aircraft control lecture 10

7/29/2019 aircraft control Lecture 10

1/41

16.333Lecture# 10StateSpaceControl

Basicstatespacecontrolapproaches


2/41

Fall2004 16.33391StateSpaceBasics

Statespacemodelsareoftheformx(t) = Ax(t)+Bu(t)y(t) = Cx(t)+Du(t)

withassociatedtransferfunctionG(s)=C(sIA)1B+D

Note:mustformsymbolicinverseofmatrix(sIA),whichishard. Time response:Homogeneouspartx =Ax, x(0)known

TakeLaplacetransformX(s)=(sIA)1x(0) x(t)=L1 (sIA)1 x(0)

I A A2Butcanshow(sIA)1 = +

s2 + s3 +...s1soL1 (sIA)1 =I+At+2!(At)2+...=eAt

Givesx(t)=eAtx(0)whereeAt isMatrixExponential1

3CalculateinMATLABRusingexpm.mandnotexp.m

Time response: ForcedSolutionMatrixcasex =Ax+Buwherexisann-vectoranduisam-vector.Camshowt

x(t) = eAtx(0)+ eA(t)Bu()d0

ty(t) = CeAtx(0)+ CeA(t)Bu()d+Du(t)

0CeAtx(0)istheinitialresponseCeA(t)B istheimpulseresponseofthesystem.

1MATLABR isatrademarkoftheMathworksInc.


3/41

Fall2004 16.33392Dynamic Interpretation

SinceA=TT1,then TeAt | | 1t . . . w..1 e =TetT1 =v1 . vn

| | nt Te wn wherewehavewritten

T w.1

.T1 = .T wn

whichisacolumnofrows. Multiplythisexpressionoutandwegetthat

nAt it Te = e viwi

i=1

AssumeAdiagonalizable,thenx =Ax,x(0)given,hassolutionx(t) = eAtx(0)=TetT1x(0)

n

= eitvi{wiTx(0)}i=1

n

it

= e viii=1

Statesolutionisalinearcombinationofthesystemmodesvieieit Determinesthenatureofthetimeresponse

viDetermines

extent

to

which

each

state

contributes

to

that

mode

i Determinesextenttowhichtheinitialconditionexcitesthemode


4/41

Fall2004 16.33393 Notethatthevi givetherelativesizingoftheresponseofeachpartofthestatevectortotheresponse.

1v1(t)= et mode100.5

v2(t)= e3t mode20.5 Clearlyeit givesthetimemodulation

i realgrowing/decayingexponentialresponsei complexgrowing/decayingexponentialdampedsinusoidal

Bottom line: The locationsoftheeigenvaluesdeterminethepolelocationsforthesystem,thus:Theydeterminethestabilityand/orperformance&transientbe-haviorofthesystem.

It istheir locationsthatwewillwanttomodifywiththecontrollers.


5/41

Fall2004 16.33394Full-stateFeedbackController

Assumethatthesingle-inputsystemdynamicsaregivenbyx = Ax+ Buy = Cx

sothatD= 0.Themulti-actuator case is quite a bitmore complicated aswewouldhavemanyextradegreesoffreedom.

RecallthatthesystempolesaregivenbytheeigenvaluesofA.Want to use the input u(t) tomodify the eigenvalues of A tochangethesystemdynamics.

r u yOO // A,B,C //

x

K

Assumeafull-statefeedbackoftheform:u= rKx

wherer issomereference inputandthegainK isR1nIfr= 0,wecallthiscontrolleraregulator


6/41

Fall2004 16.33395 Findtheclosed-loopdynamics:

x = Ax+ B(rKx)= (ABK)x+ Br= Aclx+ Br

y = Cx

Objective:PickK sothatAcl hasthedesiredproperties,e.g.,Aunstable,wantAcl stablePut2polesat2 2j

NotethattherearenparametersinKandneigenvaluesinA,soitlookspromising,butwhatcanweachieve?

Example#1:Consider:1 1

x =1 2

1x+ u

0Then

det(sIA) = (s1)(s2) 1 = s2 s+ 1 = 03sothesystemisunstable.

Defineu= k1 k2 x= Kx,then 1 1 1 1 k1 1 k2

Acl = ABK= k1 k2 =1 2 0 1 2

Sothenwehavethatdet(sIAcl) = s2+ (k13)s+ (1 2k1+ k2) = 0


7/41

Fall2004 16.33396Thus,bychoosingk1 andk2,wecanputi(Acl) anywhereinthecomplexplane(assumingcomplexconjugatepairsofpoles).

Toputthepolesats= 5, 6,comparethedesiredcharacteristicequation

(s+ 5)(s+ 6) = s2+ 11s+ 30 = 0withtheclosed-loopone

2s + (k13)x+ (1 2k1+ k2) = 0toconcludethat

k13 = 11 k1 = 141 2k1+ k2 = 30 k2 = 57

sothatK= 14 57 ,whichiscalledPolePlacement. Ofcourse, it isnotalwaysthiseasy,asthe issueofcontrollabilitymustbeaddressed.

Example#2:Considerthissystem:1 1

x =0 2

1x+ u

0withthesamecontrolapproach

1 1 1 1 k1 1 k2Acl = ABK= k1 k2 =0 2 0 0 2sothatdet(sIAcl) = (s1 + k1)(s2) = 0The feedbackcontrol canmodify thepoleat s = 1, but it cannotmovethepoleats= 2.

This system cannot be stabilizedwith full-state feedbackcontrol.

Whatisthereasonforthisproblem?


8/41

Fall2004 16.33397Itisassociatedwithlossofcontrollabilityofthee2tmode.

Basictestforcontrollability: rankMc =n B AB 1 1= 1

0 10Mc = 0 2SothatrankMc 1=


9/41

Fall2004 16.33398AckermannsFormula

Thepreviousoutlinedadesignprocedureand showedhow todo itbyhandforsecond-ordersystems.Extendstohigherorder(controllable)systems,buttedious.

AckermannsFormulagivesusamethodofdoingthisentiredesignprocessisoneeasystep.

K= 0 . . . 0 1 M1d(A)cMc = B AB . . . An1Bd(s) is the characteristic equation for the closed-loop poles,whichwethenevaluatefors= A.

Itisexplicitthatthesystemmustbecontrollablebecauseweareinvertingthecontrollabilitymatrix.

RevisitExample#1: d(s) = s2+ 11s + 30 1 1 1 1 1 1

= B AB = =Mc 0 1 2 0 0 1So 1 11 2 1 1 1 1K = 0 1 + 11 + 30I

0 1 1 2 1 2 43 14

= 0 1 = 14 5714 57

AutomatedinMatlab: place.m&acker.m(seepolyvalm.mtoo)


10/41

Fall2004 16.33399 Origins? Forsimplicity,considerathird-ordersystem(case#2),butthisextendstoanyorder. a1 a2 a3 1

A= 1 0 0 B=0 C= b1 b2 b30 1 0 0

This form is useful because the characteristic equation for the2systemisobviousdet(sIA)=s3+a1s +a2s+a3 =0

Canshowthat a1 a2 a3 1

Acl =ABK = 1 0 00 k1 k2 k30 1 0 0

a1k1 a2k2 a3k3=

1 0 0

0 1 0sothatthecharacteristicequationforthesystemisstillobvious:

2cl(s)=det(sIAcl)=s3+(a1+k1)s +(a2+k2)s+(a3+k3)=0 Wethencomparethiswiththedesiredcharacteristicequationdevel-opedfromthedesiredclosed-looppolelocations:

2d(s)=s3+(1)s +(2)s+(3)=0togetthat

a1+k1 =1 k1 =.1a1

.. .. .an+kn =n kn =nan

Pole

placement

is

avery

powerful

tool

and

we

will

be

using

it

for

mostofourstatespacework.


11/41

Fall2004 16.333910AircraftStateSpaceControl

Cannowdesignafullstatefeedbackcontrollerforthedynamics:xsp =Aspxsp+Bspe

withdesiredpolesbeingatn =3and=0.6s=1.82.4id(s)=s2+3.6s+9

Ksp=place(Asp,Bsp,[roots([1 2*0.6*3 3^2])])w Designcontrolleru= 0.0264 2.3463q

Withfullmodel,couldarrangeitsophugoidpolesremaininthesameplace,justmovetheonesassociatedwiththeshortperiodmode

s=1.82.4i, 0.00330.0672iev=eig(A);% damp short period, but leave the phugoid where it isPlist=[roots([1 2*.6*3 3^2]) ev([3 4],1)];K1=place(A,B(:,1),Plist)

uwu= 0.0026 0.0265 2.3428 0.0363 q


12/41

Fall2004 16.333911 Canalsoaddthelagdynamicstoshortperiodmodelwithincluded

xsp = Aspxsp+ a

a

=

4

c

Bsp e; e s+4 ex = 4x +4ec, a =xexsp = Asp Bsp xsp 0+ ce x 0 4 x 4

Adds=3todesiredpolelistPlist=[roots([1 2*.6*3 3^2]),-.25,-3];At2=[Asp2 Bsp2(:,1);zeros(1,3) -4];Bt2=[zeros(3,1);4];Kt=place(At2,Bt2,Plist);step(ss(At2-Bt2*Kt2,Bt2,[0 0 1 0],0),35)

wq

x

u= 0.0011 3.4617 4.9124 0.5273

0 5 10 15 20 25 30 350.25

0.2

0.15

0.1

0.05

0

Step Response

Time (sec)

r

ads

Noproblemworkingwithlargersystemswithstatespacetools Maincontrolissueisfindinggoodlocationsforclosed-looppoles


13/41

Fall2004 16.333912Estimators/Observers

Problem: So farwehaveassumedthatwehave fullaccessto thestatex(t)whenwedesignedourcontrollers.Mostoftenallofthisinformationisnotavailable.

Usually can only feedback information that is developed from thesensorsmeasurements.Couldtryoutputfeedback

u=Kx u=KySameastheproportionalfeedbackwelookedatatthebeginningoftherootlocuswork.

Thistypeofcontrolisverydifficulttodesigningeneral.

Alternative approach: Develop a replicaof thedynamic systemthatprovidesanestimateofthesystemstatesbasedonthemea-suredoutputofthesystem.

Newplan:1.Developestimateofx(t)thatwillbecalledx(t).2.Thenswitchfromu=Kx(t)tou=Kx(t).

Twokeyquestions:Howdowefindx(t)?Willthisnewplanwork?


14/41

Fall2004 16.333913EstimationSchemes

Assumethatthesystemmodelisoftheform:x = Ax+Bu, x(0)unknowny = Cx

where1.A,B,andC areknown.2.u(t)isknown3.

Measurable

outputs

are

y(t)

from

C

=

I

Goal:Developadynamicsystemwhosestatex(t)=x(t)

foralltimet0.Twoprimaryapproaches:Open-loop.Closed-loop.


15/41

Fall2004 16.333914Open-loopEstimator

Given thatweknow theplantmatricesandthe inputs,wecanjustperformasimulationthatrunsinparallelwiththesystem

x(t)=Ax +Bu(t)

Thenx(t)x(t)tprovidedthatx(0)=x(0)

MajorProblem:Wedonotknowx(0)SystemA,B,C

x(t)y(t)

//

u(t)//

//ObserverA,B,C

x(t)y(t)

//

Analysisofthiscase. Startwith:x(t) = Ax+Bu(t)x(t) = Ax +Bu(t)

Definetheestimationerror: x(t)=x(t)x(t).Nowwantx(t)=0t.Butisthisrealistic?


16/41

Fall2004 16.333915 Subtracttoget:

d(x x) = A(x x) x(t) = Axdt

whichhasthesolutionx(t) = eAtx(0)

Givestheestimationerrorintermsoftheinitialerror.

Doesthisguaranteethatx = 0 t?Oreventhatx 0 ast ? (whichisamorerealisticgoal).Responseisfineifx(0) = 0.Butwhatifx(0) = 0?

IfAstable,thenx 0 ast ,butthedynamicsoftheestima-tion errorarecompletelydeterminedby theopen-loopdynamicsofthesystem(eigenvaluesofA).Couldbeveryslow.Noobviouswaytomodifytheestimationerrordynamics.

Open-loopestimationdoesnotseemtobeaverygood idea.


17/41

Fall2004 16.333916Closed-loopEstimator

Obviouswaytofixtheproblem istousetheadditional informationavailable:Howwelldoestheestimatedoutputmatchthemeasuredoutput?

Compare: y=Cx with y =CxThenformy =yy Cx

SystemA,B,Cx(t)y(t)

//

+

u(t)

//

//

L

ObserverA,B,Cx(t) y(t)//

OO

Approach: Feedbacky toimproveourestimateofthestate.Basicformoftheestimatoris:

x(t) = Ax(t)+Bu(t)+ Ly(t)y(t) = Cx(t)

whereLisauserselectablegainmatrix. Analysis:

x =x x = [Ax+Bu] [Ax +Bu+L(yy)]= A(xx)L(CxCx)= Ax LCx =(ALC)x


18/41

Fall2004 16.333917 Sotheclosed-loopestimationerrordynamicsarenow

x =(A LC)x withsolution x(t)=e(ALC)t x(0)

Bottom line: Can select the gain L to attempt to improve theconvergenceoftheestimationerror(and/orspeeditup).Butnowmustworryaboutobservabilityofthesystemmodel.

Notethesimilarity:

RegulatorProblem: pickK forA BK3ChooseK R1n (SISO)suchthattheclosed-looppoles

det(sI A+BK)=c(s)areinthedesiredlocations.

EstimatorProblem: pickLforA LC3ChooseL Rn1 (SISO)suchthattheclosed-looppoles

det(sI A+LC)=o(s)areinthedesiredlocations.

Theseproblemsareobviouslyvery similar in fact theyarecalleddualproblems.


19/41

Fall2004 16.333918EstimationGainSelection

Forregulation,wereconcernedwithcontrollabilityof(A,B)For a controllable systemwe canplace the eigenvaluesofABK arbitrarily.

For

estimation,

were

concerned

with

observability

of

pair

(A,

C).

Foranobservable systemwe canplace the eigenvaluesofALC arbitrarily.

Testusingtheobservabilitymatrix: rankMorank

CCACA2...

CAn1

=n

The procedure for selecting L is very similar to that used for theregulatordesignprocess.


20/41

Fall2004 16.333919 Oneapproach:

Notethatthepolesof(ALC) and(ALC)T areidentical.Alsowehavethat(ALC)T = AT CTLTSodesigningLT forthistransposedsystem looks likeastandardregulatorproblem(ABK)where

A ATB CTK LT

SowecanuseKe = acker(AT,CT,P) , LKTe

NotethattheestimatorequivalentofAckermannsformulaisthat

L= e(s)M

1o

0...01


21/41

Fall2004 16.333920SimpleEstimatorExample

Simplesystem 1 1.5 1 .5A = , B= , x(0) = 0

1 2 0 1C = 1 0 , D= 0

Assumethattheinitialconditionsarenotwellknown.Systemstable,butmax(A) = 0.18Testobservability:

Crank

CA1 0

= rank1 1.5

Useopenandclosed-loopestimators. Sincetheinitialconditionsare0

notwellknown,usex(0) = 0 Open-loopestimator:

x = Ax + Buy = Cx

Closed-loopestimator:x = Ax + Bu+ Ly = Ax + Bu+ L(yy)

= (ALC)x+ Bu+ Lyy = Cx

Which isadynamicsystemwithpolesgivenbyi(ALC) andwhichtakesthemeasuredplantoutputsasaninputandgeneratesanestimateofx.


22/41

Fall2004 16.333921 Typicallysimulatebothsystemstogetherforsimplicity

Open-loopcase:x = Ax+Buy = Cxx = Ax +Buy = Cx

0.51x A 0 x

B x(0)= + u , =x(0) x 0 A x By

C 0 x

=y 0 C x

Closed-loopcase:x = Ax+Bux = (ALC)x+Bu+LCx

x A 0 x B= u

B

x LC ALC x +

Exampleusesastrongu(t)toshakethingsup

00


23/41

Fall2004 16.333922

0 0.5 1 1.5 2 2.5 3 3.5 41

0.5

0

0.5

1Open loop estimator

states

time

x1x2

0 0.5 1 1.5 2 2.5 3 3.5 41

0.5

0

0.5

1

estimationerror

time

Figure1:Open-loopestimator. Estimationerrorconvergestozero,butveryslowly.

0 0.5 1 1.5 2 2.5 3 3.5 41

0.5

0

0.5

1

Closedloop estimator

states

time

x1x2

0 0.5 1 1.5 2 2.5 3 3.5 41

0.5

0

0.5

1

estimationerror

time

Figure2: Closed-loopestimator. Convergence looksmuchbetter.


24/41

Fall2004 16.333923AircraftEstimationExample

TakeShortperiodmodelandassume thatwecanmeasure q. Canweestimatethemotionassociatedwiththeshortperiodmode?

xsp = Aspxsp+Bspuy = 0 1 xsp

Takexsp(0)=[0.5;0.05]T Systemstable,socoulduseanopenloopestimator Forclosed-loopestimator,putdesiredpolesat3,4 Forthevariousdynamicsmodelsasbefore

Csp=[0 1]; % sense qKe=place(Asp,Csp,[-3 -4]);Le=Ke;

0 2 4 6 8 1020

15

10

5

0

5

10

x1

Openloop estimator

time0 2 4 6 8 10

0.1

0.05

0

0.05

0.1

x1

time

0 2 4 6 8 108

6

4

2

0

2

4

x1

error

time0 2 4 6 8 10

0.05

0.04

0.03

0.02

0.01

0

0.01

0.02

x2

error

time

0 2 4 6 8 1020

15

10

5

0

5

10

15

x1

Closedloop estimator

time0 2 4 6 8 10

0.1

0.05

0

0.05

0.1

x1

time

0 2 4 6 8 1020

15

10

5

0

x1

error

time0 2 4 6 8 10

0.05

0.04

0.03

0.02

0.01

0

0.01

x2

error

time

Figure3: Closed-loopestimator. Convergence looksmuchbetter. Asexpected,theOLestimatordoesnotdowell,buttheclosed-looponeconvergesnicely


25/41

Fall2004 16.333924WheretoputtheEstimatorPoles?

LocationheuristicsforpolesstillapplyuseBessel,ITAE,...Maindifference: probablywanttomaketheestimatorfasterthanyou intend tomake the regulator should enhance thecontrol,whichisbasedonx(t).

ROT:Factorof23inthetimeconstantn associatedwiththeregulatorpoles.

Note: When designing a regulator, were concerned with band-width of the control getting too high often results in controlcommandsthatsaturatetheactuatorsand/orchangerapidly.Differentconcernsfortheestimator:Loopclosedinsidecomputer,sosaturationnotaproblem.However,themeasurementsyareoftennoisy,andweneedtobecarefulhowweusethemtodevelopourstateestimates.

Highbandwidthestimatorstendtoaccentuatetheeffectofsens-ingnoiseintheestimate.Stateestimatestendtotrackthemeasurements,whicharefluc-tuatingrandomlyduetothenoise.

LowbandwidthestimatorshavelowergainsandtendtorelymoreheavilyontheplantmodelEssentiallyanopen-loopestimatortendstoignorethemeasure-mentsandjustusestheplantmodel.


26/41

Fall2004 16.333925 Canalsodevelopanoptimalestimatorforthistypeofsystem.

Which isapparentlywhatKalmandidoneevening in1958whiletakingthetrainfromPrincetontoBaltimore...

Balances effect of the various types of random noise in thesystemontheestimator:

x = Ax+Bu+Bwwy = Cx+v

where:3w: processnoisemodelsuncertaintyinthesystemmodel.3v: sensornoisemodelsuncertaintyinthemeasurements.

FinalThoughts Note that the feedback gain L in the estimator only stabilizes theestimationerror.Ifthesystem isunstable,thenthestateestimateswillalsogoto

,withzeroerrorfromtheactualstates. Estimationisanimportantconceptofitsown.

NotalwaysjustpartofthecontrolsystemCriticalissueforguidanceandnavigationsystem

Morecompletediscussionrequiresthatwestudystochasticprocessesandoptimizationtheory.

Estimationisallaboutwhichdoyoutrustmore: yourmea-surementsoryourmodel.


27/41

Fall2004 16.333926CombinedRegulatorandEstimator

Asadvertised,wecanchangethepreviouscontrolu=Kxtothenewcontrolu=Kx (sameK).Wenowhavex = Ax+Buy = Cxx = Ax +Bu+L(yy)y = Cx

withclosed-loopdynamics x A BK xx

=LC ABKLC x xcl =Aclxcl

Notobviousthatthissystemwillevenbestable: i(Acl)


28/41

Fall2004 16.333927 Absolutelykeypoints:

1.i(Acl)i(Acl)why?2.Acl isblockuppertriangular,socanfindpolesbyinspection:

det(sIAcl)=det(sI(ABK))det(sI(ALC))

Theclosed-looppolesofthesystemconsistoftheunionofthe regulatorandestimatorpoles

Sowecandesigntheestimatorandregulatorseparatelywithconfi-dencethatcombinationofthetwowillworkVERYwell.

Compensatorisacombinationoftheestimatorandregulator.x = Ax +Bu+L(yy)

= (ABKLC)x+Lyu

=

Kx

xc = Acxc+Bcyu = Ccxc

Keeptrackofthisminussign.Weneedoneinthefeedbackpath,butwecanmoveitaroundtosuitourneeds.


29/41

Fall2004 16.333928 LetGc(s) bethecompensatortransferfunctionwhere

u=

Cc(sI

Ac)

1

Bc = Gc(s)y= K(sI(ABKLC))1L

sobymydefinition,u= Gcy Gc(y)

Reason formaking the definition is that when we implement thecontroller,weoftendonotjustfeedbacky(t),butinsteadhavetoincludeareferencecommandr(t)Useservoapproachandfeedbacke(t) = r(t) y(t) instead

r//

e// Gc(s) u // G(s) y//

OO

Sonowu= Gce= Gc(ry).Andifr= 0,thenwestillhaveu= Gc(y)

Importantpoints:Closed-loopsystemwillbestable,butthecompensatordynamicsneednotbe.

OftenverysimpleandusefultoprovideclassicalinterpretationsofthecompensatordynamicsGc(s).


30/41

Fall2004 16.333929 Mechanicsofclosingtheloop

G(s) :x = Ax+ Buy = Cx

Gc(s) :xc = Acxc+ Bceu = Ccxc

ande= ry,u= Gce,y= Gu.

LoopdynamicsL= GGc y= L(s)ex = Ax+ Bu= Ax+ BCcxc

xc = Acxc+ Bce

x A BCc x 0= + exc 0 Ac xc Bc x

y = C 0xc

Nowformtheclosed-loopdynamicsbyinsertinge= ry x A BCc x 0 x=xc 0 Ac xc + Bc r C 0 xc

A BCc x 0= + r

BcC Ac xc Bc

xy = C 0xc


31/41

Fall2004 16.333930Performance Issue

OftenfindwithstatespacecontrollersthattheDCgainoftheclosedloopsystemisnot1. Soy=r insteadystate.

RelativelysimplefixistomodifytheoriginalcontrollerwithscalarNu=rKxu=NrKx

Closed-loopsystemonpage5becomesx =Ax+B(NrKx)=Aclx+BNr Gcl(s)=C(sIAcl)1BNy =CxAnalyzesteadystatestepresponseyss =Gcl(0)rstep

Gcl(0)=C(Acl)1BNAndpickN sothatGcl(0)=1N = 1(C(Acl)1B)

AbitmorecomplicatedwithacombinedestimatorandregulatorOnesimpleway(notthebest)ofachievingasimilargoalistoadd

N torandforceGcl(0)=1Nowtheclosed-loopdynamicsonpage29become:x x=Acl +BclNrxc 1N =

(Ccl(Acl)1Bcl)xc xy =Ccl xc

Notethatthisfixesthesteadystatetrackingerrorproblems,but inmyexperiencecancreatestrangetransients(oftenNMP).


32/41

Fall2004 16.333931Example: CompensatorDesign

1 x =Ax+BuG(s)=s2+s+1 y=Cx

where 0 1 0

A= B= C= 1 01 1 1

Regulator: Want regulator poles to have a time constant of c =1/(n)=0.25sec(ABKr)=44jwhichcanbefoundusingplaceorackerK_r=acker(a,b,[-4+4*j;-4-4*j]);togiveKr = [ 31 7 ]

Estimator:wanttheestimatorpolestobefaster,sousee

=

1/(n)

=

0.1

sec.

Use

real

poles,

(A

LeC)

=

10

L_e=acker(a,c,[-10 -10]);

19whichgivesLe = 80

FormcompensatorGc(s)ac=a-b*K_r-L_e*c;bc=L_e;cc=K_r;dc=0;

Ac = 19 1 19 Bc = Cc = 31 7112 808(s+2.5553) u

Gc(s)=1149 =s2+27s+264 e

Lowfrequencyzero,withhigherfrequencypoles(likealead)


33/41

Fall2004 16.333932

101

100

101

102

100

101

102

Freq (rad/sec)

Mag

Plant GCompensator Gc

101

100

101

102

200

150

100

50

0

50

Freq (rad/sec)

Phase

(deg)

Plant GCompensator Gc

Figure4:Thecompensatordoesindeedlooklikeahighfrequencylead(amplificationfrom216rad/sec). Plantprettysimple looking.


34/41

Fall2004 16.333933

101

100

101

102

100

101

102

Freq (rad/sec)

Mag

Loop L

101

100

101

102

250

200

150

100

50

0

Freq (rad/sec)

Phase

(deg)

Figure5:ThelooptransferfunctionL= GcGshowsaslopechangearoundc = 5rad/secduetotheeffectofthecompensator. Significantgainandphasemargins.


35/41

Fall2004 16.333934

150

100

50

0

50

Magnitude(dB)

102

101

100

101

102

103

270

225

180

135

90

45

0

Phase

(deg)

Bode DiagramGm = 14.3 dB (at 14.9 rad/sec) , Pm = 45.8 deg (at 4.84 rad/sec)

Frequency (rad/sec)

Figure6: Quitesignificantgainandphasemargins.


36/41

Fall2004 16.333935

40 35 30 25 20 15 10 5 0 5 1025

20

15

10

5

0

5

10

15

20

25

Root Locus

Real Axis

ImaginaryAxis

Figure7: Freezethecompensatorpolesandzerosanddrawaroot locusversusanadditionalplantgain,G(s) G(s) =

(s2+

s+1).Notelocationoftheclosed-looppoles!!


37/41

Fall2004 16.333936

101

100

101

102

102

101

100

101

102

Freq (rad/sec)

Mag

Factor of N=1.0899 applied to Closed loop

Plant Gclosedloop G

cl

Figure8: Closed-looptransfersystembandwidthhas increasedsubstantially.


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Fall2004 16.333937

EstimatorDesign(est1.m)1 clear all2 close all3 figure(1);clf4 set(gcf,DefaultLineLineWidth,2)5 set(gcf,DefaultlineMarkerSize,10)6 figure(2);clf7 set(gcf,DefaultLineLineWidth,2)8 set(gcf,DefaultlineMarkerSize,10)910 load b747 % get A B Asp Bsp11 Csp=[0 1]; % sense q1213 Ke=place(Asp,Csp,[-3 -4]);Le=Ke;1415 xo=[-.5;-.05]; % start somewhere1617 t=[0:.01:10];N=floor(.15*length(t));18 % hit on the system with an input19 %u=0;u=[ones(15,1);-ones(15,1);ones(15,1)/2;-ones(15,1)/2;zeros(41,1)]/5;20 u=0;u=[ones(N,1);-ones(N,1);ones(N,1)/2;-ones(N,1)/2]/20;21 u(length(t))=0;2223 [y,x]=lsim(Asp,Bsp,Csp,0,u,t,xo);24 plot(t,y)2526 % closed-loop estimator27 % hook both up so that we can simulate them at the same time28 % bigger state = state of the system then state of the estimator29 A_cl=[Asp zeros(size(Asp));Le*Csp Asp-Le*Csp];30 B_cl=[Bsp;Bsp];31 C_cl=[Csp zeros(size(Csp));zeros(size(Csp)) Csp];32 D_cl=zeros(2,1);3334 % note that we start the estimators at zero, since that is35 % our current best guess of what is going on (i.e. we have no clue :-) )36 %37 [y_cl,x_cl]=lsim(A_cl,B_cl,C_cl,D_cl,u,t,[xo;0;0]);38 figure(1)39 subplot(221)40 plot(t,x_cl(:,[1]),t,x_cl(:,[3]),--)41 ylabel(x1);title(Closed-loop estimator);xlabel(time);grid42 subplot(222)43 plot(t,x_cl(:,[2]),t,x_cl(:,[4]),--)44 ylabel(x1);xlabel(time);grid45 subplot(223)46

plot(t,x_cl(:,[1])-x_cl(:,[3]))

47 ylabel(x1 error);xlabel(time);grid48 subplot(224)49 plot(t,x_cl(:,[2])-x_cl(:,[4]))50 ylabel(x2 error);xlabel(time);grid51 print -depsc spest_cl.eps52 jpdf(spest_cl)5354 % open-loop estimator55 % hook both up so that we can simulate them at the same time56 % bigger state = state of the system then state of the estimator57 A_ol=[Asp zeros(size(Asp));zeros(size(Asp)) Asp];58 B_ol=[Bsp;Bsp];59 C_ol=[Csp zeros(size(Csp));zeros(size(Csp)) Csp];60 D_ol=zeros(2,1);6162 [y_ol,x_ol]=lsim(A_ol,B_ol,C_ol,D_ol,u,t,[xo;0;0]);63 figure(2)64 subplot(221)65 plot(t,x_ol(:,[1]),t,x_ol(:,[3]),--)66 ylabel(x1);title(Open-loop estimator);xlabel(time);grid


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Fall2004 16.33393867 subplot(222)68 plot(t,x_ol(:,[2]),t,x_ol(:,[4]),--)69 ylabel(x1);xlabel(time);grid70 subplot(223)71

plot(t,x_ol(:,[1])-x_ol(:,[3]))

72 ylabel(x1 error);xlabel(time);grid73 subplot(224)74 plot(t,x_ol(:,[2])-x_ol(:,[4]))75 ylabel(x2 error);xlabel(time);grid76 print -depsc spest_ol.eps77 jpdf(spest_ol)78


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Fall2004 16.333939

Regular/EstimatorDesign(regest.m)1 % Combined estimator/regulator design for a simple system2 % G= 1/(s^2+s+1)3 %4 % Jonathan How5 % Fall 20046 %7 close all;clear all8 for ii=1:59 figure(ii);clf;set(gcf,DefaultLineLineWidth,2);set(gcf,DefaultlineMarkerSize,10)10 end1112 a=[0 1;-1 -1];b=[0 1];c=[1 0];d=0;13 k=acker(a,b,[-4+4*j;-4-4*j]);14 l=acker(a,c,[-10 -10]);15 %16 % For state space for G_c(s)17 %18 ac=a-b*k-l*c;bc=l;cc=k;dc=0;1920 G=ss(a,b,c,d);21 Gc=ss(ac,bc,cc,dc);2223 f=logspace(-1,2,400);24 g=freqresp(G,f*j);g=squeeze(g);25 gc=freqresp(Gc,f*j);gc=squeeze(gc);2627 figure(1);clf28 subplot(211)29 loglog(f,abs(g),f,abs(gc),--);axis([.1 1e2 .2 1e2])30 xlabel(Freq (rad/sec));ylabel(Mag)31 legend(Plant G,Compensator Gc);grid32 subplot(212)33 semilogx(f,180/pi*angle(g),f,180/pi*angle(gc),--);34 axis([.1 1e2 -200 50])35 xlabel(Freq (rad/sec));ylabel(Phase (deg));grid36 legend(Plant G,Compensator Gc)3738 L=g.*gc;3940 figure(2);clf41 subplot(211)42 loglog(f,abs(L),[.1 1e2],[1 1]);axis([.1 1e2 .2 1e2])43 xlabel(Freq (rad/sec));ylabel(Mag)44 legend(Loop L);45 grid46

subplot(212)

47 semilogx(f,180/pi*phase(L.),[.1 1e2],-180*[1 1]);48 axis([.1 1e2 -290 0])49 xlabel(Freq (rad/sec));ylabel(Phase (deg));grid50 %51 % loop dynamics L = G Gc52 %53 al=[a b*cc;zeros(2) ac];54 bl=[zeros(2,1);bc];55 cl=[c zeros(1,2)];56 dl=0;57 figure(3)58 rlocus(al,bl,cl,dl)59 %60 % closed-loop dynamics61 % unity gain wrapped around loop L62 %63 acl=al-bl*cl;bcl=bl;ccl=cl;dcl=d;6465 N=inv(ccl*inv(-acl)*bcl)66


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Fall2004 16.33394067 hold on;plot(eig(acl),d);hold off68 grid69 %70 % closed-loop freq response71

%

72 Gcl=ss(acl,bcl*N,ccl,dcl);73 gcl=freqresp(Gcl,f*j);gcl=squeeze(gcl);7475 figure(4);clf76 loglog(f,abs(g),f,abs(gcl),--);77 axis([.1 1e2 .01 1e2])78 xlabel(Freq (rad/sec));ylabel(Mag)79 legend(Plant G,closed-loop G_{cl});grid80 title([Factor of N=,num2str(N), applied to Closed loop])8182 figure(5);clf83 margin(al,bl,cl,dl)8485 figure(1);orient tall;print -depsc reg_est1.eps86 jpdf(reg_est1)87 figure(2);orient tall;print -depsc reg_est2.eps88 jpdf(reg_est2)89 figure(3);print -depsc reg_est3.eps90 jpdf(reg_est3)91 figure(4);print -depsc reg_est4.eps92 jpdf(reg_est4)93 figure(5);print -depsc reg_est5.eps94 jpdf(reg_est5)

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