Topic #13 16.31 Feedback Control State-Space Systems ? Full-state Feedback Control ? How do we change the poles of the state-space system? ? Or, even if we can change the pole locations. ? Where do we change the pole locations to? ? How well does this approach work? Copy right 2001 by Jon at h an H ow. 1 Fall 2001 16.31 13–1 Full-state Feedback Controller ? Assumethatthesingle-inputsystemdynamicsaregivenby ˙x = Ax+Bu y = Cx sothat D =0. – The multi-actuator case is quite a bit more complicated as we wouldhavemanyextradegreesoffreedom. ? Recallthatthesystempolesaregivenbytheeigenvaluesof A. – Want to use the input u(t) to modify the eigenvalues of A to changethesystemdynamics. ? Assumeafull-statefeedbackoftheform: u = r?Kx where r issome reference input andthe gain K isR 1×n – If r =0,wecallthiscontrollera regulator ? Findtheclosed-loopdynamics: ˙x = Ax+B(r?Kx) =(A?BK)x+Br = A cl x+Br y = Cx Fall 2001 16.31 13–2 ? Objective: Pick K sothat A cl hasthedesiredproperties, e.g., – A unstable,want A cl stable – Put2polesat?2±2j ? Notethatthereare n parametersin K and n eigenvaluesin A,so itlookspromising,butwhatcanweachieve? ? Example #1: Consider: ˙x = bracketleftbigg 11 12 bracketrightbigg x+ bracketleftbigg 1 0 bracketrightbigg u – Then det(sI ?A)=(s?1)(s?2)?1=s 2 ?3s+1=0 sothesystemisunstable. – De?ne u =? bracketleftbig k 1 k 2 bracketrightbig x =?Kx,then A cl = A?BK = bracketleftbigg 11 12 bracketrightbigg ? bracketleftbigg 1 0 bracketrightbigg bracketleftbig k 1 k 2 bracketrightbig = bracketleftbigg 1?k 1 1?k 2 12 bracketrightbigg – Sothenwehavethat det(sI ?A cl )=s 2 +(k 1 ?3)s+(1?2k 1 +k 2 )=0 – Thus, by choosing k 1 and k 2 , we can put λ i (A cl )anywherein thecomplexplane(assumingcomplexconjugatepairsofpoles). Fall 2001 16.31 13–3 ? Toputthepolesats =?5, ?6,comparethedesired characteristic equation (s+5)(s+6)=s 2 +11s+30=0 withtheclosed-loopone s 2 +(k 1 ?3)x+(1?2k 1 +k 2 )=0 toconcludethat k 1 ?3=11 1?2k 1 +k 2 =30 bracerightbigg k 1 =14 k 2 =57 sothat K = bracketleftbig 14 57 bracketrightbig ,whichiscalled Pole Placement. ? Ofcourse,itisnotalwaysthiseasy,astheissueofcontrollability mustbeaddressed. ? Example #2: Considerthissystem: ˙x = bracketleftbigg 11 02 bracketrightbigg x+ bracketleftbigg 1 0 bracketrightbigg u withthesamecontrolapproach A cl = A?BK = bracketleftbigg 11 02 bracketrightbigg ? bracketleftbigg 1 0 bracketrightbigg bracketleftbig k 1 k 2 bracketrightbig = bracketleftbigg 1?k 1 1?k 2 02 bracketrightbigg sothat det(sI ?A cl )=(s?1+k 1 )(s?2)=0 Sothefeedbackcontrolcanmodifythepoleats =1,butitcannot movethepoleat s =2. ? The system cannot be stabilized with full-state feed- back control. Fall 2001 16.31 13–4 ? Whatisthereasonforthisproblem? – Itisassociatedwithlossofcontrollabilityofthe e 2t mode. ? Considerthebasiccontrollabilitytest: M c = bracketleftbig B AB bracketrightbig = bracketleftbiggbracketleftbigg 1 0 bracketrightbigg bracketleftbigg 11 02 bracketrightbiggbracketleftbigg 1 0 bracketrightbiggbracketrightbigg Sothat rank M c =1< 2. ? Considerthe modal test todevelopalittlemoreinsight: A = bracketleftbigg 11 02 bracketrightbigg ,decomposeas AV = VΛ ?Λ=V ?1 AV where Λ= bracketleftbigg 10 02 bracketrightbigg V = bracketleftbigg 11 01 bracketrightbigg V ?1 = bracketleftbigg 1 ?1 01 bracketrightbigg Convert ˙x= Ax+Bu z=V ?1 x ?→ ˙z =Λz+V ?1 Bu where z = bracketleftbig z 1 z 2 bracketrightbig T .But: V ?1 B = bracketleftbigg 1 ?1 01 bracketrightbiggbracketleftbigg 1 0 bracketrightbigg = bracketleftbigg 1 0 bracketrightbigg sothatthedynamicsinmodalformare: ˙z = bracketleftbigg 10 02 bracketrightbigg z+ bracketleftbigg 1 0 bracketrightbigg u ? WiththiszerointhemodalB-matrix,caneasilyseethatthemode associatedwiththe z 2 stateis uncontrollable. – Must assume that the pair (A, B) are controllable. Fall 2001 16.31 13–5 Ackermann’s Formula ? Thepreviousoutlinedadesignprocedureandshowedhowtodoit byhandforsecond-ordersystems. – Extendstohigherorder(controllable)systems,buttedious. ? Ackermann’s Formula gives us a method of doing this entire designprocessisoneeasystep. K = bracketleftbig 0 ... 01 bracketrightbig M ?1 c Φ d (A) where –M c = bracketleftbig BAB...A n?1 B bracketrightbig – Φ d (s) is the characteristic equation for the closed-loop poles, whichwethenevaluatefors = A. – Itisexplicitthatthesystem must be controllablebecause weareinvertingthecontrollabilitymatrix. ? Revisitexample # 1: Φ d (s)=s 2 +11s+30 M c = bracketleftbig B AB bracketrightbig = bracketleftbiggbracketleftbigg 1 0 bracketrightbigg bracketleftbigg 11 12 bracketrightbiggbracketleftbigg 1 0 bracketrightbiggbracketrightbigg = bracketleftbigg 11 01 bracketrightbigg So K = bracketleftbig 01 bracketrightbig bracketleftbigg 11 01 bracketrightbigg ?1 parenleftBigg bracketleftbigg 11 12 bracketrightbigg 2 +11 bracketleftbigg 11 12 bracketrightbigg +30I parenrightBigg = bracketleftbig 01 bracketrightbig parenleftbiggbracketleftbigg 43 14 14 57 bracketrightbiggparenrightbigg = bracketleftbig 14 57 bracketrightbig ?AutomatedinMatlab: place.m& acker.m(see polyvalm.mtoo) Fall 2001 16.31 13–6 ? Wheredidthisformulacomefrom? ? For simplicity, consider a third-order system (case #2), but this extendstoanyorder. A = ? ? ?a 1 ?a 2 ?a 3 100 010 ? ? B = ? ? 1 0 0 ? ? C = bracketleftbig b 1 b 2 b 3 bracketrightbig – Seekeybene?tofusing control canonical state-spacemodel – This form is useful because the characteristic equation for the systemisobvious?det(sI ?A)=s 3 +a 1 s 2 +a 2 s+a 3 =0 ? Canshowthat A cl = A?BK = ? ? ?a 1 ?a 2 ?a 3 100 010 ? ? ? ? ? 1 0 0 ? ? bracketleftbig k 1 k 2 k 3 bracketrightbig = ? ? ?a 1 ?k 1 ?a 2 ?k 2 ?a 3 ?k 3 100 01 ? ? sothatthecharacteristicequationforthesystemisstillobvious: Φ cl (s)=det(sI ?A cl ) = s 3 +(a 1 +k 1 )s 2 +(a 2 +k 2 )s+(a 3 +k 3 )=0 Fall 2001 16.31 13–7 ? Wethencomparethis withthe desiredcharacteristicequationde- velopedfromthedesiredclosed-looppolelocations: Φ d (s)=s 3 +(α 1 )s 2 +(α 2 )s+(α 3 )=0 togetthat a 1 +k 1 = α 1 . . . a n +k n = α n ? ? ? k 1 = α 1 ?a 1 . . . k n = α n ?a n ? Togetthespeci?csoftheAckermannformula,wethen: – TakeanarbitraryA,B andtransformittothecontrolcanonical form(x?z = T ?1 x) – Solve for the gains ? K using the formulas above for the state z (u = ? Kz) – Thenswitchbacktogainsneededforthestate x,sothat K = ? KT ?1 (u = ? Kz= Kx) ? Pole placement is a very powerful tool and we will be using it for mostofthiscourse.