Synthesis, Nonlinear design. Whynonlineardesign methods? Relative degree. Example (Zero dynamics for linear systems) ξ 1 = ξ 2

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1 Synthesis, Nonlinear design Whynonlineardesign methods? Introdction Relative degree & zero-dynamics (rev) Exact Linearization (intro) Control Lyapnov fnctions Lyapnov redesign Nonlinear damping Backstepping Control Lyapnov fnctions (CLFs) passivity robst/adaptive Linear design degraded by nonlinearities (eg satrations) Linearization not controllable (eg pocket parking) Long state transitions (eg satellite orbits) Inherently nonlinear Ch 3-32, 4-43 Nonlinear Systems, Khalil The Joy of Feedback, P V Kokotovic Relative degree A system s relative degree: How many times yo need to take the derivative of the otpt signal before the inpt shows p Note: A nonlinear system may have state-dependent relative degree Example: The ball and beam process (see process homepage for more information) If nothing else stated we assme a fixed relative degree in the seqel For a nonlinear system with relative degreed we have ẋ=f(x)+ (x) y=h(x) ẏ = d dt h(x)= h(x) ẋ= h x x f(x)+ h x (x) = L f h(x)+ L h(x) =0if d> y (k) = L k fh(x) ifk<d (2) y (d) = L d f h(x)+l L (d ) f h(x) () Using the same kind of coordinate transformations as for the feedback linearizable systems above, we can introdce new state space variables, ξ, where the firstd coordinates are chosen as ξ =h(x) ξ 2 =L f h(x) (3) ξ d =L (d ) f h(x) Under some conditions on involtivity, the Frobenis theorem garantees the existence of another(n d) fnctions to provide a local state transformation of fll rank Sch a coordinate change transforms the system to the normal form ξ = ξ 2 ξ d = ξ d ξ d =L d f h(ξ,z)+l L d f h(ξ,z) ż= ψ(ξ,z) (4) y= ξ whereż= ψ(ξ,z) represent the zero dynamics of ordern d Byrnes+Isidori 99] Example (Zero dynamics for linear systems) Consider the linear system y= s s 2 +2s+ (5) with the following state-space description ẋ = 2x +x 2 + ẋ 2 = x (6) y = x y 0 x 0 ẋ 0 x 2 +=0 ẋ 2 = =x 2 (7) The remaining dynamics is an nstable system corresponding to the zeros=in the transfer fnction (??) We have the relative degree = Find the zero-dynamics, by assigningy 0

2 Exact (feedback) Linearization Idea: Transform the nonlinear system into a linear system by means of feedback and/or a change of variables After this, a stabilizing state feedback is designed For general nonlinear systems feedback linearization comprises state transformation inversion of nonlinearities linear feedback r v y + =β ( ) Σ z Simple example ẍ= l sin(x)+cos(x) x=t(z) L x Inner feedback linearization and oter linear feedback control Pt gives (locally) = cos(x) ( l sin(x)+v) ẍ=v Design linear controllerv= l x+ l 2 ẋ, etc State transformation More difficlt example, where we need a state transformation ẋ =asin(x 2 ) ẋ 2 = x 2 + Can not cancelasin(x 2 ) Introdce z =x z 2 =asinx 2 so that ż =z 2 ż 2 =( z 2 +)acosx 2 Feedback linearization ( nonlinear version of pole-zero cancellation ) Feedback linearization can be interpreted as a nonlinear version of pole-zero cancellations which can not be sed if the zero-dynamics are nstable, i e, for nonminimm-phase system Linear systems: See paper Middleton (999) Atomatica 35(5), "Slow stable open-loop poles: to cancel or not to cancel"] Then feedback linearization is (locally) possible by =z 2 +v/(acos(z 2)) When to cancel nonlinearities? Matching ncertainties ẋ = x 3 + ẋ 2 =x Nonrobst and/or not necessary However, note the difference between tracking or reglation!! Will see later how optimal criteria will give hints (8) ẋ =x 2 ẋ n =x d ẋ n =L d f h(x,z)+l L d f h(x,z) ż= ψ(x,z) y=x Integrator chain and nonlinearities (+ zero-dynamics) Note that ncertainties de to parameters etc are collected in L d f h(x,z)+l L d f h(x,z) (9) Achieving passivity by feedback ( Feedback passivation ) Need to have relative degree one weakly minimm phase NOTE! (Nonlinear) relative degree and zero-dynamics invariant nder feedback! Two major challenges: avoid non-robst cancellations make it constrctive by finding matching inpt-otpt pairs Exact Linearization Often sefl in simple cases Important intition may be lost Nonlinear version of "pole-zero cancellations" Related to Lie brackets and flatness 2

3 Fromanalysis to synthesis Lyapnov criterion Search for(v,) sch that x f+ ]<0 IQC criterion Search forq(s) and τ,,τ m sch that» " # T+T 2QT 3](iω) X» T+T τ 2QT 3](iω) I kπ k(iω) I for ω 0, ] In both cases, the problem is non-convex and hard Heristic idea: Iterate between the argments k <0 Convexity for state feedback Problem Sppose α φ(v)/v β Given the system ẋ=f (x):=ax+eφ(fx)+b find= Lx andv(x)=x T Px sch that x f (x)<0 Soltion Solve forp,l (A+ αef BL) T P+P(A+ αef BL)<0 (A+βEF BL) T P+P(A+βEF BL)<0 or eqivalently convex in(q,k)=(p,lp ) (AQ+ αefq BK) T +(AQ+ αefq BK)<0 (AQ+βEFQ BK) T +(AQ+βEFQ BK)<0 ControlLyapnov Fnction (CLF) A positive definite radially nbondedc fnctionv is called a CLF for the systemẋ=f(x,) if for eachx =0, there exists sch that x (x)f(x,)<0 (Notation:L fv(x)<0) Example Check ifv(x,y)=x 2 +(y+x 2 ) 2 ] 2 /2 is a CLF for the system ẋ=xy ẏ= y+ Whenf(x,)=f(x)+ (x),v is a CLF if and only if L f V(x)<0for allx =0sch that L V(x) =0 L f V(x,y)=x 2 y+(y+x 2 )( y+2x 2 y) L V(x,y)=2(y+x 2 )x 2 +(y+x 2 ) 2 ] L V(x,y)=0 y= x 2 L f V(x,y)= x 4 <0 if(x,y) Sontag s formla Backstepping idea IfV is a CLF for the systemẋ=f(x)+ (x), then a continos asymptotically stabilizing feedback is defined by 0 (x):= ifl V(x)=0 L fv+ (L f V) 2 +((L V)(L V) T ) 2 L (L V)(L V) T V] T ifl V(x) =0 Note: Can cancel factorl V =0if scalar 0 (x):= ifl V(x)=0 L fv+ (L f V) 2 +(L V) 4 (x) ifl V(x) =0 L V Problem Given a CLF for the system ẋ=f(x,) find one for the extended system ẋ=f(x,y) ẏ=h(x,y)+ Idea Usey to control the first system Usefor the second Note potential for recrsivity motivation: Feedback Linearization Lyapnov Redesign Consider the nominal system One of the drawbacks with feedback linearization is that exact cancellation of nonlinear terms may not be possible de to e g, parameter ncertainties A sggested soltion: stabilization via feedback linearization arond a nominal model consider known bonds on the ncertainties to provide an additional term for stabilization ( Lyapnov redesign ) with the known control law ẋ=f(x,t)+g(x,t) = ψ(x,t) so that the system is niformly asymptotically stable Assme that a Lyapnov fnctionv(x,t) is known st α ( x ) V(x,t) α 2 ( x ) t + x f(t,x)+gψ] α 3( x ) 3

4 Pertrbed system LyapnovRedesign cont distrbance δ= δ(t,x,) ẋ=f(x,t)+g(x,t)+ δ] (0) Assme the distrbance satisfies the bond δ(t,x,ψ+v) ρ(x,t)+ κ 0 v If we know ρ and κ 0 how do we design additional controlv sch that= ψ(x,t)+v stabilizes (??)? The matching condition: pertrbation enters at same place as control signal Apply= ψ(x,t)+v ẋ=f(x,t)+g(x,t)ψ+g(x,t)v+ δ(t,x,ψ+v)] () V= t + x f(t,x)+gψ]+ x Gv+ δ] α 3( x )+ x Gv+ Introdcew= x G] V α 3 ( x )+w T v+w T δ Choosevsch thatw T v+w T δ 0: Two alternatives presented in Khalil ( 2 -norm / -norm) Note:v appears at same place as δ de to the matching condition LyapnovRedesign cont w T v+w T δ w T v+ w T 2 δ 2 w T v+w T δ w T v+ w T δ Alternative : If δ(t,x,ψ+v) 2 ρ(x,t)+ κ 0 v 2, 0 κ 0 < take w v= η(t,x) w 2 where η ρ/( κ 0 ) Alternative 2: If take δ(t,x,ψ+v) ρ(x,t)+ κ 0 v, 0 κ 0 < v= η(t,x)sgnw where η ρ/( κ 0 ) Restriction on κ 0 <bt not on growth of ρ Alt and alt 2 coincide for single-inpt systems Note: control laws are discontines fcn ofx (risk of chattering) Example: Matched ncertainty Example cont + x Example: Exponentially decaying distrbance (t)= (0)e kt linear feedback= cx, c>0 ϕ(x)=x 2 ϕ( ) ẋ= cx+ (0)e kt x 2 Similar to peaking problem in the first lectre: Finite escape of soltion to infinity if (0)x(0)>c+k ẋ=+ ϕ(x) (t) We want to garantee that x(t) stay bonded for all initial vales x(0) and all bonded distrbances (t) Nonlinear damping Modify the control law in the previos example as: where = cx s(x)x s(x)x will be denoted nonlinear damping Use the Lyapnov fnction candidatev= x2 2 V=x+xϕ(x) = cx 2 x 2 s(x)+xϕ(x) Choose to complete the sqares! s(x)= κϕ 2 (x) V= cx 2 x 2 s(x)+xϕ(x) = cx 2 κ xϕ ] κ 2κ 4κ 2 cx κ Note! V is negative whenever x(t) 2 κc How to proceed? 4

5 Yong s ineqality Can show thatx(t) converges to the set R= x: x(t) 2 κc i e, x(t) stays bonded for all bonded distrbances Letp>,q>st(p )(q )=, then for all ǫ>0and all(x,y) R 2 xy< ǫp p x p + qǫ q y q Standard case: (p=q=2, ǫ 2 /2= κ ) Remark: The nonlinear damping κxϕ 2 (x) renders the system Inpt-To-State Stable (ISS) with respect to the distrbance Or example: xy<κ x 2 + 4κ y 2 xϕ(x) (t)< κx 2 ϕ 2 (x)+ 2 (t) 4κ Backstepping idea Problem Given a CLF for the system ẋ=f(x,) find one for the extended system ẋ=f(x,y) ẏ=h(x,y)+ Idea Usey to control the first system Usefor the second Note: potential for recrsivity Backstepping LetV x be a CLF for the systemẋ=f(x)+ (x)ȳ with corresponding asymptotically stabilizing control lawȳ= φ(x) ThenV(x,y)=V x (x)+y φ(x)] 2 /2 is a CLF for the system with corresponding control law Proof ẋ=f(x)+ (x)y ẏ=h(x,y)+ = φ x f(x)+ (x)y] x (x) h(x,y)+ φ(x) y x V=( x / x)(f+ y)+(y φ)h+ ( φ/ x) (f+ y)] =( x / x)(f+ φ)+(y φ)( x / x) ( φ/ x) (f+ y)+h =( x / x)(f+ φ) (y φ) 2 <0 Backstepping Example Example again (step by step) For the system ẋ=x 2 +y ẏ= we can choosev x (x)=x 2 and φ(x)= x 2 xto get the control law with Lyapnov fnction = φ (x)f(x,y) h(x,y)+ φ(x) y = (2x+)(x 2 +y) x 2 x y V(x,y)=V x (x)+y φ(x)] 2 /2 =x 2 +(y+x 2 +x) 2 /2 Find (x) which stabilizes (??) ẋ =x 2 +x 2 ẋ 2 =(x) Idea : Try first to stabilize thex -system withx 2 and then stabilize the whole system with We know that ifx 2 = x x 2 thenx 0asymptotically ( exponentially ) ast (2) We can t expect to realizex 2 = α(x ) exactly, bt we can always try to get the error 0 Introdce the error states z =x z 2 =x 2 α (x ) where α (x )= x x 2 x2 ż = ẋ =z 2 + z 2 + α (z )= = z 2 +z 2 z 2 z = z +z 2 ż 2 = ẋ 2 α =(x) known α α = d dt ( z2 z )= z ż ż = z ( z +z 2 ) ( z +z 2 )= = z 2 z z 2 z 2 z (3) Start with a Lyapnov for the first sbsystem (z -dynamics): V = 2 z2 0 V = z ż = z 2 +z z 2 Note : Ifz 2 =0we wold achievev = z 2 0 with α (x ) 5

6 Now look at the agmented Lyapnov fcn for the error system Backward propagation of desired control signal V 2 = V + 2 z2 2 0 V 2 = V +z 2 ż 2 = = z 2 +z z 2 +z 2 ( z 2 +z z 2 ) = z 2 +z 2( z 2 +z z 2 +z 2 +z ) choose= z 2 = z 2 z2 2 0 x 2 + f( ) x so if=z 2 z z 2 z 2 z (z,z 2 ) 0asymptotically (exponentially) (x,x 2 ) 0asymptotically Asz =x andz 2 =x 2 α =x 2 +x 2 +x, we can expressas a ( nonlinear ) state feedback fnction of x andx 2 If we cold sex 2 as control signal, we wold like to assign it to α(x ) to stabilize thex -dynamics x α f+ α Move the control backwards throgh the integrator z 2 =x 2 α + + dα/dt f+ α Note the change of coordinates! x x Adaptive Backstepping System : ẋ =x 2 + θ(x ) ẋ 2 =x 3 ẋ 3 =(t) where is a known fnction ofx and θ is an nknown parameter Introdce new (error) coordinates z (t)=x (t) z 2 (t)=x 2 (t) α (z,ˆ θ) where α is sed as a control to stabilize thez - system wrt a certain Lyapnov-fnction (4) (5) Lyapnov fnction : V = 2 z θ 2 where θ=(ˆ θ θ) is the parameter error (Back-) Step : ż (t) = x 2 z 2 (t)+ α (z,ˆ θ)+θ(z (t)) Note: If we sed θ= ˆ τ as pdate law and ifz 2 =0then V = z 2 0 Step 2: Introdcez 3 =x 3 α 2 (z,z 2,ˆ θ) and se α 2 as control to stabilize the(z,z 2 )-system V = z z + θ ˆ θ=z (z 2 + α + θ)+ θ ˆ θ= = z z 2 + α +ˆ θ ]+ θ( ˆ θ z ) z τ Choose α = z ˆ θ V = z 2 +z z 2 + θ( ˆ θ τ ) Agmented Lyapnov fnction : V 2 =V + 2 z 2 2 ż 2 = ẋ 2 α = x 3 = z 3 + α 2 α (x 2 + θ) α θ ˆ ˆ θ V 2 = V +z 2 ż 2 = = = z 2 +z 2z 3 + α 2 +z + α (z z 2 ) α θ ˆ ]+ z ˆ θ z 2 + θ θ (τ ˆ α +z 2 ) ] z τ 2 Choose α 2 = z 2 z α (z z 2 )+ α ˆ θ τ 2 Note : Ifz 3 =0and we sed ˆ θ= τ 2 as pdate law we wold get V 2 = z 2 z2 2 0 Reslting sbsystem ż ż 2 ] ] = Hrwitz z z 2 ] ] + α θ+ ] ] τ 2 = α z x V 2 = z 2 z2 2 +z α 3z 2 + θ( ˆ θ τ 2 ) z ( ˆ 2 θ τ ˆ θ 2 ) Step 3 : ż 3 = ẋ 3 α 2 ] 0 z 3 α ( ˆ θ τ ˆ θ 2 ) = α 2 ż α 2 ż 2 α 2 θ= ˆ = z 2 ˆ θ = ph z 2 6

7 We now want to choose=(z,z 2,ˆ θ) sch that the whole system will be stabilized wrtv 3 Jst to simplify the expressions, introdce, and choose = z 2 z 3 + α 2 (z 2 + α 2 +ˆ θ)+ α 2 x 3 + α 2 θ+ ˆ z 2 ˆ θ =? Agmented Lyapnov fnction : V 3 = V z 3 2 = 2 z 2 + θ 2 2 V 3 = z 2 z2 2 z2 3 +z α 2 3+ θ ˆ θ (τ 2 z 3 )] z τ 3 α z 2 ˆ ( ˆ θ τ 2 ) θ θ=(τ ˆ α 3 )= τ 2 z 2 3 V 3 = z 2 +z 3 (+z 2 σ) We are almost there : 0 0 ż= α z+ α θ+ ˆ (τ θ 2 ˆ θ) 0 α 2 ] θ= ˆ α α 2 z z 2 z 3 V 3 = z 2 α +z 3 +z 2 ˆ (τ θ 2 ˆ θ) Crcial : α ˆ θ (τ 2 ˆ θ)=z α α 2 3 ˆ θ z known def =σ 0 0 ż= + σ z+ α 0 + θ 0 α 2 Observer backstepping Choose= z 2 σ Finally : V 3 = z 2 GS ofz=0,ˆ θ= θ andx 0(by La Salle s Theorem ) Closed-loop system : 0 ż= + σ α z+ θ 0 σ α 2 skew symmetric I Observer backstepping is based on the following steps: A (nonlinear) observer is designed which provides (exponentially) convergent estimates 2 Backstepping is applied to a system where the states have been replaces by their estimates The observation errors are regarded as (bonded) distrbances and handled by nonlineardamping ] τ 3 = α α 2 z Backstepping applies to systems in strict-feedback form ẋ =f (x )+x 2 Compare with Strict-feedforward systems ẋ 2 =f 2 (x,x 2 )+x 3 ẋ n =f n (x,x 2,x n,x n )+ ẋ =x 2 +f (x 2,x 3,,x n,) ẋ 2 =x 3 +f 2 (x 3,,x n,) ẋ n =x n +f n (x n,) ẋ n = 7