problem3-HW6

# problem3-HW6 - Problem 3 Exercise problem 3 Homework set...

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(4) (2) O (1) (5) (7) (3) O O O O O O O O (6) with('linalg'): Define the system matrices A , B 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])); d 1 0 1 1 2 1 0 1 0 1 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 colspace(R); 1 0 0 0 0 , 0 0 0 1 1 , 0 0 1 0 0 , 0 1 0 0 0 T1:=transpose(stackmatrix(Matrix(1, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0}), Matrix(1, 5, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 1, (1, 5) = 1}), Matrix(1, 5, { Problem 3, Exercise set 6

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(13) (11) O O (9) O (12) (10) O (8) O (1, 1) = 0, (1, 2) = 0, (1, 3) = 1, (1, 4) = 0, (1, 5) = 0}), Matrix(1, 5, {(1, 1) = 0, (1, 2) = 1, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0}))); T1 d 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 The controllable subspace has dimension 4, and it is spanned by the columns of the matrix T1. Let us find a complementary subspace of the controllable subspace. For simplicity, we compute the orthogonal complement: nullspace(transpose(T1)); 0 0 0 K 1 1 T2:=matrix(5,1,[0,0,0,-1,1]); T2 d 0 0 0 K 1 1 Collecting all the vectors, we form a change of basis adapted to the controllable subspace. The corresponding transformation matrix T is
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problem3-HW6 - Problem 3 Exercise problem 3 Homework set...

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