Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp A Symbolic Approach for Solving Algebraic Riccati Equations G. Rance, Y. Bouzidi, Al. Quadrat, Ar. Quadrat Journées Nationales de Calcul Formel Monday, January 22 th 2018 JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 1 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp Overview 1 Algebraic Riccati Equations for the optimal control problem 2 A new algebraic description 3 The case of 3 order systems 4 A practical example 5 Conclusion and perspectives JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 2 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp Overview 1 Algebraic Riccati Equations for the optimal control problem 2 A new algebraic description 3 The case of 3 order systems 4 A practical example 5 Conclusion and perspectives JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 3 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp The linear optimal control problem Input : a linear dynamical system � x ( t ) = A x ( t ) + B u ( t ) ˙ x (0) = x 0 where x ( t ) ∈ R n the state vector, u ( t ) ∈ R m the control vector A (resp. B ) is an n × n (resp. n × m ) real matrix Output : a control u that stabilizes the system and minimizes a quadratic cost functional � + ∞ 1 [ x ( t ) T Q x ( t ) + u ( t ) T R u ( t )] d t 2 0 where Q (resp. R ) is a positive semi-definite (resp. positive definite) symmetric real matrix. Goal : Achieve a control reference using the minimum energy JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 4 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp Optimal control : mathematical simplifications Let introduce the Lagrange multiplier λ = ( λ 1 , . . . , λ n ) and the following functional + ∞ � 1 [ x ( t ) T Q x ( t ) + u ( t ) T R u ( t ) − λ ( t )(˙ x ( t ) − A x ( t ) − B u ( t ))] d t 2 0 By a variation computation, the problem is reduced to solving the following OD systems λ ( t ) T + A T λ ( t ) T + Q x ( t ) = 0 , ˙ � x ( t ) = A x ( t ) − B R − 1 B T λ ( t ) T , � ˙ u = − R − 1 BT λ ( t ) T x ( t ) − A x ( t ) − B u ( t ) = 0 , ˙ λ ( t ) T = − Q x ( t ) − A T λ ( t ) T . ˙ R u ( t ) + B T λ ( t ) T = 0 . − − − − − − − − − − → If we seek for a solution of the form λ ( t ) T = P ( t ) x ( t ), P ( t ) must satisfy the differential equation P = A P + A T P + P B R − 1 B T P T + Q ˙ If we consider a constant matrix P , this yields the following algebraic equation A P + A T P + P B R − 1 B T P T + Q = 0 The optimal control is then given as u ( t ) = − R − 1 B T P x ( t ) JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 5 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp Algebraic Riccati Equations An Algebraic Riccati Equation is the following quadratic matrix equation A T X + X A + X B R − 1 B T X + Q = 0 (1) where A is a real n × n matrix and Q , R are real symmetric n × n matrices Solving Algebraic Riccati Equations • Computing all the solutions X of (1) • Computing specific solutions of (1) : real, hermitian, positive definite... • A positive definite solution is stabilizing Algebraic Riccati Equations are fundamental in many linear control theory problems (Estimation, Filtering, Robust control,...) JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 6 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp Riccati Equations and invariant subspaces Solutions of (1) can be constructed in term of the invariant subspaces of the following 2 n × 2 n Hamiltonian matrix � A − B R − 1 B T � H := − A T − Q Theorem [Zhou et al. (1996)] Let V ⊂ C 2 n be an n -dimensional invariant subspace of H and let X 1 , X 2 ∈ C n × n be two complex matrices such that � X 1 � V = Im X 2 If X 1 is invertible, then X := X 2 X − 1 is a solution of the Riccati Equation (1). 1 Invariant subspaces can be obtained via eigenvalues and eigenvectors computation JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 7 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp The spectral factorization problem The spectrum of H is symmetric with respect to the real and imaginary axis If we consider the characteristic polynomial of H f ( λ ) = det( H − λ I 2 n ) Then f ( λ ) = f ( − λ ) Invariant subspaces can be obtained by computing factorizations of the form f ( λ ) = g ( λ ) g ( − λ ) where g ( λ ) ∈ C [ λ ] This problem is known as the spectral factorization problem JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 8 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp The problem under consideration n th order Single Input ( u ) Single Output ( y ) systems � x = A x + B u ˙ y = C x 0 1 0 0 . . . . ... . B := (0 . . . 0 1) T 0 0 1 . A := . . ... ... . . C := ( c 0 c n − 1 ) . . 0 . . . 0 0 0 1 . . . − a 0 − a 1 . . . − a n − 2 − a n − 1 where a := ( a 0 , · · · , a n − 1 ) , c := ( c 0 , · · · , c n − 1 ) are unknown parameters . Goal : Compute a closed loop control u that stabilizes y and minimizes � + ∞ 1 [ y ( t ) 2 + u ( t ) 2 ] dt 2 0 This control will depend on the parameters a,c � observe the effect of parameters on the optimization problem ! JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 9 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp The problem under consideration This yields the following Algebraic Riccati Equation R := X A + A T X − X B B T X + C T C = 0 (2) where X is a symmetric matrix Theorem [Zhou et al. (1996)] If the pair ( A , C ) is observable, then The positive definite solution X of (2) is unique The positive definite solution X of (2) is a stabilizing solution Goal : Compute the positive definite solution of (2) JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 10 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp Overview 1 Algebraic Riccati Equations for the optimal control problem 2 A new algebraic description 3 The case of 3 order systems 4 A practical example 5 Conclusion and perspectives JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 11 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp Algebraic description R = 0 ⇔ n ( n +1) polynomial equations of n ( n +1) unknowns 2 2 Noting X = ( x i , j ) then n ( n − 1) elements of R yields 2 x i , j = x i − 1 , j +1 + f ( a k , c k , x k , n | k = 1 · · · n ) Recursion → x i , j = f ( a k , c k , x k , n | k = 1 · · · n ) Two halting conditions : • Strictly above the anti-diagonal → First row • Below the anti-diagonal → Last column JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 12 / 27
Algebraic Riccati Equations for the optimal control problem A new algebraic description The case of 3 order systems A practical example Conclusion and persp Algebraic description For k = 1 . . . n , we set x k , 0 = x 0 , k := 0, and for ( i , j ) ∈ N 2 , we define : � N ( i , j ) := i − 1 , 2 ≤ i + j ≤ n + 1 (stly. above anti-diag.) N ( i , j ) := n − j + 1 , n + 1 < i + j ≤ 2 n + 1 (below anti-diag.) The elements of X solution of R = 0 are determined only by the b k ’s x k , n = b k − 1 − a k − 1 (last column of X) N ( i , j ) � ( − 1) k b i − 1 − k b j − 1+ k − θ N ( i , j ) x i , j − 1 = k =0 where 1 ≤ k ≤ n , 1 ≤ i < j ≤ n , and θ m is defined by : m � ( − 1) k � � θ m := a i − 1 − k a j − 1+ k + c i − 1 − k c j − 1+ k k =0 The number of variables is now equal to n JNCF 2018 Yacine Bouzidi A Symbolic Approach for Solving Algebraic Riccati Equations 13 / 27
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