Robust Control 9 Design Summary & Examples

Transcription

1 Robust Control 9 Design Summary & Examples Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/6/003

2 Outline he H Problem Solution via γ-iteration Robust stability via coprime factorization Robust stability with loop shaping Example: non-collocated revisited

3 he H Problem We will formulate a problem similar to above, but with two changes: 1) the disturbances are not stochastic. wt may be any member of the class () wt = 1 ) Choose K so that the maximum of z t over all admissible disturbances () wt is a minimum, i.e. min max zt ( ) K w = 1 () ()

4 he H Problem, Cont d Recall: 1 z () t z() t dt = Z ( jω) Z( jω) dω π 1 = W ( jω) F ( jω) F( jω) W( jω) dω π he maximum performance energy over all disturbances with unit norm occurs when W is alligned with the maximum eigenvalue max ( ) ( ) ( ) ( ) ( ) w = 1 = 1 z t z t dt W jωσ ( F jω ) W jωdω π ( ω ) = max σ F( j ) = F min F ω K * of F F :

5 Solution of the H Problem A stabilizing controller that satisfies Fs ( ) < γ is A ZL K = F 0 where ( ) A = A+ B + L D W + B F + Z L C + Z L D F uu ( uu ) 1 ( xy ) yy F = R R + B X L = Y C + V V 1 1 W = B 1 X, Z = I Y X γ γ X 0, Y 0 satisfy the Riccati equations 1 X A + A X X B R B BB X = R + R R R γ 1 1 r r uu 1 1 xx xu uu xu 1 AY + Y A Y C V C C C Y = V + V V V γ 1 1 e e yy 1 1 xx xy yy xy If γ is fixed, these Riccati equations are decoupled

6 Solution of the H Problem, Cont d And the following 3 conditions are satisfied 1) he Hamiltonial matrix A BRuu Rxu BRuu B + BB 1 1 γ 1 1 Rxx RxuRuu Rxu ( A BRuu Rxu ) + has no eigenvalues on the imaginary axis. Equivalently, A + BW + B F 1 is stable. ) he Hamiltonian matrix ( A VxyVyy C) CVyy C + C 1C1 γ 1 1 Vxx + VxyVyy Vxy A+ VxyVyy C has no eigenvalues on the imaginary axis. Equivalently, 1 A+ LC + Y C C γ 1 1 is stable. () () ρ YX < γ ρ = λ 3) ( ) where max i i is the spectral radius.

7 Strategy for Solving H ypically there is a minimum value of γ for which the above conditions are satisfied. his is the optimal control. he usual method is to find the optimal control by γ-iteration. Begin with a large γ and reduce it until one of the conditions fail. hen use a bisection method, until within an acceptable tolerance. Usually, we don t need to get that close - a few iterations will suffice.

8 Robust Controller Design with Coprime Uncertainty Consider the family of perturbed plants { 1 ( ) ( ) [ ] ε} G = D + N + < p D N N D with some stability margin ε > 0. he system is robustly stabilized by the controller u = Ky if and only if γ K ( I GK ) D I ε Robust Stabilizer Design Problem: Find the lowest achievable γ and the corresponding maximum stability margin ε and the corresponding controller K.

9 Solution to RSDP min 1 max ( 1 ( XZ )) 1/ where X, Z are the unique positive definite sol'ns of the ARE's ( ) ( ) A BS D C Z + Z A BS D C ZC R CZ + BS B = ( ) ( ) A BS D C X + X A BS D C XBS B X + C R C = and γ = ε = + ρ ( ) minimal realization ( A, B, C, D) G s R= I + DD, S = I + D D 0 0

10 Solution to RSDP, Cont d a solution that guarantees K 1 1 ( I GK ) D γ for some γ > γ I is given by K A + BF + γ ( L ) ZC ( C+ DF) γ ( L ) ZC ( 1 γ ) ( ) 1 F S D C B X min 1 1 = B X D = + L= I + XZ

11 Loop Shaping for Performance: Closed Loop ransfer Functions Assume the closed loop is stable. hen: σ ( S ) << σ ( ) σ ( ) σ ( ) ( KS ) 1. disturbance rejection 1. noise attenuation << 1 3. reference tracking 1 4. control energy σ << 1 ( KS ) σ ( ) 5. robust stability with additive uncertainty σ << 1 6. robust stability with mult output uncertainty << 1

12 Loop Shaping for Performance: Open Loop ransfer Function Assume the closed loop is stable. hen: σ ( L) σ ( L) σ ( L) ( K ) 1. disturbance rejection >> 1. noise attenuation << 1 3. reference tracking >> 1 4. control energy σ << 1 ( K ) 5. robust stability with additive uncertainty σ << 1, if σ ( L) << 1 ( L) 6. robust stability with mult output uncertainty σ << 1

13 Example: Collocated vs Noncollocated he goal is to position the load. A position sensor can be placed on motor or load. θ1 θ L motor load flexible shaft

14 Example: d dt θ θ1 0 0 ω k/ J c / J k/ J 0 ω 0 1/ J = L θ θ 0 0 ω k/ J 0 k/ J c / Jω 1/ J 0 z = [ θ1 θ1 ω 1 ω 1 0 ] or z = [ ] θ θ ω ω y = z J = J = 1, k = 1, c = c = 0 1 1

15 Example P = or P= Non-collocated Collocated

16 Simplified Collocated G c ( s) = ( s 0.916) ( s s 1.969) ( s ) ( s+.49) ( s s ) ( s ) 3.65 ( s +.49) + + +

17 Simplified Noncollocated G c ( s) = + + s s s s ( s 0.44) ( s s.089) ( ) ( + ) ( ) + + s s s ( s 0.44) ( s ) ( + ) ( )

18 Noncollocated

19 Robust Stability gammin = 3.80 Zero/pole/gain: (s+0.363) (s^ s +.00) (s+3.15) (s+.35) (s^ s ) gammin = Zero/pole/gain: (s+.001) (s ) (s^ s ) (s ) (s^ s +.744) (s^ s +.733)

20 Step Responses Red: Shaped robust stability Blue: γ-iteration Green: Robust stability

21 Robust Stability

22 Shaped Robust Stability