# ps7-soln - Problem Set 7 Solutions March 7 2010 Problem 1...

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Problem Set 7 Solutions March 7, 2010 Problem 1 First, we calculate the contribution to the steady-state error due only to disturbance: Rearranging the block diagram to show D(s) as the input, D ( s ) E ( s ) + K 1 ( s +2) s +3 K 2 s ( s +4) Figure 1: Rearranged block diagram. Therefore, E ( s ) = D ( s ) K 2 s ( s +4) 1 + K 1 K 2 ( s +2) s ( s +3)( s +4) = D ( s ) K 2 ( s + 3) s ( s + 3)( s + 4) + K 1 K 2 ( s + 2) Next, we calculate the contribution to the steady-state error due to the reference input R ( s ): E R ( s ) = R ( s ) 1 1 + K 1 K 2 ( s +2) s ( s +3)( s +4) For D ( s ) = 1 s , e D ( ) = lim s 0 sE D ( s ) = 3 2 K 1 . This identity holds provided E D ( s ) has poles in the LHP or at most one pole in the origin. For R ( s ) = 1 /s 2 , e R ( ) = lim s 0 sE R ( s ) = 1 K 1 K 2 6 = 6 K 1 K 2 . This identity holds provided E R ( s ) has poles in the LHP or at most one pole in the origin. Design: e D ( ) = 0 . 000012 = 3 2 K 1 , or K 1 = 125000. Similarly, e R ( ) = 0 . 003 = 6 K 1 K 2 , or K 2 = 0 . 016 Note that, for these values of K 1 and K 2 to be acceptable, both E R ( s ) and E D ( s ) must satisfy the property above (have poles in the LHP and at most one pole in the origin). It is easy to check that this is the case. Problem 2 a. E ( s ) = R ( s ) C ( s ). But, C ( s ) = [ R ( s ) C ( s ) H ( s )] G 1 ( s ) G 2 ( s ) + D ( s ). 1

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Solving for C ( s ), C ( s ) = R ( s ) G 1 ( s ) G 2 ( s ) 1 + G 1 ( s ) G 2 ( s ) H ( s ) + D ( s ) 1 + G 1 ( s ) G 2 ( s ) H ( s ) Substituting into E ( s ), E ( s ) = b 1 G 1 ( s ) G 2 ( s ) 1 + G 1 ( s ) G 2 ( s ) H ( s ) B R ( s ) 1 1 + G 1 ( s ) G 2 ( s ) H ( s ) D ( s ) b. For R ( s ) = D ( s ) = 1
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ps7-soln - Problem Set 7 Solutions March 7 2010 Problem 1...

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