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141_1_Final_sol

# 141_1_Final_sol - EE141 Principles of Feedback...

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EE141 Principles of Feedback Control (Winter 2010) Solutions to Final Problem 1 (1) sX 1 = - 0 . 4 X 1 + 0 . 2 X 2 + 0 . 02 X 3 + U sX 2 = 0 . 2 X 1 - 0 . 4 X 2 sX 3 = 0 . 2 X 1 - 0 . 02 X 3 Then we can eliminate X 2 , X 3 and get X 1 U = ( s + 0 . 4)( s + 0 . 02) s 3 + 0 . 82 s 2 + 0 . 132 s + 0 . 0008 (2) With Routh Test, s 3 1 0 . 132 s 2 0 . 82 0 . 0008 s 1 0 . 131 s 0 0 . 0008 The first column is positive. Thus the system is stable. (3) U = 0 . 5 s and since the system is stable. The steady state concentration of insulin in the plasma upon an injection is lim s 0 sX 1 ( s ) = lim s 0 s ( s + 0 . 4)( s + 0 . 02) s 3 + 0 . 82 s 2 + 0 . 132 s + 0 . 0008 0 . 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 + 0 . 606 s + 0 . 0036 1

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We can use a PD controller on the feedback path here, then the close-loop transfer function is H 1 + DH = 1 s 2 + 0 . 606 s + 0 . 0036 1 + ( k D s + k P ) 1 s 2 + 0 . 606 s + 0 . 0036 = 1 s 2 + ( k D + 0 . 606) s + k p + 0 . 0036 Since t s = 4 . 6 σ 1 σ 4 . 6 t r = 1 . 8 ω n 0 . 9 ω n 2 we have k D + 0 . 606 = 2 σ 9 .
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141_1_Final_sol - EE141 Principles of Feedback...

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