Multivariable Control Lecture 3 Description of Linear Time Invariant Systems John T. Wen September 7, 26
Outline Mathematical description of LTI Systems Ref: 3.1-3.4 of text September 7, 26Copyrighted by John T. Wen Page 1
Description of MIMO LTI Systems Z t x(t) = e At x o + e A(t τ) Bu(τ)dτ. September 7, 26Copyrighted by John T. Wen Page 2
State Space State equation: ẋ = Ax+Bu, x() = x ; y = Cx+Du Complete solution: Matrix Exponential: e At. Z t x(t) = e At x + y(t) = Ce At x +C e A(t τ) Bu(τ)dτ Z t e A(t τ) Bu(τ)dτ+Du(t). How does one calculate e At? Try A = 1 2, A = 2 1 1 2 2, A = α β β α. Internal Stability: A is Hurwitz. September 7, 26Copyrighted by John T. Wen Page 3
Impulse Response When input is δ(t), output is g(t) = Ce At B1(t)+δ(t). Output y(t) can also be written as the sum of zero-input and zero-state responses: y(t) = e At x +(g u)(t). September 7, 26Copyrighted by John T. Wen Page 4
Transfer Function Laplace transform: ˆf(s) = R f(t)e st dt. Well defined if f is of exponential order: there exist α, β so that f(t) βe αt. Domain of convergence: Re(s) > α. Laplace transform of the state equation (with x() = ): ŷ(s) = Ĝ(s)û(s). G(s) = C(sI A) 1 B+D = A B. C D G(s) is also the Laplace transform of the impulse response g(t). September 7, 26Copyrighted by John T. Wen Page 5
SISO vs. MIMO MIMO: order is important (G(s)K(s) K(s)G(s) in general)! G(s) = s+1 1 2 s+2 1 s+2 1 s+1, K(s) = s+2 s 2 s+1 s 2 1. What do you think the zeros and poles of these transfer functions are? We shall see that there is an unstable pole/zero cancellation though it is not obvious by inspection alone. September 7, 26Copyrighted by John T. Wen Page 6
MATLAB tools Try it for the system sys1 = ss(a, B, C, D) sys2 = tf(num, den) sys3 = zpk(z, p, k) sys1 = tf(sys2) [num, den] = tfdata(sys1) s+1 s+3 s 1 s+4 1 (s+1)(s+3). To define this LTI object in MATLAB, use G=tf({[1 1],[1-1];,1},{[1 3],[1 4];1,conv([1 1],[1 3])}) Valid LTI operations: sys1*sys2,sys1+sys2,inv(sys1) September 7, 26Copyrighted by John T. Wen Page 7
Controllability ẋ = Ax+Bu; x() = x. At time T : Z 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] R n. (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 7, 26Copyrighted by John T. Wen Page 8
Controllability (cont.) How do we computel T? Given a linear operatora :V W, wherev andw are Hilbert Spaces (complete normed linear spaces equipped with inner product), thena, the adjoint ofa, is defined as the operator that satisfies ThenL T : Rn L2 m [,T] is < w,a v > 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 = Z T W T,c satisfies the linear differential equation: e A(T t) BB T e AT (T t) dt = Z 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 7, 26Copyrighted by John T. Wen Page 9
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) inr (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) T C +. September 7, 26Copyrighted by John T. Wen Page 1
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 7, 26Copyrighted by John T. Wen Page 11
Another Eigenvalue Assignment Algorithm We look for F to satisfy (A+BF)e i = λ i e i, i = 1,...,n. Define f i = Fe i, then [ Let F 1 = [ 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 7, 26Copyrighted by John T. Wen Page 12
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 7, 26Copyrighted by John T. Wen Page 13
Observability (cont.) l T : R n L p 2 [,T], so l T : Lp 2 [,T] Rn and Observability grammian: l T w = Z T e AT t C T w(t)dt. W T,o = l Tl T = W T,o satisfies the linear differential equation: Z 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 7, 26Copyrighted by John T. Wen Page 14
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) inn (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) T C +. C 1 ] where September 7, 26Copyrighted by John T. Wen Page 15
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 7, 26Copyrighted by John T. Wen Page 16
Kalman Decomposition Kalman Decomposition: represent (A, B,C, D) in R n =V co V co 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 7, 26Copyrighted by John T. Wen Page 17