Robust Control 5 Nominal Controller Design Continued
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1 Robust Control 5 Nominal Controller Design Continued Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 4/14/2003
2 Outline he LQR Problem A Generalization to LQR Min-Max Control Solving the Riccati Equation Classical LQG Stability Margins & LR
3 he Linear Quadratic Regulator Problem (state feedback) x = Ax + Bu, 1 J( x) = lim x ( ) Qx( ) u ( ) Ru( ) d 0 τ τ + τ τ τ heorem: If, is controllable (stabilizable) the optimal controller and X = ( AB) for any initial state is u t K = X R B X () = Kx() t, where: 0 is the unique positive semidefinite solution of the algebraic Riccati equation: + + = 0 A X XA XBR B X Q
4 Stability of LQR An autonomous linear system x = Ax is asymptotically stable if there exists P> 0, Q> 0 that satisfy the Lyapunov equation + = PA A P Q Furthermore, if det A 0, the system is asymptotically if and only if there exist, P> 0, Q 0 that satisfy the Lyapunov equation. ( ) Basis of proof: set V x = x Px, P > 0 and compute V V = x = x PAx + x A Px x If V = x Ax, Q > 0 we have stability by the Lyapunov stability heorem. If det A 0, use LaSalle's hm for sufficiency, Chetaev Instability hm for necessity.
5 Stability of LQR ~ 2 R B X XA + A X XBR B X = Q ( ) Now consider, x = Ax+ Buu, = Kx x = A+ BK xwhere K = ( ) ( ) ( ) ( R B X BK ) + XBK XB + X B R B X X X A+ BK + A+ BK X = Q XBR B ( ) ( ) X Lyapunov Equation
6 Generalization of LQR 1 J( x) = lim x ( ) Qx( ) 2 x ( ) Su( t) u ( ) Ru( ) d τ τ τ τ τ τ R> Q SR S 0, 0 Notice that (complete the square) ( ) ( ) ( ) x Qx + 2x Su + u Ru = u + R Sx R u + R Sx + x Q SR S x Define a new control ( ) = +, = + = + Bu 1 u u R S x x Ax Bu x A BR S x 1 J( x) = lim x ( τ) ( Q SR S ) x( τ) u ( τ) Ru( τ) dτ + 0 u = R ( B P+ S) x ( ) + ( ) + ( ) P A BR S A BR S P PBR B P Q SR S
7 RE & Associated Hamiltonian Matrix A BR B A X + XA XBR B X + Q = 0 H = if λ is an eigenvalue of H, so is λ ( ) ( ) A, B controllable, A, C observable where Q= C C there are no eigenvalues on the imaginary axis = such that H Q Λ 0 = 0 Λ A ( ) Λ is composed of real Jordan blocks, with Reλ Λ > 0 he positive definite solution is X =
8 Computational ools - 1 CARE Solve continuous-time algebraic Riccati equations. [X,L,G,RR] = CARE(A,B,Q,R,S,E) computes the unique symmetric stabilizing solution X of the continuous-time algebraic Riccati equation -1 A'XE + E'XA - (E'XB + S)R (B'XE + S') + Q = 0 or, equivalently, F'XE + E'XF - E'XBR B'XE + Q - SR S' = 0 with F:=A-BR S'. When omitted, R,S and E are set to the default values R=I, S=0, and E=I. Additional optional outputs include the gain matrix -1 G = R (B'XE + S'), the vector L of closed-loop eigenvalues (i.e., EIG(A-B*G,E)), and the Frobenius norm RR of the relative residual matrix.
9 Computational ools 2 LQR Linear-quadratic regulator design for continuous-time systems. [K,S,E] = LQR(A,B,Q,R,N) calculates the optimal gain matrix K such that the state-feedback law u = -Kx minimizes the cost function J = Integral {x'qx + u'ru + 2*x'Nu} dt. subject to the state dynamics x = Ax + Bu. he matrix N is set to zero when omitted. Also returned are the Riccati equation solution S and the closed-loop eigenvalues E: -1 SA + A'S - (SB+N)R (B'S+N') + Q = 0, E = EIG(A-B*K).
10 Min-Max Control x = Ax+ Bu+ Ew, y = Cx ( AC) 2 (, ) = ( ) ( ) + ρ ( ) ( ) γ ( ) ( ), ρ, γ > 0 0 Juw y tyt u tut w twt dt objective: min max Juw (, ) u () ( )( ) heorem: Suppose wt has bounded energy, AB,, AE, stabilizable,, detectable, then w 1 the optimal control i s u( t) = Kx( t), K = B S ρ 1 the worst case disturbance is wt ( ) = E Sxt ( ) 2 γ S is the unique, symmetric, nonnegative solution of the Riccati equation: 1 1 AS+ SA S BB EE S CC 2 = ρ γ
11 Min-Max Hamiltonian Matrix γ ρ 2 = = ρ γ AS SA S BB EE S CC H stabilizability/detectability γ such that there are no pure eigenvalues of H if γ > γ min γ produces LQR solution γ = γ min is the full state feedback H solution all γ γ < are valid min-max controllers min 1 1 A EE BB CC min A
12 Min-Max, Continued 1 1 Reλ( H) 0 Reλ A+ EE S BB S 0 2 < γ ρ 1 1 he closed loop system matrix is A BB S, since EE S 2 ρ γ is destabilizing, there is some stability margin
13 Linear Quadratic Gaussian (LQG) Problem Consider the linear system x = Ax+ Bu+ w y = Cx+ v where wv, are independent, zero-mean white noise processes having covariances { } { } { τ δ τ τ δ τ τ } E w() t w ( ) = W ( t ), E v() t v ( ) = V ( t ), E w() t v ( ) = 0 () Find u t that minimizes 1 J = E lim 0 xqx+ uru dt, Q= Q 0, R= R > 0
14 LQG Solution ( ) u = Kxˆ( t), K = R B P PA + A P PBR B P = Q, P 0 x ˆ = Axˆ+ Bu+ L y Cxˆ, L= SC V SA + AS SC V CS = W, S 0
15 LQG Robustness: State Feedback Loop break point x = 0 K ( ) 1 si A B x ( ) = ( ) = Φ, Φ= ( ) ( ) = ( ) 1 G s K si A B K B si A S s I K si A B ( ) ( ) For R diagonal, σ S jω 1 ω 1 π GM :, PM : (in each channel) 2 3
16 LQG Robustness: State Feedback 2 When K is determined via the Ricatti equation, it is known { ( ) } σ min S = min I K jωi A B v 1, ω > 0 v = 1 Similarly, R { ( ) } I C jωi A L v min 1, ω > 0 v = 1 V V R
17 LQG Robustness: Output Feedback y = 0 G ( s ) G ( s) c p y p ( ) = ( ) G s C si A B ( ) = ( + + ) Gc s K s A BK LC L No guaranteed margins!
18 Output Feedback-a closer look ŷ C v u xˆ = Axˆ+ Bu+ Lv K x = Ax + Bu ˆx y = Cx Φ ( s) [ si A] 1 Loop transfer functions: y v L Φ ˆx C ŷ (A) (B) (C) Φ Φ CΦL C BK BK L KΦB C L u Plant is invisible B Φ x K (D) 1 K Φ BK LC LCΦB observer is invisible Loop breaking at (A) & (D) do not produce the excellent margins of state feedback.
19 Loop ransfer Recovery Recall L= Let SA + AS SC V CS = W 2 W = W0 + BB phase plant SC V, with ( ) ( ) = ( ), ρ a scalar. It can be shown, for a minimum lim G s G s K si A B ρ c Correspondingly, p ( ) = ( ) ( ) lim G s K si A BG s ρ c ρ complete recovery! some of the estimator poles tend to the plant zeros For nonminimum phase plants, estimator poles tend to LHP plant zeros, recovery is not complete. p
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