1. LQR formulation 2. Selection of weighting matrices 3. Matlab implementation. Regulator Problem mm3,4. u=-kx

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1 MM8.. LQR Reglator 1. LQR formlation 2. Selection of weighting matrices 3. Matlab implementation Reading Material: DC: p , , Matlab fnctions: lqr, lqry, dlqr, lqrd, care, dare 3/26/2008 Introdction to Process Control 1 Reglator Problem mm3,4 Objective: Drive all states to zeros with satisfactory natral mode transients to initial conditions or distrbances X=AX+B C Y =-KX. Plant X = AX + B Sensor Y=CX Y Controller -K Estimator X = AXˆ + B + L( y CXˆ ) ˆ. 3/26/2008 Introdction to Process Control 2

2 Reference Inpt for the Estimation Case-mm6 r N Plant X & = AX + B Sensor Y=CX M Controller -K Estimator X = ( A BK LC) Xˆ ˆ. + Ly + Mr plant controller : : X& = AX + B y = CX & Xˆ = ( A BK = K Xˆ + N r Determine parameter M,N: 3/26/2008 Introdction to Process Control 3 LC ) Xˆ = + N r Ly + K Xˆ Zero assignment estimator: all zeros of the closed loop can be assigned Atonomos estimator: state estimator error is independent of r racking-error estimator: the tracking error r-y is sed in control Mr 1. LQR Problem - Formlation Consider a LI system Cost fnction: dx () t = AX() t+ Bt () dt Yt () = CXt () J ( ) = ( XQX+ Rdt ) Optimal problem: minimize the linear qadratic cost fnction by choosing the control inpt nder the constraint of state space model 0 or J ( ) = ( Y QY + R) dt 0 3/26/2008 Introdction to Process Control 4

3 LQR Problem - Soltion Pre-condition: system is stabilizable (controllable) Linear state feedback: t () = KXt () with K = 1 R B P where P is the soltion of the algebraic Riccati eqation: -1 A P PA-PBR B P Q = Optimal cost: min J ( ) = X (0) PX(0) 3/26/2008 Introdction to Process Control 5 LQR Problem - Benefits Balance of control effort and system performance Fixed fll-state feedback soltion t () = KXt () with K = 1 R B P he soltion leads to a stable closed-loop system Xt & () = ( A BKXt ) () he minimal cost only depends on the initial condition and P min J ( ) = X (0) PX(0) 3/26/2008 Introdction to Process Control 6

4 LQR Problem Why? See the distribted paper and DC textbook X=AX+B C Y =-KX 3/26/2008 Introdction to Process Control 7 LQR Problem - Attentions Selection of weighting matrices Q, R Q positive semi-definite, R positive definite Soltions of Riccati eqation t 1 () = R BPX() t where P is the soltion of the algebraic Riccati eqation: -1 A P + PA-PBR B P + Q = 0 here is only one positive semi-definite soltion 3/26/2008 Introdction to Process Control 8

5 2. Selection of Weighting Matrices A certain amont of trial and error is sally reqired Bryson s rles (1975): (DC textbook: p ) Step 1: find ot the maximm deviations of otpts and inpts, e.g., for otpts:m1,m2,m3, for inpts: 1, 2 Step 2: constrct Q_1 as a diagonal matrix with elements Q_1(1,1)=1/m1^2, Q_1(2,2)=1/m2^2, Q_1(3,3)=1/m3^2; Step 3: constrct R as a diagonal matrix with elements R(1,1)=1/1^2, R(2,2)=1/2^2; Step 4: constrct the cost fnction as following and design controller based on it. Step 5: select \rho by trial and error J = ρ ( X C Q CX + R) dt 3/26/ Matlab Implementation [K,S,e] = lqr(sys,q,r,n), [K,S,e] = LQR(A,B,Q,R,N) [K,S,e] = lqry(sys,q,r,n) [K,S,e] = dlqr(a,b,q,r,n) [Kd,S,e] = lqrd(a,b,q,r,s), [Kd,S,e] = lqrd(a,b,q,r,n,s) care: Solve continos-time algebraic Riccati eqation dare: Solve discrete-time algebraic Riccati eqations 3/26/2008 Introdction to Process Control 10

6 Example 1: consider a system Example A=[0 1; -1 0]; B=[0; 2]; C=[1 0]; D=0; Select Q=[1 0; 0 3], R=[2], [K,S,E]=lqr(A,B,Q,R) K=[ ], E=[ ] check the closed loop system: eig(a-b*k) Adding the reference inpt: =-KX+Nr*r N=[A B; C D]^-1*[0; 0; 1]; Nr=N(3)+K*N(1:2); Closed loop system: Ac=A-B*K; Bc=B*Nr; Cc=C; Dc=D; syscl=ss(ac,bc,cc,dc); step(syscl,syscl) Download LQRcontrol.m 3/26/2008 Introdction to Process Control 11 LQG Optimal Control RLQG = LQGREG(KES,K) 3/26/2008 Introdction to Process Control 12