Hw6_f09 - ME 132 Fall 2009 Solutions to Homework 6 1 There are several different ways to find state-space realizations of single linear ordi nary

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Unformatted text preview: ME 132 Fall 2009 Solutions to Homework 6 1. There are several different ways to find state-space 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 non-zero 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 state-space 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.

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Hw6_f09 - ME 132 Fall 2009 Solutions to Homework 6 1 There are several different ways to find state-space realizations of single linear ordi nary

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