Inner outer and spectral factorizations
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Inner outer and spectral factorizations
- 1. Inner-Outer and Spectral Factorizations SOLO HERMELIN Updated: 27.05.11 1
- 2. Table of Content SOLO Inner-Outer and Spectral Factorizations 2
- 3. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations 3 Given the Linear Time-Invariant System ( ) 00 xxuBxAx =+= and the Quadratic Cost Function: [ ] TT T TT c PPRRdt u x RS SP uxJ =>= = ∫ ∞ &0 2 1 0 The Optimal Regulator that Minimizes the Cost Function Jc is given by the following procedure: Define the Hamiltonian of the Optimization Problem: ( ) [ ] ( )uBxA u x RS SP uxuxH T T TT ++ = λλ 2 1 ,, The Euler-Lagrange Equations are: ( )λλ λ λ λ TTTT T T T BxSRuBxSuR u H uSxPA x H td d uBxA H td xd +−=⇒++= ∂ ∂ = −−−= ∂ ∂ −= += ∂ ∂ = −1 0
- 4. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations 4 ( )λλ λ λ λ TTTT T T T BxSRuBxSuR u H uSxPA x H td d uBxA H td xd +−=⇒++= ∂ ∂ = −−−= ∂ ∂ −= += ∂ ∂ = −1 0 or: ( ) ( ) ( ) = −−−− −− = −− −− λλλ x A x SRBASRSP BRBSRBAx HTTT TT 11 11 ( ) ( ) ( ) −−−− −− = −− −− TTT TT H SRBASRSP BRBSRBA A 11 11 : We want to find X (t) such that: , then( ) ( ) ( )txtXt =λ ( ) ( )XBSRKxKxXBSRu TT cc TT +=⇒−=+−= −− 11 :
- 5. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations 5 ( ) ( ) ( ) = −−−− −− = −− −− λλλ x A x SRBASRSP BRBSRBAx HTTT TT 11 11 We want to find X (t) such that: , then( ) ( ) ( )txtXt =λ ( ) ( ) ( ) ( ) ( )txtXtxtXt +=λ ( ) ( ) ( )[ ]xXBRBxSRBAXxXxXSRBAxSRSP TTTTT 1111 −−−− −−+=−−−− ( ) ( ) ( )[ ] ( )txxSRSPXBRBXSRBAXXSRBAX TTTTT ∀=−+−−+−+ −−−− 01111 ( ) ( ) ( )TTTTT SRSPXBRBXSRBAXXSRBAX 1111 −−−− −−+−−−−= We obtain the following Differential Riccati Equation for :X ( ) ( ) ( ) [ ] ( ) ( ) ( ) [ ] ( ) ( ) 0 0 1 11 11 1111 =+++−+= −= −−−− −− −= −−+−−−−= − −− −− −−−− PXBSRXBSAXXA X I AIX X I SRBASRSP BRBSRBA IX SRSPXBRBXSRBAXXSRBA TTTTTT HTTT TT TTTTT If a Steady-state Solution exists for t→∞ then and:0→X Continuous Algebraic Riccati Equation (CARE)
- 6. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations Theorem 1 A Unique Stabilizing ( A+BKc=[A-BR-1 (ST +BT X)] is Stable) Solution of CARE, X=XT , is obtained iff: a.(A,B) is Stabilizable b.Re[λi(AH)]≠0 for all i=1,2,…,2n The Unique Stabilizing Solution is denoted X=Ric [AH] Proof a. It is clear that the condition that (A,B) is Stabilizable is a necessary condition to stabilize A+BKc. b. Rewrite ( ) ( ) ( ) ( ) ( )[ ] − +−− −+− = −−−− −− = − −− −− −− IX I XBSRBA BRBXBSRBA IX I SRBASRSP BRBSRBA A TTT TTT TTT TT H 0 0 0 : 1 11 11 11 If λi is an eigenvalue of AH, - λi is also; hence AH has n eigenvalues λi s.t. Re[λi] ≤ 0 and n eigenvalues λj s.t. Re[λj] ≥ 0. If AH has no eigenvalue of jω axis, then we can find a solution X s.t. all the eigenvalues of A-BR-1 (ST +BT X)=A+BKc are the stable eigenvalues of AH. ( ) ( ) 01 1 1 =+++−+ − PXBSRXBSAXXA TTTTTT
- 7. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations Theorem 1 A Unique Stabilizing ( A+BKc=[A-BR-1 (ST +BT X)] is Stable) Solution of CARE is obtained iff: a.(A,B) is Stabilizable b.Re[λi(AH)]≠0 for all i=1,2,…,2n The Unique Stabilizing Solution is denoted X=Ric [AH] Proof (continue -1) c. Assume that there are two Stabilizing Solutions X1=X1 T and X2 =X2 T that satisfy the CARE ( ) ( ) ( ) ( ) 0 0 2 1 222 1 1 111 =+++−+ =+++−+ − − PXBSRXBSAXXA PXBSRXBSAXXA TTTTTT TTTTTT Subtracting those two equations we obtain: ( ) ( ) ( ) ( ) 0 0 1 1 21 1 2 2 1 21 1 1 1 2121 1 2121 =−+ −−−−−−−+− −− −−−− XBRBXXBRBX XBRBXXBRBXSRBXXXXBRSAXXXXA TT TTTTT or: ( )[ ] ( ) ( ) ( )[ ] 01 1 21212 1 =+−−+−+− −− XBSRBAXXXXXBSRBA TTTTT
- 8. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations Theorem 1 A Unique Stabilizing ( A+BKc=[A-BR-1 (ST +BT X)] is Stable) Solution of CARE is obtained iff: a.(A,B) is Stabilizable b.Re[λi(AH)]≠0 for all i=1,2,…,2n The Unique Stabilizing Solution is denoted X=Ric [AH] Proof (continue -2) c. Assume that there are two Stabilizing Solutions X1=X1 T and X2 =X2 T that satisfy the CARE This is a Sylvester Matrix Equation and since both X1 and X2 are Stabilizing Solutions we must have: ( )[ ] ( ) ( ) ( )[ ] 01 1 21212 1 =+−−+−+− −− XBSRBAXXXXXBSRBA TTTTT ( )[ ]{ } ( )[ ]{ } ( )[ ]{ } ( )[ ]{ } jiXBSRBAXBSRBA njXBSRBA niXBSRBA TT j TT iTT j TT i ,0ReRe ,,10Re ,,10Re 2 1 1 1 2 1 1 1 ∀≠+−++−⇒ =∀<+− =∀<+− −− − − λλ λ λ Therefore the solution of the Sylvester Matrix Equation is Unique, and by substitution we can see that X1 = X2 is the Unique Solution. q.e.d.
- 9. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations Proof (1) (1) The Matrix YT Z is Symmetric. (2) If Y-1 exists, the X=ZY-1 is the Solution of CARE such that the Matrix ((A-BR-1 ST )-BR-1 BT X) has the Eigenvalues λ1,λ2,…,λn. X is Symmetric. (3) If (A,B) Stabilizable and Re[λi(AH)]≠0 for all i=1,2,…,2n (Theorem 1) Y-1 exists. Theorem 2 Let the columns of the Matrix be the Eigenvectors of AH corresponding to the Stable Eigenvalues λ1,λ2,…,λn, and J the corresponding Jordan Matrix. Then ( )nxnnxn RZYR Z Y ∈∈ ,2 Define TTT SRSPQBRBWSRBAE 111 :,:,: −−− −==−= ( ) ( ) ( ) −− − = −−−− −− = −− −− TTTT TT H EQ WE SRBASRSP BRBSRBA A 11 11 : =−− =− ⇒ = −− − ⇒ = JZZEQY JYZWYE J Z Y Z Y EQ WE J Z Y Z Y A TTHWe have ( ) ZYJZWZZEYZYJWZEYJYZWYE TTTTTTTTTTT =−⇒=−⇒=− JZYQYYZEYJZYZEYQYYJZZEQYY TTTTTTTTTT −−=⇒=−−⇒=−−
- 10. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations Proof (1) (continue – 1) Since –YT QT Y-ZT WZ is symmetric so is (YT Z)J+JT (YT Z) or ( ) ( )ZYJJZYZWZQYYZWZZYJJZYQYY JZYQYYZEY ZYJZWZZEY TTTTTTTTTT TTTT TTTTT +=−−⇒+=−−⇒ −−= =− ( ) ( ) ( ) ( )[ ] ( ) ( )TTTTTTTTTTT ZYJJZYZYJJZYZYJJZY +=+=+ Because J represents the stable Eigenvalues Re[λi(J)+λj(J)]≠0 for all I,j=1,…,n, the Unique Solution of the Sylvester Matrix Equation is ( ) ( )[ ] ( ) ( )[ ] [ ]0=−+− TTTTTTT ZYZYJJZYZY Sylvester Matrix Equation ( ) ( ) [ ] ( ) ( ) SymmetricisZYZYZYZYZY TTTTTTT =⇒=− 0 (1) The Matrix YT Z is Symmetric. (2) If Y-1 exists, the X=ZY-1 is the Solution of CARE such that the Matrix ((A-BR-1 ST )-BR-1 BT X) has the Eigenvalues λ1,λ2,…,λn. X is Symmetric. (3) If (A,B) Stabilizable and Re[λi(AH)]≠0 for all i=1,2,…,2n (Theorem 1) Y-1 exists. Theorem 2 Let the columns of the Matrix be the Eigenvectors of AH corresponding to the Stable Eigenvalues λ1,λ2,…,λn, and J the corresponding Jordan Matrix. Then ( )nxnnxn RZYR Z Y ∈∈ ,2
- 11. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations Proof (2) =−− =− JZZEQY JYZWYE T 111 −−− =−⇒=− YJYYZWEYJYZWYE 1111111 −−−−−−− =−⇒=− YJZYZWYZEYZYJYYZWEYZ 111 −−− =−−⇒=−− YJZYZEQYJZZEQY TT ( ) ( ) ( ) ( ) [ ]01111 =+−+ −−−− QYZWYZEYZYZET Define TTT SRSPQBRBWSRBAE 111 :,:,: −−− −==−= Therefore X=ZY-1 is the Unique Stabilizing Solution of the CARE ( ) ( ) ( ) [ ]01111 =−+−−+− −−−− TTTTT SRSPXBRBXSRBAXXSRBA q.e.d. (1) The Matrix YT Z is Symmetric. (2) If Y-1 exists, the X=ZY-1 is the Solution of CARE such that the Matrix ((A-BR-1 ST )-BR-1 BT X) has the Eigenvalues λ1,λ2,…,λn. X is Symmetric. (3) If (A,B) Stabilizable and Re[λi(AH)]≠0 for all i=1,2,…,2n (Theorem 1) Y-1 exists. Theorem 2 Let the columns of the Matrix be the Eigenvectors of AH corresponding to the Stable Eigenvalues λ1,λ2,…,λn, and J the corresponding Jordan Matrix. Then ( )nxnnxn RZYR Z Y ∈∈ ,2 Assume Y-1 exists:
- 12. Linear Quadratic Regulator (LQR) Problem SOLO Inner-Outer and Spectral Factorizations Proof (2) (continue – 1) q.e.d. From (1) YT Z is Symmetric, so ( ) YZZYZY TTTT == TTTTTTTT ZYXZXYZYZYYYZZY −−− =⇒=⇒=⇒= 11 1− = YZX ( ) TTTT XYZZYX === −− 1 X is Symmetric (1) The Matrix YT Z is Symmetric. (2) If Y-1 exists, the X=ZY-1 is the Solution of CARE such that the Matrix ((A-BR-1 ST )-BR-1 BT X) has the Eigenvalues λ1,λ2,…,λn. X is Symmetric. (3) If (A,B) Stabilizable and Re[λi(AH)]≠0 for all i=1,2,…,2n (Theorem 1) Y-1 exists. Theorem 2 Let the columns of the Matrix be the Eigenvectors of AH corresponding to the Stable Eigenvalues λ1,λ2,…,λn, and J the corresponding Jordan Matrix. Then ( )nxnnxn RZYR Z Y ∈∈ ,2
- 13. References SOLO 13 S. Hermelin, “Robustness and Sensitivity Design of Linear Time-Invariant Systems”, PhD Thesis, Stanford University, 1986 G.Stein, J.Doyle, B.Francis, “Advances in Multivariable Control”, ONR/Honeywell Workshop, 1984 Inner-Outer and Spectral Factorizations T. Kailath, A.H. Sayed, B. Hassibi, “Linear Estimation”. Prentice Hall, 2000
- 14. 14 SOLO Technion Israeli Institute of Technology 1964 – 1968 BSc EE 1968 – 1971 MSc EE Israeli Air Force 1970 – 1974 RAFAEL Israeli Armament Development Authority 1974 – 2013 Stanford University 1983 – 1986 PhD AA
- 15. Sylvester Matrix Equation : Anxn Xnxm+Xnxm Bmxm=Cnxm SOLO 15 Consider the Sylvester Matrix Equation where Anxn, Bmxm, Cnxm are given matrices. Then there exists a Unique Solution Xnxm if and only if Matrices James Joseph Sylvester (1814 – 1887) nxmmxmnxmnxmnxn CBXXA =+ ( )[ ] ( )[ ] njiAA nxnjnxni ,,1,0ReRe =∀≠+ λλ ∫ ∞ = 0 dteCeX tB nxm tA nxm mxmnxn Note : If B=AT : Lyapunov Equation The Necessary and Sufficient Condition for the existence of a Unique Solution is nxn T nxnnxnnxnnxn CAXXA =+ In particular if λi(A)=- λj(A) = j ω a Unique Solution does not exist. Aleksandr Mikhailovich Lyapunov 1857 - 1918 ( ) ( ) mjniBA mxmjnxni ,,1,,,10 ==∀≠+ λλ the Unique Solution is given by If ( )[ ] ( )[ ] mjniBA mxmjnxni ,,1,,,10ReRe ==∀<+ λλ
- 16. Sylvester Matrix Equation : Anxn Xnxm+Xnxm Bmxm=Cnxm SOLO Matrices Proof Let rewrite [ ]0 21 22221 11211 21 22221 11211 21 22221 11211 21 22221 11211 21 22221 11211 = − + = −+ nmnn m m mmmm m m nmnn m m nmnn m m nnnn n n nxmmxmnxmnxmnxn ccc ccc ccc bbb bbb bbb xxx xxx xxx xxx xxx xxx aaa aaa aaa CBXXA = nmnn m m nnnn n n nxmnxn xxx xxx xxx aaa aaa aaa XA 21 22221 11211 21 22221 11211 ( ) ( ) ( ) [ ]nxmnxnc mnxn nxn nxn Xvec m nm m n n nxn nxn nxn nxmcnxnm XAvec cA cA cA c x x c x x c x x A A A XvecAI nxmc = = =⊗ 2 1 1 2 2 12 1 1 11 00 00 00 ( ) [ ] ( ) ( ) nxmc Xvec xmn m nm m n n mcnxmc c x x c x x c x x cccvecXvec 1 1 2 2 12 1 1 11 21 : ⋅ == Define the Vectorization Operator vec:
- 17. Sylvester Matrix Equation : Anxn Xnxm+Xnxm Bmxm=Cnxm SOLO Matrices Proof (continue – 1) Let rewrite [ ]0 21 22221 11211 21 22221 11211 21 22221 11211 21 22221 11211 21 22221 11211 = − + = −+ nmnn m m mmmm m m nmnn m m nmnn m m nnnn n n nxmmxmnxmnxmnxn ccc ccc ccc bbb bbb bbb xxx xxx xxx xxx xxx xxx aaa aaa aaa CBXXA ( ) ( ) ( ) [ ]mxmnxmc mmmmm mm mm Xvec m nm m n n nmmnmnm nmnm nmnn nxmcn T mxm BXvec cbcbcb cbcbcb cbcbcb c x x c x x c x x IbIbIb IbIbIb IbIbIb XvecIB nxmc = +++ +++ +++ = =⊗ 2211 2222112 1221111 1 2 2 12 1 1 11 21 22212 12111 = mmmm m m nmnn m m mxmnxm bbb bbb bbb xxx xxx xxx BX 21 22221 11211 21 22221 11211
- 18. Sylvester Matrix Equation : Anxn Xnxm+Xnxm Bmxm=Cnxm SOLO Matrices Proof (continue – 2) We have [ ] [ ] [ ] [ ] [ ]0=−+=−+ nxmcmxmnxmcnxmnxncnxmmxmnxmnxmnxnc CvecBXvecXAvecCBXXAvec ( ) ( ) [ ]nxmcnxmcn T mxmnxnm CvecXvecIBAI =⊗+⊗or Let use the Jordan decomposition for Anxn and Bmxm 11 & −− == TTT BBB T mxmAAAnxn SJSBSJSA ( ) ( ) ( ) ( ) ( ) ( ) n TTT m TT TTT I AABBBAAA I BB nBBBAAAmn T mxmnxnm SSSJSSJSSS ISJSSJSIIBAI 1111 11 −−−− −− ⊗+⊗= ⊗+⊗=⊗+⊗ ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) 111 111 1111 −−− −−− ⊗ −− ⊗⊗⊗⊗ −− ⋅⊗⋅=⊗⊗ ⊗=⊗ ⊗⊗+⊗⊗= ATB T nTBATB TT ATBAm TT SS AB IJSS ABB SSJI AABAB DBCADCBA BABA SSSJSSJSSS ( )( )( ) 1− ⊗⊗+⊗⊗=⊗+⊗ ABnBAmABn T mxmnxnm SSIJJISSIBAI T T T This equation has a Unique Solution iff is Nonsingular.( )n T mxmnxnm IBAI ⊗+⊗
- 19. Sylvester Matrix Equation : Anxn Xnxm+Xnxm Bmxm=Cnxm SOLO Matrices Proof (continue – 3) where ( )( )( ) 1− ⊗⊗+⊗⊗=⊗+⊗ ABnBAmABn T mxmnxnm SSIJJISSIBAI T T T [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] = lJ J J J 00 00 00 2 1 [ ] = i i i i i xkki ii J λ λ λ λ λ 0000 1000 0000 0010 0001 Ji are Upper-Triangular Matrices ( )nBAm IJJI T ⊗+⊗Therefore is also Upper Triangular with diagonal elements λi[A] + λj[B] (i=1,…,n, j=1,…,m), that are the Eigenvalues of and therefore to the Similar Matrix . This Matrix is Nonsingular iff ( )nBAm IJJI T ⊗+⊗ ( )n T mxmnxnm IBAI ⊗+⊗ ( ) ( ) mjniBA mxmjnxni ,,1,,,10 ==∀≠+ λλ
- 20. Sylvester Matrix Equation : Anxn Xnxm+Xnxm Bmxm=Cnxm SOLO Matrices Since ( )[ ] ( )[ ] [ ] [ ]0lim,0lim,,1,,,10ReRe ====∀<+ ∞→∞→ tB t tA t mxmjnxni eemjniBA λλ Proof (continue – 4) Let rewrite [ ] [ ] = − =−+= nxm m Snnxm nxm m nxnnxm mxm nnxmnxmmxmnxmnxmnxn X I AIX X I AC B IXCBXXA 0 0 ( ) ( ) nxmnxm tB nxm tA nxm CPeCetP mxmnxn −=⇒−= 0:Define By differentiation ( ) mxmnxmnxmnxnmxm tB nxm tAtB nxm tA nxn nxm BPPABeCeeCeA td tPd mxmnxnmxmnxn −−=−−= Let Integrate the Differential Equation ( ) ( ) ( ) mxmnxmnxmnxn nxm nxmnxm BdtPdtPAdt td tPd PP −+ −==−∞ ∫∫∫ ∞∞∞ 000 0 We have and where ∫ ∞ = 0 dteCeX tB nxm tA nxm mxmnxn ( ) [ ]0=∞nxmP mxmnxmnxmnxnmxmnxmnxmnxnnxm BXXABdtPdtPAC += −+ −= ∫∫ ∞∞ 00
- 21. Matrix Differential Riccati Equation SOLO Inner-Outer and Spectral Factorizations ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tDtXtCtXtBtXtXtAtX +++= where all matrices are nxn and A (t),B (t), C(t), D(t) are integrable over an interval t0 ≤ t ≤ tf. This Matrix Differential Equation can be decompose in the following 2 Linear Differential Equations: The Nonlinear Matrix Differential Riccati Equation is given by ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tZtAtYtDtZ tZtCtYtBtY += −−= If in the interval t0 ≤ t ≤ tf the Matrix Y(t) is nonsingular, then the solution of the Riccati Differential Equation is ( ) ( ) ( ) 10 1 ttttYtZtX ≤≤= − To verify we have ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tY td tYd tYtY td d tY td d tZtY td tZd tX td d 11111 & −−−−− −=+= ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tXtCtXtBtXtXtAtD tYtZtCtYtBtYtZtYtZtAtYtDtX td d +++= −−−+= −−− 111
- 22. Matrix Differential Riccati Equation SOLO Inner-Outer and Spectral Factorizations ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tDtXtCtXtBtXtXtAtX +++= If yhe matrices are nxn and A ,B , C, D are constant, then This Matrix Differential Equation can be decompose in the following 2 Linear Differential Equations: The Nonlinear Matrix Differential Riccati Equation is given by ( ) ( ) ( ) ( ) ( ) ( )tZAtYDtZ tZCtYBtY += −−= If in the interval t0 ≤ t ≤ tf the Matrix Y(t) is nonsingular, then the solution of the Riccati Differential Equation is ( ) ( ) ( ) 10 1 ttttYtZtX ≤≤= − Define ( ) ( ) ( ) −− = = AD CB G tZ tY tW :&: to obtain which have the solution( ) ( )tGWtW = ( ) ( ) ( )0 0 tWetW ttG − =
Editor's Notes
- K. Ogata, “State Space Analysis of Control Systems”, Prentice Hall, Inc., 1967 , pp.356-357
- K. Ogata, “State Space Analysis of Control Systems”, Prentice Hall, Inc., 1967 , pp.356-357
- K. Ogata, “State Space Analysis of Control Systems”, Prentice Hall, Inc., 1967 , pp.356-357
- K. Ogata, “State Space Analysis of Control Systems”, Prentice Hall, Inc., 1967 , pp.356-357
- K. Ogata, “State Space Analysis of Control Systems”, Prentice Hall, Inc., 1967 , pp.356-357
- K. Ogata, “State Space Analysis of Control Systems”, Prentice Hall, Inc., 1967 , pp.321-322
- K. Ogata, “State Space Analysis of Control Systems”, Prentice Hall, Inc., 1967 , pp.321-322