problem_11_15

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

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Unformatted text preview: Problem 11.15 of the Palm textbook (MODIFIED) Plant TF: G P s ( 29 = 2 σ 2- σ- 26 σ σ + 2 ( 29 σ + 3 ( 29 • 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 τ = 2 / 3 . SOLUTION Closed-loop characteristic equation: = 1 + Κ 2 σ 2- σ- 26 σ σ + 2 ( 29 σ + 3 ( 29 ÷ = 1+ Κ 2 σ 2- σ- 26 σ 3 + 5 σ 2 + 6 σ ÷ = 1 + Κ Ν σ ( 29 ∆ σ ( 29 ÷ 1. Number of branches : max N P , N Z { } = μ αξ 3,2 { } = ΤΗΡΕΕ βρανχηεσ 2. Start/end points : OL zeros : z 1 = -3.36, ζ 2 = 3.86 ΟΛ πολεσ π 1 = 0, π 2 = -2, π 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:...
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problem_11_15 - Problem 11.15 of the Palm textbook(MODIFIED...

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