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Unformatted text preview: ME 132 Fall 2009 Solutions to Homework 6 1. There are several different ways to find statespace realizations of single, linear, ordi nary, differential equations. One approach is listed below. (a) We can write the transfer function form of the ODE as: ( s 4 + 4 s 3 + 10 s 2 + 7 s 2) Y ( s ) = ( s 3 + 5 s 2 2 s 1) U ( s ) Y ( s ) = s 3 + 5 s 2 2 s 1 s 4 + 4 s 3 + 10 s 2 + 7 s 2 U ( s ) Define X ( s ) as X ( s ) := 1 s 4 + 4 s 3 + 10 s 2 + 7 s 2 U ( s ) so Y ( s ) = ( s 3 + 5 s 2 2 s 1) X ( s ). Now, define x 1 := x , x 2 := ˙ x 1 = ¨ x , x 3 := ˙ x 2 = x (3) , and x 4 := ˙ x 3 = x (4) , and we get the state space representation of the differential equation: ˙ x 1 ˙ x 2 ˙ x 3 ˙ x 4 = 1 1 1 2 7 10 4 x 1 x 2 x 3 x 4 + 1 u y = bracketleftbig 1 2 5 1 bracketrightbig x 1 x 2 x 3 x 4 (b) We can almost repeat the same procedure as above, except that we have to make the transfer function from u to y strictly proper. If we left the transfer function as is, then the above procedure would not work. The result is a nonzero D matrix, as we’ll see below. The transfer function can be made strictly proper by simply polynomial division: Y ( s ) = 3 s 3 2 s + 1 s 3 + 5 s 2 4 s U ( s ) = parenleftbigg 15 s 2 + 10 s + 1 s 3 + 5 s 2 4 s + 3 parenrightbigg U ( s ) From here, we can see that the D matrix is D = 3. Now, the procedure the same as before, and the statespace representation is...
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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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