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20095ee141_1_EE141_hw4_sol

20095ee141_1_EE141_hw4_sol - EE141 Principles of Feedback...

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EE141 Principles of Feedback Control (Fall 2009) Solutions to Homework 4 Problem 4.12 (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. Alternatively, if we use the result from type-1 system, we can observe that the open loop transfer function is T ol = D ( s ) 1 s 2 + K . If we need to track a ramp reference input with constant steady-state error, the system has to be type-1 system, which implies T ol has a pole at the origin. In other words, 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. (c) If D ( s ) = k p + k I s 1
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we have Y ( s ) R ( s ) = ( k p + k I s ) 1 s 2 + K 1 + ( k p + k I s ) 1 s 2 + K = k p s + k I s 3 + ( K + k p ) s + k I .
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