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makeSoln10

# makeSoln10 - ME 132 Solutions 10 1 State-space realizations...

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Unformatted text preview: ME 132 Solutions # 10 1 State-space 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 State-space 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 state-space realizations for each subsystem into the overall system realization. Using the methodology presented in class, this yields the state-space 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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makeSoln10 - ME 132 Solutions 10 1 State-space realizations...

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