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

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ECE521 Linear Systems Fall 2009 Homework 6 Solutions Problem 1 : Consider a continuous system defined by A and B , A = 0 1 0 0 0 0 1 0 0 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 0 parenrightbigg and A - b 2 k T 2 = A - B parenleftbigg 0 k T 2 parenrightbigg Hence, operating on B and K = parenleftbigg k T 1 0 parenrightbigg or B and K = parenleftbigg 0 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 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 . We also know (Lecture 13) that W = Q R ( A, b 1 ) Q R ( ˆ A, ˆ b 1 ) - 1

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2 will transform ( A, b 1 ) to ( ˆ A, ˆ b 1 ) where W - 1 AW = ˆ A and W - 1 b 1 = ˆ b 1 From the latter equations we see that the last column of W is equal to b 1 . In addition as ˆ A = A the first equation show that AW = WA . Let A = [ a 1 a 2 a 3 a 4 ] and W = [ w 1 w 2 w 3 w 4 ].
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sol6.09 - ECE521 Linear Systems Fall 2009 Homework 6...

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