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20112ee141_1_EE141_hw4_sol

# 20112ee141_1_EE141_hw4_sol - EE141 Principles of Feedback...

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EE141 Principles of Feedback Control (Winter 2010) Solutions to Homework 4 Problem 4.9 (a) Y = 1 s 2 ( D ( R - Y ) + W - KY ) ( s 2 + K + D ) Y = DR + W Y = D s 2 + K + D R + 1 s 2 + K + D W In order to track a ramp reference input with constant steady-state error, R ( s ) = 1 s 2 Y = D s 2 + K + D R E ( s ) = R ( s ) - Y ( s ) = s 2 + K s 2 + K + D R e ss = lim s 0 sE ( s ) = lim s 0 s s 2 + K s 2 + K + D 1 s 2 = lim s 0 s 2 + K s 3 + Ks + Ds Thus lim s 0 sD ( s ) is a constant, which implies D ( s ) must have a pole at the origin. (b) Y ( s ) = 1 s 2 + K + D ( s ) W ( s ) = 1 s 2 + K + D ( s ) 1 s l lim s 0 sY ( s ) = lim s 0 1 s l - 1 ( s 2 + K + D ( s )) = 0 iff lim s 0 s l - 1 D ( s ) = Because D ( s ) has one pole at the origin, the above equation is satisfied iff l = 1. Thus system will reject step disturbances. Problem 4.23 (a) m ˙ v = K a U - bv msV ( s ) = K a U ( s ) - bV ( s ) V ( s ) U ( s ) = K a ms + b = 10 1000 s + 10 = 0 . 01 s + 0 . 01 1

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where m = 1000 , K a = 10 , b = 10. (b) V ( s ) = 1 s + 0 . 02 U ( s ) + 0 . 05 s + 0 . 02 W ( s ) = 1 s + 0 . 02 ( - k p V ( s )) + 0 . 05 s + 0 . 02 W ( s ) V ( s ) W ( s ) = 0 . 05 s + 0 . 02 + k p Since the reference velocity is 0 and disturbance W is 2 percent, E ( s ) = - V ( s ) = - 0 . 05 s + 0 . 02 + k p W ( s ) = - 0 . 05 s + 0 . 02 + k p 2 s The steady state error is e ss = lim s 0
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