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problem_11_15

# problem_11_15 - Problem 11.15 of the Palm textbook(MODIFIED...

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Problem 11.15 of the Palm textbook (MODIFIED) Plant TF: G P s ( ) = 2 s 2 ! s ! 26 s s + 2 ( ) s + 3 ( ) Sketch the root locus. Determine the range of proportional control gains K for which the system response is stable. Determine the value of K required for a time constant of ! = 2 / 3 . SOLUTION Closed-loop characteristic equation: 0 = 1 + K 2 s 2 ! s ! 26 s s + 2 ( ) s + 3 ( ) " # \$ % & ' = 1 + K 2 s 2 ! s ! 26 s 3 + 5 s 2 + 6 s " # \$ % & ' = 1 + K N s ( ) D s ( ) " # \$ % & ' 1. Number of branches : max N P , N Z { } = max 3,2 { } = THREE branches 2. Start/end points : OL zeros : z 1 = ! 3.36, z 2 = 3.86 OL poles : p 1 = 0, p 2 = ! 2, p 3 = ! 3 Therefore, the RL START at poles p 1 , p 2 , p 3 . Two RL branches ENDS at z 1 and z 2 while one branch ends at ! . 4. Since the RL can exist on the real axis only to the left of an ODD number of real OL zeros/poles, the regions on the real axis containing the RL are: p 1 < s < z 2 , p 3 < s < p 2 and s < z 2 . 5. Here we have one branch of the RL going to ! ; we need to determine the asymptote for this branch:

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! 0 = p i " # z i " N P # N z = 0 # 2 # 3 [ ] # # 3.36 + 3.86 [ ] 3 # 2 = # 5.5 \$ intercept with the real axis ( ) ! k = 2 k " 1 ( ) 180 ° N P " N z = 360 ° k " 180 ° = 180 ° 6. BA/RE points d ds N s ( ) D s ( ) ! " # \$ % & = d ds 2 s 2 ' s ' 26 s 3 + 5 s 2 + 6 s ! " # \$ % & 0 = 4 s ' 1 ( ) s 3 + 5 s 2 + 6 s ( ) ' 2 s 2 ' s ' 26 ( ) 3 s 2 + 10 s + 6 ( ) s 3 + 5 s 2 + 6 s ( ) 2 Solving: s 2 + 16 s + 13 = 0 ! s 1 = " 0.86 and s 2 = " 15.1 7. Departure/arrival angles We need to determine the departure angles from the four OL poles from which the RL branches start. To this end, we will use the general equation: ! s " z 1 ( ) # \$ % & " ! s " p 1 ( ) + ! s " p 2 ( ) # \$ % & = 180 ° Departure angle for p 1 : ! p 1 " z 1 ( ) # \$ % & "

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problem_11_15 - Problem 11.15 of the Palm textbook(MODIFIED...

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