aircraft control lecture 9

7/29/2019 aircraft control Lecture 9

1/33

16.333Lecture# 9BasicLongitudinalControl

Basicaircraftcontrolconcepts Basiccontrolapproaches


2/33

Fall2004 16.33381BasicLongitudinalControl

Goal: analyzeaircraft longitudinaldynamicstodetermine ifthebe-haviorisacceptable,andifnot,thenmodifyitusingfeedbackcontrol.

Note that we could (and will) work with the full dynamics model,butfornow,letsfocusontheshortperiodapproximatemodelfromlecture75.

xsp =Aspxsp+Bspewheree istheelevatorinput,and

w Zw/m U0xsp = q , Asp = I1(Mw +MwZw/m) I1(Mq +MwU0)yy yyZe/mBsp = I1(Me +MwZe/m)yy

Addthat =q,sos=q

Taketheoutputas,inputise,thenformthetransferfunction(s) 1q(s)

= = 0 1 (sIAsp)1Bspe(s) s e(s)

Asshowninthecode,forthe747(40Kft,M=0.8)thisreducesto:(s) 1.1569s+0.3435

=e(s) s(s2+0.7410s+0.9272)Ge(s)

so that the dominant roots have a frequency of approximately 1rad/secanddampingofabout0.4


3/33

Fall2004 16.33382

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.11

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

0.09

0.95

0.090.20.30.420.540.68

0.84

0.95

0.20.30.420.540.68

0.84

0.2

0.4

0.6

0.8

0.8

0.2

0.4

0.6

PoleZero Map

Real Axis

ImaginaryAxis

Figure1: Pole-zeromapforGqe Basic problem is that there are vast quantities of empirical data to

showthatpilotsdonotliketheflyingqualitiesofanaircraftwiththiscombinationoffrequencyanddampingWhatdotheyprefer?

Acceptable

0.1

1

0

2

3

4

5

6

7

0.2 0.4 2 4

s

s

Unacceptable

Poor

Satisfactory

0.6 0.8 1

Unda

mpednaturalfrequency

rad/sec

Damping ratio

Figure2: ThumbPrintcriterion Thiscriterionhasbeenaroundsincethe1950s,butitisstillvalid.

Goodtarget:

frequency

3rad/sec

and

damping

of

about

0.6

Problem is that the short period dynamics are no where near these

numbers,sowemustmodifythem.


4/33

Fall2004 16.33383Coulddoitbyredesigningtheaircraft,butitisabitlateforthat...

3.5 3 2.5 2 1.5 1 0.5 03

2

1

0

1

2

3

0.84

0.7 0.56 0.10.20.320.44

0.7 0.44 0.32

0.95

0.10.2

2.5

0.84

0.56

3

0.95

0.511.52

PoleZero Map

Real Axis

ImaginaryAxis

Figure3: Pole-zeromapandtargetpole locations

Ofcoursethereareplentyofotherthingsthatwewillconsiderwhenwe

design

the

controllers

Smallsteadystateerrortocommandse withinlimitsNooscillationsSpeedcontrol


5/33

Fall2004 16.33384FirstShortPeriodAutopilot

Firstattempttocontrolthevehicleresponse: measureandfeed itbacktotheelevatorcommande.Unfortunatelytheactuator isslow,sothere isanapparent lag in

theresponsethatwemustmodelce

//4

s+ 4

ae

// Ge(s) //

ook

OO

coo

Dynamics: a is the actual elevator deflection, ec is the actuatorecommandcreatedbyourcontroller

4= Ge(s)ea; a = H(s)ec; H(s) =e s+ 4Thecontrolisjustbasicproportionalfeedback

c = k(c)eWhichgivesthat

= Ge(s)H(s)k(c)orthat

(s) Ge(s)H(s)k=c(s) 1 + Ge(s)H(s)k

Looksgood,buthowdoweanalyzewhatisgoingon?

Need

to

be

able

to

predict

where

the

poles

are

going

as

afunction

ofk RootLocus


6/33

Fall2004 16.33385RootLocusBasics

r e u y//OO //

Gc(s) // Gp(s) //

AssumethattheplanttransferfunctionisoftheformGp = Kp

Np

= Kpi(szpi)Dp i(sppi)

andthecontrollertransferfunctionisNc

Gc(s) = Kc = Kci(szci)Dc i(spci)

Signalsare:u controlcommandsy output/measurementsr referenceinpute responseerror

This istheunity feedback form. WecouldaddthecontrollerGc inthefeedbackpathwithoutchangingthepolelocations.

Willadd

disturbances

later.


7/33

Fall2004 16.33386 Basicquestions:

Analysis: GivenNc andDc,wheredotheclosed looppolesgoasafunctionofKc?

Synthesis: GivenKp,NpandDp,howshouldwechoseKc,Nc,Dctoputtheclosedlooppolesinthedesiredlocations?

Block diagram analysis: Sincey = GpGceande= ry,theneasytoshowthat

y GcGp=

r 1 + GcGpGcl(s)

whereKcKpNcNp

Gcl(s) =DcDp+ KcKpNcNp

istheclosed looptransfer function

Thedenominator iscalledthecharacteristic equationc(s) andtherootsofc(s) = 0 arecalledtheclosed-looppoles(CLP).

TheCLPareclearly functionsofKc foragivenKp,Np,Dp,Nc,Dcalocusofroots[Evans,1948]


8/33

Fall2004 16.33387RootLocusAnalysis

GeneralrootlocusishardtodeterminebyhandandrequiresMatlabtoolssuchasrlocus(num,den)toobtainfullresult,butwecangetsomeimportantinsightsbydevelopingashortsetofplottingrules.FullrulesinFPE,page260.

Basicquestions:1.Whatpointsareontherootlocus?2.Wheredoestherootlocusstart?3.Wheredoestherootlocusend?4.When/whereisthelocusontherealline?5.Giventhats0 is foundtobeonthe locus,whatgain isneed for

thattobecometheclosed-looppolelocation?

Question#1: ispoints0 ontherootlocus? AssumethatNc andDcareknown,let

Ld =NcDc

NpDp

andK= KcKp thenc(s) = 1 + KLd(s) = 0

orequivalentlyforvaluesofsforwhichLd(s) = 1/K,withKreal.ForK positive,s0 isontherootlocusif

Ld(s0) = 180 l 360, l= 0,1,...

IfK negative,s0 isontherootlocusif [0 locus]Ld(s0) = 0

l 360, l= 0,1,...Theseareknownasthephaseconditions.


9/33

Fall2004 16.33388 Question#2: Wheredoestherootlocusstart?

NcNpc = 1 + K = 0DcDp

DcDp+ KNcNp = 0So ifK 0,then locusstartsatsolutionsofDcDp = 0 whicharethepolesoftheplantandcompensator.

Question#3: Wheredoestherootlocusend?Alreadyshownthatfors0 tobeonthelocus,musthave

1Ld(s0) =

KSoifK ,thepolesmustsatisfy:

NcNpLd = = 0DcDp

Thereareseveralpossibilities:

1.Polesare locatedatvaluesofs forwhichNcNp = 0, whicharethezerosoftheplantandthecompensator

2.IfLoopLd(s) hasmorepolesthanzerosAs s , Ld(s)| 0,butwemustensurethatthephase| | |

conditionis

still

satisfied.


10/33

Fall2004 16.33389 MoredetailsasK :

AssumetherearenzerosandppolesofLd(s)Thenforlarge|s|,

1Ld(s)

(s )pnSotherootlocusdegeneratesto:

11+ =0

n

(s )

p

SonpolesheadtothezerosofLd(s)Remainingp n poles head to s = along asymptotes| |

definedbytheradiallines180+360 (l 1)

l = l=1,2,...p n

sothatthenumberofasymptotes isgovernedbythenumberofpolescomparedtothenumberofzeros(relativedegree).Ifzi arethezeros ifLd andpj arethepoles,thenthecentroid

oftheasymptotesisgivenby:p n

zipj =

p n

Example: L(s)=s4

0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.80.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

Root Locus

Real Axis

ImaginaryAxis


11/33

Fall2004 16.333810 Question#4: When/whereisthelocusontherealline?

Locuspointsonthereal linearetothe leftofanodd numberofrealaxispolesandzeros[K positive].

Explanationabittoodetailedandnotthatrelevant

Question#5: Giventhats0 is foundtobeonthe locus,whatgainisneededforthattobecometheclosed-looppolelocation?

Need

1 =K

Ld(s0)

Dp(s0)Dc(s0)Np(s0)Nc(s0)| |

SinceK=KpKc,signofKc dependsonsignofKp3

e.g.,assume

that

Ld(s0)

=

180

,then

need

Kc and Kp tobesamesignsothatK>0


12/33

Fall2004 16.333811RootLocusExamples

Figure4: Basic Figure5: Twopoles

Figure6: Addzero Figure7: Threepoles

Figure8: Addzero

Examplessimilar

to

control

design

process:

add

compensator

dynamics

to

mod-

ifyrootlocusandthenchosegaintoplaceCLPatdesiredlocationonthelocus.


13/33

Fall2004 16.333812

Figure9: Complexcase

5 4 3 2 1 0 1 2 3 41

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

Root Locus

Real Axis

ImaginaryAxis

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.55

4

3

2

1

0

1

2

3

4

5

Root Locus

Real Axis

ImaginaryAxis

Figure10: Verycomplexcase


14/33

Fall2004 16.333813DynamicCompensation

For a given plant, can draw a root locus versus K. But if desiredpole locations are not on that locus, then need to modify it usingdynamiccompensation.Basicrootlocusplotsgiveusanindicationoftheeffectofadding

compensatordynamics. Butneedtoknowwhattoaddtoplacethepoleswherewewantthem.

Newquestions:Whattypeofcompensationisrequired?Howdowedeterminewheretoputtheadditionaldynamics?

Therearethreeclassictypesofcontrollersu= Gc(s)e

1.Proportionalfeedback: Gc Kg again,sothatNc = Dc = 1Samecasewehavebeenlookingat.

2.Integralfeedbackt Ki

u(t) = Ki e()d Gc(s) =0 s

Usedtoreduce/eliminatesteady-stateerrorIfe() isapproximatelyconstant,thenu(t) willgrowtobevery

largeandthushopefullycorrecttheerror.


15/33

Fall2004 16.333814Considererrorresponseof

Gp

(s) = 1/(s+ a)(s+ b)(a>0,b>0)toastep,

r(t) = 1(t) r(s) = 1/swhere

er =

11 + GcGp e(s) =

r(s)(1 + GcGp)

Toanalyzeerror,useFVTlime(t) = lim se(s)

s 0t sothatwithproportionalcontrol,

s 1 1limess =lim =

t s0 s 1 + KgGp(s) 1 +Kgabsocanmakeess small,butonlywithaverylargeKg

Withintegralcontrol,lims 0Gc(s) = ,soess 0

Integral control improves the steady state, but this is attheexpenseofthetransientresponse(typicallygetsworsebecausethesystemislesswelldamped)


16/33

Fall2004 16.3338151

Example#1: G(s)= ,addintegralfeedbackto(s+a)(s+b)

improvethesteadystateresponse.

6 5 4 3 2 1 0 1 24

3

2

1

0

1

2

3

4

Root Locus

Real Axis

ImaginaryAxis

Figure11: RLafteradding integralFBIncreasingKitoincreasespeedoftheresponsepushesthepoles

towardstheimaginaryaxismoreoscillatoryresponse.

Combineproportionaland integral(PI)feedback:Gc =K1+K2 =K1s+K2

s swhichintroducesapoleattheoriginandzeroats=K2/K1PIsolvesmanyoftheproblemswithjustintegralcontrol(I).

3 2.5 2 1.5 1 0.5 05

4

3

2

1

0

1

2

3

4

5

Root Locus

Real Axis

ImaginaryAxis

Figure12: RLwithproportionaland integralFB


17/33

Fall2004 16.3338163.DerivativeFeedbacku= Kde sothatGc(s) = Kds

DoesnothelpwiththesteadystateProvides feedback on the rate of change of e(t) so that the

controlcananticipatefutureerrors.

1Example#2G(s) = ,(a>0,b>0)

(sa)(sb)withGc(s) = Kds

2 1.5 1 0.5 0 0.5 1 1.5 21.5

1

0.5

0

0.5

1

1.5

Root Locus

Real Axis

ImaginaryAxis

Figure13: RLwithderivativeFB

DerivativefeedbackisveryusefulforpullingtherootlocusintotheLHP- increasesdampingandmorestableresponse.

Typicallyusedincombinationwithproportionalfeedbacktoformproportional-derivativefeedbackPD

Gc(s) = K1+ K2swhichmovesthezerofromtheorigin.


18/33

Fall2004 16.333817ControllerSynthesis

Firstdeterminewherethepolesshouldbelocated Willproportionalfeedbackdothejob? Whattypesofdynamicsneedtobeadded? Typicallyusemainbuild-

ingblock(s+ z)

GB(s) = Kc(s+ p)

Canbemadeto look likevariouscontrollers,dependingonhowwepickKc,p,andzIfwepickz>p,withpsmall,then

(s+ z)GB(s) Kc

s

whichisessentiallyaPIcompensator,calledalag.Ifwepickpz,thenatlowfrequency,theimpactofp/(s+ p)

issmall,soGB(s) Kc(s+ z)

whichisessentiallyPDcompensator,calledalead. Various algorithms exist to design the components of the lead and

lagcompensators(see16.31coursenotesonline)


19/33

Fall2004 16.333818PolePlacement

Oneoptionforsimplesystemsiscalledpoleplacement. Consider a simple system Gp = s2 for which we want the closed

looppolestobeat1 2i Proportionalcontrolclearlynotsufficient,souseacompensatorwith

1pole.(s+ z)

Gc = K(s+ p)

Sothereare3CLP. Knowthatthedesiredcharacteristicequationis

d(s) = (s2+ 2s+ 5)(s+ ) = 0 Theactualclosedlooppolessolve:

c(s) = 1 + GpGc = 02s (s+ p) + K(s+ z) = 03 2s + s p+ Ks+ Kz = 0

Clearlyneedtopullthepolesattheorigin intotheLHP,soneedalead compensator Good rule of thumb is to takep = 10z as astartingpoint.

Comparethecharacteristicequations:2c(s) = s2+ 10zs + Ks+ Kz= 0

d(s) = (s2+ 2s+ 5)(s+ )3 2= s + s (+ 2) + s(2+ 5) + 5= 0

gives2s + 2=10z

s 2+ 5=K0s 5=zK


20/33

Fall2004 16.333819solvefor,z,K

25 5zK

=

52z; =52z

z=2.23,=20.25,K=45.5

25 20 15 10 5 020

15

10

5

0

5

10

15

20

Root Locus

Real Axis

ImaginaryAxis

Figure14: CLPwithpoleplacement


21/33

Fall2004 16.333820Autopilot

Backtotheproblemon94: feedbacktoe tocontrolshortperiodmodel

c(s)=1+Ge(s)H(s)k =0with

1.1569s+0.3435 4Ge(s)=s3+0.7410s2+0.9272s, H(s)=s+4

Pole/zeromap:

Figure15: Pole/zeromapoftheshortperiodautopilotwith feedback Note: Kp 0,thenwewouldhavetodrawa0 locus

Poleatoriginwouldmovetorightalongrealaxisunstable.

4 3 2 1 0 1 2 3 4

4

3

2

1

0

1

2

3

4

SP Autopilot with FB: k

> 0

Real Axis

Imagina

ryAxis

Figure16: Withk >0


22/33

Fall2004 16.333821 Somustusek


23/33

Fall2004 16.333822 Couldusearategyroasthesensorinsteadandfeedbackq

1.1569s+0.3435 4Gqe(s)=s2+0.7410s+0.9272, H(s)=s+4

Figure18: PZmapwithqFB Again,pickthegainkq


24/33

Fall2004 16.333823 Cangetanevenbetterresult ifwecombinetheqand feedback

likePDonthemeasurement.ce

//4

s+ 4ae

// Gqeq

//

oo

1s

//

oo

kq

OO

k

OO

coo

ce = kqqk(c) ae = H(s)ceq = s

=1sGqeae = Geae

= GeH[(k + kqs)kc] GeHk=c 1 + GeH(k + kqs)

Nowpickkq = k,anddoRLversuskc(s) = 1 + GeH(1 + s)k = 0

Placesazeroats= 1/

Stepresponse

is

reasonable,

but

now

need

to

pick

the

k

and

kq

to

getthedesiredsp andsp


25/33

Fall2004 16.333824

4.5 4 3.5 3 2.5 2 1.5 1 0.5 0

4

3

2

1

0

1

2

3

4

SP PD Autopilot with =1 FB: k

< 0

Real Axis

ImaginaryAxis

Figure20: PDfeedbackwith= 1.aretheclosed looppoles.

0 5 10 15 20 25 30 35

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

step response to c

Time (sec)

(t)

r

ads

Figure21: PDfeedbackwith= 1 - step


26/33


27/33

Fall2004 16.333826Summary

Presentedonlybasicsofcontroldesignandanalysis,butthisisenoughforustogetstarted

Workingwithsmallermodelsisgoodfordesign,butneedtoconfirmthatthefullmodelstillstable

Could work with the full model, but that is much easier with statespacetools


28/33

Fall2004 16.333827

SPcontrolcodes(lect9a.m)1 %2 % Fall 20043 % 16.3334 % Codes to investigate control for the SP5 %6 close all7 figure(1);clf8 set(gcf,DefaultLineLineWidth,2)9 set(gcf,DefaultlineMarkerSize,10)10 rlocus([1],conv([1 0],[1 3 2]))11 print -depsc rl_pi11213 load b747 Asp Bsp14 figure(10);pzmap(ss(Asp,Bsp,[0 1],0));grid on;15 hh=get(10,children);hhh=get(hh(1),children)16 set(hhh(1),MarkerSize,24);set(hhh(1),LineWidth,2)17 set(hhh(2),MarkerSize,24);set(hhh(2),LineWidth,2)18 axis([-1 0.1 -1 1])19 print -depsc pz_sp20 jpdf(pz_sp)2122 figure(10);clf23 test=[% thumbnail data from Nelson 16724 .5 .4;25 .8 .4;26 1 .55;27 .7 .6]28 test(:,2)=test(:,2)*2*pi;29 stest=[-test(:,1).*test(:,2) test(:,2).*sqrt(1-test(:,1).^2)]30 fill(stest(:,1),stest(:,2),blue,stest(:,1),-stest(:,2),blue)31 hold on32 pzmap(ss(Asp,Bsp,[0 1],0));grid on;33 hh=get(10,children);hhh=get(hh(1),children)34 set(hhh(1),MarkerSize,24);set(hhh(1),LineWidth,2)35 set(hhh(2),MarkerSize,24);set(hhh(2),LineWidth,2)36 hold off37 print -depsc pz_sp238 jpdf(pz_sp2)3940 figure(1);clf41 set(gcf,DefaultLineLineWidth,2)42 set(gcf,DefaultlineMarkerSize,10)43 rlocus([1],conv([1 0],[1 3 2]))44 print -depsc rl_pi14546

figure(2);

47 set(gcf,DefaultLineLineWidth,2)48 set(gcf,DefaultlineMarkerSize,12)49 rlocus([1 3],conv([1 0],[1 3 2]))50 print -depsc rl_pi25152 figure(2);53 set(gcf,DefaultLineLineWidth,2)54 set(gcf,DefaultlineMarkerSize,12)55 rlocus([1 0],conv([1 -2],[1 -1]))56 print -depsc rl_d15758 %Example: G(s)=1/2^259 %Design Gc(s) to put the clp poles at -1 + 2j60 z=roots([-20 49 -10]);z=max(z),k=25/(5-2*z),alpha=5*z/(5-2*z),61 num=1;den=[1 0 0];62 knum=k*[1 z];kden=[1 10*z];63 rlocus(conv(num,knum),conv(den,kden));64 hold;plot(-alpha+eps*j,d);plot([-1+2*j,-1-2*j],d);hold off65 r=rlocus(conv(num,knum),conv(den,kden),1)66 print -depsc rl_pp


29/33

Fall2004 16.3338286768 jpdf(rl_pi1)69 jpdf(rl_pi2)70 jpdf(rl_d1)71

jpdf(rl_pp)

7273 load b747 Asp Bsp74 G=tf(ss(Asp,Bsp,[0 1],0))*tf(1,[1 0]);75 H=tf(4,[1 4]);76 rlocus(G*H)77 axis([-3 3 -3 3]*1.5)78 title(SP Autopilot with \theta FB: k_\theta79 print -depsc sp_ap180 jpdf(sp_ap1)8182 rlocus(-G*H)83 axis([-3 3 -3 3]*1.5)84 title(SP Autopilot with \theta FB: k_\theta85 print -depsc sp_ap286 jpdf(sp_ap2)8788 load b747 Asp Bsp89 G=tf(ss(Asp,Bsp,[0 1],0));90 H=tf(4,[1 4]);91 rlocus(-G*H)92 rr=rlocus(-G*H,1)93 hold on

> 0,FontSize,16)


30/33

Fall2004 16.333829

SPcontrolcodes: Part2(lect9b.m)1 % 16.333 Fall 20042 % SP design meeting the performance goals a3 %4 load b747 Asp Bsp5 G=tf(ss(Asp,Bsp,[0 1],0))*tf(1,[1 0]);6 H=tf(4,[1 4]);78 zeta=1.95;9 k_th=1;10 PD=tf([zeta 1],1);1112 rlocus(-G*H*PD)13 rr=rlocus(-G*H*PD,k_th)14 hold on

bit better

15 plot(rr+sqrt(-1)*eps,d,MarkerFaceColor,b)16 plot(-1.8+2.4*sqrt(-1),s,MarkerFaceColor,r)17 plot(-1.8-2.4*sqrt(-1),s,MarkerFaceColor,r)18 hold off19 axis([-3 0 -3 3]*1.5)20 title([SP PD Autopilot with \zeta=,num2str(zeta),21 print -depsc sp_pd222 jpdf(sp_pd2)2324 figure(2);25 set(gcf,DefaultLineLineWidth,2)26 set(gcf,DefaultlineMarkerSize,12)27 G_cl=G*H*(-k_th)/(1+G*H*PD*(-k_th));28 step(G_cl,35)

FB: k_\theta


31/33

Fall2004 16.333830

SPcontrol: BasicDynamics(lect9.m)1 clear all2 prt=1;3 %4 % B747 longitudinal dynamics5 % 16.333 Fall 20046 %7 % B747 at Mach 0.8, 40,000ft, level-flight8 % From Etkin and Reid9 %10 % metric11 Xu=-1.982e3;Xw=4.025e3;12 Zu=-2.595e4;Zw=-9.030e4;Zq=-4.524e5;Zwd=1.909e3;13 Mu=1.593e4;Mw=-1.563e5;Mq=-1.521e7;Mwd=-1.702e4;1415 g=9.81;theta0=0;S=511;cbar=8.324;16 U0=235.9;Iyy=.449e8;m=2.83176e6/g;cbar=8.324;rho=0.3045;17 Xdp=.3*m*g;Zdp=0;Mdp=0;18 Xde=-3.818e-6*(1/2*rho*U0^2*S);Zde=-0.3648*(1/2*rho*U0^2*S);19 Mde=-1.444*(1/2*rho*U0^2*S*cbar);;2021 A=[Xu/m Xw/m 0 -g*cos(theta0);[Zu Zw Zq+m*U0 -m*g*sin(theta0)]/(m-Zwd);22 [Mu+Zu*Mwd/(m-Zwd) Mw+Zw*Mwd/(m-Zwd) Mq+(Zq+m*U0)*Mwd/(m-Zwd) ...23 -m*g*sin(theta0)*Mwd/(m-Zwd)]/Iyy;24 [ 0 0 1 0]];25 B=[Xde/m Xdp/m;Zde/(m-Zwd) Zdp/(m-Zwd);(Mde+Zde*Mwd/(m-Zwd))/Iyy ...26 (Mdp+Zdp*Mwd/(m-Zwd))/Iyy;0 0];2728 % Short-period Approx29 Asp=[Zw/m U0;30 [Mw+Zw*Mwd/m Mq+U0*Mwd]/Iyy];31 Bsp=[Zde/m;(Mde+Zde/m*Mwd)/Iyy];32 [nsp,dsp]=ss2tf(Asp,Bsp,eye(2),zeros(2,1));33 [Vsp,evsp]=eig(Asp);evsp=diag(evsp);34 %rifd(evsp)3536 %37 % CONTROL38 %39 % actuator dynamics are a lag at 440 actn=4;actd=[1 4]; % H(s) in notes41 %42 % use the short period model since that is all we are trying to control43 %44 [nsp,dsp]=ss2tf(Asp,Bsp,eye(2),zeros(2,1));45 [nfull,dfull]=ss2tf(A,B(:,1),eye(4),zeros(4,1));46

%

47 % form num and den for the "plant" = act_dyn*G == N/D48 %49 % short period model50 N=conv(actn,nsp(2,:));% q is second state51 D=conv(actd,dsp);52 % full model53 Nqfull=conv(actn,nfull(3,:));% q is third state54 Ntfull=conv(actn,nfull(4,:));% theta is fourth state55 Dfull=conv(actd,dfull);56 %57 % add an extra pole to get the integrator for the \dot theta = q58 %59 Ntheta=conv(N,1);Dtheta=conv(D,[1 0]);6061 figure(1);clf;62 K_th0=-1;63 rlocus(sign(K_th0)*Ntheta,Dtheta);64 r_th0=rlocus(Ntheta,Dtheta,K_th0);hold on;plot(r_th0+eps*sqrt(-1),v);65 hold off66 sgrid(.6,3); % gives target damping and freqs


32/33

Fall2004 16.33383167 axis([-5 .5 -4 4]);grid68 title(Closed-loop poles using only theta FB)69 if prt70 print -depsc ac_th071

end

7273 figure(2);clf;74 K_q=-1;75 rlocus(sign(K_q)*N,D);76 r_q=rlocus(N,D,K_q);hold on;plot(r_q+eps*sqrt(-1),v);77 hold off78 sgrid(.6,3);79 axis([-5 .5 -4 4]);grid80 title(Closed-loop poles using q FB)81 if prt82 print -depsc ac_q83 end84 %85 %closed-loop dynamics with q FB in place (inner loop)86 % GH/(1+k_q GH) = (N/D) / (1+k_q N/D) = N/(D+k_q N)87 %88 Nq=N;Dq=K_q*[0 N]+D;89 %90 % add integrator for q ==> theta91 %92 K_th=-1;93 %94 figure(3);clf;95 rlocus(sign(K_q)*Nq,conv(Dq,[1 0]));grid96 r_th=rlocus(Nq,conv(Dq,[1 0]),K_th);hold on;97 plot(r_q+eps*sqrt(-1),v);plot(r_th+eps*sqrt(-1),^);98 hold off99 sgrid(.6,3);100 axis([-5 .5 -4 4]);grid101 title(Closed-loop poles also using theta FB)102 if prt103 print -depsc ac_th104 end105 %106 % as a final check, form the closed-loop dynamics107 % using the original short period model108 Ncl=Ntheta*K_th;109 Dcl=Dtheta+conv([0 N],[K_q K_th]);110 roots(Dcl)111 %112 SPcl=tf(Ncl,Dcl);113 [y,t]=step(SPcl);114 figure(4)115 plot(t,y);xlabel(time);ylabel(\theta rad,FontSize,12)116 if prt117 print -depsc ac_th_step118 end119120 %121 % as a final check, form the closed-loop dynamics122 % using the full dynamics from del_e to theta123 Nfcl=Ntfull*K_th;124 Dfcl=Dfull+conv(Ntfull,[K_q K_th]);125 roots(Dfcl)126 figure(3)127 hold on;plot(roots(Dfcl)+eps*sqrt(-1),r*);hold off128 axis([-3 .5 -2 2]);grid129 title(Closed-loop poles with q and th FB on FULL model)130 if prt131 print -depsc ac_full132 end133134

%return

135136 %137 % add theta state to the short period approx model138 %


33/33

1

2

3

4

5

6

7

8

9

Fall2004 16.333832139 Ksp=place(Asp,Bsp,[roots([1 2*.6*3 3^2])])140 Asp2=[Asp [0 0];0 1 0];Bsp2=[Bsp;0];141 Ksp2=place(Asp2,Bsp2,[roots([1 2*.6*3 3^2]),-.25])142 %143

%

add

actuatormodel

to

SP

with

theta

144 %145 Plist=[roots([1 2*.6*3 3^2]),-.25,-3];146 At2=[Asp2 Bsp2(:,1);zeros(1,3) -4];Bt2=[zeros(3,1);4];147 Kt2=place(At2,Bt2,[Plist])148 step(ss(At2-Bt2*Kt2,Bt2,[0 0 1 0],0),35)149 hh=get(gcf,children);hhh=get(hh(1),children)150 set(hhh(1),MarkerSize,24);set(hhh(1),LineWidth,2)151 ylabel(\theta rads,FontSize,12)152 print -depsc fsfb_step.eps153 jpdf(fsfb_step)154 %155 ev=eig(A);156 % dampe short period, but leave the phugoid where it is157 Plist=[roots([1 2*.6*3 3^2]) ev([3 4],1)];158 K1=place(A,B(:,1),Plist)159 %160 % add actuator model16 %16 At=[A B(:,1);zeros(1,4) -4];Bt=[zeros(4,1);4];16 Kt=place(At,Bt,[Plist -3])1616 save b747 A B Asp Bsp1616 %!eps2pdf /f="ac7_fig1.eps"16 %!mv c:\ac7_fig1.pdf ./ac7_fig1.pdf16170

aircraft control lecture 9

Documents