# hw4 - EML 4312 Fall 2009 MC/AC Proportional and Lead/Lag...

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EML 4312 Fall 2009 MC/AC, Proportional and Lead/Lag Compensation Due Monday November 2nd. 1. Consider the following block diagram. K in the block diagram could represent a constant gain or polynomial (e.g., lead or lead/lag compensator). a. (10 points) For the following given plant model P ( s ) = s + a ( s + 1) 2 ( s + 5) develop a controller to place a closed-loop pole at s 1 = ± 2 : 5 + 5 j: sol. Coe¢ cient Matching A = ( s + 1) 2 ( s + 5) + K ( s + a ) = s 3 + 7 s 2 + (11 + K ) s + 5 + Ka D = ( s + 2 : 5 + 5 j )( s + 2 : 5 ± 5 j )( s + c ) = s 3 + (5 + c ) s 2 + (31 : 25 + 5 c ) s + 31 : 25 c By matching coe¢ cients c = 2 K = 41 : 25 ± 11 = 30 : 25 a = 31 : 25(2) 30 : 25 = 2 : 0661 : So this can be done with a pure constant K = 30 : 25 sol. MC/AC K s + a ( s + 1) 2 ( s + 5) s = 2 : 5+5 j = 1 K ± 2 : 5 + 5 j + a ( ± 2 : 5 + 5 j + 1) 2 ( ± 2 : 5 + 5 j + 5) = 1 K q ( ± 2 : 5 + a ) 2 + 25 ( p ± 1 : 5 2 + 25) 2 ± p 2 : 5 2 + 25 ² = 1 From the AC 6 s + a ± 6 s + 1 ± 6 s + 1 ± 6 s + 5 j s = 2 : 5+5 j = ² 180 6 ( a ± 2 : 5 + 5 j ) ± 6 ( ± 1 : 5 + 5 j ) ± 6 ( ± 1 : 5 + 5 j ) ± 6 (2 : 5 + 5 j ) = ² 180 tan 1 ( 5 a ± 2 : 5 ) ± ( ± 73 : 3) ± ( ± 73 : 3) ± 63 : 435 = ² 180 tan 1 ( 5 a ± 2 : 5 ) = 96 : 8 5 a ± 2 : 5 = ± 8 : 345 a = 1 : 9 1

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Now go back to MC K q (0 : 6) 2 + 25 ( p 1 : 5 2 + 25) 2 ± p 2 : 5 2 + 25 ² = 1 K 30 : 25 = 1 K = 30 : 25 ± b. (10 points) For the following given plant model P ( s ) = ( s + 1) s 2 ( s + b ) develop a controller to place a closed-loop pole at s 1 = 2 : 0 + 15 j: sol. Coe¢ cient Matching A = s 2 ( s + b ) + K = s 3 + bs 2 + Ks + K D = ( s + 2 + 15 j )( s + 2 15 j )( s
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hw4 - EML 4312 Fall 2009 MC/AC Proportional and Lead/Lag...

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