Chapter 3. LQ, LQG and Control System Design. Dutch Institute of Systems and Control

Transcription

1 Chapter 3 LQ, LQG and Control System H 2 Design

2 Overview LQ optimization state feedback LQG optimization output feedback H 2 optimization non-stochastic version of LQG Application to feedback system design

3 LQ theory State space description xt &() = Axt () But (), x(0) = x o zt () = Dxt () Use state feedback to control the system such that is minimal 0 J = [ z ( t) Qz( t) u ( t) Ru( t)] dt ut () x(0) xt () zt () control error control effort

4 Criterion Control error: nonnegative-definite quadratic expression such as 2 i i z () t Qz() t = q z () t Control effort: similar quadratic expression i rade-off form J = [ z ( t) Qz( t) ρu ( t) Rou( t)] dt 0 ρ large: low gain ρ small: high gain

5 Choice of Q and R A sensible initial choice of the weighting matrices is z u () t Qz() t () t Ru() t = = i ρ i 2 i () t 2 i,max z z 2 i () t 2 i,max u u Fine tune Q and R by trial and error

6 Solution Optimal state feedback law x(0) ut () = Fxt () State feedback gain ut () x& = Ax Bu z = Dx xt () zt () 1 F = R B X F Solve X from the Matrix Algebraic Riccati Equation 1 A X XA D QD XBR B X = 0

7 Properties of the solution Assume that the system xt &() = Axt () But (), zt () = Dxt () is stabilizable and detectable, and that Q and R are positive-definite hen he ARE has finitely many solutions here is a unique nonnegative-definite solution X he closed-loop optimal system xt &() = ( A BF) xt () is stable J = x (0) Xx(0)

8 Guaranteed stability margin the Nyquist plot of the loop gain stays outside this disc with radius 1 Conclusions the gain margin is infinite the phase margin is at least 60 o he modulus margin is at least 1

9 What can we achieve with LQ control? We may achieve But Good transient response from nonzero initial conditions Good stability Good robustness State feedback is seldom possible What about disturbances?

10 LQG control Modified plant description u xt &() = Axt () But () Gvt () yt () = Cxt () wt () zt () = Dxt () v Plant z y w White noise with intensity V White noise with intensity W Error output Measured output Controller

11 Observers Observer equation u Plant y z xt & ˆ() = Axt ˆ() But () K[ y( t) Cxˆ ( t)] K Plant model Observation error et () = xt ˆ() xt () Error equation (no noise) et &() = ( A KCet ) () Observer If A KC is stable then e(t) converges to 0

12 In the presence of the noises v and w et &() = ( A KCet ) () Gvt () Kwt () Kalman filter If the error system is stable then the error variance matrix Y() t = Ee() t e () t converges to a constant matrix Y that satisfies the equation ( A KC) Y Y( A KC) GVG KWK = 0 he variance is minimal if K = YC W where Y satisfies the ARE 1 AY YA GVG YC W CY = 0 1 Kalman filter

13 Properties of the Kalman filter Assume that the system xt &() = Axt () Gvt () yt () = Cxt () is stabilizable and detectable and that V and W are positive-definite hen he ARE has finitely many symmetric solutions he ARE has a unique nonnegative-definite symmetric solution Y he corresponding error system and the filter itself are stable lim ete ( ) ( t) = Y t

14 Optimal output feedback u Plant z y LQG control: Interconnect the optimal state feedback law with F Observer the Kalman filter to minimize Separation principle Certainty equivalence lim E z ( t) Qz( t) u ( t) Ru( t) t

15 Properties of the LQG solution he eigenvalues of the interconnected system are the list union of the eigenvalues of A BF and A KC Hence, the closed-loop system is stable if both the Kalman filter and the state feedback solution are stable

16 Exact state reconstruction-1 Assumptions: he disturbance v is additive to the control input u, that is, xt &() = Axt () But [ () vt ()] he open-loop plant H( s) = C( si A) B is square he zeros of the plant are all in the left-half plane, 1 ( ) that is, H s has all its poles in the left-half complex plane 1

17 Exact state reconstruction-2 hen if there is no measurement noise the state may be reconstructed with zero error v u y Hs () H 1() s uv ( si A) - 1 B x Hence, if W = σ W then o Y σ K σ σ 0 0 σ 0 differentiating filter

18 Loop transfer recovery-1 Consider the broken LQG loop u Plant z y Denote the loop gain as L σ F ^x Observer Assume that the conditions for exact state reconstruction hold and W = σ W o hen σ 0 1 L ( s) F( si A) B σ

19 Loop transfer recovery-2 Key assumption: he plant has left-half plane zeros only hen May approach the state feedback loop gain arbitrarily closely Several of the closed-loop poles, compensator poles and compensator zeros go to as we get closer he robustness properties of the LQ solution are recovered Caveat: If the loop is broken differently then robustness may not apply

20 H 2 Optimization H 2 optimization amounts to removing the stochastic interpretation from LQG optimization u Hs () y White noise with intensity V Output signal has spectral density φy ( ω) = H( jω) VH ( jω)

21 H 2 Norm u Hs () y Variance of the output signal Variance matrix of y Ey () t y() t = tr Ey() t y () t = tr φ ( ω) df Set V = I = tr H( jω) VH ( jω) df = tr H( jω) H ( jω) df y f = ω /2π Square of the 2-norm of the system transfer matrix H 2

22 Re-interpretation of the LQG paradigm-1 u v Ce P y z w Write z v = H u w Assume Q = I R = I hen zt () zt () E z () t z() t u () t u() t = E ut () ut () V 0 = tr H ( jω) H ( jω) df H = 0 W 2 2 Let V = I, W = I

23 Re-interpretation of the LQG paradigm-2 Hence, solution of the LQG problem amounts to minimization of the 2-norm H 2 of the transfer matrix H from the white noise inputs v and w to the outputs z and u Notes We have gone from a signal interpretation to a systems interpretation he norm interpretation is non-stochastic V and W now simply are tuning parameters

24 he standard H 2 optimization problem Open-loop system z G11 G12 w y = G21 G 22 u w u G z y Closed-loop system 1 z = G 11 G12K( I G22K) G 12 w H K Minimize H 2

25 Solution of the H 2 problem-1 State space representation xt &() = Axt () Bwt () But () zt () = Cxt () D ut () yt () = C xt () D wt () D ut () Assumptions the systems xt &() = Axt () But 2 (), zt () = Cxt 1 () xt &() = Axt () Bwt (), y() t = C x() t 1 2 are stabilizable and detectable D D, D D are nonsingular

26 Solution of the H 2 problem-2 By suitable substitutions the problem may be reduced to an LQG problem Its solution involves the solution of two AREs he compensator consists of an observer interconnected with a state feedback law

27 Design by the LQG method-1 System equations x& = Ax Bu Gv y = Cx w, z = Dx Parameter selection Let G = B so that we may apply loop recovery Preferably D = C, that is, y(t) = z(t)w(t). Otherwise, we deal with inferential control, which is less robust Choose Q and R = ρro by scaling arguments as explained Likewise choose V and W = σ Wo by scaling arguments

28 Design by the LQG method-2 Design Determine the state feedback gain F by varying ρ and tuning Q and Ro o Assess the parameter selection by considering the closed-loop pole locations and transient responses Design the observer by making σ small enough to achieve loop recovery, and by tuning V and Wo o Assess the design by considering the closed-loop pole locations and the sensitivity functions

29 H 2 interpretation of the LQG problem-1 Block diagram 1 Ps () = CsI ( A) B u v R z 1 Rs () = DsI ( A) B Ce P w In the case of noninferential control y Rs () = Ps ()

30 H 2 interpretation of the LQG problem-2 Redraw the diagram as 1 Wo ( s) = R( s) P ( s) v u - Ce P W o w y z In the case of non-inferential control Wo ( s) = I

31 H 2 interpretation of the LQG problem-3 It is not difficult to find that G ( s) = G ( s) 2 2 tr o o o o ( ) H = W SPVP S W Q W W W Q V R UWU R df where S = ( I PC ) = ( I PC ) PC 1 = ( I C P) C P e e e U = C ( I PC ) e e e 1 e Weighted combination of sensitivity functions

32 A more flexible configuration z 2 W 2 v V 1 V W, V 1 2, W 1 2 shaping filters weighting filters Ce u P W 1 z 1 V 2 w V W Q R = I = I = I = I 2 2 ( ) H = tr W SPVV P S W WV V W W VV W W UV V W df y

33 Example-1 MIMO plant u 1 Ps () s s 2 = 1 0 s 2 1 s 2 1 y 1 u 2 y s2 2 Poles at 0, 0, 2 No zeros Required bandwidth: 1 rad/s on both channels

34 Example-2 Choice of weighting matrices Q = I, R = ρ I, V = I, W = σ I Loci of the regulator poles as B0 Closed-loop poles for ρ = , 0.72 ± ϕ 0.70 he loci of the observer poles are identical For σ = the observer poles are 200, 7.1 ± j 7.1

35 Example-3 Sensitivity matrix he sensitivity for disturbances at the second output is unsatisfactory

36 Example-4 Add integral action in the second channel by the integrator-in-the-loop method. o this end, modify the plant to u 1 1 s 2 y 1 u u2 ' 2 1 y 2 s1 s s2 Open-loop poles at 0, 0, 0, 2. Open-loop zero at 1

37 Example-5 Root loci 2 1/ρ Im 1 1/ρ 0-1 1/ρ Re Choose ρ = 0.5, σ =

38 Example-6 Sensitivity matrix

39 Example-7 Closed-loop step response matrix r C P z