State Feedback Control, Lecture Notes - Linear System And Control, Study notes of Production Planning and Control

Download State Feedback Control, Lecture Notes - Linear System And Control and more Study notes Production Planning and Control in PDF only on Docsity! Linear Control Systems Lecture # 12 State Feedback Control – p. 1/30 The system ẋ = Ax+Bu y = Cx+Du is asymptotically stable if and only if Re[λi] < 0 for all eigenvalues of A This condition also guarantees that the system is BIBO stable The transient response is determined by exponential modes of the form tk−1eλit – p. 2/30 Consider the state feedback control u = Fx+ v F is an m× n gain matrix and v(t) is an m-dimensional external input Closed-loop system ẋ = (A+BF )x+Bv y = (C +DF )x+Dv The stability and transient response of the closed-loop system are determined by the eigenvalues of (A+BF ) – p. 5/30 The eigenvalues of (A+BF ) are called the closed-loop eigenvalues, while those of A are called the open-loop eigenvalues Can we choose F to arbitrarily assign the eigenvalues of (A+BF )? Lemma: The uncontrollable eigenvalues of (A,B), if any, cannot be relocated by feedback – p. 6/30 Proof: Suppose (A,B) is not controllable. Then there is a nonsingular n× n matrix P such that  = P−1AP = [ Â11 Â12 0 Â22 ] , B̂ = P−1B = [ B̂1 0 ] where Â11 is q × q, B̂1 is q ×m, the pair (Â11, B̂1) is controllable, and the eigenvalues of Â22 are the uncontrollable eigenvalues Partition FP as FP = [ F̂1 F̂2 ] where F̂1 is m× q and F̂2 is m× (n− q) – p. 7/30 Single-input Systems Since (A,B) is controllable, there is a nonsingular matrix P such that Ac = P −1AP, Bc = P −1B Ac =       0 1 · · · 0 ... ... . . . ... 0 0 · · · 1 −α0 −α1 · · · −αn−1       , Bc =       0 ... 0 1       det(sI −A) = sn + αn−1s n−1 + · · ·+ α1s+ α0 – p. 10/30 Let F = FcP−1, where Fc = [ f0 f1 · · · fn−1 ] Ac +BcFc =       0 1 · · · 0 ... ... . . . ... 0 0 · · · 1 −(α0 − f0) −(α1 − f1) · · · −(αn−1 − fn−1)       det[sI − (Ac +BcFc)] = sn + (αn−1 − fn−1)s n−1 + · · ·+ (α1 − f1)s+ (α0 − f0) – p. 11/30 If {λ1, . . . , λn} are the desired eigenvalues, then the desired characteristic polynomial is αd(s) = (s− λ1)× · · · × (s− λn) def = sn + dn−1s n−1 + · · ·+ d0 Choose fi = αi − di, for 0 ≤ i ≤ n− 1 Hence, (Ac +BcFc) has the desired eigenvalues {λ1, . . . , λn} and so does (A+BF ) = P (Ac +BcFc)P −1 because the eigenvalues are invariant under state (similarity) transformations – p. 12/30 αd(Ac) = n−1 ∑ i=1 (di − αi)A i−1 c Let ei be the unit vector with one in the ith element eT1 = [ 1 0 · · · 0 ] , eTn = [ 0 · · · 0 1 ] Cc = [Bc, AcBc, . . . , A n−1 c Bc] =         0 · · · · · · 0 1 0 · · · 0 1 ∗ ... ... ... 0 1 ∗ · · · ∗ 1 ∗ · · · · · · ∗         eT1 Cc = e T n ⇒ e T 1 = e T nC −1 c – p. 15/30 Ac =       0 1 · · · 0 ... ... . . . ... 0 0 · · · 1 −α0 −α1 · · · −αn−1       eT1Ac = e T 2 = [ 0 1 0 · · · 0 ] eT1A 2 c = e T 2Ac = e T 3 eT1A n−1 c = e T n eT1αd(Ac) = e T 1 n−1 ∑ i=0 (di − αi)A i−1 c = n−1 ∑ i=0 (di − αi)e T i+1 – p. 16/30 eT1αd(Ac) = n−1 ∑ i=0 (di − αi)e T i+1 = −Fc Fc = −e T nC −1 c αd(Ac) Recall Cc = P −1C ⇒ C−1c = C −1P Ac = P −1AP ⇒ Aic = P −1AiP, for i ≥ 0 ⇒ αd(Ac) = P −1αd(A)P F = FcP −1 – p. 17/30 αd(s) = (s+ 1) 3 = s3 + 3s2 + 3s+ 1 αd(A) = A 3 + 3A2 + 3A+ I =    6 −2 −5 10 −2 −3 4 1 −2    eT3 = [ 0 0 1 ] F = −eT3 C −1αd(A) = [ −0.8 −1.4 −1.2 ] – p. 20/30 Multi-input systems Suppose B has full rank. Since (A,B) is controllable, there is a nonsingular matrix P such that P−1AP = Ac = Āc + B̄cAm, P −1B = Bc = B̄cBm Bm is nonsingular. Let F = FcP−1 A+BF = P (Ac +BcFc)P −1 Ac +BcFc = Āc + B̄cAm + B̄cBmFc Take Fc = B −1 m (Adm −Am) where Adm is an m× n matrix to be chosen – p. 21/30 Ac +BcFc = Āc + B̄cAdm Āc = Block diag                  1 . . . 1 0 · · · · · · 0       µi×µi , i = 1, . . . ,m            B̄c = Block diag                  0 ... 0 1       µi×1 , i = 1, . . . ,m            – p. 22/30 Example: A =    −1 0 1 −2 2 −2 −1 0 3    , B =    1 0 0 2 −1 1    Eigenvalues of A are 2, −0.7321, 2.7321 Desired eigenvalues are −1, −1± j αd(s) = s 3 + 3s2 + 4s+ 2 – p. 25/30 Matlab Calculations: W = ctrb(A,B); rank(W), rank(B) ans = 3 ans = 2 rank(W(:,1:3)) ans = 3 µ1 = 2, µ2 = 1 Cbar = [W(:,1) W(:,3) W(:,2)]; M = inv(Cbar); Q = [M(2,:); M(2,:)*A; M(3,:)]; P = inv(Q); PI = Q; Ac = PI*A*P; Bc = PI*B; Am = [Ac(2,:);Ac(3,:)]; Bm = [Bc(2,:);Bc(3,:)]; Adm = [0 0 1;-2 -4 -3]; Fc = inv(Bm)*(Adm - Am); F = Fc*PI; – p. 26/30 Ac =    0 1 0 1 3 −0.5 6 0 1    , Bc =    0 0 1 −0.5 0 1    Am = [ 1 3 −0.5 6 0 1 ] , Bm = [ 1 −0.5 0 1 ] F = [ 0 −1.5 5 0.6667 −3.3333 4.6667 ] – p. 27/30