problem4-HW6

# problem4-HW6 - Exercise set 6 set 6, problem 3 Homework -...

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O O (6) (7) (5) O (2) O (3) O (4) O (1) O O with('linalg'): Define the system matrices A , B , C A:=diag(2,2,2,-3,-3); d 2 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 K 3 0 0 0 0 0 K 3 A[1,2]:=1; 1, 2 d 1 print(A); 2 1 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 K 3 0 0 0 0 0 K 3 B:=transpose(matrix(2,5,[1,1,2,0,0,0,1,1,1,1]));C:=matrix(2,5, [1,0,0,1,-1,0,0,1,0,0]); d 1 0 1 1 2 1 0 1 0 1 d 1 0 0 1 K 1 0 0 1 0 0 Compute the controllability matrix R:=evalm(augment(B,A&*B,A&*A&*B,A&*A&*A&*B,A&*A&*A&*A&*B)); R d 1 0 3 1 8 4 20 12 48 32 1 1 2 2 4 4 8 8 16 16 2 1 4 2 8 4 16 8 32 16 0 1 0 K 3 0 9 0 K 27 0 81 0 1 0 K 3 0 9 0 K 27 0 81 rank(R); 4 The pair (A,B) is not controllable. Let us compute the controllable subspace imageR:=colspace(R); Exercise set 6 - Problem 4

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O (8) O (7) O (12) O (13) O O (11) (10) (9) imageR d 0 0 1 0 0 , 1 0 0 0 0 , 0 0 0 1 1 , 0 1 0 0 0 Compute the observability matrix Theta:=evalm(stackmatrix(C,C&*A,C&*A&*A,C&*A&*A&*A,C&*A&*A&*A&* A)); ! d 1 0 0 1 K 1 0 0 1 0 0 2 1 0 K 3 3 0 0 2 0 0 4 4 0 9 K 9 0 0 4 0 0 8 12 0 K 27 27 0 0 8 0 0 16 32 0 81 K 81 0 0 16 0 0 rank(Theta); 4 The system is not observable. Let us compute the unobservable subspace: kernel_Theta:=nullspace(Theta); kernel_Theta d 0 0 0 1 1 The intersection between the controllable and unobservable subspaces is given by the unobservable subspace itself: kerTheta_int_imR:=intbasis(imageR,kernel_Theta); kerTheta_int_imR d 0 0 0 1 1 Let us find a complementary subspace of the controllable subspace. For simplicity, we compute the orthogonal complement: imageR_complement:=nullspace(transpose(T1)); imageR_complement d 0 0 0 K 1 1 We find the Kalman decomposition. Construct a change of basis adapted to the intersection of the unobservable and controllable subspace, the rest of the controllable subspace, and the uncontrollable subspace (the intersection between the unobservable and uncontrollable subspace is empty): T:=transpose(matrix(5,5,[kerTheta_int_imR[1],imageR[1],imageR[3] ,imageR[4],imageR_complement[1]])); T d 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 K 1 1 0 0 0 1 The system expressed in the new basis is given by
O (16) (14) O (15) O Ak:=evalm(inverse(T)&*A&*T);Bk:=evalm(inverse(T)&*B);Ck:=evalm (C&*T); Ak d K 3 0 0 0 0 0 2 1 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 K 3 Bk d 0 1 1 0 1 1 2 1 0 0 Ck d 0 1 0 0 K 2 0 0 0 1 0 It is readily seen that the system is stabilizable, since the eigenvalue associated with the matrix A 33 of the uncontrollable subsystem is " = K 3. It is also clear that the system is detectable, as the eigenvalue associated with the matrix 11 of the unobservable subsystem is " = K 3. The controllable and observable subsystem 22 , B 2 , C 2 is A22:=submatrix(Ak,2. .4,2. .4);B2:=submatrix(Bk,2. .4,1. .2);C2:= submatrix(Ck,1. .2,2. .4); A22 d 2 1 0 0 2 0 0 0 2 B2 d 1 0 1 1 2 1 C2 d 1 0 0 0 0 1 Let us compute the transfer matrix of the system 22 , 2 , 2 : W2:=simplify(evalm(C2&*inverse(diag(s,s,s)&+(-A22))&*B2)); W2 d s K 1 K 2 2 1 K 2 2 2 K 2 1 K 2 Note the unstable poles, repeated with multiplicity 2. This is consistent with the structure of the matrix

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O O O (19) (20) (22) O (17) O O (18) O (21) A 22 , which is already in Jordan form. Let us compare it with the transfer matrix of the full system: W:=simplify(evalm(C&*inverse(diag(s,s,s,s,s)&+(-A))&*B)); W d s K 1 2 K 4 C 4 1 2 K 4 C 4 2 K 2 1 K 2 which shows that the two transfer matrices are indeed equal. Next, we compute a state-feedback matrix
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## This note was uploaded on 02/21/2011 for the course ECE 750 taught by Professor Serrani during the Fall '10 term at Ohio State.

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problem4-HW6 - Exercise set 6 set 6, problem 3 Homework -...

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