ps10-soln - Problem Set 10 Solutions March 28, 2010 Problem...

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Unformatted text preview: Problem Set 10 Solutions March 28, 2010 Problem 1 Let K = bracketleftbig k 1 k 2 bracketrightbig . We need to show that the eigenvalues of A + BK can be assigned to be the eigenvalues of any real 2x2 matrix by a suitable choice of K . To do this, we simply need to show that with a suitable K , the characteristic polynomial of A + BK can made into the characteristic polynomial of any real 2x2 matrix. The characteristic polynomial of a real 2x2 matrix will have the form x 2 + a 1 x + a for real numbers a and a 1 . We have A + BK = bracketleftbigg 3 + 3 k 1 4 + 3 k 2- 2 + k 1 6 + k 2 bracketrightbigg The characteristic polynomial of A + BK is x 2 + (- 9- 3 k 1- k 2 ) x + (26 + 14 k 1 + 9 k 2 ). Equating this polynomial to the general polynomial given above, we get a linear system of equations: 3 k 1 + k 2 =- 9- a 1 14 k 1 + 9 k 2 =- 26 + a which is solvable for any real a , a 1 since the matrix bracketleftbigg 3 1 14 9 bracketrightbigg is invertible. Hence, the eigenvalues of A + BK can be arbitrarily assigned. Problem 2 Let K = bracketleftbig k 1 k 2 k 3 bracketrightbig . We have A + BK = - 2 + k 1 k 2 k 3 3 + 2 k 1- 5 + 2 k 2 2 k 3- 9 + k 1 k 2 7 + k 3 The characteristic polynomial of A + BK is x 3 + (- k 1- 2 k 2- k 3 ) x 2 + (- 39 + 2 k 1 + 7 k 2 + 2 k 3 ) x + (- 70 + 35 k 1 + 49 k 2 + 35 k 3 ). If we evaluate this polynomial at x = 7 we get 0 for any k 1 , k 2 , k 3 , which implies that 7 is an eigenvalue of A + BK for any K . Hence, no linear state feedback can stabilize this system. Another way to see why we cannot change the eigenvalue 7 is to change the state variable to z with z = T 1 x where T = 1 1 1 1 1 We have the new system given by z ( t ) = T 1...
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ps10-soln - Problem Set 10 Solutions March 28, 2010 Problem...

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