Lecture 10: Linear Matrix Inequalities Dr.-Ing. Sudchai Boonto

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1 Dr-Ing Sudchai Boonto Department of Control System and Instrumentation Engineering King Mongkuts Unniversity of Technology Thonburi Thailand

2 Linear Matrix Inequalities A linear matrix inequality (LMI) has the form M(p) = M 0 + p 1 M p N M N < 0 where M 0, M 1,, M N are given symmetric matrices, p = [ p 1 p 2, p N ] T is a column vector of real scalar variables the matrix inequality M(p) < 0 means that the left hand side is negative definite An important property of LMIs is that the set of all solutions p is convex LMIs can be used as constraints for the minimization problem min p c T p subject to M(p) < 0 where the elements of the vector c in the linear cost function are weights on the individual decision variables 2/25

3 Linear Matrix Inequalities the convex problem can be solved by efficient, polynomial-time interior-point methods Several LMI constraints can be combined into a single constraint of type for example the constraint M 1 (p) < 0 and M 2 (p) < 0 is equivalent to the single LMI constraint [ ] M 1 (p) 0 < 0 0 M 2 (p) 3/25

4 Pole Region Constraints the condition that the poles of a system are located within a given region in the complex plane can be formulated as an LMI constraint the dynamic system ẋ(t) = Ax(t) This system is stable if an only if the matrix A has all eigenvalues in the left half plane, which is true iff there exists a positive definite, symmetric matrix P that satisfies the Lyapunov inequality PA T + AP < 0 This inequality is linear in the matrix variable P, and one can use LMI solvers to search for solutions 4/25

5 Pole Region Constraints assume that A is a 2 by 2 matrix and write the symmetric matrix variable P as [ ] [ ] [ ] [ ] p 1 p P = = p 1 + p 2 + p 3 p 2 p the LMI represents a necessary and sufficient condition for the matrix A to have all eigenvalues in the left half plane one can express an arbitrary region D in the complex plane in terms of two matrix L = L T and M as the set of all complex numbers that satisfy and LMI constraint D = {s C : L + Ms + M T s < 0} where s denotes the complex conjugate of s 5/25

6 Pole Region Constraints Such a region is called an LMI region 6/25

7 Pole Region Constraints Example: poles region constraint Im α l α r β Re From Re s < α r, we have s + s < α r 2 s + s 2α r < 0 7/25

8 Pole Region Constraints Example: poles region constraint and Im s > α l s + s > α l 2 s s + 2α l < 0 Thus [ ] 2αl 0 L v =, M v = 0 2α r [ ] For the conic sector, a complex number s = x + jy lies in the conic sector if and only if y x < tan β = sin β cos β Rewrite the above conditions in the form y 2 x 2 < sin2 β cos 2 β and x sin β > 0 8/25

9 Pole Region Constraints Example: poles region constraint we get x 2 sin 2 β y 2 cos 2 β > 0 x sin β y2 cos 2 β x sin β > 0 By Schur s complement we have [ M c s + M 2x sin β c s = 2jy cos β ] 2jy cos β > 0 2x sin β Thus L c = [ ] 0 0, M c = 0 0 [ sin β ] cos β cos β sin β The two constraints can be combined as [ ] [ ] Lc 0 Mc 0 L =, M = 0 L v 0 M v 9/25

10 Constraints on the H Norm Consider the system with transfer function T(s) as state space realization ẋ(t) = Ax(t) + Bw(t), x(0) = 0 z(t) = Cx(t) + Dw(t) Assuming that T(s) is stable, the H norm of the system is T 2 0 z T (t)z(t)dt = max w 0 0 w T, x(0) = 0 (t)w(t)dt It follows that T < γ is equivalent to 0 (z T (t)z(t) γ 2 w T (t)w(t))dt < 0 Holding true for all square integrable, non-zero w(t) 10/25

11 Constraints on the H Norm Introduce a Lyapunov function V(x) = x T Px with P = P T > 0 Since x(0) = x( ) = 0, the constraint T < γ is enforced by the existence of a matrix P = P T > 0 such that dv(x) dt + 1 γ zt (t)z(t) γw T (t)w(t) < 0 for all x(t), w(t); to turn into a LMI, substitute dv(x) dt To obtain = x T (A T P + PA)x + x T PBw + w T B T Px, z = Cx + Dw [ ] T x AT P + PA + 1 γ CT C w B T P + 1 γ DT C PB + 1 [ ] γ CT D x < 0 γi + 1 γ DT D w 11/25

12 Constraints on the H Norm For T < γ the above must hold for all x and w, thus the block matrix must be negative definite The condition can be rewritten as [ ] [ ] A T P + PA PB B T + 1 C T [ ] P γi γ D T C D < 0 By Schur complement, we have Theorem (Bound real lemma) T < γ if and only if there exists a positive definite, symmetric matrix P that satisfies the linear matrix inequality A T P + PA PB C T B T P γi D T < 0 C D γi 12/25

13 Constraints on the H Norm Using cvx sys = rss(3,3); A = sysa; B = sysb; C = sysc; D = sysd; n = size(a,1); nu = size(b,2); ny = size(d,1); cvx_begin sdp variable P(n,n) symmetric variable gamma; minimize gamma; subject to P > 0; [A'*P + P*A, P*B, C'; B'*P, -gamma*eye(nu), D'; C, D, -gamma*eye(ny)] < 0; cvx_end display(p); 13/25

14 Controller Design Using LMIs Generalized Plant w P z u K v The generalized plant P(s) has a state space realization ẋ(t) = Ax(t) + B w w(t) + B u u(t) z(t) = C z x(t) + D zw w(t) + D zu u(t) v(t) = C v x(t) + D vw w(t) 14/25

15 Controller Design Using LMIs Controller dynamics The controller dynamics are represented by a state space model ζ(t) = A K ζ(t) + B K v(t) u(t) = C K ζ(t) + D K v(t) The state space realization of the closed loop-system is where [ A + B u D K C v A c = B K C v C c = ẋ c (t) = A c x c (t) + B c w(t) z(t) = C c x c (t) + D c w(t) B u C K A K ], B c = [ B w + B u D K D vw B K D vw [C z + D zu D K C v D zu C K ], D c = D zw + D zu D K D vw ], 15/25

16 Controller Design Using LMIs H State Feedback State feedback u = Fx yields the closed-loop system ẋ(t) = (A + B u F)x(t) + B w w(t) z(t) = (C z + D zu F)x(t) + D zw w(t) Replacing the system matrices in Bounded real lemma by the closed-loop matrices and using the variable transformation Y = FP leads to the following result: a necessary and sufficient condition for a state feedback controller to achieve a H -norm less than γ i s the existence of matrices P = P T > 0 and Y that satisfy AP + PA T + B u Y + Y T B T u B w PC T z + Y T D T zu B T w γi D T zw C z P + D zu Y D zw γi < 0, F = YP 1 16/25

17 Controller Design Using LMIs H State Feedback G(s) = A B w B u C z D zw D zu I 0 0 with (A, B u) assumed to be stabilizable minimize subject to γ P = P T > 0 AP + PA T + B uy + Y T B T u B w PC T z + Y T D T zu B T w γi D T zw < 0 C z P + D zu Y D zw γi If this has a solution then F = YP 1 17/25

18 H State Feedback Design Using cvx G = ss(a, [Bw, Bu], [Cz; Cv], [Dzw, Dzu; Dvw, zeros(nv,nu)]); cvx_begin sdp variable P(n,n) symmetric; variable Y(nu,n); variable gamma; minimize gamma; subject to P > 0; [A*P + P*A' + Bu*Y + Y'*Bu', Bw, P*Cz' +Y'*Dzu'; Bw', -gamma*eye(nu,nu), Dzw'; Cz*P + Dzu*Y, Dzw, -gamma*eye(nv,nv)] < 0; cvx_end F = Y*inv(P); % check closed-loop poles Aclp = A + Bu*F; disp(eig(aclp)); 18/25

19 Controller Design Using LMIs H Output Feedback P(s) = A B w B u C z D zw D zu C v D vw 0 for output feedback (assume D K = 0): [ ] u = K(s)y = A K B K C K 0 G(s) = F l (P(s), K(s)) = G(s) = [ A c C c B c D c ] with (A, B u ) assumed to be stabilizable and (C v, A) assumed to be detectable y A B u C K B w B K C v A K B K D vw C z D zu C K D zw 19/25

20 Controller Design Using LMIs H Output Feedback The LMI condition is : A T c P + PA c PB c C T c B T cp γi D T c < 0 C c D c γi Partition P as: [ ] X R P = R T [ ] and P 1 Y S = S T Define an inertia-preserving transform via: PT Y = T X where T Y = [ ] [ ] Y I I X S T, T 0 X = 0 R T 20/25

21 Controller Design Using LMIs H Output Feedback The matrices T X and T Y can be used to transform the nonlinear constraint into a linear one This transformation is based on the fact that where [ ] T T Y PAcT Y = T T X AcT AY + B u CK A Y = à K XA + B K C v [ ] T T Y PB B ] w c = XB w + B, C c T Y = [C z Y + D zu CK C z K D vw à K = RA K S T + RB K C v Y + XB u C K S T + XAY B K = RB K C K = C K S T D K = D K [ ] T T Y I Y PT Y = I X 21/25

22 Controller Design Using LMIs H Output Feedback The LMI condition is : A T c P + PA c PB c C T c B T cp γi D T c < 0, and P > 0 C c D c γi T T Y 0 0 A T c P + PA c PB c C T c T Y I 0 B T cp γi D T c 0 I 0 = 0 0 I C c D c γi 0 0 I AY + YA T + B u CK + C T K BT u à T + A B w YC T z + C T K DT zu A T X + XA + B K C v + C T v B T K XB w + B K D vw C T z γi D T < 0 zw γi 22/25

23 Controller Design Using LMIs H Output Feedback minimize subject to : γ [ ] Y I > 0 I X AY + YA T + B u CK + C T K BT u à T + A B w YC T z + C T K DT zu A T X + XA + B K C v + C T v B T K XB w + B K D vw C T z γi D T < 0 zw γi If this has a solution γ, X, Y, Ã, B and C then PP 1 = I = RS T = I YX (solve for R and S) Solve for A K, B K and C K from: à = RA K S T + RB K C vy + XB uc K S T + XAY B = RB K C = C K S T 23/25

24 H Output Feedback Design Using cvx G = ss(a, [Bw, Bu], [Cz; Cv], [Dzw, Dzu; Dvw, zeros(nv,nu)]); cvx_begin sdp variable X(n,n) symmetric; variable Y(n,n) symmetric; variable AK(n,n); variable BK(n,n); variable CK(nu,n); variable gamma; minimize gamma; subject to [Y, eye(n,n); eye(n,n), X] > 0; cvx_end [A*Y + Bu*CK + Y*A' + CK'*Bu', A+AK', Bw, Y*Cz' + CK'*Dzu'; A'+AK, X+A + A'*Y + BK*Cv + Cv'*BK', X*Bw + BK*Dvw, Cz'; Bw', Bw'*X + Dvw'*BK', -gamma*eye(nw,nw), Dzw'; Cz*Y + Dzu*CK, Cz, Dzw, -gamma*eye(nz,nz)] < 0; 24/25

25 Reference 1 Herbert Werner Lecture note on Optimal and Robust Control, /25