sol6.09 - ECE521 Linear Systems Fall 2009 Homework 6...

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Unformatted text preview: ECE521 Linear Systems Fall 2009 Homework 6 Solutions Problem 1 : Consider a continuous system defined by A and B , A = 0 1 0 0 1 0 0 1 1 1- 3 4 , B = 1 0 0 0 0 0 0 1 Find two different matrices K so the eigenvalues of A- BK are- 1 ± i and- 2 ± i . Let B = [ b 1 b 2 ], and k T 1 and k T 2 be two row vectors. Note that A- b 1 k T 1 = A- B parenleftbigg k T 1 parenrightbigg and A- b 2 k T 2 = A- B parenleftbigg k T 2 parenrightbigg Hence, operating on B and K = parenleftbigg k T 1 parenrightbigg or B and K = parenleftbigg k T 2 parenrightbigg is the same as operating on b 1 and k T 1 or b 2 and k T 2 . As long as ( A,b i ) is controllable, we can place poles at any desired locations. This can be done by transforming ( A,b i ) to the cannonical form. Once we have the cannonical we need to choose free parameters so they match the coefficients of the characteristic polynomial. In our case, the desired polynomial is ( λ- (- 1 + i))( λ- (- 1- i))( λ- (- 2 + i))( λ- (- 2- i)) = λ 4 + 6 λ 3 + 15 λ 2 + 18 λ + 10 Note that ( A,b 1 ) as well as ( A,b 2 ) are controllable pairs. Indeed Q B ( A,b 1 ) = 1 0 0 0 0 0 1 0 0 1 4 0 1 4 13 is nonsingular and ( A,b 2 ) is already in the cannonical form (which is always controllable). (a) We first work with the pair ( A,b 1 ). We need to transform the pair ( A,b 1 ) to the cannonical form, say ( ˆ A, ˆ b 1 ) where ˆ b 1 = [0 0 0 1] T . As A is in the cannonical form we must have ˆ A = A ....
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This note was uploaded on 02/09/2010 for the course MAE 123 taught by Professor 123 during the Spring '10 term at École Normale Supérieure.

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sol6.09 - ECE521 Linear Systems Fall 2009 Homework 6...

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