Multivariable Control Lecture 4 Controllability, Observability, Full State Feedback, Observer Based Control John T. Wen September 13, 24 Ref: 3.2-3.4 of Text
Controllability ẋ = Ax + Bu; x() = x. At time T : T x(t ) = e AT x + e A(T τ) Bu(τ)dτ. Or we can write it as a linear operator equation: b = x(t ) e AT x = L T u; L T : L m 2,T ] Rn. (A,B) is controllable L T is onto L T is full (row) rank N (L T ) = {} C AB := si A W c,t = L T L T is nonsingular ] B AB... A n 1 B is full rank ] B is full rank for all s σ(a). (Popov-Belevitch-Hautus, PBH, Test) September 13, 24Copyrighted by John T. Wen Page 1
Controllability (cont.) How do we compute LT? Given a linear operator A : V W, where V and W are Hilbert Spaces (complete normed linear spaces equipped with inner product), then A, the adjoint of A, is defined as the operator that satisfies Then LT : Rn L2 m,t ] is < w,av > W =< A w,v > V, v V,w W. (L T x)(t) = B T e AT (T t) x. It follows that the controllability grammian is W T,c = L T L T = T W T,c satisfies the linear differential equation: e A(T t) BB T e AT (T t) dt = T e At BB T e AT t dt. dw T,c dt = AW T,c +W T,c A T + BB T. If A is Hurwitz, then W T,c W c as T and solves the Lyapunov equation AW c +W c A T + BB T =. September 13, 24Copyrighted by John T. Wen Page 2
Controllability (cont.) R (L T ) = R (L T L T ) = R (C AB). Controllability is invariant under coordinate transformation: (A,B) is controllable iff (T 1 AT,T 1 B) is controllable. W T,c is not invariant under coordinate transformation. MATLAB tools: ctrb, gram(sys, c ). Proposition: R (L T ) is A-invariant. Represent (A,B) in R (L T ) R (L T ) : A where (A 11,B 1 ) is controllable. A 11 A 12 A 22, B B 1 Stabilizablity: A 22 is Hurwitz. PBH Test for Stabilizability: si A B ] is full rank for all s σ(a) C +. September 13, 24Copyrighted by John T. Wen Page 3
Eigenvalue Assignment (A,B) is controllable iff the eigenvalues of (A + BF) can be arbitrarily assigned. Proof: (if) SI case: Put (A,B) in controllable canonical form. MI case: Let B 1 be any column of B, then there exists F 1 so that (A + BF 1,B 1 ) is controllable (generic property, so a random F 1 may be used). (only if) Use PBH test. (A,B) is stabilizable if and only if there exists F so that A + BF is Hurwitz. September 13, 24Copyrighted by John T. Wen Page 4
Another Eigenvalue Assignment Algorithm We look for F to satisfy Define f i = Fe i, then Let F 1 = (A + BF)e i = λ i e i, i = 1,...,n. A λ i I Fe 1... Fe n ], E = B ] e i f i =. e 1... e n ]. Then F = F 1 E 1 ; E is invertible iff (A,B) is controlable. Note that F is completely determined for SISO case, but there are multiple solutions for MIMO, which can be used to assign some of the eigenvectors. September 13, 24Copyrighted by John T. Wen Page 5
Observability W.l.o.g, let D =. Write the output as y(t) Du(t) = Ce At x, t,t ]. y = l T x; l : R n L p 2,T ]. (C,A) is observable l T is 1-1 l T is full (column) rank N (l T ) = {} W o,t = l T l T is nonsingular C CA O CA := is full rank si A. C CA n 1 is full rank for all s σ(a). September 13, 24Copyrighted by John T. Wen Page 6
Observability (cont.) l T : R n L p 2,T ], so l T : Lp 2,T ] Rn and Observability grammian: l T w = T e AT t C T w(t)dt. W T,o = l T l T = W T,o satisfies the linear differential equation: T e AT t C T Ce At dt. dw T,o dt = AT W T,o +W T,o A +C T C. If A is Hurwitz, then W T,o W o as T and solves the Lyapunov equation A T W o +W o A T +C T C =. September 13, 24Copyrighted by John T. Wen Page 7
Observability (cont.) N (l T ) = N (l T l T ) = N (O CA ). Observability is invariant under coordinate transformation: (A,C) is observable iff (T 1 AT,CT ) is observable. W T,o is not invariant under coordinate transformation. Duality: (A,B) controllable iff (B T,A T ) observable. MATLAB tools: obsv, gram(sys, o ). N (l T ) is A-invariant. Represent (C,A) in N (l T ) N (l T ): (C 1,A 11 ) is observable. Detectability: A 22 is Hurwitz. PBH Test for Stabilizability: si A C A A 11 A 21 A 22, C is full rank for all s σ(a) C +. C 1 ] where September 13, 24Copyrighted by John T. Wen Page 8
Observer Based Control ˆx = (A + LC + BF + LDF) ˆx Ly, u = F ˆx. (F,L) chosen to place eigenvalues of (A + BF) and (A + LC). Closed loop eigenvalues are the eigenvalues of (A + BF) and (A + LC). September 13, 24Copyrighted by John T. Wen Page 9
Kalman Decomposition Kalman Decomposition: represent (A,B,C,D) in R n = V co Vco V c o V c o where V co = unobserv subspace for (A c,b c ) V co = unobserv subspace for (A c,b c ) V c o = unobserv subspace for (A c,b c ) V c o = unobserv subspace for (A c,b c ). A = A c,o A 13 A 21 A c,o A 23 A 24 A c,o A 43 A c,o, B = B co B c,o, C = C c,o C c,o ]. September 13, 24Copyrighted by John T. Wen Page 1