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Unformatted text preview: ME 132 Solutions # 10 1 Statespace realizations The controllable canonical form can be calculated from the method ology presented in class/previous homework solutions. (a) x 1 x 2 x 3 x 4 x 5 y = 1 1 1 1 3 0 2 7 6 1 1 3 2 x 1 x 2 x 3 x 4 x 5 u (b) x 1 x 2 x 3 y = 1 1 4 3 2 1 13 10 6 3 x 1 x 2 x 3 u 2 Statespace realizations (a) There may be several ways to solve this problem. One way to do it is to recognize that H ( s ) = bracketleftBigg 2 s +2 s s 2 +3 s +2 2 s s +1 bracketrightBigg = bracketleftBigg H 11 H 12 H 21 H 22 bracketrightBigg corresponds to the picture given below. Then, from the previous homework, we can combine the statespace realizations for each subsystem into the overall system realization. Using the methodology presented in class, this yields the statespace realization  2 2 1 2 3 1 1 1 1 1 1 2 1 (b) In the above realization, there are four states....
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This note was uploaded on 10/20/2010 for the course ME 132 taught by Professor Tomizuka during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Tomizuka

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