141_1_Final_sol

141_1_Final_sol - EE141 Principles of Feedback Control...

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Unformatted text preview: EE141 Principles of Feedback Control (Winter 2010) Solutions to Final Problem 1 (1) sX 1 =- . 4 X 1 + . 2 X 2 + . 02 X 3 + U sX 2 = . 2 X 1- . 4 X 2 sX 3 = . 2 X 1- . 02 X 3 Then we can eliminate X 2 , X 3 and get X 1 U = ( s + . 4)( s + . 02) s 3 + . 82 s 2 + . 132 s + . 0008 (2) With Routh Test, s 3 1 . 132 s 2 . 82 . 0008 s 1 . 131 s . 0008 The first column is positive. Thus the system is stable. (3) U = . 5 s and since the system is stable. The steady state concentration of insulin in the plasma upon an injection is lim s sX 1 ( s ) = lim s s ( s + . 4)( s + . 02) s 3 + . 82 s 2 + . 132 s + . 0008 . 5 s = 5 (4) G 3 has the fastest response due to two most negative poles. Thus G 3 is corresponding to the middle graph. G 1 has a relative larger steady state response than G 2 . Thus G 1 is corresponding to the left graph and G 2 corresponding to the right graph. (5) H ( s ) = 1 s 2 + . 606 s + . 0036 1 We can use a PD controller on the feedback path here, then the close-loop transfer function...
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This note was uploaded on 07/02/2011 for the course EE 141 taught by Professor Balakrishnan during the Spring '07 term at UCLA.

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141_1_Final_sol - EE141 Principles of Feedback Control...

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