,likealogarithmicfunctionofn,orlikeanexponentialfunctionofn?Decidingamongtheseoptionsrequiresmoredata. Herearemanyvaluesoff[n] (startingwithmonth0):
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,...
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51 Differenceequations
Thesamplesgrowquickly.Theirgrowthistoorapidtobelogarithmic,unlessf[n]isanunusualfunctionlike(logn)20.Theirgrowth
is
probably
also
too
rapid
for
f[n]
tobe a polynomial in n, unless f[n] is na high-degree polynomial. A likely alternative isexponentialgrowth. To testthathypothesis,usepictorialreasoningbyplottinglnf[n]versusn. Theplottedpointsoscillateaboveandbelowabest-fitstraightline. Thereforelnf[n]growsalmostexactlylinearlywithn,andf[n]isapproximatelyanexponentialfunctionofn:
f[n]Azn,wherezandAareconstants.
lnf[n]
Howcanzbeestimatedfromf[n]data?n f[n]/f[n1]10 1.6181818181818
best-fitlineasngrows,theexponentialapprox-Becausetheplottedpointsfalleverclosertothe 20 1.6180339985218imation f[n] Aznbecomes more exact as n 30 1.6180339887505
40 1.6180339887499grows.Iftheapproximationwereexact,thenf[n]/f[n1]wouldalwaysequalz, so f[n]/f[n1]becomes 50 1.6180339887499anevercloserapproximationtozasnincreases.Theseratiosseemtoconvergetoz=1.6180339887499.Itsfirstfewdigits1.618mightbefamiliar. Foramemoryamplifier,feedtheratiototheonlineInverseSymbolicCalculator(ISC).Givenanumber,it guesses its mathematical source. When given the Fibonacci z, the InverseSymbolicCalculatorsuggeststwoequivalentforms: thatzisarootof1xx2 orthatitis (1+ 5)/2. Theconstantisthefamousgoldenratio[5].Therefore,f[n] An. Tofindtheconstantof
n f[n]/f[n1]proportionality A, takeout thebigpartbydividing f[n]by n. These ratios hover around0.723...,soperhapsAis31. Alas,exper 10
20 0.72362506926472
0.72360679895785
imentswithlargervaluesofnstronglysuggestthatthedigitscontinue0.723606... whereas3
3040
0.723606797750060.72360679774998
1 = 0.73205.... Abit of experimentation or 50 0.72360679774998the Inverse Symbolic Calculator suggests that0.72360679774998probablyoriginatesfrom/ 5.
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6 1.1 Rabbits
Thisconjecturehasthemeritofreusingthe 5alreadycontainedinthedefinitionof,soitdoesnotintroduceanewarbitrarynumber. WiththatconjectureforA,theapproximationforf[n] becomes
n+1f[n] .5
Howaccurateisthisapproximation?Totesttheapproximation,takeoutthebig
n r[n]/r[n 1]partbycomputingtheresidual:
2 0.61803398874989601r[n] = f[n] n+1/ 5. 3 0.61803398874989812
4 0.61803398874988624The
residual
decays
rapidly,
perhaps
expo 5 0.61803398874993953nentially.Thenrhasthegeneralform 6 0.61803398874974236
r[n] Byn, 7 0.618033988750294148 0.61803398874847626
wherey and B are constants. To findy, 9 0.61803398875421256computetheratiosr[n]/r[n 1].Theycon 10 0.61803398873859083verge to 0.61803..., which isalmostex-
nactly1 or1/.Thereforer[n] B(1/) .WhatistheconstantofproportionalityB?
nTocomputeB,divider[n] by(1/) . Thesevalues,ifnisnottoolarge(Problem1.2), almost instantlysettleson0.27639320225. With luck, thisnumbercanbeexplainedusingand 5. Afewnumericalexperimentssuggesttheconjecture
1 1B= .
5 Theresidualbecomes
n+11 1r[n] = .
5 ThenumberofrabbitpairsisthesumoftheapproximationAzn andtheresidualByn:
f[n] = 1 n+1 ()(n+1) .5
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71 Differenceequations
Howbizarre! TheFibonaccisignalfsplitsintotwosignalsinatleasttwoways. First, it is thesumof theadult-pairssignal aand thechild-pairssignalc.Second,itisthesumf1 + f2 wheref1 andf2 aredefinedby
1f1[n] n+1;51
f2[n] (1/)n+1.5
Theequivalenceofthesedecompositionswouldhavebeendifficulttopredict. Instead, many experiments and guesses were needed to establishtheequivalence. Anotherkindofquestion,alsohardtoanswer,arisesbychangingmerelytheplussignintheFibonaccidifferenceequationintoaminussign:
g[n] = g[n 1] g[n 2].Withcorrespondinginitialconditions,namelyg[0] = g[1] = 1,thesignalgrunsasfollows(staringwithn= 0):
1,1,0,1,1,0,1,1,0,1,1, 0 , . . . . oneperiod
Ratherthangrowingapproximatelyexponentially,thissequenceisexactlyperiodic. Why? Furthermore,ithasperiod6. Why? Howcanthisperiod
bepredictedwithoutsimulation?Arepresentationsuitedforsuchquestionsisintroducedin??. Fornow,letscontinueinvestigatingdifferenceequationstorepresentsystems.
Problem 1.2 Actualresiduallnr[n]
nHereisasemiloggraphoftheabsoluteresidual|r[n]|computednumericallyupton=80. (Theabsoluteresidualisusedbecausethe residual is often negative and wouldhaveacomplexlogarithm.) Itfollowsthepredictedexponentialdecayforawhile,butthenmisbehaves.Why?
1.2 LeakytankIntheFibonaccisystem,therabbitschangedtheirbehaviorgrewuporhadchildrenonlyonceamonth.TheFibonaccisystemisadiscrete-time
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8 1.2 Leakytank
system. Thesesystemsaredirectlysuitableforcomputationalsimulationandanalysisbecausedigitalcomputers themselvesact likediscrete-timesystems. However, manysystems in the world such as pianostrings,earthquakes,
microphones,
or
eardrums
are
naturally
described
as
continuous-timesystems.
Toanalyzecontinuous-timesystemsusingdiscrete-timetoolsrequiresapproximations.Theseapproximationsareillustratedinthesimplestinterestingcontinuous-timesystem: aleakytank. Imagineabathtuborsinkfilledtoaheighthwithwater. Attimet= 0,thedrainisopenedandwaterflowsout. Whatisthesubsequentheightofthewater?Att= 0,thewaterlevelandthereforethepressureisatitshighest,sothewaterdrainsmostrapidlyatt= 0.Asthewaterdrainsandthelevelfalls,thepressureandtherateofdrainagealsofall.Thisbehavioriscapturedbythefollowingdifferentialequation:
h
leak
dh= f(h),
dtwheref(h) isanas-yet-unknownfunctionoftheheight.Finding f(h) requires knowing the geometry of the tub and piping andthen
calculating
the
flow
resistance
in
the
drain
and
piping.
The
simplest
modelforresistanceisaso-calledlinearleak: thatf(h) isproportionaltoh.Thenthedifferentialequationsimplifiesto
dhh.dt
Whatarethedimensionsofthemissingconstantofproportionality?Thederivativeontheleftsidehasdimensionsofspeed(heightpertime),sothemissingconstanthasdimensionsofinversetime. Calltheconstant1/,whereisthetimeconstantofthesystem.Then
dh h= .
dt
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91 Differenceequations
Analmost-identicaldifferentialequationdescribesthevoltageVonacapacitordischargingacrossaresistor:
dV 1= V.
dt RC
R
C
V
Itistheleaky-tankdifferentialequationwithtimeconstant= RC.
Problem 1.3 KirchoffslawsUseKirchoffslawstoverifythisdifferentialequation.
Approximatingthecontinuous-timedifferentialequationasadiscrete-timesystemenables thesystem tobesimulatedbyhandandcomputer. Inadiscrete-timesystem,timeadvancesinlumps.Ifthelumpsize,alsoknownasthetimestep,isT,thenh[n] isthediscrete-timeapproximationofh(nT). Imaginethatthesystemstartswithh[0] = h0. Whatish[1]? Inotherwords,whatisthediscrete-timeapproximationforh(T)?Theleaky-tankequationsaysthat
dh h= .
dt Att= 0thisderivativeish0/. Ifdh/dtstaysfixedforawholetimestepaslightlydubiousbutsimpleassumptionthentheheightfallsbyapproximatelyh0T/ inonetimestep.Therefore
T Th[1] = h0 h0 = 1 h[0].
Usingthesameassumptions,whatish[2] and,ingeneral,h[n]?The reasoning to compute h[1] from h[0] applies when computing h[2]fromh[1]. Thederivativeatn = 1equivalently,att = T is h[1]/.Thereforebetweenn = 1andn = 2equivalently,betweent = T andt= 2Ttheheightfallsbyapproximatelyh[1]T/tau,
T Th[2] = h1 h1 = 1 h[1].
Thispatterngeneralizestoaruleforfindingeveryh[n]:
Th[n] = 1 h[n 1].
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10 1.2 Leakytank
Thisimplicitequationhastheclosed-formsolutionn
Th[n] = h0 1 .
Howcloselydoesthissolutionreproducethebehavioroftheoriginal,continuous-timesystem?The original, continuous-time differential equation dh/dt = htau issolvedby
h(t) = h0et/.Atthediscretetimest= nT,thissolutionbecomes
nh(t) = h0enT/ = h0 eT/ .Thediscrete-timeapproximationreplaceseT/ with1 T/. ThatdifferenceisthefirsttwotermsintheTaylorseriesforeT/ : 2 3
eT/ = 1T +1 T 1 T + . . . . 2 6
Therefore the discrete-time approximation is accurate when the higher-ordertermsintheTaylorseriesaresmallnamely,whenT/
1.
Thismethodofsolvingdifferentialequationsbyreplacingthemwithdiscrete-timeanalogsisknownastheEulerapproximation,anditcanbeusedtosolveequationsthatareverydifficultorevenimpossibletosolveanalytically.
Problem 1.4 Whichistheapproximatesolution?
n
Here are unlabeled graphs showing the discrete-time samplesh[n] and thecontinuous-timesamplesh(nT), forn =0 . . . 6.Whichgraphshowsthediscrete-timesignal?Problem 1.5 LargetimestepsSketchthediscrete-timesamplesh[n] inthreecases:(a.)T =/2 (b.)T = (c.)T = 2 (d.)T = 3Problem 1.6 TinytimestepsShowthatasT0,thediscrete-timesolution
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111 Differenceequations
T nh[n] = h0 1 .
approachesthecontinuous-timesolution
h(t) = h0enT/.HowsmalldoesT havetobe,asafunctionofn,inorderthatthetwosolutionsapproximatelymatch?
1.3 FallofafogdropletTheleakytank(Section1.2)isafirst-ordersystem,anditsdifferentialequationanddifferenceequationarefirst-orderequations.However,thephysicalworldisoftensecondorderbecauseNewtonssecondlawofmotion,F= ma,containsasecondderivative.Forsuchsystems,howapplicableistheEulerapproximation?ToillustratetheissuesthatariseinapplyingtheEulerapproximationtosecond-ordersystems,letssimulatethefallofafogdropletactedonbygravity(F= mg)andairresistance. Afogdropletissmallenoughthatitsairresistanceisproportional to velocity rather than to the more usual velocity squared.Then the net downward force on the droplet is mg bv, wherev is itsvelocityandbisaconstantthatmeasuresthestrengthofthedrag.Intermsofpositionx,withthepositivedirectionasdownward,Newtonssecondlawbecomes
d2x dxm = mg b .
dt2 dtDividingbothsidesbymgives
d2x b dx= g .
dt2 m dtWhat
are
the
dimensions
of
b/m?
Theconstantb/mturnsthevelocitydx/dtintoanacceleration,sob/mhasdimensionsofinversetime.Thereforerewriteitas1/,wherem/bisatimeconstant.Then
d2x 1 dx= g .
dt2 dt
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12 1.3 Fallofafogdroplet
Whatisadiscrete-timeapproximationforthesecondderivative?Intheleaky-tankequation,
dh h= ,dt thefirst derivative at t = nT had the Eulerapproximation (h[n+ 1] h[n])/Tandh(t= nT) becameh[n].Thesecondderivatived2x/dt2 isthelimitofadifferenceoftwofirstderivatives.UsingtheEulerapproximationprocedure,approximatethefirstderivativesatt= nTandt= (n+ 1)T:
dx
x[n+ 1] x[n] ;
dt t=nT Tdx x[n+ 2] x[n+ 1] .dt t=(n+1)T T
Thend2x 1x[n+ 2] x[n+ 1] x[n+ 1] x[n] .dt2 t=nT T T T
Thisapproximationsimplifiestod2x
1 (x[n+ 2] 2x[n+ 1] + x[n]) .dt2 t=nT T2
The Euler approximation for the continuous-time equation of motion isthen
1 1 x[n+ 1] x[n](x[n+ 2] 2x[n+ 1] + x[n]) = g
T2 Tor
Tx[n+ 2] 2x[n+ 1] + x[n] = gT2 (x[n+ 1] x[n]).
Ouroldfriendfromtheleakytank, theratioT/,hasreappearedinthisproblem. To simplify the subsequent equations, define T/. Thenaftercollecting the like terms, thedifferenceequation for the falling fogdropletis
x[n+ 2] = (2 )x[n+ 1] (1 )x[n] + gT2.
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131 Differenceequations
Asexpected, thisdifferenceequation issecondorder. Like theprevioussecond-orderequation,theFibonacciequation,itneedstwoinitialvalues.Letstryx[1] = x[0] = 0.Physically,thefogdropletstartsfromrestatthereference
height
0,
and
at
t= 0
starts
feeling
the
gravitational
force
mg.
Foratypicalfogdropletwithradius10m,thephysicalparametersare:
m4.21012kg;b2.8109kgs1;1.5103s1.
Relativeto,thetimestepTshouldbesmall,otherwisethesimulationerrorwilllarge.Atimestepsuch as 0.1ms is a reasonable compromisebe- x[n]
(m)
2010
tweenkeepingreducingtheerrorandspeeding 0upthesimulation. Thenthedimensionlessratio 0 10 20 nis0.0675.Asimulationusingtheseparametersshows x initially rising faster than linearly, probably quadratically, thenrisinglinearlyatarateofroughly1.5mpertimestepor1.5cms1.Thissimulationresultexplainsthelongevityoffog.Fogis,roughlyspeaking,acloudthathassunktotheground. Imaginethatthiscloudreachesupto500m(atypicalcloudthickness). Then,tosettletotheground,thecloudrequires
500mtfall 9hours.
1.5cms1Inotherwords,fogshouldlastovernightinagreementwithexperience!
Counting timestepsHowmanytimestepswould the fog-dropletsimulationrequire(withT = 0.1ms)inorderforthedroplettofall500minthesimulation?Howlongwouldyourcomputer,oranothereasilyavailablecomputer,requiretosimulatethatmanytimesteps?
Problem 1.7 TerminalvelocityBysimulatingthefogequation
x[n+ 2] = (2 )x[n+ 1] (1 )x[n] + gT2.withseveralvaluesofTandtherefore,guessarelationbetweeng,T,,andtheterminalvelocityoftheparticle.
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14 1.4 Springs
1.4 SpringsNowletsextendoursimulationstothemostimportantsecond-ordersystem: thespring. Springsareamodelforavastnumberofsystemsinthenaturalandengineeredworlds: planetaryorbits,chemicalbonds,solids,electromagneticradiation,andevenelectronprotonbonds.Sincecolorresultsfromelectromagneticradiationmeetingelectronprotonbonds,grassisgreenandtheskyisbluebecauseofhowspringsinteractwithsprings.Thesimplestspringsystemisamassconnectedtoaspringandfreetooscillateinjustonedimension.Itsdifferentialequationis
d2xm + kx= 0,
dt2
wherexistheblocksdisplacementfromtheequilibriumposition,mistheblocksmass,andkisthespringconstant.Dividingbymgives
d2x k+ x= 0.
dt2 mDefiningtheangularfrequency k/mgivesthecleanequation:
d2x+ 2x= 0.
dt2NowdividetimeintouniformstepsofdurationT,andreplacethesecondderivatived2x/dt2 withadiscrete-timeapproximation:
d2x x[n+ 2] 2x[n+ 1] + x[n] ,dt2 t=nT T2
whereasusualthesamplex[n] correspondstothecontinuous-timesignalx(t) att= nT.Then
x[n+ 2] 2x[n+ 1] + x[n]+ 2x[n] = 0
T
2oraftercollectingliketerms,
x[n+ 2] = 2x[n+ 1] 1+ (T)2 x[n].DefiningT,
x[n+ 2] = 2x[n+ 1] 1+ 2 x[n].
km
x
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151 Difference equations
This second-order difference equation needs two initial values. A simplepair is x[0] = x[1] = x0. This choice corresponds to pulling the mass right-wards by x0, then releasing it at t = T. What happens afterward?To simulate the system numerically,one shouldchoose T to make small. As a reasonable small, try 100 samples per oscillation period: = name: dummy
file: shm-forward2/100 or roughly 0.06. Alas, the simulation pre-
state: unknowndicts that the oscillations grow to infinity.What went wrong?Perhaps , even 0.06, is too large. Here are two simulations with smallerat values of:
x x
t t
0.031 0.016These oscillations also explode. The only difference seems to be the rate ofgrowth (Problem 1.8).
Problem 1.8 Tiny values of Simulate
x[n + 2] = 2x[n + 1] 1 + 2 x[n]using very small values for . What happens?
An alternative explanation is that the discrete-time approximation of thederivative caused the problem. If so, it would be surprising, because thesame approximation worked when simulating the fall of a fog droplet. Butlets try an alternative definition: Instead of defining
dx x[n + 1] x[n] ,dt t=nT T
try the simple change todx x[n] x[n 1] .dt T
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Usingthesameprocedureforthesecondderivative,d2x
x[n] 2x[n 1] + x[n 2] .dt2 t=nT T2
Thediscrete-timespringequationisthen x(1+ 2)x[n] = 2x[n 1] x[n 2],
ort
2x[n 1] x[n 2]x[n] = .
1+ 2Usingthesameinitialconditionsx[0] = x[1] = 1,whatisthesubsequenttimecourse?Thebadnewsisthattheseoscillationsdecaytozero!However,thegoodnewsisthatchangingthede- xrivativeapproximationcansignificantlyaffectthe
behavior of thediscrete-time system. Lets try asymmetricsecondderivative: t
d2x x[n+ 1] 2x[n] + x[n 1] .dt2 t=nT T2
Thenthedifferenceequationbecomesx[n+ 2] = (2 2)x[n+ 1] x[n].
Nowthesystemoscillatesstably,justasaspringwithoutenergy lossorinputshouldbehave.Why did the simple change to a symmetric second derivative solve theproblem of decaying or growing oscillations? The representation of thealternativediscrete-timesystemsasdifferenceequationsdoesnothelpanswerthatquestion. Itsanswerrequiresthetwomost important ideas insignalsandsystems:operators(??)andmodes(??).
Problem 1.9 Differentinitialconditionsx
t
Herearethesubsequentsamplesusingthesymmetricsecondderivativeandinitialconditionsx[0] = 0,x[1] = x0. Theamplitude is, however, much largerthan x0. Is thatbehavior physically reasonable? Ifyes,explainwhy.Ifnot,explainwhatshouldhappen.
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2Differenceequationsandmodularit
2.1 Modularity:Makingtheinputliketheoutput 172.2 Endowmentgift 212.3
Rabbits
25
Thegoalsofthischapterare: toillustratemodularityandtodescribesystemsinamodularway; totranslateproblemsfromtheirrepresentationasaverbaldescrip
tionintotheirrepresentationasdiscrete-timemathematics(differenceequations);and
tostartinvestigatingthesimplestsecond-ordersystem,thesecond-simplestmoduleforanalyzinganddesigningsystems.
Thethemesofthischapteraremodularityandtherepresentationofverbaldescriptionsasdiscrete-timemathematics. Weillustratethesethemeswith twoexamples, money inahypotheticalMITendowment fundandrabbitsreproducinginapen,settingupdifferenceequationstorepresentthem. Therabbitexample,whichintroducesanewmoduleforbuildingandanalyzingsystems,isafrequentvisitortothesechapters.Inthischapterwebegintostudyhowthatmodulebehaves. Before introducingtheexamples,weillustratewhatmodularityisandwhyitisuseful.
2.1 Modularity:MakingtheinputliketheoutputAcommonbutalasnon-modularwaytoformulatedifferenceanddifferentialequationsusesboundaryconditions. Anexamplefrompopulation
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192 Differenceequationsandmodularity
meaningofphilosophicalasirrelevantbutansweringithelpsustounderstandanddesignsystems. HerethesystemistheUnitedStates. Theinputtothesystemisonenumber,theinitialpopulationp[2007];however,the
output
is
a
sequence
of
populations
p[2008], p[2009], . . ..
In
this
formulation,thesystemsoutputcannotbecometheinputtoanothersystem.
Therefore we cannot design large systemsby combining small, easy-tounderstand systems. Nor wecan we analyze large, hard-to-understandsystemsbybreakingthemintosmallsystems.Instead, wewould likeamodular formulation inwhich the input is thesame kind of object as the output. Here is the US-population questionreformulatedalongthoselines:Ifx[n] peopleimmigrateintotheUnitedstatesinyearn,andtheUSpopulationgrowsat1%annually,whatisthepopulationin
yearn
?The
input
signal
is
the
number
of
immigrants
versus
time,
so
it
is
asequenceliketheoutputsignal. Includingtheeffectofimmigration,therecurrenceis
p[n] = (1+ r)p[n 1] + x[n] . output reproduction immigration
Theboundaryconditionisnolongerseparatefromtheequation! Insteaditispartoftheinputsignal.Thismodularformulationisnotonlyelegant;itisalsomoregeneralthanistheformulationwithboundaryconditions,forwecanrecasttheoriginalquestionintothisframework. Therecastinginvolvesfindinganinputsignalheretheimmigrationversustimethatreproducestheeffectoftheboundaryconditionp[2007] = 3108.
Pausetotry 2. What input signal reproduces the effect of theboundarycondition?
Theboundaryconditioncanbereproducedwiththisimmigrationschedule(theinputsignal):
3108 ifn= 2007;x[n] = 0 otherwise.
ThismodelimaginesanemptyUnitedStatesintowhich300millionpeoplearriveintheyear2007. Thepeoplegrow(innumbers!) atanannualrate
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20 2.1Modularity:Makingtheinputliketheoutput
of1%,andwewanttoknowp[2077],theoutputsignal(thepopulation)intheyear2077.Thegeneralformulationwithanarbitraryinputsignalishardertosolvedirectlythanisthefamiliarformulationusingboundaryconditions,whichcanbesolvedbytricksandguesses.Forourinputsignal,theoutputsignalis
31081.01n2007 forn2007;p[n] = 0 otherwise.
Exercise 2. Checkthatthisoutputsignalsatisfiestheboundaryconditionandthepopulationequation.
In later chapters you learn how to solve the formulation with an arbitraryinputsignal. Hereweemphasizenotthemethodofsolutionbutthemodularformulationwhereasystemturnsonesignalintoanothersignal.Thismodulardescriptionusingsignalsandsystemshelpsanalyzecomplexproblemsandbuildcomplexsystems.Toseehowithelps,firstimagineaworldwithtwocountries:IrelandandtheUnitedStates.SupposethatpeopleemigratefromIrelandtotheUnitedStates,areasonablemodelinthe1850s. SupposealsothattheIrishpopulationhasanintrinsic10annualdeclineduetofaminesandthatanother10%ofthepopulationemigrateannuallytotheUnitedStates.IrelandandtheUnitedStatesaretwosystems,withonesystemsoutput(Irishemigration)feeding intotheothersystems input(theUnitedStatess immigration).Themodulardescriptionhelpswhenprogrammingsimulations.Indeed,giantpopulation-growthsimulationsareprogrammedinthisobject-orientedway. Eachsystemisanobjectthatknowshowitbehaveswhatitoutputswhen fed inputsignals. Theuserselectssystemsandspecifiesconnectionsamongthem. Fluid-dynamicssimulationsuseasimilarapproach
by
dividing
the
fluid
into
zillions
of
volume
elements.
Each
elementisasystem,andenergy,entropy,andmomentumemigratebetween
neighboringelements.Our one- or two-component population systems are simpler than fluid-dynamicssimulations, thebetter to illustratemodularity. Using twoexamples,wenextpracticemodulardescriptionandhowtorepresentverbaldescriptionsasmathematics.
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2 Differenceequationsandmodularity 21
2.2 EndowmentgiftThefirstexampleforrepresentingdescriptionsasmathematicsinvolvesahypotheticalendowmentgifttoMIT.Adonorgives107 dollarstoMITto support projects proposed and chosenby MIT undergraduates! MITwould like touse this fund fora long timeanddraw 0.5106 everyyearforaso-called5%drawdown. Assumethatthemoneyisplacedinareliableaccountearning4%interestcompoundedannually.HowlongcanMITanditsundergraduatesdrawonthefundbeforeitdwindlestozero?Nevermakeacalculationuntilyouknowroughlywhattheanswerwillbe! ThismaximisrecommendedbyJohnWheeler,abrilliantphysicistwhosemostfamousstudentwasMITalumRichardFeynman [9]. WehighlyrecommendWheelersmaximasawaytobuildintuition.Sohereareafewestimationquestionstogetthementaljuicesflowing. Startwiththebroadestdistinction,whetheranumberisfiniteorinfinite.Thisdistinctionsuggeststhefollowingquestion:
Pausetotry 3. Willthefundlastforever?
Alas, the fund will not last forever. In the first year, the drawdown isslightlygreaterthantheinterest,sotheendowmentcapitalwilldwindleslightly. Asaresult,thenextyearsinterestwillbesmallerthanthefirstyearsinterest. Sincethedrawdownstaysthesameat$500,000annually(whichis5%oftheinitialamount),thecapitalwilldwindlestillmoreinlateryears,reducingtheinterest,leadingtoagreaterreductionininterest,leadingtoagreaterreductionincapital. . . Eventuallythefundevaporates.Giventhatthe lifetime isfinite,roughlyhowlongisit? Canyourgreat-grandchildrenuseit?
Pauseto
try
4.
Will
the
fund
last
longer
than
or
shorter
than
100
years?
The figure of 100 years comes from the differencebetween the outflow theannualdrawdownof5%of thegiftand the inflowproducedbythe interest rate of 4%. The differencebetween 5% and 4% annually is
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22 2.2 Endowmentgift
= 0.01/year. Thedimensionsofareinversetime,suggestinganendowmentlifetimeof1/,whichis100years. Indeed, ifeveryyearwerelikethefirst,thefundwouldlastfor100years. However,theinflowfrominterest
decreases
as
the
capital
decreases,
so
the
gap
between
outflow
and
inflow increases. Thus this1/method,basedonextrapolating thefirstyearschangetoeveryyear,overestimatesthelifetime.Havingwarmedupwithtwoestimates, letsdescribethesystemmathematicallyandsolveforthetrue lifetime. Indoingso,wehavetodecidewhatistheinputsignal,whatistheoutputsignal,andwhatisthesystem.Thesystemistheleasttrickypart:Itisthebankaccountpaying4interest.Thegiftof$10millionismostlikelypartoftheinputsignal.
Pausetotry 5. Isthe$500,000annualdrawdownpartoftheoutputortheinputsignal?
Thedrawdownflowsoutof theaccount, and theaccount is thesystem,soperhapsthedrawdown ispartoftheoutputsignal. No!! Theoutputsignaliswhatthesystemdoes,whichistoproduceoratleasttocomputeabalance.Theinputsignaliswhatyoudotothesystem.Here,youmovemoneyinoroutofthesystem:
bankaccountmoneyinorout balance
Theinitialendowmentisaone-timepositiveinputsignal,andtheannualdrawdownisarecurringnegativeinputsignal. Tofindhowlongtheendowment lasts, find when the output signal crossesbelow zero. Theseissuesofrepresentationarehelpfultofigureoutbeforesettingupmathematics.Otherwisewithgreateffortyoucreateirrelevantequations,whereuponnoamountofcomputingpowercanhelpyou.Now lets represent the description mathematically. First represent theinputsignal. Tominimize the largenumbers and dollar signs, measuremoneyinunitsof$500,000.Thischoicemakestheinputsignaldimensionless:
X=20,1,1,1,1,...
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232 Differenceequationsandmodularity
Weusethenotationthatacapitalletterrepresentstheentiresignal,whilealowercaseletterwithanindexrepresentsonesamplefromthesignal. Forexample, P is thesequenceofpopulationsandp[n] is thepopulation inyear
n.
Theoutputsignalis
Y= 20,?,?,?, . . . Pausetotry 6. Explainwhyy[0] = 20.
Theproblemistofillinthequestionmarksintheoutputsignalandfindwhenitfallsbelowzero.Thedifferenceequationdescribingthesystemis
y[n] = (1+ r)y[n 1] + x[n],whereristheannualinterestrate(here,r= 0.04).Thisdifferenceequationisafirst-orderequationbecauseanyoutputsampley[n] dependsontheoneprecedingsampley[n 1].Thesystemthattheequationrepresentsissaidtobeafirst-ordersystem. Itisthesimplestmoduleforbuildingandanalyzingcomplexsystems.
Exercise 3. ComparethisequationtotheoneforestimatingtheUSpopulationin2077.Nowwehaveformulatedtheendowmentproblemasasignalprocessed
byasystem toproduceanothersignalallhailmodularity! andrepresentedthisdescriptionmathematically. However,wedonotyetknowhowtosolvethemathematicsforanarbitraryinputsignalX.Buthereweneedtosolveitonlyfortheparticularinputsignal
X= 20,
1,
1,
1,
1,
. . . .
Withthatinputsignal,therecurrencebecomes
y[n] = 1.04y[n 1] 1 n > 0;20 n= 0.
They[0] = 20reflectsthat thedonorseeds theaccountwith20unitsofmoney, which is the $10,000,000 endowment. The 1 in the recurrence
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24 2.2 Endowmentgift
reflectsthatwedraw1uniteveryyear. Withoutthe1term,thesolutiontotherecurrencewouldbey[n] 1.04n,wherethesymbolmeansexceptfor a constant. The 1 means that simple exponential growth is not asolution.
However,
1
is
a
constant
so
it
may
contribute
only
a
constant
to
thesolution. Thatreasoningisdubiousbutsimple,sotryitfirst. Usinga
bitofcourage,hereisaguessfortheformofthesolution:y[n] = A1.04n + B (guess),
whereAandBareconstantstobedetermined. BeforefindingAandB,figureoutthemostimportantcharacteristic,theirsigns.So:
Pausetotry 7. Assumethatthisformiscorrect.WhatarethesignsofAandB?
Sincetheendowmenteventuallyvanishes,thevariabletermA 1.04nmustmakeanegativecontribution; soA < 0. Since the initialoutputy[0] ispositive,BmustovercomethenegativecontributionfromA; so B > 0.
Pausetotry 8. FindAandB.SolvingfortwounknownsAandBrequirestwoequations.Eachequationwillprobablycomefromonecondition. Somatchtheguesstotheknown
balancesattwotimes.Thetimes(valuesofn)thatinvolvetheleastcalculationaretheextremecasesn = 0andn = 1. Matchingtheguesstothe
behavioratn= 0givesthefirstequation:20= A+ B (n= 0condition).
Tomatch theguess to thebehaviorat n = 1, firstfindy[1]. At n = 1,whichisoneyearafterthegift,0.8unitsofinterestarrivefrom4%of20,and1unitleavesasthefirstdrawdown.So
y[1] = 20+ 0.8 1= 19.8.Matchingthisvaluetotheguessgivesthesecondequation:
19.8= 1.04A+ B (n= 1condition).
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252 Differenceequationsandmodularity
Bothconditionsaresatisfiedwhen A = 5and B = 25. Aspredicted,A < 0 andB > 0.Withthatsolutiontheguessbecomes
y[n] = 25 5
1.04n.Thissolutionhasastrangebehavior. Afterthebalancedropsbelowzero,the1.04n growsevermorerapidlysothebalancebecomesnegativeeverfaster.
Exercise 4. Doesthatbehaviorofbecomingnegativemoreandmore rapidly indicate an incorrect solution to therecurrencerelation,oranincompletemathematicaltranslationofwhathappensinreality?
Exercise 5. Theguess,withthegivenvaluesforA andB,worksforn = 0andn = 1. (Howdoyouknow?) Showthatitisalsocorrectforn > 1.
Nowwecananswertheoriginalquestion: Whendoesy[n] falltozero?nAnswer: When1.04 > 5,whichhappensatn = 41.035.... SoMITcan
drawon the fund inyears1 , 2 , 3 , . . . , 4 1, leaving loosechange in theaccountforalargegraduationparty.Theexactcalculationisconsistentwiththeargumentthatthelifetimebelessthan100years.
Exercise 6. HowmuchloosechangeremainsafterMITdrawsitslastpayment?Converttorealmoney!
2.3 RabbitsThesecondsystemtorepresentmathematicallyisthefecundityofrabbits.The Encyclopedia Britannica (1981 edition) states this population-growthproblemasfollows[6]:
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272 Differenceequationsandmodularity
Letc[n] bethenumberofchildpairsatmonthnanda[n] bethenumberofadultpairsatmonthn.Theseintermediatesignalscombinetomaketheoutputsignal:
f[n] = a[n] + c[n] (outputsignal).Pausetotry 11. What equation contains the rule that childrenbe
comeadultsinonemonth?
Becausechildrenbecomeadultsinonemonth,andadultsdonotdie,thepoolofadultsgrowsbythenumberofchildpairsinthepreviousmonth:
a[n] = a[n 1] + c[n 1] (growing-upequation).The two terms on the right-hand side represent the two ways tobe anadult:1. Youwereanadultlastmonth(a[n 1]),or2. youwereachildlastmonth(c[n 1])andgrewup.Thenextequationsaysthatalladults,andonlyadults,reproducetomakenewchildren:
c[n] = a[n 1].However, this equation is not completebecause immigration also contributes child pairs. The number of immigrant pairs at month n is theinputsignalx[n].Sothefullstoryis:
c[n] = a[n 1] + x[n] (childequation)Ourgoalisarecurrenceforf[n],thetotalnumberofpairs.Soweeliminatethenumberofadultpairsa[n] andthenumberofchildpairsc[n] infavoroff[n]. Doitintwosteps. First,usethegrowing-upequationtoreplacea[n 1] inthechildequationwitha[n 2] + c[n 2]. Thatsubstitutiongives
c[n] = a[n 2] + c[n 2] + x[n].
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28 2.3 Rabbits
Sincef[n] = c[n] + a[n],wecanturntheleftsideintof[n] byaddinga[n].Thegrowing-upequationsaysthata[n] isalsoa[n 1] + c[n 1],soaddthosetermstotherightsideandprayforsimplification.Theresultis
c[n] + a[n] = a[n 2] + c[n 2] +x[n] + a[n 1] + c[n 1] . f[n] f[n2] f[n1]
Theleftsideisf[n]. Therightsidecontainsa[n 2] + c[n 2],whichisf[n 2];anda[n 1] + c[n 1],whichisf[n 1].Sothesumofequationssimplifiesto
f[n] = f[n 1] + f[n 2] + x[n].The Latin problem description is from Fibonaccis LiberAbaci [10], published in1202, and thisequation is the famousFibonaccirecurrencebutwithaninputsignalx[n] insteadofboundaryconditions.Thismathematicalrepresentationclarifiesonepointthatisnotobviousintheverbalrepresentation: Thenumberofpairsofrabbitsatmonthndependsonthenumberinmonthsn1 andn2.Becauseofthisdependenceontwoprecedingsamples,thisdifferenceequationisasecond-orderdifferenceequation.Sinceallthecoefficientsareunity,itisthesimplestequation of that category, and ideal as a second-order system to understandthoroughly.Tobuildthatunderstanding,weplaywiththesystemandseehowitresponds.
2.3.2 TryingtherecurrenceToplaywiththesystemdescribedbyFibonacci,weneedtorepresentFi
bonaccisboundaryconditionthatonepairofchildrabbitsenterthewallsonlyinmonth0. ThecorrespondinginputsignalisX=1 , 0 , 0 , 0 , . . .. UsingthatX,knownasanimpulseoraunitsample,therecurrenceproduces(leavingouttermsthatarezero):
f[0] = x[0] = 1,f[1] = f[0] = 1,f[2] = f[0] + f[1] = 2,f[3] = f[1] + f[2] = 3,. . .
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2 Differenceequationsandmodularity 29
Whenyoutryafewmorelines,yougetthesequence:F=1,1,2,3,5,8,13,21,34,..Whenyou tireofhandcalculation, askacomputer tocontinue. Here isslowPythoncodetoprintf[0],f[1],. . .,f[19]:def f(n):
i f n < 2 : return 1return f(n-1) + f(n-2)
print [f(i) for i in range(20)]Exercise 7. Write the corresponding Matlab or Octave code,
then rewrite the code in one of the languages Python,Matlab,orOctavetobeefficient.
Exercise 8. WriteMatlab, Octave, orPythoncode tofindf[n]whentheinputsignalis1 , 1 , 1 , . . ..Whatisf[17]?
2.3.3 RateofgrowthTosolvetherecurrenceinclosedformmeaninganexplicitformulaforf[n]thatdoesnotdependonprecedingsamplesitishelpfultoinvestigateitsapproximategrowth.Evenwithoutsophisticatedtechniquestofindtheoutputsignal,wecanunderstandthegrowthinthiscasewhentheinputsignalistheimpulse.
Pausetotry 12. Whentheinputsignalistheimpulse,howfastdoesf[n]grow? Is itpolynomial, logarithmic,orexponential?
Fromlookingatthefirstfewdozenvalues,itlookslikethesequencegrowsquickly. Thegrowth isalmostcertainly toorapidtobe logarithmicand,almostascertain,toofasttobepolynomialunlessitisahigh-degreepolynomial.Exponentialgrowthisthemostlikelycandidate,meaningthatanapproximationforf[n]is
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30 2.3 Rabbits
f[n] znwherez isaconstant.Toestimatez,playwiththerecurrencewhenn > 0,whichiswhentheinputsignaliszero. Thef[n] areallpositiveand,sincef[n] = f[n 1] + f[n 2] whenn > 0,thesamplesareincreasing: f[n] >f[n 1].Thisboundturnsf[n] = f[n 1] + f[n 2] intotheinequality
f[n] < f[n 1] + f[n 1].Sof[n] < 2f[n 1] orf[n]/f[n 1] < 2;thereforetheupperboundonzisz < 2.Thisboundhasacounterpartlowerboundobtainedbyreplacingf[n 1] byf[n 2] in theFibonaccirecurrence. Thatsubstitutionturnsf[n] = f[n 1] + f[n 2] into
f[n] > f[n 2] + f[n 2].Therightsideis2f[n 2] sof[n] > 2f[n 2].Thisboundleadstoalower
bound:z2 > 2 orz > 2.Therangeofpossiblez isthen
2 < z < 2. Letschecktheboundsbyexperiment.Hereisthesequenceofratiosf[n]/f[n1] forn = 1 , 2 , 3 , . . .:
1.0, 2.0, 1.5, 1.666 . . . , 1.6, 1.625, 1.615 . . . , 1.619 . . . , 1.617 . . .
Theratiosseemtooscillatearound1.618,whichliesbetweenthepredictedbounds 2and2. Inlaterchapters,usingnewmathematicalrepresentations,youlearnhowtofindtheclosedfromforf[n].Wehavewalkedtwostepsinthatdirectionbyrepresentingthesystemmathematicallyandbyinvestigatinghowf[n] grows.
Exercise 9. Useamorerefinedargumenttoimprovetheupperboundtoz <
3.
Exercise 10. Doesthenumber1.618lookfamiliar?
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312 Differenceequationsandmodularity
Exercise 11. [Hard!]Considerthesamesystembutwithonerabbit pair emigrating into the system every month,not only in month 0. Compare the growth withFibonaccisproblem, whereonepairemigrated inmonth0only. Isitnowfasterthanexponential? Ifyes,howfastisit? Ifno,doestheorderofgrowthchangefromz1.618?
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3Blockdiagramsandoperators:Twonewrepresentations
3.1
Disadvantagesof
difference
equations
34
3.2 Blockdiagramstotherescue 353.3 Thepowerofabstraction 403.4 Operationsonwholesignals 413.5 Feedbackconnections 453.6 Summary 49
Thegoalsofthischapterare: tointroducetworepresentationsfordiscrete-timesystems:block
diagramsandoperators; tointroducethewhole-signalabstractionandtoexhortyoutouse
abstraction; tostartmanipulatingoperatorexpressions; tocompareoperatorwithdifference-equationandblock-diagram
manipulations.
Thepreceding chaptersexplained the verbal-description anddifference-equation representations. This chapter continues the theme of multiplerepresentationsbyintroducingtwonewrepresentations:blockdiagramsand operators. New representations are valuablebecause they suggestnewthoughtsandoftenprovidenewinsight;anexpertengineervaluesherrepresentationsthewayanexpertcarpentervalueshertools.Thischapterfirstintroducesblockdiagrams,discussesthewhole-signalabstractionand
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34 3.1 Disadvantagesofdifferenceequations
thegeneralvalueofabstraction,thenintroducestheoperatorrepresentation.
3.1 DisadvantagesofdifferenceequationsChapter2illustratedthevirtuesofdifferenceequations. Whencomparedtotheverbaldescriptionfromwhichtheyoriginate,differenceequationsarecompact,easytoanalyze,andsuitedtocomputerimplementation.Yetanalyzingdifferenceequationsofteninvolveschainsofmicro-manipulationsfromwhichinsightishardtofind.Asanexample,showthatthedifferenceequation
d[n] = a[n] 3a[n 1] + 3a[n 2] a[n 3]isequivalenttothissetofequations:
d[n] = c[n] c[n 1]c[n] = b[n] b[n 1]b[n] = a[n] a[n 1].
Asthefirststep,usethelastequationtoeliminateb[n] andb[n 1] fromthec[n] equation:
c[n] = (a[n] a[n 1]) (a[n 1] a[n 2]) = a[n]2a[n1]+a[n2]. b[n] b[n1]Usethatresulttoeliminatec[n] andc[n 1] fromthed[n] equation:
d[n] = (a[n] 2a[n 1] + a[n 2]) (a[n 1] 2a[n 2] + a[n 3]) c[n] c[n1]
= a[n] 3a[n 1] + 3a[n 2] a[n 3].Voil:Thethree-equationsystemisequivalenttothesingledifferenceequation.
But
what
a
mess.
Each
step
is
plausible
yet
the
chain
of
steps
seems
random.Ifthelaststephadproduced
d[n] = a[n] 2a[n 1] + 2a[n 2] a[n 3],it would not immediately look wrong. We would like a representationwhereitwouldlookwrong,perhapsnotimmediatelybutatleastquickly.Blockdiagramsareonesuchrepresentation.
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353 Blockdiagramsandoperators:Twonewrepresentations
Exercise 12. Althoughthissectionpointedoutadisadvantageofdifferenceequations,itisalsoimportanttoappreciatetheirvirtues.Therefore,inventaverbaldescription(astory)torepresentthesingleequation
d[n] = a[n] 3a[n 1] + 3a[n 2] a[n 3]and then a verbal description to represent theequivalent set of three equations. Now have funshowing, without converting to difference equations,thatthesedescriptionsareequivalent!
3.2 BlockdiagramstotherescueBlockdiagramsvisuallyrepresentasystem.Toshowhowtheywork,hereareafewdifferenceequationswithcorrespondingblockdiagrams:
Delay1/2+ y[n] = (x[n] + x[n 1])/2
averagingfilter+
Delayy[n] = y[n 1] + x[n]
accountwith0%interest
Pausetotry 13. Draw theblock diagram for the endowment accountfromSection2.2.
Theendowmentaccountisabankaccountthatpays4%interest,soitneedsagainelement in the loop, withgainequal to 1.04. Thediagram isnotunique. Youcanplacethegainelementbeforeorafterthedelay. Hereisonechoice:
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36 3.2 Blockdiagramstotherescue
+
1.04Delay
y[n] = 1.04y[n 1] + x[n]endowmentaccountfromSection2.2
Amazingly,allsystemsinthiscoursecanbebuiltfromonlytwoactionsandonecombinator:
A action1:multiplybyDelay
+action2:delayonetickcombinator:addinputs
3.2.1 BlockdiagramfortheFibonaccisystemTopracticeblockdiagrams,wetranslate(represent)theFibonaccisystemintoablockdiagram.
Pausetotry 14. RepresenttheFibonaccisystemofSection1.1usingablockdiagram.
WecouldtranslateFibonaccisdescription(Section1.1)directlyintoablockdiagram,butweworkedsohardtranslatingthedescriptionintoadifferenceequationthatwestartthere.Itsdifferenceequationis
f[n] = f[n 1] + f[n 2] + x[n],where the inputsignalx[n]ishowmanypairsofchildrabbitsenterthesystematmonthn,andtheoutputsignalf[n] ishowmanypairsofrabbitsareinthesystematmonthn. Intheblockdiagram,itisconvenienttoletinputsignalsflowinfromtheleftandtoletoutputsignalsexitattherightfollowingtheleft-to-rightreadingcommontomanylanguages.
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373 Blockdiagramsandoperators:Twonewrepresentations
Exercise 13. Do signals-and-systems textbooks in Hebrew orArabic,whicharewrittenrighttoleft,putinputsignalsontherightandoutputsignalsontheleft?
TheFibonaccisystemcombinestheinputsample,thepreviousoutputsample,andthesecond-previousoutputsample.Thesethreesignalsarethereforeinputstothepluselement. Thepreviousoutputsampleisproducedusingadelayelementtostoresamplesforonetimetick(onemonth)beforesendingthemonward.Thesecond-previousoutputsampleisproducedbyusingtwodelayelementsinseries. SotheblockdiagramoftheFibonaccisystemis
+ DelayDelayDelay
f[n]x[n]
3.2.2 ShowingequivalenceusingblockdiagramsWe introducedblock diagrams in the hope of finding insight not easilyvisiblefromdifferenceequations.Souseblockdiagramstoredotheproofthat
d[n] = a[n] 3a[n 1] + 3a[n 2] a[n 3]isequivalentto
d[n] = c[n] c[n 1],c[n] = b[n] b[n 1],b[n] = a[n] a[n 1].
Thesystemofequationsisacascadeofthreeequationswiththestructureoutput= thisinput previousinput.
Theblockdiagramforthatstructureis
-1 Delay+
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38 3.2 Blockdiagramstotherescue
wherethegainof1producesthesubtraction.Thecascadeofthreesuchstructureshastheblockdiagram
Delay+
-1 Delay+
-1 Delay-1
Thisdiagramhasadvantagescomparedtothesetofdifferenceequations.First,thediagramhelpsusdescribethesystemcompactly. Eachstageinthecascade isstructurally identical, and thestructural identity isapparentbylookingatit.Whereasinthedifference-equationrepresentation,thecommonstructureofthethreeequationsishiddenbythevaryingsignalnames.
Each
stage,
it
turns
out,
is
a
discrete-time
differentiator,
the
simplestdiscrete-timeanalogofacontinuous-timedifferentiator.Sotheblock
diagrammakesapparentthatthecascadeisadiscrete-timetripledifferentiator.Second,theblockdiagramhelpsrewritethesystem,whichweneedtodotoshowthat it is identical to thesingledifferenceequation. Sofollowasignalthroughthecascade. Thesignalreachesaforkthreetimes(markedwithadot),andeachforkoffersachoiceofthebottomortopbranch.Threetwo-waybranchesmeans23 or8pathsthroughthesystem.Letsexaminea
few
of
them.
Three
paths
accumulate
two
delays:
1. lowroad,lowroad,highroad:
Delay+
-1 Delay+
-1 Del-1
2. lowroad,highroad,lowroad:
-1 Delay+
-1 Delay+
-1 Del
3. highroad,lowroad,lowroad:
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3 Blockdiagramsandoperators:Twonewrepresentations 39
-1 Delay+
-1 Delay+
-1 Del
Besidesthetwodelays,eachpathaccumulatestwogainsof1,makingagainof1.Sothesumofthethreepathsisagainof3andadoubledelay.
Exercise 14. Show the other five pathsare: threepaths withasingledelayandagainof1,onepathwiththreedelays and a gain of 1, and one path that goesstraightthrough(nogain,nodelay).
Ablockdiagramrepresentingthosefourgroupsofpathsis
3 Delay
3 Delay Delay
1 Delay Delay Delay
+
Thesingledifferenceequationd[n] = a[n] 3a[n 1] + 3a[n 2] a[n 3].
alsohasthisblockdiagram.Thepictorialapproachisanadvantageofblockdiagramsbecausehumansare sensorybeings and vision is an important sense. Brains, over hundreds of millions of years of evolution, have developed extensive hardwaretoprocesssensoryinformation. However,analyticalreasoningandsymbolmanipulationoriginatewithlanguage,skillperhaps100,000yearsold,soourbrainshavemuch lesspowerfulhardware in thosedomains.
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40 3.3 Thepowerofabstraction
Notsurprisingly,computersarefarmoreskilledthanarehumansatanalytical tasks likesymbolic algebra and integration, and humans are farmoreskilledthanarecomputersatperceptualtaskslikerecognizingfacesor
speech.
When
you
solve
problems,
amplify
your
intelligence
with
a
visualrepresentationsuchasblockdiagrams.
Ontheotherside,exceptbytracingandcountingpaths,wedonotknowtomanipulateblockdiagrams;whereasanalyticrepresentationslendthemselves to transformation, an important property when redesigning systems. So we need a grammar forblock diagrams. To find the rules ofthisgrammar,weintroduceanewrepresentationforsystems,theoperatorrepresentation. Thisrepresentationrequiresthewhole-signalabstractionin which all samples of a signal combine into one signal. It is a subtlechange
of
perspective,
so
we
first
discuss
the
value
of
abstraction
in
general,thenreturntotheabstraction.
3.3 ThepowerofabstractionAbstraction is a great tools of human thought. All language isbuilt onit: Whenyouuseaword,youinvokeanabstraction. Theword,evenanordinarynoun,standsforarich,subtle,complexidea. Takecowandtrytoprogramacomputertodistinguishcowsfromnon-cows;thenyoufindhowdifficultabstractionis. Orwatchachildsabilitywithlanguagedevelopuntilshelearnsthatredisnotapropertyofaparticularobjectbutisanabstractpropertyofobjects. Nooneknowshowthemindmanagestheseamazingfeats,norinwhatamountstothesameignorancecananyoneteachthemtoacomputer.Abstraction is so subtle that even Einstein once missed its value. Einstein formulated the theory of special relativity [7] with space and timeasseparateconceptsthatmingleintheLorentztransformation.Twoyearslater, the mathematician Hermann Minkowskijoined the two ideas intothespacetimeabstraction:
TheviewsofspaceandtimewhichIwishtolaybeforeyouhavesprungfromthesoilofexperimentalphysics,andthereinliestheirstrength.Theyareradical.Henceforth spaceby itself, and timeby itself, are doomed to fade away intomereshadows,andonlyakindofunionofthetwowillpreserveanindependentreality.
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413 Blockdiagramsandoperators:Twonewrepresentations
See the English translation in [11] or the wonderful textbook SpacetimePhysics [1], whosefirstauthorrecentlyretired from theMITphysicsdepartment. Einsteinthoughtthatspacetimewasapreposterousinventionof
mathematicians
with
time
to
kill.
Einstein
made
a
mistake.
It
is
perhaps the fundamentalabstraction ofmodernphysics. Themoral is that
abstractionispowerfulbutsubtle.
Exercise 15. Findafewabstractionsinchemistry,biology,physics,andprogramming.
IfwelackEinsteinsphysicalinsight,weoughtnottocompoundtheabsence
with
his
mistake.
So
look
for
and
create
abstractions.
For
example,
inaprogram,factoroutcommoncodeintoaprocedureandencapsulatecommonoperationsintoaclass. Ingeneral,organizeknowledgeintoabstractionsorchunks[15].
3.4 OperationsonwholesignalsForsignalsandsystems,thewhole-signalabstractionincreasesourabilitytoanalyzeandbuildsystems. Theabstractionistakeallsamplesofasignalandlumpthemtogether,operatingontheentiresignalatonceandasoneobject.Wehavenotbeenthinkingthatwaybecausemostofourrepresentationshinderthisview. Verbaldescriptionsanddifferenceequationsusuallyimplyasample-by-sampleanalysis.Forexample,fortheFibonaccirecurrenceinSection2.3.2,wefoundthezerothsamplef[0],usedf[0]tofindf[1],usedf[0]andf[1]tofindf[2],foundafewmoresamples,thengottiredandaskedacomputertocarryon.Blockdiagrams,thethirdrepresentation,seemtoimplyasample-by-sampleanalysisbecausethedelayelementholdsontosamples,spittingoutthesampleafteronetimetick.Butblockdiagramsliveinbothworldsandcanalsorepresentoperationsonwholesignals.Justreinterprettheelementsinthewhole-signalview,asfollows:
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42 3.4 Operationsonwholesignals
A action1:multiplywholesignalbyDelay
+action
2:
shift
whole
signal
right
one
tick
combinator:addwholesignals
Tobenefitfromtheabstraction, compactlyrepresent theprecedingthreeelements. Whenasignalisasingleobject,thegainelementactslikeordinarymultiplication,andthepluselementactslikeadditionofnumbers.Ifthedelayelementcouldalsoactlikeanarithmeticoperation,thenallthreeelementswouldact ina familiarway, andblockdiagramscouldbemanipulatedusingtheordinaryrulesofalgebra. Inordertobringthedelayelementintothisfamiliarframework,weintroducetheoperatorrepresentation.
3.4.1 OperatorrepresentationInoperatornotation, thesymbolRstandsfortheright-shiftoperator. Ittakesasignalandshiftsitonesteptotheright. HereisthenotationforasystemthatdelaysasignalX byoneticktoproduceasignalY:
Y=R{X}.Nowforgetthecurlybraces,tosimplifythenotationandtostrengthentheparallelwithordinarymultiplication.Thecleannotationis
Y=RX.Pausetotry 15. Convinceyourselfthatright-shiftoperatorR,rather
thantheleft-shiftoperatorL,isequivalenttoadelay.
LetstesttheeffectofapplyingRtothefundamentalsignal,theimpulse.Theimpulseis
I=1 , 0 , 0 , 0 , . . . ApplyingRtoitgives
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44 3.4 Operationsonwholesignals
Withthesemappings,thedifferenceequationturnsintothecompactformD= (1 3R+ 3R2 R3)A.
Toshowthatthisformisequivalenttothesystemofthreedifferenceequations,translatethemintoanoperatorexpressionconnectingtheinputsignalAandtheoutputsignalD.
Pausetotry 16. Whataretheoperatorversionsofthethreedifferenceequations?
Thesystemofequationsturnsintotheseoperatorexpressionsd[n] = c[n] c[n 1] D= (1 R)C,c[n] = b[n] b[n 1] C= (1 R)B,b[n] = a[n] a[n 1] B= (1 R)A.
EliminateBandCtogetD= (1 R)(1 R)(1 R)A= (1 R)3A.
ExpandingtheproductgivesD= (1 3R+ 3R2 R3)A,
which matches the operator expression corresponding to the single difference equation. The operator derivation of the equivalence is simplerthantheblock-diagramrewriting,andmuchsimplerthanthedifference-equationmanipulation.Nowextendtheabstractionbydividingouttheinputsignal:
D= 1 3R+ 3R2 R3.
ATheoperatorexpressionontheright,beingindependentoftheinputandoutputsignals,isacharacteristicofthesystemaloneandiscalledthesystemfunctional.The moral of the example is that operators help you efficiently analyzesystems. Theyprovideagrammarforcombining,forsubdividing,andingeneral forrewritingsystems. It isa familiargrammar, thegrammarof
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453 Blockdiagramsandoperators:Twonewrepresentations
algebraicexpressions.Letsseehowextensivelyoperatorsfollowthese.Inthenextsectionwestretchtheanalogyandfindthatitdoesnotbreak.
Exercise 16. What is theresultofapplying1 Rto thesignal1, 2, 3, 4, 5, . . .?
Exercise 17. Whatistheresultofapplying(1 R)2 tothesignal1,4,9,16,25,36,...?
3.5 FeedbackconnectionsThesystemwith(1 R)3 asitssystemfunctionalusedonlyfeedforwardconnections:Theoutputcouldbecomputeddirectlyfromafixednumberofinputs. However,manysystemssuchasFibonacciorbankaccountscontainfeedback,wheretheoutputdependsonpreviousvaluesoftheoutput. Feedback produces new kinds of system functionals. Lets testwhethertheyalsoobeytherulesofalgebra.
3.5.1 AccumulatorHereisthedifferenceequationforthesimplestfeedbacksystem,anaccumulator:
y[n] = y[n 1] + x[n].Itisabankaccountthatpaysnointerest. Theoutputsignal(thebalance)isthesumoftheinputs(thedeposits,whetherpositiveornegative)uptoandincludingthattime.Thesystemhasthisblockdiagram:
+
Delay
NowcombinethevisualvirtuesofblockdiagramswiththecompactnessandsymbolicvirtuesofoperatorsbyusingRinsteadofDelay.Theoperatorblockdiagramis
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46 3.5 Feedbackconnections
X +
R
Y
Pausetotry 17. Whatisitssystemfunctional?Either from this diagram or from the difference equation, translate intooperatornotation:
Y= RY+ X.CollecttheYtermsononeside,andyoufindendupwiththesystemfunctional:
Y 1= .
X 1 RItisthereciprocalofthedifferentiator.ThisoperatorexpressionisthefirsttoincludeRinthedenominator. Onewaytointerpretdivisionistocomparetheoutputsignalproducedbythedifferenceequationwiththeoutputsignalproducedbythesystemfunctional
1/(1 R).
For
simplicity,
test
the
equivalence
using
the
impulse
I= 1 , 0 , 0 , 0 , . . .
asthe inputsignal. Sox[n]is1forn = 0and is0otherwise. Thenthedifferenceequation
y[n] = y[n 1] + x[n]producestheoutputsignal
Y= 1 , 1 , 1 , 1 , . . . .
Exercise 18. Checkthisclaim.
Theoutputsignalisthediscrete-timestepfunction.Nowapply1/(1R)to the impulseIby importing techniques fromalgebraorcalculus. Use
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47
1
3 Blockdiagramsandoperators:Twonewrepresentations
syntheticdivision,Taylorseries,orthebinomialtheoremtorewrite1/(1 R)as
=1
+R+R2 +R3 + .
1 R
Toapply1/(1 R)totheimpulse,applytheeachofterms1,R,R2,. . . totheimpulseI:
1I=1, 0, 0, 0, 0, 0, 0, . . . , RI=0, 1, 0, 0, 0, 0, 0, . . . ,
R2I=0, 0, 1, 0, 0, 0, 0, . . . , R3I=0, 0, 0, 1, 0, 0, 0, . . . , R
4
I=0, 0, 0, 0, 1, 0, 0, . . . , . . .
AddthesesignalstogettheoutputsignalY.
Pausetotry 18. WhatisY?
Forn0,they[n]samplegetsa1 fromtheRnI term,andfromnootherterm.
So
the
output
signal
is
all
1s
from
n
=
0 onwards.
The
signal
with
thosesamplesisthestepfunction:
Y=1 , 1 , 1 , 1 , . . . . Fortunately,thisoutputsignalmatchestheoutputsignalfromrunningthedifferenceequation.So,foranimpulseinputsignal,theseoperatorexpressionsareequivalent:
1and 1 +R+R2+R3 + .
1 RExercise 19. Ifyouaremathematicallyinclined,convinceyour
selfthatverifyingtheequivalenceforthe impulseissufficient. Inotherwords,wedonotneedtotryallotherinputsignals.
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48 3.5 Feedbackconnections
ThemoralisthattheRoperatorfollowstherulesofalgebraandcalculus.Sohavecourage:Useoperatorstofindresults,anddonotworry.
3.5.2 FibonacciTakingourownadvice,wenowanalyzetheFibonaccisystemusingoperators.Therecurrenceis:
output=delayedoutput + twice-delayedoutput + input.
Pausetotry 19. Turnthisexpressionintoasystemfunctional.
TheoutputsignalisF,andtheinputsignalisX.ThedelayedoutputisRX,andthetwice-delayedoutputisRRXorR2X. So
F=RF+R2F+X.CollectallFtermsononeside:
FRFR2F=X.ThenfactortheF:
(1RR2)F=X.ThendividebothsidesbytheRexpression:
1F= X.
1RR2Sothesystemfunctionalis
1.
1RR2
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4Modes
4.1 GrowthoftheFibonacciseries 524.2 TakingoutthebigpartfromFibonacci 554.3 Operatorinterpretation 574.4 Generalmethod:Partialfractions 59
Thegoalsofthischapterare: to illustrate the experimental way that an engineer studies sys
tems,evenabstract,mathematicalsystems; toillustratewhatmodesarebyfindingthemfortheFibonaccisys
tem;and to decompose second-order systems into modes, explaining the
decompositionusingoperatorsandblockdiagrams.Thefirstquestioniswhatamodeis.Thatquestionwillbeansweredaswedecompose theFibonaccisequence intosimplersequences. Eachsimplesequencecanbegeneratedbyafirst-ordersystemliketheleakytankandiscalledamodeofthesystem.BydecomposingtheFibonaccisequenceintomodes,wedecomposethesystemintosimpler,first-ordersubsystems.The plan of the chapter is to treat the Fibonacci system first as ablack
boxproducinganoutput signal Fand to developcomputationalprobestoexaminesignals. Thisexperimentalapproachishowanengineerstudiesevenabstract,mathematicalsystems. Theresultsfromtheprobeswillshowushow todecompose thesignal into itsmodes. Thesemodesarethenreconciledwithwhattheoperatormethodpredictsfordecomposingthesystem.
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52 4.1 GrowthoftheFibonacciseries
Whydescribetheexperimental,andperhapsharder,methodforfindingthemodesbeforegivingtheshortcutsusingoperators? WeknowtheoperatorexpressionfortheFibonaccisystem,andcouldjustrewriteitusingalgebra.
The
answer
is
that
the
operator
method
has
meaning
only
after
you feelmodes inyourfingertips, a feelingdevelopedonlyasyouplaywithsignals. Withoutfirstplaying, wewouldbe teachingyouamazingfeatsofcalculationonmeaninglessobjects.Furthermore,theexperimentalapproachworksevenwhennodifferenceequationisavailabletogeneratethesequence. Engineersoftencharacterizesuchunknownorpartiallyknownsystems.Thesystemmightbe: computational: Imaginedebuggingsomeoneelsesprogram. Yousend
intestinputstofindouthowitworksandwhatmakesitfail. electronic: ImaginedebuggingaCPUthatjustreturnedfromthefabri
cationrun,perhapsinquantitiesofmillions,butthatdoesnotcorrectlydividefloating-pointnumbers[12]. Youmightgiveitnumberstodivide until you find the simplest examples that give wrong answers.Fromthatdatayoucanoftendeducetheflawinthewiring.
mathematical:Imaginecomputingprimestoinvestigatethetwin-primeconjecture[16],oneoftheoutstandingunsolvedproblemsofnumbertheory.[Theconjecturestatesthatthereareaninfinitenumberofprimepairs
p,p+2
,such
as
(3,5),(5,7)
,etc.]
The
new
field
of
experimental
mathematics,whichusescomputationaltoolstoinvestigatemathematicsproblems,islively,growing,andafertilefieldforskilledengineers[4,14,8].
Sowehopethat,throughexperimentalprobesoftheFibonaccisequence,youlearnageneralapproachtosolvingproblems.
4.1 GrowthoftheFibonacciseriesSection1.1.2estimatedhowfastthesequencef[n]grew.Itseemedtogrowgeometricallywithanorderofgrowthbetween 2and2.Ourfirstprojectistoexperimentallynarrowthisrangeandtherebytoguessaclosedformfortheorderofgrowth.Oneprobetofindtheorderofgrowthistocomputethesuccessiveratiosf[n]/f[n1]. Theratiososcillatedaround1.618,butthisestimate isnot
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4Modes 53
accurateenoughtoguessaclosedform. Sincetheoscillationsintheratiodieoutasngrows,letsestimatetheratioaccuratelybycomputingitforlargen.Ourtoolfortheseexperimentsourprobeisacomputerthatweprogram
in
Python,
a
clean,
widely
available
language.
Use
any
tool
that
fitsyou,perhapsanotherlanguage,agraphingcalculator,oraspreadsheet.Section2.3.2offeredthisPythoncodetocomputef[n]:def f(n):
i f n < 2 : return 1return f(n-1) + f(n-2)
Butthecodeisslowwhennislarge.Herearetherunningtimestoevaluatef[n]onaPentiumCoreDuo1.8GHzprocessor:
n 10 15 20 25 30 time(ms) 0.17 1.5 21 162 1164
Thetimesgrowrapidly!
Exercise 21. Whatistherunningtimeofthisimplementation?
Thetimesmightseem lowenoughtobeusable,but imaginebeingonadesert
island
with
only
a
graphing
calculator;
then
the
times
might
be
a
factorof10orof100longer. Wewouldliketobuildanefficientcomputationalprobesothatitiswidelyusable.Anefficientfunctionwouldstorepreviouslycomputedanswers,returningthestoredanswerwhenpossibleratherthanrecomputingoldvalues. InPython,onecanstorethevaluesinadictionary,whichisanalogoustoahashinPerloranassociativearrayinawk. ThememoizedversionoftheFibonaccifunctionis:
memo = {}def f(n):
if n < 2 : return 1if n in memo : return memo[n]
memo[n] = f(n-1) + f(n-2)return memo[n]
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54 4.1 GrowthoftheFibonacciseries
Pausetotry 20. Whatistherunningtimeofthememoizedfunction,inbig-Ohnotation?
Thenewfunctionrunsinlineartimeafasterprobe!sowecaninexpensivelycomputef[n]forlargen.Herearetheratiosf[n]/f[n1]:
n f[n]/f[n1]5 1.60000000000000009
10 1.6181818181818181715 1.6180327868852459920 1.6180339985218033025 1.6180339886704431230
1.61803398875054083
35 1.6180339887498895740 1.6180339887498949045 1.61803398874989490
Thesevaluesareverystablebyn =45,perhapslimitedinstabilityonlybytheprecisionofthefloating-pointnumbers.Letsseewhatclosedformwouldproducetheratio1.61803398874989490atn =45. Onesourceforclosedformsisyourintuitionandexperience.AnotherwonderfulsourceistheInverseSymbolicCalculatorByusingtheInverseSymbolicCalculator,youincreaseyourrepertoireofclosedformandtherebyenhanceyourintuition.
Pausetotry 21. AsktheInverseSymbolicCalculatorabout1.6180339887498949TheInverseSymbolicCalculatorthinksthat1.61803398874989490ismostlikelythepositiverootofx2 x1or,equivalently,isthegoldenratio:
1+ 5
2
Letsusethathypothesis.Thenf[n]n.
Butwedonotknowtheconstanthiddenbythesymbol. FindthatconstantbyusingtheInverseSymbolicCalculatoronemoretime. Hereisa
.
http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.htmlhttp://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.htmlhttp://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.htmlhttp://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.htmlhttp://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.htmlhttp://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.htmlhttp://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.htmlhttp://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html8/8/2019 Course Note Signals
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4Modes 55
table of the ratio f[n]/n. With luck it converges to a constant. And itdoes:
n f[n]/n0 1.00000000000000000
10 0.7236250692647178120 0.7236067989578528530 0.7236067977500580940 0.7236067977499780550 0.7236067977499778360 0.7236067977499774970 0.7236067977499772780 0.7236067977499770590 0.72360679774997672
100 0.72360679774997649 Aroundn= 10,theratioslooklike 3 10.732butlaterratiosstabilizearoundavalueinconsistentwiththatguess.
Pausetotry 22. AsktheInverseSymbolicCalculatorabout0.7236067977499764Whichofthealternativesseemmostreasonable?
TheInverseSymbolicCalculatorprovidesmanyclosedformsfor0.723606797749976
Achoicethatcontains 5isreasonablesincecontains 5. Theclosed formnearestto0.72360679774997649andcontaining 5is(1+ 1/ 5)/2,whichisalso/ 5.SotheFibonaccisequenceisroughly
f[n] n.
54.2 TakingoutthebigpartfromFibonacci
Nowletstakeoutthebigpartbypeelingawaythe n contributionto5
seewhatremains.DefinethesignalF1by
f1[n] = n.5
ThissignalisonemodeoftheFibonaccisequence.Theshapeofamodeisitsorderofgrowth,whichhereis.Theamplitudeofamodeistheprefactor,whichhereis/ 5.Themodeshapeisacharacteristicofthesystem,
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56 4.2 TakingoutthebigpartfromFibonacci
whereastheamplitudedependsontheinputsignal(forthisexample,theinputsignalwastheimpulse).Soweoftenhavemoreinterestintheshapethan in theamplitude. However, hereweneedshapeandamplitude inorder
to
determine
the
signal
and
peel
it
away.
SotabulatetheresidualsignalF2 = F F1:
n f2[n] = f[n] f1[n]0 +0.276393202250021061 0.170820393249936812 +0.105572809000084263 0.065247584249852774 +0.040325224750231045 0.024922359499623076
+0.01540286525060708
7 0.009519494249015998 +0.005883371001587539 0.00363612324743201
10 +0.00224724775415552Theresidualsignalstartssmallandgetssmaller,sothemainmodeF1 isanexcellentapproximationtotheFibonaccisequenceF.TofindaclosedformfortheresidualsignalF2,retrythesuccessive-ratiosprobe:
n f2[n]/f2[n 1]1 0.618033988749894462 0.618033988749896013 0.618033988749893904 0.618033988749890465 0.618033988749939536 0.618033988749742367 0.618033988750294148 0.618033988748476269 0.61803398875421256
10 0.61803398873859083Thesuccessiveratiosarealmostconstantandlooksuspiciouslylike1 ,whichisalso1/.
Exercise 22. Showthat1 = 1/.
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4Modes 57
Sof2[n] ()n. Toevaluatetheamplitude,dividef2[n]bythemodeshape()n.Hereisatableofthoseresults:
n f2[n]/()n1 0.276393202250020902 0.276393202250021403 0.276393202250021014 0.276393202250019015 0.276393202250038996 0.276393202249970837 0.276393202250149418 0.276393202249514979 0.27639320225144598
10 0.27639320224639063Thosevaluesstabilizequicklyandlooklikeoneminustheamplitudeofthen mode.Sotheamplitudeofthe()n modeis1/ 5,whichisalso1/( 5).Thustheresidualsignal,combiningitsshapeandamplitude,is
1f2[n] = ()n.
5NowcombinetheF1 andF2 signalstogettheFibonaccisignal:
f[n] = f1[n] + f2[n] 1
= n + ()n.5 5Thisclosedform,deducedusingexperiment,isthefamousBinetformulaforthenth Fibonaccinumber.
Exercise 23. Usepeelingawayandeducatedguessingtofindaclosedformfortheoutputsignalwhentheimpulseisfedintothefollowingdifferenceequation:
y[n] = 7y[n1] 12y[n2] + x[n].
4.3 OperatorinterpretationNextwe interpretthisexperimentalresultusingoperatorsandblockdiagrams. Modes are the simplest persistent responses that a system can
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58 4.3 Operatorinterpretation
make,andarethebuildingblocksofallsystems,sowewouldliketofindtheoperatororblock-diagramrepresentationsforamode.The Fibonacci signal decomposed into two simpler signals F1 and F2 whicharealsothemodesandeachmodegrowsgeometrically. Geometricgrowthresultsfromonefeedbackloop.Sothen modeisproducedbythissystem
+
R
withthe
system
functional
(1
R)
1.
The()n modeisproducedbythissystem+
1R
withthesystemfunctional(1+ R/)1.TheFibonaccisystemisthesumofthesesignalsscaledbytherespectiveamplitudes,soitsblockdiagramisaweightedsumoftheprecedingblockdiagrams. Thesystem functionalfortheFibonaccisystem isaweightedsumofthepure-modesystemfunctionals.Soletsaddtheindividualsystemfunctionalsandseewhatturnsup:
F(R) = F1(R) + F2(R) 1 1 1
= + 51 R 51+ R/
1= .
1 R R2ThatfunctionalisthesystemfunctionalfortheFibonaccisystemderiveddirectlyfromtheblockdiagram(Section3.5.2)! Sotheexperimentalandoperatorapproachesagreethattheseoperatorblockdiagramsareequivalent:
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4Modes 59
15
11+R/
11RR2
5
11R
+=
where,tomakethediagrameasiertoparse,systemfunctionalsstandforthefirst- andsecond-ordersystemsthattheyrepresent.
Exercise 24. Write the system of difference equations that corresponds to the parallel-decompositionblock diagram. Show that the system is equivalent to theusualdifferenceequation
f[n] = f[n 1] + f[n 2] + x[n].Theequivalenceisobviousneitherfromtheblockdiagramsnorfromthedifferenceequationsdirectly.Makingtheequivalenceobviousneedseitherexperimentortheoperatorrepresentation.Havingexperimented,youarereadytousetheoperatorrepresentationgenerallytofindmodes.
4.4 Generalmethod:PartialfractionsSowewouldlikeawaytodecomposeasystemwithoutpeelingawayandguessing.Andwehaveone:themethodofpartialfractions,whichshowsthevalueof theoperatorrepresentationandsystemfunctional. Becausethesystemfunctionalbehaveslikeanalgebraicexpressionoronemightsay,becauseitisanalgebraicexpressionitisofteneasiertomanipulatethanistheblockdiagramorthedifferenceequation.Havinggonefromthedecomposedfirst-ordersystemstotheoriginalsecond-ordersystemfunctional,letsnowgotheotherway:fromtheoriginalsystem functional to the decomposed systems. To do so, first factor the Rexpression:
1 1 1= .
1 R R2 1 R 1+ R/
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60 4.4 Generalmethod:Partialfractions
Thisfactoring,aseriesdecomposition,willhelpusstudypolesandzerosina later chapter. Here we use it to find theparallel decompositionbyusingthetechniqueofpartialfractions.Thepartialfractionsshouldusethetwofactorsindenominator,soguessthisform:
1 a b= + ,
1 R R2 1 R 1+ R/where a and b are unknown constants. After adding the fractions, thedenominatorwillbetheproduct(1 R)(1+ R/) andthenumeratorwillbetheresultofcrossmultiplying:
a(1+ R/) + b(1 R) = a+ (a/)R+ b bR.Wewantthenumeratortobe1. If we set a= andb= 1/,thenatleasttheRtermscancel,leavingonlytheconstanta+b. Sowechoseaandbtoolargebythesuma+ b,whichis+ 1/or 5.Soinsteadchoose
a= / 5,b= 1/( 5).
Ifyouprefersolvinglinearequationstotheguess-and-checkmethod,herearethelinearequations:
a+ b= 1,a/ b= 0,
whosesolutionsaretheonesdeducedusingtheguess-and-checkmethod.Themoral:Tofindhowasystembehaves,factoritssystemfunctionalandusepartialfractionstodecomposethatfactoredformintoasumoffirst-ordersystems.Withthatdecomposition,youcanpredicttheoutputsignal
becauseyouknowhowfirst-ordersystemsbehave.Youcanpracticethenewskillofdecompositionwiththefollowingquestion:
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4Modes 61
Exercise 25. Lookagainatthesystemy[n] = 7y[n 1] 12y[n 2] + x[n].
Decomposetheoperatorrepresentationintoasumof two modes and draw the correspondingblockdiagram (usingblock diagram elements). Whenthe input signal X is the impulse, do the operatorandblock-diagramdecompositionsproducethesame closed form that you findby peeling awayandguessing?
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5Repeatedroots
5.1 Leaky-tankbackground 645.2 Numericalcomputation 655.3
Analyzing
the
output
signal
67
5.4 Deformingthesystem:Thecontinuityargument 685.5 Higher-ordercascades 70
Afterreadingthischapteryoushouldbeable touseacontinuityargument toexplainthenon-geometricoutputofamodewithadoubleroot.
Modesgeneratepersistentoutputs. Sofarourexamplesgeneratepersistentgeometricsequences. Butamodefromarepeatedroot,suchasfromthesystemfunctional(1R/2)3,producesoutputsthatarenotgeometricsequences. Howdoesrootrepetitionproduce thisseeminglystrange
behavior?Theanalysisdependsontheideathatrepeatedrootsareanunlikely,specialsituation. Ifrootsscatterrandomlyonthecomplexplane, theprobability iszero that tworoots landexactlyon thesameplace. Ageneric,decentsystemdoesnothaverepeatedroots,andonlythroughspecialcontrivancedoesaphysicalsystemacquirerepeatedroots. Thisfactsuggestsdeformingarepeated-rootsystemintoagenericsystembyslightlymovingonerootsothatthemodesofthedeformedsystemproducegeometricsequences. Thisnewsystemisthereforequalitativelyeasiertoanalyzethanistheoriginalsystem,anditcanapproximatetheoriginalsystemascloselyasdesired. Thiscontinuityargumentdependsontheideathattheworld
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655 RepeatedrootsVout R
= .Vin 1 (1 )R
Exercise 27. Whatblock diagram corresponds to this systemfunctional?
Adoublerootarisesfromcascadingtwoidenticalsystems.Hereisitshigh-levelblockdiagramshowingtheinput,intermediate,andoutputsignals:
leakytankorRCcircuit leakytankorRCcircuitV0
V1V2
Itssystemfunctionalisthesquareofthefunctionalforonesystem: 2V2 R
= .V0 1 (1 )R
Thenumerator(R)2 doesnotaddinterestingfeaturestotheanalysis,sosimplifylifebyignoringit.Tosimplifythealgebrafurther,define= 1.WiththatdefinitionandwithouttheboringRfactor,thepurifiedsystemis
V2 1= .V0 (1 R)2
5.2 NumericalcomputationBydesign,thiscascadesystemhasadoublerootat= 1 .Letssimulateitsimpulseresponseandfindpatternsinthedata.Simulationrequireschoosingnumericalvaluesfortheparameters,andheretheonlyparameteris= T/.AnaccuratediscretizationusesatimestepTmuchshorterthanthesystemtimeconstant;otherwisethesystemchangessignificantlybetweensamples,obviatingthediscrete-timeapproximation.Souse1.
Pausetotry 24. Writeaprogramtosimulatetheimpulseresponse,choosingareasonableor.
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66 5.2 Numericalcomputation
The following simulation uses = 0.05 or = 0.95 in computing theimpulseresponse:from scipy import *N = 100impulse = zeros(N)impulse[0] = 1beta = 0.95# return the output signal from feeding INPUT signal througha system# with a feedback loop containing a delay and the given GAIN.def onestage(input, gain):output = input * 0
output[0] = input[0]for i in range(1,len(output)): # 1..n-1
output[i] = input[i] + gain*output[i-1]return output
signal = impulse # start with the impulsefor gain in [beta, beta]: # run it through each system
signal = onestage(signal, gain)print signalTheimpulseresponseis:
n y[n]0 1.0000001 1.9000002 2.7075003 3.4295004 4.0725315 4.6426866 5.1456437 5.5866988 5.9707849 6.302494
10 6.586106. . .
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5 Repeatedroots 67
5.3 AnalyzingtheoutputsignalThe impulse response contains a pattern. To find it, play with the dataandmakeconjectures.Thefirstfewsampleslookliken+ 1.However,byn= 10thatconjecturelooksdubious.Solookforanotherpattern.Asinglesystem(1 R)1wouldhaveageometric-sequenceoutputwherethenthsampleisn. Maybethatgeometricdecayappearsinthedoublesystemandswampstheconjecturedn+ 1growth.Therefore,takeoutthebigpartfromtheimpulseresponsebytabulatingthesignaly[n]/0.95n. Todoso,addonelineofcodetothepreviousprogram:print signal/0.95**arange(N)Thedataare
n y[n]/0.95n0 1.0000001 2.0000002 3.0000003 4.0000004 5.0000005 6.0000006 7.0000007 8.0000008 9.0000009 10.000000
10 11.000000Nowy[n] = n+ 1isexact!Theimpulseresponseofthedoublecascadeisthesignal
y[n] = (n+ 1) 0.95n forn0.The factor of 0.95n makes sensebecause a single system (1 0.95R)1wouldhave0.95nasitsimpulseresponse.Buthowdoesthefactorofn+ 1arise? To understand its origin, one method is convolution, which wasdiscussedinthelecture.Hereweshowanalternativemethodusingacontinuityargument.
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68 5.4 Deformingthesystem:Thecontinuityargument
5.4 Deformingthesystem:ThecontinuityargumentThe cascade is hard to analyzebecause its roots are replicated. So deformthecascadebymakingthesecondrootbe0.951insteadof0.95.Thatslightlydeformedsystemhasthefunctional
1 1 .1 0.951R 1 0.95R
Sincetheroothardlymoved,theimpulseresponseshouldbealmostthesameastheimpulseresponseoftheoriginalsystem. Thisassumptionisthe essence of the continuity argument. We could find the responsebyslightlymodifyingtheprecedingprogram. However,reachingforaprogramtoooftendoesnotaddinsight.Alternatively,nowthatthesystemsrootsareunequal,wecaneasilyusepartialfractions.Thefirststepinpartialfractionsistofindthemodes:
1 1M1 = and M2 = .
1 0.951R 1 0.95RThesystemfunctionalisalinearcombinationofthesemodes:
1 1 C1 C2 = .1 0.951R 1 0.95R 1 0.951R 1 0.95R
Exercise 28. ShowthatC1 = 951andC2 = 950.Thepartial-fractionsdecompositionis
1 1 1 0.951 0.95 = .1 0.95R 1 0.951R 0.001 1 0.951R 1 0.95R
The0.951/(1 0.951R) systemcontributestheimpulseresponse0.951n+1,andthe0.95/(1 0.95R) systemcontributestheimpulseresponse0.95n+1.
Exercise 29. Checktheseimpulseresponses.
Sotheimpulseresponseofthedeformedsystemisy[n] = 1000(0.951n+1 0.95n+1).
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695 Repeatedroots
Since 0.951 0.95, the difference in parentheses is tiny. However, thedifferenceismagnifiedbythefactorof1000outsidetheparentheses. Theresultingsignalisnottiny,andmightcontainthenon-geometricfactorofn
+ 1
in
the
impulse
response
of
atrue
double
root.
Toapproximatethedifference0.951n+1 0.95n+1,usethebinomialtheorem,keepingonlythetwolargestterms:
0.951n+1 = (0.95+ 0.001)n+1 0.95n+1 + (n+ 1)0.95n 0.001+ .
Thustheapproximateimpulseresponseisy[n] 1000 (n+ 1) 0.95n 0.001.
Thefactorof1000cancelsthefactorof0.001toleavey[n] (n+ 1) 0.95n,
whichiswhatweconjecturednumerically!Thusthelinearprefactorn+ 1comesfromsubtractingtwogarden-variety,geometric-sequencemodesthatarealmost identical. Thesignreflectsthatwekeptonlythefirsttwotermsinthebinomialexpansionof0.951n+1.However, as the deformation shrinks, the shifted root at 0.951becomesinstead0.9501or0.95001,etc. Astherootapproaches0.95,thebinomialapproximationbecomesexact,asdoestheimpulseresponse(n+ 1) 0.95n.Theresponse(n+ 1) 0.95n istheproductofanincreasingfunctionwithadecreasingfunction,witheachfunctionfightingforvictory. Insuchsituations,onefunctionusuallywinsatthen 0extreme, and theotherfunctionwinsatthen extreme,withamaximumproductwherethetwofunctionsarrangeadraw.
Exercise 30. Sketchn+ 1,0.95n,andtheirproduct.
Pausetotry 25. Whereisthemaximumof(n+ 1) 0.95n?Thisproductreachesamaximumwhentwosuccessivesamp
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