{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

20112ee141_1_EE141_hw4_sol

20112ee141_1_EE141_hw4_sol - EE141 Principles of Feedback...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}