e483h6s_07

# e483h6s_07 - 1 ENSC 483 HW#6 Solutions by M. Saif Simon...

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ENSC 483 HW#6 Solutions by M. Saif 1 Simon Fraser University School of Engineering Science ENSC 483 - Modern Control Systems 1. Transformation into OCF. Consider the SISO system described by ˙ x = Ax + B u y = Cx Prove that the system can be transformed by x = P ˆ x , with P being a nonsingular transformation into an equivalent system with: ˆ A = 0 0 ··· 0 0 - a 0 1 0 ··· 0 0 - a 1 . . . . . . . . . . . . . . . . . . 0 0 ··· 1 0 - a n - 2 0 0 ··· 0 1 - a n - 1 ˆ c = ( 0 0 0 ··· 0 1 ) given that ρ c cA cA 2 . . . cA n - 1 = n Hint: To prove this, follow along the same lines as the proof for transformation into CCF which we did in the class, and take P = ( v Av A 2 v ··· A n - 1 v ) Solution: Deﬁning ˆ x = P - 1 x = ˆ A = P - 1 AP . Let P = [ v Av A 2 v ··· A n - 1 v ] where v is a n column vectors selected such that ρ ( P ) = n . Let P - 1 = q 1 q 2 ··· q n = P - 1 AP = q 1 Av q 1 A 2 v ··· q 1 A n v q 2 Av q 2 A 2 v ··· q 2 A n v . . . . . . . . . . . . q n Av q n A 2 v ··· q n A n v Note also that P - 1 P = I = q 1 v q 1 Av ··· q 1 A n - 1 v q 2 v q 2 Av ··· q 2 A n - 1 v . . . . . . . . . . . . q n v q n Av ··· q n A n - 1 v

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ENSC 483 HW#6 Solutions by M. Saif 2 Notice now that the i th column of P - 1 AP is the i + 1 th column of P - 1 P ( i = 1 , 2 , ··· ,n - 1) . The last column of P - 1 AP is a i = - q i +1 A n v i = 0 , 1 , ··· ,n - 1 Now ˆ C = CP which implies that Cv = 0 CAv = 0 . . . CA n - 2 v = 0 CA n - 1 v = 1 = C CA . . . CA n - 2 CA n - 1 v = 0 0 . . . 0 1 As we know, the above equation has a unique solution so long that ρ C CA . . . CA n - 2 CA n - 1 = n which we assumed to be the case. Finally, to show that the matrix P is nonsingular, we need to prove that its columns are linearly independent. Assuming otherwise, then α 1 v + α 2 Av + ··· + α n A n - 1 v = 0 for some α i s not all zero. However, pre-multiplying this expression by C results in α n = 0 . Similarly, pre-multiplying by CA results in α n - 1 = 0 . Continuing in this manner we can show that all the α s are zero which of course is a contradiction and therefore we conclude that the columns of P are linearly independent and therefore P is nonsingular. 2. Consider the system
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## This note was uploaded on 10/09/2009 for the course ENSC 1166 taught by Professor Mehrdadsaif during the Spring '07 term at Simon Fraser.

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e483h6s_07 - 1 ENSC 483 HW#6 Solutions by M. Saif Simon...

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