Modern Optimal Control

Transcription

1 Modern Optimal Control Matthew M. Peet Arizona State University Lecture 21: Optimal Output Feedback Control

2 connection is called the (lower) star-product of P and Optimal Output Feedback ansformation (LFT). Recall: Linear Fractional Transformation z y P w u K to z is given by Plant: [z ] [ P11 P = 12 ] where P = ] [ w D y P S(P, K) =P 11 + P 12 K(I P 22 K) 1 21 P 22 u 11 D 12 C 2 D 21 DP A B 1 B 2 Controller: AK B u = Ky where K = K M. Peet Lecture 21: 2 / 28 C K C 1 D K

3 Optimal Control Choose K to minimize P 11 + P 12 (I KP 22 ) 1 KP 21 AK B Equivalently choose K to minimize [ A 0 0 A K C K ] [ B B K D K [ C1 0 ] + [ D 12 0 ] [ I D K D 22 I where Q = (I D 22 D K ) 1. ] 1 I DK 0 CK D 22 I C 2 0 ] 1 0 CK C 2 0 B 1 + B 2D KQD 21 B KQD 21 D 11 + D 12D KQD 21 H M. Peet Lecture 21: 3 / 28

4 Optimal Control Recall the Matrix Inversion Lemma: Lemma 1. 1 A B C D A = 1 + A 1 B(D CA 1 B) 1 CA 1 A 1 B(D CA 1 B) 1 (D CA 1 B) 1 CA 1 (D CA 1 B) 1 [ ] (A BD = 1 C) 1 (A BD 1 C) 1 BD 1 D 1 C(A BD 1 C) 1 D 1 + D 1 C(A BD 1 C) 1 BD 1 M. Peet Lecture 21: 4 / 28

5 Optimal Control Recall that 1 I DK I + DK QD = 22 D K Q I Q D 22 QD 22 where Q = (I D 22 D K ) 1. Then 1 A 0 B2 0 I DK 0 CK A cl := + 0 A K 0 B K D 22 I C 2 0 A 0 B2 0 I + DK QD = + 22 D K Q 0 CK 0 A K 0 B K QD 22 Q C 2 0 A + B2 D = K QC 2 B 2 (I + D K QD 22 )C K B K QC 2 A K + B K QD 22 C K Likewise C cl := [ C 1 0 ] + [ D 12 0 ] I + D K QD 22 D K Q 0 CK QD 22 Q C 2 0 = C 1 + D 12 D K QC 2 D 12 (I + D K QD 22 )C K M. Peet Lecture 21: 5 / 28

6 Thus we have A + B 2 D K QC 2 B 2 (I + D K QD 22 )C K B 1 + B 2 D K QD 21 [C1 B K QC 2 A K + B K QD 22 C K ] B K QD 21 + D 12 D K QC 2 D 12 (I + D K QD 22 )C K D 11 + D 12 D K QD 21 where Q = (I D 22 D K ) 1. This is nonlinear in (A K, B K, C K, D K ). Hence we make a change of variables (First of several). A K2 = A K + B K QD 22 C K B K2 = B K Q C K2 = (I + D K QD 22 )C K D K2 = D K Q M. Peet Lecture 21: 6 / 28

7 This yields the system A + B2 D K2 C 2 B 2 C K2 [C1 B K2 C 2 A K2 ] + D 12 D K2 C 2 D 12 C K2 [ AK2 B Which is affine in K2 C K2 D K2 ]. B 1 + B 2 D K2 D 21 B K2 D 21 D 11 + D 12 D K2 D 21 M. Peet Lecture 21: 7 / 28

8 Hence we can optimize over our new variables. However, the change of variables must be invertible. If we recall that (I QM) 1 = I + Q(I MQ) 1 M then we get I + D K QD 22 = I + D K (I D 22 D K ) 1 D 22 = (I D K D 22 ) 1 Examine the variable C K2 C K2 = (I + D K (I D 22 D K ) 1 D 22 )C K = (I D K D 22 ) 1 C K Hence, given C K2, we can recover C K as C K = (I D K D 22 )C K2 M. Peet Lecture 21: 8 / 28

9 Now suppose we have D K2. Then D K2 = D K Q = D K (I D 22 D K ) 1 implies that D K = D K2 (I D 22 D K ) = D K2 D K2 D 22 D K or (I + D K2 D 22 )D K = D K2 which can be inverted to get D K = (I + D K2 D 22 ) 1 D K2 M. Peet Lecture 21: 9 / 28

10 Once we have C K and D K, the other variables are easily recovered as B K = B K2 Q 1 = B K2 (I D 22 D K ) A K = A K2 B K (I D 22 D K ) 1 D 22 C K To summarize, the original variables can be recovered as 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 M. Peet Lecture 21: 10 / 28

11 Or [ Acl B cl C cl D cl ] := Acl B cl = A 0 B C cl D cl C 1 0 D 11 A + B2 D K2 C 2 B 2 C K2 [C1 B K2 C 2 A K2 ] + D 12 D K2 C 2 D 12 C K2 + B 1 + B 2 D K2 D 21 B K2 D 21 D 11 + D 12 D K2 D 21 0 B 2 I 0 AK2 B K2 0 I 0 C 0 D K2 D K2 C 2 0 D A 0 0 B2 AK2 B A cl = + K2 0 I 0 0 I 0 C K2 D K2 C 2 0 B1 0 B2 AK2 B B cl = + K2 0 0 I 0 C K2 D K2 D 21 C cl = [ C 1 0 ] + A 0 D K2 B K2 0 I 12 C K2 D K2 C 2 0 D cl = A D K2 B K2 0 D12 C K2 D K2 D 21 M. Peet Lecture 21: 11 / 28

12 Lemma 2 (Transformation Lemma). Suppose that Y1 I > 0 I X 1 Then there exist X 2, X 3, Y 2, Y 3 such that 1 X1 X X = 2 Y1 Y X2 T = 2 X 3 Y2 T = Y 1 > 0 Y 3 Y1 I where Y cl = Y2 T has full rank. 0 M. Peet Lecture 21: 12 / 28

13 Proof. Since Y1 I > 0, I X 1 by the Schur complement X 1 > 0 and X1 1 Y 1 > 0. Since I X 1 Y 1 = X 1 (X1 1 Y 1 ), we conclude that I X 1 Y 1 is invertible. Choose any two square invertible matrices X 2 and Y 2 such that X 2 Y T 2 = I X 1 Y 1 Because Y 2 is non-singular, Y T cl = are also non-singular. [ Y1 Y 2 I 0 ] I 0 and X cl = X 1 X 2 M. Peet Lecture 21: 13 / 28

14 Proof. Now define X and Y as X = Y T cl X cl and Y = X 1 cl Ycl T. Then XY = Y 1 cl X cl X 1 Y cl = I cl M. Peet Lecture 21: 14 / 28

15 Lemma 3 (Converse Transformation Lemma). X1 X Given X = 2 X2 T > 0 where X X 2 has full column rank. Let 3 X 1 Y1 Y = Y = 2 Y2 T Y 3 then Y1 I > 0 I X 1 Y1 I and Y cl = Y2 T has full column rank. 0 M. Peet Lecture 21: 15 / 28

16 Proof. I 0 Since X 2 is full rank, X cl = also has full column rank. Note that X 1 X 2 XY = I implies Ycl T Y1 Y X = 2 X1 X 2 I 0 I 0 X2 T = = X X 3 X 1 X cl. 2 Hence Y T cl = Y1 Y 2 I 0 = Y = X I 0 X 1 X cl Y 2 has full column rank. Now, since XY = I implies X 1 Y 1 + X 2 Y2 T = I, we have [ ] I 0 Y1 I Y X cl Y cl = X 1 X 2 Y2 T = 1 I Y1 I 0 X 1 Y 1 + X 2 Y2 T = X 1 I X 1 Furthermore, because Y cl has full rank, Y1 I = X I X cl Y cl = X cl Y Xcl T = Ycl T XY cl > 0 1 M. Peet Lecture 21: 16 / 28

17 Theorem 4. The following are equivalent. There exists a ˆK AK B = K such that S(K, P ) C K D H < γ. K X1 I There exist X 1, Y 1, A n, B n, C n, D n such that > 0 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 < 0 γi C 1Y 1 + D 12C n C 1 + D 12D nc 2 D 11 +D 12D nd 21 γi Moreover, [ AK2 B K2 C K2 D K2 ] [ X2 X = 1 B 2 0 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 ] [ X1 AY 1 0 Y T 0 0 ] C 2 Y 1 I ] 1 M. Peet Lecture 21: 17 / 28

18 Proof: If. Suppose there exist X 1, Y 1, A n, B n, C n, D n such that the LMI is feasible. Since X1 I > 0, I Y 1 by the transformation lemma, there exist X 2, X 3, Y 2, Y 3 such that 1 X X2 Y Y2 X := X2 T = X 3 Y2 T > 0 Y 3 Y I AK B where Y cl = Y2 T has full row rank. Let K = K where 0 C K D K 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. M. Peet Lecture 21: 18 / 28

19 Proof: If. and where [ AK2 B K2 C K2 D K2 ] = 1 [ X2 X 1 B 2 An B n 0 I C n D n ] [ X1 AY 1 0 Y T C 2 Y 1 I ] 1. M. Peet Lecture 21: 19 / 28

20 Proof: If. As discussed previously, this means the closed-loop system is A 0 B Acl B 1 0 B 2 cl = I 0 AK2 B K2 0 I 0 C cl D cl C C 1 0 D 11 0 D K2 D K2 C 2 0 D A 0 B 1 0 B 2 = I 0 C 1 0 D 11 0 D 12 1 [ ] X2 X 1 B 2 An B n X1 AY 1 0 Y T I 0 0 I C n D n 0 0 C 2 Y 1 I C 2 0 D 21 We now apply the KYP lemma to this system M. Peet Lecture 21: 20 / 28

21 Proof: If. Expanding out, we obtain Y cl T 0 0 A T cl X + XA cl XB cl Ccl T Y cl I 0 Bcl T X γi DT cl 0 I 0 = 0 0 I C cl D cl γi 0 0 I 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 Acl B Hence, by the KYP lemma, S(P, K) = cl satisfies S(P, K) H < γ. C cl D cl < 0 M. Peet Lecture 21: 21 / 28

22 Proof: Only If. [ AK B Now suppose that S(P, K) H < γ for some K = K S(P, K) H < γ, by the KYP lemma, there exists a X1 X X = 2 X2 T > 0 X 3 such that C K AT cl X + XA cl X Ccl T Bcl T X γi DT cl < 0 C cl D cl γi D K ]. Since Because the inequalities are strict, we can assume that X 2 has full row rank. Define Y1 Y Y = 2 Y2 T = X 1 Y1 I and Y Y cl = 3 Y2 T 0 Then, according to the converse [ transformation ] lemma, Y cl has full row rank and X1 I > 0. I Y 1 M. Peet Lecture 21: 22 / 28

23 Proof: Only If. Now, using the given A K, B K, C K, D K, define the variables An B n X2 X = 1 B 2 AK2 B K2 Y T C n D n 0 I C K2 D K2 C 2 Y 1 I X1 AY where A K2 = A K + B K (I D 22 D K ) 1 D 22 C K C K2 = (I + D K (I D 22 D K ) 1 D 22 )C K B K2 = B K (I D 22 D K ) 1 D K2 = D K (I D 22 D K ) 1 Then as before A 0 B Acl B 1 0 B 2 cl = I 0 C cl D cl C 1 0 D 11 0 D 12 1 [ ] X2 X 1 B 2 An B n X1 AY 1 0 Y T I 0 0 I C n D n 0 0 C 2 Y 1 I C 2 0 D 21 M. Peet Lecture 21: 23 / 28

24 Proof: Only If. Expanding out the LMI, we find 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 = Y cl T 0 0 A T cl X + XA cl X Ccl T Y cl I 0 Bcl T X cl γi Dcl T 0 I 0 < I C cl D cl γi 0 0 I M. Peet Lecture 21: 24 / 28

25 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 > 0 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 < 0 γi C 1Y 1 + D 12C n C 1 + D 12D nc 2 D 11 +D 12D nd 21 γi M. Peet Lecture 21: 25 / 28

26 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 0 I C n D n ] [ X1 AY 1 0 Y T 0 0 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 0 C 2 Y 1 I ] 1. M. Peet Lecture 21: 26 / 28

27 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 21: 27 / 28

28 Alternative Formulation Theorem 5. Let N o and N c be full-rank matrices whose images satisfy Im N o = Ker C 2 D 21 Im N c = Ker B2 T D12 T Then the following are equivalent. There exists a ˆK AK B = K such that S(K, P ) C K D H < 1. K X1 I There exist X 1, Y 1 such that > 0 and I Y 1 T No 0 X 1A + A T X 1 T T B1 0 I T X 1 I T No 0 < 0 0 I C 1 D 11 I T AY Nc Y 1 A T T C 0 I 1 Y 1 I T Nc 0 < 0 B1 T D11 T 0 I I] M. Peet Lecture 21: 28 / 28