Lecture 3 Nonlinear Control. Center Manifold Theorem Robustness analysis and quadratic inequalities

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1 Lecture 3 Nonlinear Control Center Manifold Theorem Robustness analysis and quadratic inequalities

2 Material Khalil, Ch 8. Center manifold theorem lecture notes A. Megretski and A. Rantzer, System Analysis via Integral Quadratic Constraints, IEEE Transactions on Automatic Control, 47:6, 1997 U. Jönsson, Lecture Notes on Integral Quadratic Constraints User s guide to µ toolbox, Matlab

3 The Center Manifold Theorem [Khalil ch 8] What can we do if the linearizationa= f x imaginary axis? has zeros on the

4 Center Manifold Theory Assume that a system ( possibly via a state space transformation[x] [y T,z T ] T ) can be written as ẏ = A 0 y+f 0 (y,z) ż = A z+f (y,z) A : asymptotically stable A 0 : eigenvalues on imaginary axis f 0 andf second order and higher terms.

5 Center Manifold Theorem Assume[y T,z T ] T =0is an equilibrium point. For everyk 2 there exists a δ k >0andC k mappinghsuch thath(0)=0and h (0)=0and the surface z=h(y) y δ k is invariant under the dynamics above.

6 Proof Outline For any continuously differentiable functionh k, globally bounded together with its first partial derivative and with h k (0)=0,h (0)=0, leth k+1 be defined by the equations ẏ = A 0 y+f 0 (y,h k (y)) ż = A z+f (y,h k (y)) h k+1 (y) = z Under suitable assumptions, it can be verified that this defines h k+1 uniquely. Furthermore, the sequence{h i } is contractive in the normsup y h i (y) and the limith satisfies the conditions for a center manifold.

7 Usage 1) Determine z = h(y), at least approximately. (E.g., do a series expansion and identify coefficients...) 2) The local stability for the entire system can be proved to be the same as for the dynamics restricted to a center manifold: ẏ=a 0 y+f 0 (y,h(y))

8 Usage cont d In the case of using series expansion ofh(y)=c 2 y 2 +c 3 y , you would need to continue (w.r.t the order of the terms) until you have been able to determined the local behavior. (Low order terms dominate locally). Identify the coefficients from the boundary condition [Khalil (8.8, 8.11)] h y (y)[a0 y+f 0 (y,h(y))] A h(y) f (y,h(y))=0

9 Example ẏ = z ż = z+ay 2 +byz HereA 0 =0andA = 1.z=h(y) gives h+ay 2 +byh h h=0 hence h(y)=ay 2 +O( y 3 ) Substituting into the dynamics we get ẏ=ay 2 +O( y 3 ) sox=(0,0) is unstable fora =0.

10 Non-uniqueness The center manifold need not be unique Example ẏ = y 3 ż = z z=h(y) gives h y 3 =z=h(y) which has the solutions h(y)=ce 1/(2y2 ) for all constants C.

11 Robustness analysis and quadratic inequalities µ-analysis S-procedure Multipliers Integral Quadratic Constraints Performance analysis

12 Preview Example A linear system of equations { x=y y= x x=y=1 Equations with uncertainty { (x y) 2 < ǫ 1 x 2 (y+0.1x 1.1) 2 < ǫ 2 (x 1) 2 +(y 1) 2 < ǫ 3 Given ǫ 1 and ǫ 2, how do we find a valid ǫ 3?

13 Example v C(sI A) 1 B w + r Question: For what values of is the system stable? Note: May be large differences if we consider complex or real uncertainties. A formula for Computation of the Real Stability Radius, L. Qiu, B. Bernhardsson, A. Rantzer, E.J. Davison, and P.M. Young. Automatica, pp , vol 31(6), 1995.

14 Parametric Uncertainty in Linear Systems LetD R n n contain zero. The systemẋ=(a+b C)x is then exponentially stable for all D if and only if A is stable det [ I C(iωI A) 1 B ] =0for ω R, D v C(sI A) 1 B w + r

15 Use quadratic inequalities at each frequency! [ 1r w= I C(iωI A) B] 1 { w= v+r v=c(iωi A) 1 Bw For example, if {[ ] δ1 0 D= 0 δ 2 } : δ k [ 1,1] Then a bound of the form w 2 < γ 2 r 2 can be obtained using w 1 r 1 2 < v 1 2 [ ] [ ] v1 w 2 r 2 2 < v 2 2 =C(iωI A) 1 w1 B This verifies thatdet [ I C(iωI A) 1 B ] =0. v 2 w 2

16 Structured Singular Values GivenM C n n and a perturbation set D={diag[δ 1 I r1,...,δ m I rm, 1,..., p ]: δ k R, l C m l m l } the structured singular value µ D (M) is defined by µ D (M)=sup{ σ( ) 1 : D,det(I M )=0} See Matlab s µ toolbox

17 Reformulated Definition The following two conditions are equivalent (i) (ii) 0 =det[i M(iω)] for all D and ω R µ D (M(iω))<1for ω R

18 Bounds on µ IfD consists of full complex matrices, then µ D (M)= σ(m). where σ(m) is the largest singular value of M = the larges eigenvalue of the matrixm M. IfD consists of perturbations of the form = δi with δ [ 1,1], then µ D (M) is equal to the magnitude ρ R (M) of the largest real eigenvalue of M ( the spectral radius ). In general ρ R (M) µ D (M) σ(m)

19 Computation of µ Define U D ={U D:U U=I} D D ={D=D C n : D = D for all D} G D ={G=G C n : G = G for all D} Then sup ρ R (UM) µ D (M) inf µ(d,g) inf U U D D D D D G G D σ(dmd 1 ) D where µ(d,g)=inf{µ>0: M D DM+j(GM M G)<µ 2 D D}

20 S-procedure LetM 0,M 1,...M p be quadratic functions ofz R n wheret i =T T i. M i =z T T i z+2u i z+v i, i=0,...,p Consider the following condition onm 0,M 1,...M p : M 0 (z) 0 forallzsuchthat M i (z) 0, i=1,...,p (1)

21 Consider the following condition onm 0,M 1,...M p : M 0 (z ) 0 for allz such that (1) M i (z ) 0, i=1,...,p Obviously, if there exists τ 1 0,...τ p 0such that for allz M 0 (z)+ then (1) holds. p τ i M i (z) 0 (2) i=1 Nontrivial fact, that whenp=1,(1) implies (2), provided that there exist somez 0 such thatm 0 (z 0 )<0.

22 S-procedure for quadratic inequalities The inequality x y 1 T M 0 x y 0 1 follows from the inequalities x y 1 T M 1 x y 0 1 x y 1 T M 2 x y 0 1 if there exist τ 1,τ 2 0such that M 0 + τ 1 M 1 + τ 2 M 2 0 Numerical algorithms in available (e.g. in Matlab), see also [Boyd et al]

23 S-procedure in general The inequality σ 0 (h) 0 follows from the inequalities σ 1 (h) 0,...,σ n (h) 0 if there exist τ 1,...,τ n 0such that σ 0 (h)+ τ k σ k (h) 0 h k

24 S-procedure losslessness by Megretski/Treil Let σ 0,σ 1,...,σ n be time-invariant quadratic forms onl m 2. Suppose that there existsz such that σ 1 (z )>0,...,σ m (z )>0 Then the following statements are equivalent σ 0 (z) 0for allz such that σ 1 (z) 0,...,σ n (z) 0 There exist τ 1,...,τ n 0such that σ 0 (z)+ k τ k σ k (z) 0 z

25 Integral Quadratic Constraint An IQC expresses information of a subsystem. Should be convenient to use for analysis of a larger system. Unifies Multiplier (Zames-Falb) Passivity Absolute stability µ

26 Passivity Theorem is a Small Phase Theorem r 1 e 1 S 1 y 1 y 2 e 2 r 2 S 2 φ 2 φ 1 A passive operator can also be viewed as a sector condition[0, ). Compare circle criterion: Nyquist curve should avoid "circle" ( 1 α, 1 β ) whole LHP as α 0and β.

27 Multipliers Cut the loops in smart ways or introduce multipliers (Bounded operatorsm andm 1 ). Same idea as loop transformations: should be easier to prove stability for transformed system. G 1 G 1 M G 2 M 1 G 2

28 Multipliers, cont d Example: Negative feedback of linear system G(s)=C(sI A) 1 B with nonlinearity with positivity property ϕ(y) y 0. Circle criterion assures exponential stability if Re{G(iω)}>0 ω R Compare (strict) passivity conditions

29 Zames-Falb (1968): Multipliers, cont d Circle criterion can be improved if there are additional assumptions on the nonlinearity as e.g., monotonicity or bounds on slope. If dϕ(y) dy 0, Z-F introduced extra freedom with H RL and H L1 1 such that absolute stability is assured if Re{G(iω) 1 (1+H(iω))}>0, ω R{0} Compare with Popov-criterion conditions

30 Integral Quadratic Constraint v v The causal bounded operator onl2 m is said to satisfy the IQC defined by the matrix function Π(iω) if [ ] [ ] v(iω) v(iω) ( v)(iω) Π(iω) ( v)(iω) dω 0 for allv L 2. Trivial for Π>0, but almost all interesting cases have non-positive Π.

31 Integral Quadratic Constraint v v The causal bounded operator onl2 m is said to satisfy the IQC defined by the matrix function Π(iω) if [ ] [ ] v(iω) v(iω) ( v)(iω) Π(iω) ( v)(iω) dω 0 for allv L 2. Trivial for Π>0, but almost all interesting cases have non-positive Π.

32 Example Gain and Passivity Suppose the gain of is at most one. Then [ ] [ ] [ ] v(iω) 0 ( v 2 v 2 I 0 v(iω) )dt= ( v)(iω) 0 I ( v)(iω) dω 0 Suppose instead that is passive. Then [ ] [ ] [ ] v(iω) 0 I v(iω) 0 v(t)( v)(t)dt= ( v)(iω) I 0 ( v)(iω) dω 0 Note: Scaling in Parseval s formula neglected here (does not affect sign of IQC).

33 Exercise Show that a nonlinearity satisfying the sector condition αy 2 ϕ(t,y)y βy 2 satisfies the IQC, ϕ IQC(Π) given by Π(jω)=Π= [ ] 2α β α+ β α+ β 2 Note: Satisfies a quadratic inequality (for every frequency)= satisfies integral quadratic inequality

34 IQC s for Coulomb Friction Zames/Falb s property f(t)= 1 ifv(t)<0 f(t) [ 1,1] ifv(t)=0 f(t)=1 ifv(t)>0 0 0 v(t)[f(t)+(h f)(t)]dt, h(t) dt 1 0 ] [ ][ v f] 0 1+H(iω) dω [ v f 1+H( iω) 0

35 structure Π(iω) Condition passive (iω) 1 δ [ 1,1] δ(t) [ 1,1] ( v)(t)=sgn(v(t)) [ [ ] 0 I I 0 [ ] x(iω)i 0 0 x(iω)i [ ] X(iω) Y(iω) Y(iω) X(iω) [ ] X Y Y T X 0 1+H(iω) 1+H(iω) 0 ] x(iω) 0 X=X 0 Y= Y H L1 1

36 Well-posed Interconnection f v τ G(s) w e The feedback interconnection { v =Gw+f w = (v)+e is said to well-posed if the map(v,w) (e,f) has a causal inverse. It is called BIBO stable if the inverse is also bounded.

37 Use as many IQCs as possible to characterize the nonlinearity/uncertainty IQC(Π 2 ) IQC(Π 1 ) IQC(Π 3 ) In this caseiqc(π 3 ) does not help to restrict the complete set that satisfy the IQCs.

38 IQC Stability Theorem τ G(s) LetG(s) be stable and proper and let be causal. For all τ [0,1], suppose the loop is well posed and τ satisfies the IQC defined by Π(iω). If [ G(iω) I ] [ G(iω) Π(iω) I then the feedback system is BIBO stable. ] <0 for ω [0, ]

39 Computations via LMI s [ G(iω) I ] k [ G(iω) τ k Π(iω) I ] <0 for ω [0, ] [ (iωi A) 1 B I ] (M+ k [ (iωi A) τ k M k ) 1 B I ] <0 for ω [0, ] n i=1 Solve for τ 1,...,τ n 0andP. [ A τ k M k + T P+PA PB B T P 0 ] <0.

40 Relation to Passivity and Gain Theorems τ G(s) [ ] I 0 A stability theorem based on gain is recovered with. 0 I [ ] 0 I A passivity based stability theorem is recovered with. I 0

41 Special Case µ Analysis Note that =diag{δ 1,...,δ m }, with δ k 1satisfies the IQC defined by [ ] X(iω) 0 Π(iω)= 0 X(iω) wherex(iω)=diag{x 1 (iω),...,x m (iω)}>0. Feedback loop stability follows if there existsx(iω)>0with G(iω) X(iω)G(iω)<X(iω) ω [0, ] or equivalently, withd(iω) D(iω)=X(iω) sup D(iω)G(iω)D(iω) 1 <1 ω

42 Combination of Uncertain and Nonlinear Blocks The operator (v 1,v 2 )=(δv 1,φ(v 2 )) where δ [ 1,1] α φ(v 2 )/v 2 β satisfies all IQC s defined by matrix functions of the form X(iω) 0 Y(iω) 0 Π(iω)= 0 2α β 0 α+ β Y(iω) 0 X(iω) 0 0 α+ β 0 2 wherex(iω)=x(iω) andy(iω)=y(iω).

43 Proof idea of IQC Theorem Combination of the IQC for with the inequality for G gives existence ofc 0 >0such that v c 0 v τg (v) v L 2,τ [0,1] If(I τg ) 1 is bounded for some τ [0,1] then the above inequality gives boundedness of(i νg ) 1 for all ν with c 0 G τ ν <1 Hence, boundedness for τ = 0 gives boundedness for τ<(c 0 G ) 1. This, in turn, gives boundedness for τ<2(c 0 G ) 1 and so on. Finally the whole interval[0,1] is covered.

44 A toolbox for IQC analysis Copy /home/kursolin/matlab/lmiinit.m to the current directory or download and install the IQCbeta toolbox from G(iω) e(t) y(t) 10s 2 s 3 +2s 2 +2s >> abst_init_iqc; >> G = tf([10 0 0],[ ]); >> e = signal >> w = signal >> y = -G*(e+w) >> w==iqc_monotonic(y) >> iqc_gain_tbx(e,y)

45 A simulation model

46 An analysis model defined graphically The text version (i.e., NOT the gui) is strongly recommended by the IQCbeta author(s) at present version!!

47 z iqc_gui( fricsystem ) extracting information from fricsystem... scalar inputs: 5 states: 10 simple q-forms: 7 LMI #1 size = 1 states: 0 LMI #2 size = 1 states: 0 LMI #3 size = 1 states: 0 LMI #4 size = 1 states: 0 LMI #5 size = 1 states: 0 Solving with 62 decision variables... ans = The text version (i.e., NOT the gui) is strongly recommended by the IQCbeta author(s) at present version!!

48 A library of analysis objects

49 Bounds on Auto Correlation u system The auto correlation bound u(t) u(t T)dt α corresponds to Ψ(iω)=2α e iωt e iωt. u(t) u(t)dt,

50 Dominant Harmonics u system For small ǫ>0, the constraint 0 û(iω) 2 dω (1+ǫ) b a û(iω) 2 dω means that the energy of u is concentrated to the interval[a, b].

51 Incremental Gain and Passivity v v A causal nonlinear operator onl2 m incremental gain less than γ if is said to have (v 1 ) (v 2 ) γ v 1 v 2 v 1,v 2 L 2 It is called incrementally passive if 0 T 0 [ (v 1 ) (v 2 )][v 1 v 2 ]dt T>0,v 1,v 2 L 2

52 Incremental Stability f v τ G(s) w e The feedback interconnection { v =Gw+f w = (v)+e is called incrementally stable if there is a constant C such that any two solutions(e 1,f 1,v 1,w 1 ),(e 2,f 2,v 2,w 2 ) satisfies v 1 v 2 + w 1 w 2 C e 1 e 2 +C f 1 f 2

53 Robust Performance G 11 (iω) G 12 (iω) G 21 (iω) G 22 (iω)

54 A Converse Small Gain Theorem The static case A matrixm satisfies σ(m)<1if and only if0 =det(i M) for all matrices with σ( ) 1. The dynamic case A stable transfer matrixg(s) satisfies G <1if and only if [I (s)g(s)] 1 is stable for every stable (s) with <1.

55 Proof in the Static Case Suppose thatdet(i M)=0and σ( ) 1. Then there existsx =0such thatx Mx=0and so σ(m) 1. x = Mx Mx On the other hand, if σ(m) 1, there existsx =0with Mx x. Let = xm x Mx 2 Then σ( ) 1and(I M)x=0, sodet(i M)=0.

56 Robust Performance Theorem Suppose that σ(m 11 ) 1and thatd is a connected set of matrices with0 D. Let D 1 ={diag( 1, ): σ( 1 ) 1, D}. Then the following conditions are equivalent. (i) σ(m 11 +M 12 [I M 22 ] 1 M 21 )<1for D ( [ ][ ]) 1 0 M11 M (ii) 0 =det I 12 for 0 M 21 M { 22 D σ( 1 ) 1 ([ ]) M11 M (iii) µ 12 D1 1for ω R M 21 M 22

57 Proof The conditions (ii) and (iii) are equivalent by the definition of µ. Moreover, we showed on the previous slide that (i) fails if and only if there exists 1 with σ( 1 ) 1andx 1 =0such that 0={I 1 (M 11 +M 12 [I M 22 ] 1 M 21 )}x 1 Introducex 2 =[I M 22 ] 1 M 21 x 1. Then [ ] [ ][ x1 1 0 M11 M = 12 x 2 0 M 21 M 22 ][ x1 This is possible if and only if (ii) fails, so the equivalence of (i) and (ii) is proved. x 2 ]

58 Example Compute maxsup δ k 1 ω δ 1 (iω) 2 +(2+ δ 2 )iω+2+ δ 1 δ 2 This is the worst case gain of the system { ÿ= (2+ δ 2 )ẏ (2+ δ 1 δ 2 )y+ δ 1 u= 2ẏ 2y δ 1 v 1 δ 2 v 2 v 1 = δ 2 v 3 +u, v 2 =ẏ, v 3 =y ẏ y ÿ ẏ v δ 1 v 1 v δ 2 v 2 v δ 2 v 3 y u

59 Performance Analysis via S-procedure The performance criterion σ 0 (h) 0 h N follows from the IQC s σ 1 (h) 0,...,σ n (h) 0 h N if there exist τ 1,...,τ n 0such that σ 0 (h)+ k τ k σ k (h) 0 h L n 2

60 Performance Bounds from IQC s v y G 11 G 12 G 21 G 22 w u Suppose that satisfies the IQC defined by Π. Then the gain bound y γ u holds provided that the system is stable and [ ] Π 11 0 Π 12 0 [ ] G(iω) 0 0 I 0 0 G(iω) I Π 21 0 Π 22 0 I γ 2 I ω R

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