Modern Optimal Control

Transcription

1 Modern Optimal Control Matthew M. Peet Arizona State University Lecture 22: H 2, LQG and LGR

2 Conclusion To solve the H -optimal state-feedback problem, we solve min γ such that γ,x 1,Y 1,A n,b n,c n,d n [ ] X1 I > I Y 1 AY 1 +Y 1A T +B 2C n +Cn T B2 T T T T A T + A n + [B 2D nc 2] T X 1A+A T X 1 +B nc 2 +C2 T Bn T T T [B 1 + B 2D nd 21] T [XB 1 + B nd 21] T < γi C 1Y 1 + D 12C n C 1 + D 12D nc 2 D 11 +D 12D nd 21 γi M. Peet Lecture 22: 2 / 21

3 Conclusion Then, we construct our controller using where [ AK2 B K2 C K2 D K2 ] = D K = (I + D K2 D 22 ) 1 D K2 B K = B K2 (I D 22 D K ) C K = (I D K D 22 )C K2 A K = A K2 B K (I D 22 D K ) 1 D 22 C K. [ ] 1 [[ ] X2 X 1 B 2 An B n I C n D n [ ]] [ X1 AY 1 Y T and where X 2 and Y 2 are any matrices which satisfy X 2 Y 2 = I X 1 Y 1. e.g. Let Y 2 = I and X 2 = I X 1 Y 1. The optimal controller is NOT uniquely defined. Don t forget to check invertibility of I D 22 D K 2 C 2 Y 1 I ] 1. M. Peet Lecture 22: 3 / 21

4 Conclusion The H -optimal controller is a dynamic system. [ ] Transfer Function ˆK(s) AK B = K C K D K Minimizes the effect of external input (w) on external output (z). z L2 S(P, K) H w L2 Minimum Energy Gain M. Peet Lecture 22: 4 / 21

5 Motivation H 2 -optimal control minimizes the H 2 -norm of the transfer function. The H 2 -norm has no direct interpretation. G 2 H 2 = 1 Trace(Ĝ (ıω)ĝ(ıω))dω 2π Motivation: Assume external input is Gaussian noise with spectral density S w Theorem 1. E[w(t) 2 ] = 1 2π Trace(Ŝw(ıω))dω For an LTI system P, if w is noise with spectral density Ŝw(ıω) and z = P w, then z is noise with density Ŝ z (ıω) = ˆP (ıω)ŝ(ıω) ˆP (ıω) M. Peet Lecture 22: 5 / 21

6 Motivation Then the output z = P w has signal variance (Power) E[z(t) 2 ] = 1 Trace(Ĝ (ıω)s(ıω)ĝ(ıω))dω 2π S H G 2 H 2 If the input signal is white noise, then Ŝ(ıω) = I and E[z(t) 2 ] = G 2 H 2 M. Peet Lecture 22: 6 / 21

7 Colored Noise If the noise is colored, then we can still use the same approach to H 2 optimal control. Let Ĥ(ıω)Ĥ(ıω) = Ŝw(ıω). Then design an H 2 -optimal controller for the plant [ ˆP11 ˆP s (s) = (s)ĥ(s) ] ˆP12 (s) ˆP 21 (s)ĥ(s) ˆP22 (s) Now, using the controller and filtered plant, S(P s, K)(s) = P 11 H + P 12 (I KP 22 ) 1 KP 21 H = S(P, K)H For noise with spectral density S w, let z = S(P, K)w. Then if S z is the spectral density of z, we have S z (s) = S(P, K)(s)Ŝw(s)S(P, K)(s) = S(P, K)(s)Ĥ(s)Ĥ(s) S(P, K)(s) = S(P s, K)(s)Ŝ(P s, K)(s) and hence E[z(t) 2 ] = Ŝ(P s, K) 2 H 2 = γ. Thus minimization of Ŝ(P s, K) 2 H 2 achieves the optimal system response to noise with density Ŝw. M. Peet Lecture 22: 7 / 21

8 Theorem 2. Suppose ˆP (s) = C(sI A) 1 B. Then the following are equivalent. 1. A is Hurwitz and ˆP H2 < γ. 2. There exists some X > such that trace CXC T < γ AX + XA T + BB T < M. Peet Lecture 22: 8 / 21

9 Proof. Suppose A is Hurwitz and ˆP H2 < γ. Then the Controllability Grammian is defined as X c = Now recall the Laplace transform e At BB T e AT dt ( Λe At ) (s) = e At e ts dt = e (si A)t dt = (si A) 1 e (si A)t dt = (si A) 1 Hence ( ΛCe At B ) (s) = C(sI A) 1 B. t= t= M. Peet Lecture 22: 9 / 21

10 Proof. ( ΛCe At B ) (s) = C(sI A) 1 B implies ˆP 2 H 2 = C(sI A) 1 B 2 H 2 = 1 Trace((C(ıωI A) 1 B) (C(ıωI A) 1 B))dω 2π = 1 Trace((C(ıωI A) 1 B)(C(ıωI A) 1 B) )dω 2π = Trace = TraceCX c C T Ce At BB e A t C dt Thus X c and TraceCX c C T = ˆP 2 H 2 < γ. M. Peet Lecture 22: 1 / 21

11 Proof. Likewise TraceB T X o B = ˆP 2 H 2. To show that we can take strict the inequality X >, we simply let X = e At ( BB T + ɛi ) e AT dt for sufficiently small ɛ >. Furthermore, we already know the controllability grammian X c and thus X ɛ satisfies the Lyapunov inequality. A T X ɛ + X ɛ A + BB T < These steps can be reversed to obtain necessity. M. Peet Lecture 22: 11 / 21

12 Full-State Feedback Lets consider the full-state feedback problem A B 1 B 2 Ĝ(s) = C 1 D 12 I D 12 is the weight on control effort. D 11 = is neglected as the feed-through term. C 2 = I as this is state-feedback. ˆK(s) = [ K ] M. Peet Lecture 22: 12 / 21

13 Full-State Feedback Theorem 3. The following are equivalent. 1. S(K, P ) H2 < γ. 2. K = ZX 1 for some Z and X > where [ A B2 ] [ X Z ] + [ X Z T ] [ A T B T 2 ] + B 1 B T 1 < Trace [ C 1 X + D 12 Z ] X 1 [ C 1 X + D 12 Z ] < γ However, this is nonlinear, so we need to reformulate using the Schur Complement. M. Peet Lecture 22: 13 / 21

14 Applying the Schur Complement gives the alternative formulation convenient for control. Theorem 4. Suppose ˆP (s) = C(sI A) 1 B. Then the following are equivalent. 1. A is Hurwitz and ˆP H2 < γ. 2. There exists some X, Z > such that [ ] [ ] A T X + XA XB X C T B T <, >, TraceZ < γ X γi C Z M. Peet Lecture 22: 14 / 21

15 Full-State Feedback Theorem 5. The following are equivalent. 1. S(K, P ) H2 < γ. 2. K = ZX 1 for some Z and X > where [ ] [ ] X A B2 + [ X Z ] [ ] Z T A T B2 T + B 1 B1 T < [ ] X (C 1 X + D 12 Z) T > C 1 X + D 12 Z W TraceW < γ Thus we can solve the H 2 -optimal static full-state feedback problem. M. Peet Lecture 22: 15 / 21

16 Relationship to LQR Thus minimizing the H 2 -norm minimizes the effect of white noise on the power of the output noise. This is why H 2 control is often called Least-Quadratic-Gaussian (LQG). LQR: Full-State Feedback Choose K to minimize the cost function subject to dynamic constraints x(t) T Qx(t) + u(t) T Ru(t)dt ẋ(t) = Ax(t) + Bu(t) u(t) = Kx(t), x() = x M. Peet Lecture 22: 16 / 21

17 Relationship to LQR To solve the LQR problem, let [ 1 ] Q 2 C 1 = [ ] D 12 = R 1 2 B 2 = B and B 1 = I. So that [ ] A + BK S( ˆP, ˆK) A + B2 K B = 1 = Q 1 2 C 1 + D 12 K D 11 R 1 2 K And solve the H 2 full-state feedback problem. Then if ẋ(t) = A CL x(t) = (A + BK)x(t) = Ax(t) + Bu(t) u(t) = Kx(t), x() = x I Then x(t) = e A CLt x M. Peet Lecture 22: 17 / 21

18 Relationship to LQR If then x(t) = e A CLt x and = Trace ẋ(t) = A CL x(t) = (A + BK)x(t) = Ax(t) + Bu(t) u(t) = Kx(t), x() = x x(t) T Qx(t) + u(t) T Ru(t)dt = = x 2 Trace [ 1 ] T [ 1 ] x T e AT CL t Q 2 Q 2 e ACLt x R 1 2 K R 1 dt 2 K = x 2 S(K, P ) 2 H 2 x T e AT CL t (Q + K T RK)e A CLt x dt B 1 e AT CL t (C 1 + D 12 K) T (C 1 + D 12 K)e A CLt B T 1 dt Thus LQR reduces to a special case of H 2 static state-feedback. M. Peet Lecture 22: 18 / 21

19 H 2 -optimal output feedback control Theorem 6 (Lall). The following are equivalent. [ ] There exists a ˆK AK B = K such that S(K, P ) C K D H2 < γ. K [ ] X1 I There exist X 1, Y 1, Z, A n, B n, C n, D n such that > I Y 1 AY 1 +Y 1 A T +B 2 C n +Cn T B2 T T T A T + A n + [B 2 D n C 2 ] T X 1 A+A T X 1 +B n C 2 +C2 T Bn T T <, [B 1 + B 2 D n D 21 ] T [XB 1 + B n D 21 ] T γi X 1 I T I Y 1 T >, C 1 Y 1 + D 12 C n C 1 + D 12 D n C 2 Z D 11 + D 12 D n D 21 =, trace(z) < γ M. Peet Lecture 22: 19 / 21

20 H 2 -optimal output feedback control As before, the controller can be recovered as [ AK2 B K2 C K2 D K2 ] = [ X2 X 1 B 2 I ] 1 [[ ] An B n C n D n for any full-rank X 2 and Y 2 such that [ ] [ X1 X 2 Y1 Y X2 T = 2 X 3 Y2 T Y 3 To find the actual controller, we use the identities: D K = (I + D K2 D 22 ) 1 D K2 B K = B K2 (I D 22 D K ) C K = (I D K D 22 )C K2 [ ]] [ X1 AY 1 Y T ] 1 A K = A K2 B K (I D 22 D K ) 1 D 22 C K 2 C 2 Y 1 I ] 1 M. Peet Lecture 22: 2 / 21

21 Robust Control Before we finish, let us briefly touch on the use of LMIs in Robust Control. p M q Questions: Is S(, M) stable for all? Determine sup S(, M) H. M. Peet Lecture 22: 21 / 21