CE4.1 case 3

CE4.1 case 3 - %- %- % CE 4.1 (case 3) %- A = [0 1 0 0 0...

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Sheet1 Page 1 %------------------------------------------------- %------------------------------------------------- % CE 4.1 (case 3) %------------------------------------------------- A = [0 1 0 0 0 0 B = [0 C = [0 0 0 0 1 0] D = [0] JbkR = ss(A,B,C,D) Q = obsv(JbkR) Q1=[C CharPoly = poly(A) % characteristic polynomial if (rank(Q) == size(A,1)) % Logic to assess % observability disp('System is observable.') else disp('System is NOT observable.') end P = ctrb(JbkR) % controllability % matrix P a1 = CharPoly(2) Pccfi = [a1 1 % matrix Pccf Tccf = P*Pccfi % transformation matrix Accf = inv(Tccf)*A*Tccf % formula Bccf = inv(Tccf)*B Cccf = C*Tccf Dccf = D Qocf = inv(Pccfi) Tocf=inv(Q)*Qocf % transformation %matrix Aocf=inv(Tocf)*A*Tocf % via formula Bocf = inv(Tocf)*B Cocf = C*Tocf Docf = D
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Sheet1 Page 2 [JbkROCF,TOCF] = canon(JbkR,'companion') % using canon AOCF = JbkROCF.a BOCF = JbkROCF.b COCF = JbkROCF.c DOCF = JbkROCF.d %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Answer from Matlab
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CE4.1 case 3 - %- %- % CE 4.1 (case 3) %- A = [0 1 0 0 0...

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