582hw6Solutions

582hw6Solutions - EEE 582 HW # 6 SOLUTIONS Problem 14.4 To...

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Unformatted text preview: EEE 582 HW # 6 SOLUTIONS Problem 14.4 To verify the stated identity, multiply from left and right with the quantities in the inverses: P ( I ¡ QP ) ¡ 1 = ( I ¡ PQ ) ¡ 1 P , ( I ¡ PQ ) P ( I ¡ QP ) ¡ 1 ( I ¡ QP ) = ( I ¡ PQ )( I ¡ PQ ) ¡ 1 P ( I ¡ QP ) , ( I ¡ PQ ) P = P ( I ¡ QP ) Now, the transfer function from r to y is found by substituting u = Kx + Nr in the state equation: G r ! y = C ( sI ¡ A ¡ BK ) ¡ 1 BN = C ( sI ¡ A ) ¡ 1 [( sI ¡ A )( sI ¡ A ) ¡ 1 ¡ BK ( sI ¡ A ) ¡ 1 ] ¡ 1 BN Now, apply the identity with P = B and Q = K ( sI ¡ A ) ¡ 1 . G r ! y = C ( sI ¡ A ) ¡ 1 [ I ¡ PQ ] ¡ 1 PN = C ( sI ¡ A ) ¡ 1 P [ I ¡ QP ] ¡ 1 NC ( sI ¡ A ) ¡ 1 B ( I ¡ K ( sI ¡ A ) ¡ 1 B ] ¡ 1 N Note: This expression can help in visualizing the feeddback/feedforward interpretation of the controller. However, state feedback is not equivalent to feedforward control. The realization of the above transfer function requires 2n states whereas the origina with state feedback requires only n states. The extra n states involverequires 2n states whereas the origina with state feedback requires only n states....
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This document was uploaded on 04/28/2011.

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582hw6Solutions - EEE 582 HW # 6 SOLUTIONS Problem 14.4 To...

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