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# 5r1 x4 10x4 r2 x5 4r2 x6 x5 4x6 c1 x 2 x 4 c2

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Unformatted text preview: �� � C2 +�� The input, output, and state vectors are: �� r �= 1 u r2 �� c �= 1 y c2 � ⎡ x1 �⎢ �x2 ⎢ �x ⎢ � = � 3⎢ x �⎢ �x4 ⎢ � 5 ⎣ x x6 From the block diagram, we can write down the state equations and the output equation: x1 = 2r1 − x1 ˙ x2 = x1 − 2x2 ˙ x3 = −6x3 + 0.5r1 ˙ x4 = −10x4 + r2 ˙ x5 = 4r2 ˙ x6 = x5 − 4x6 ˙ c1 = x 2 + x 4 c2 = x 3 + x 6 In matrix form: � −1 �1 � � ˙ = � 0 � x � 0 � � 0 0 � 01 � = y 00 ⎡ � 0 0 0 00 2 ⎢ �0 −2 0 0 0 0⎢ � � 0 −6 0 0 0⎢ ⎢ � + �0.5 ⎢x � 0 0 0 −10 0 0 ⎢ � �0 0 0 0 0 0⎣ 0 0 0 1 −4 0 � � � 0100 00 �+ x � u 1001 00 2 ⎡ 0 ⎢ 0⎢ ⎢ 0⎢ u ⎢� 1⎢ 4⎣ 0 Problem 2 1. sI − A = � s+1 0 −1 s + 2 � � �� 1 1 s+2 0 s+1 � (sI − A) = = 1 1 s+1 (s + 1)(s + 2) (s+1)(s+2) � � e−t 0 −1 −1 � �(t) = L [(sI − A) ] = −t e − e−2t e−2t −1 2. 0 1 s+2 ⎤ �� � e−0 0 10 = �(0) = −0 01 e − e−2(0) e−2(0) � 3. � −1 e−2t 0 (t) = −3t −2t e − e−t e−t e � � et 0 = et − e2t e2t = �(−t) � 1 � 4. �(t2 )�(t1 ) = � e−t2 −t2 − e−2t2 e 0...
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## This note was uploaded on 11/06/2011 for the course AERO 100 taught by Professor Willcox during the Fall '03 term at MIT.

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