# hw5 - Aaron Marburg E105 Feedback Control of Dynamic...

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Unformatted text preview: Aaron Marburg E105: Feedback Control of Dynamic Systems Homework #5 Solutions Each problem worth 5 pts Total: 35 pts 1 FPE 5.1 b) and d)iii Set up the given characteristic equations in the form suited to Evan’s root-locus method. Give L ( s ) ,a ( s ) and b ( s ) and the parameter K in terms of the original parameters in each case. Be sure to select K so that a ( s ) and b ( s ) are monic in each case and the degree of b ( s ) is not greater than that of a ( s ). 1.1 Part b) s 2 + cs + c + 1 = 0 versus parameter c Start by rearranging like so: s 2 + cs + c + 1 = 0 ( s 2 + 1 ) + c ( s + 1) = 0 Divide through by s 2 + 1 to get: 1 + c s + 1 s 2 + 1 = 0 Which means L ( s ) = s + 1 s 2 + 1 b ( s ) = s + 1 which is monic a ( s ) = s 2 + 1 which is monic K = c and the order of b ( s ) is less than the order of a ( s ) (1 < 2). 1.2 FPE 5.1diii 1 + bracketleftbigg k P + k I s + k D s τs + 1 bracketrightbigg G ( s ) = 0 Assume that G ( s ) = A c ( s ) d ( s ) where c ( s ) and d ( s ) are monic polyno- mials with the degree of d ( s ) greater than that of c ( s ). Solve versus k D . 1 Sometimes there’s no hope but to brute force it. Here’s one approach. Write it out as: parenleftbigg 1 + bracketleftbigg k P + k I s bracketrightbigg G ( s ) parenrightbigg + k D bracketleftbigg s τs + 1 bracketrightbigg G ( s ) = 0 And just divide that in to get: 1 + k D bracketleftBig s τs +1 bracketrightBig G ( s ) ( 1 + bracketleftbig k P + k I s bracketrightbig G ( s ) ) = 0 Start simplifying: 1 + k D parenleftbigg sG ( s ) τs + 1 parenrightbigg s s + k P G ( s ) s + k I G ( s ) = 0 Expand out G ( s ) 1 + k D sA c ( s ) d ( s ) τs + 1 s s + k P A c ( s ) d ( s ) s + Ak I c ( s ) d ( s ) = 0 1 + k D parenleftbigg sAc ( s ) d ( s )( τs + 1) parenrightbigg d ( s ) s d ( s ) s + k P Ac ( s ) s + Ak I c ( s ) = 1 + k D Ac ( s ) s 2 ( τs + 1)( d ( s ) s + k P Ac ( s ) s + Ak I c ( s )) = In the interest of making everything monic, turn the k D term into a k D A τ term, which gives 1 + parenleftbigg k D A τ parenrightbigg c ( s ) s 2 ( s + 1 τ )( d ( s ) s + Ac ( s ) k P s + Ak I c ( s )) = 0 I’ll just stop there. In this case L ( s ) = c ( s ) s 2 ( s + 1 τ )( d ( s ) s + Ac ( s ) k P s + Ak I c ( s )) b ( s ) = c ( s ) s 2 a ( s ) = ( s + 1 τ )( d ( s ) s + Ac ( s ) k P s + Ak I c ( s )) K = k D A τ The numerator is of order (order of c )+2, and it’s monic because c ( s ) are monic....
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hw5 - Aaron Marburg E105 Feedback Control of Dynamic...

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