HW2_Solu - HW 2 Solution 1) - - - 2) 25.4% 3) 4) 5. Poles:...

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HW 2 Solution 1) 2) 3) 25.4% - - -
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4)
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5. Poles: 0, -6, -9; no finite zeros. Breakaway point: K = - σ ( σ + 6) ( σ + 9)= - σ 3 - 15 σ 2 - 54 σ dK d = -3 σ 2 - 30 σ - 54 = 0 = σ 2 + 10 σ + 18 = 0 σ = - 10 ± 100 - 72 2 = -5 ± 7 = -2.35 or -7.65 and –2.35 is the break away. or Using transition method + = + i i p z 1 1
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9 1 6 1 1 0 + + + + = After simplifying 0 18 10 2 = + + solving it, = - 2.35 Angle of asymptotes: Real-axis intercept: a = poles - zeros n - m =(0 –6 –9)/(3-0)=-15/3=-5 Angle: a = (2 k + 1) n - m = 3 ) 1 2 ( + k = 3 5 , , 3 Root locus crosses the jw-axis at j7.348 with a gain of 810.
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6. Angles of Departure ∑θ z -∑θ p =- 180˚ ∑θ z =∠ (S + 3) = tan - 1 2 0  = 90˚ ∑θ p =∠ (s + 1) + (s + 3 + 2j) + (s + 5) + θ d = tan - 1 2 - 2  + tan - 1 4 0  + tan - 1 2 2 + θ d = 135˚ + 90˚ + 45˚ + θ d = 270˚ + θ d 90˚ - (270˚ + θ d) = - 180˚ - 180˚ - θ d = - 180˚ θ d
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HW2_Solu - HW 2 Solution 1) - - - 2) 25.4% 3) 4) 5. Poles:...

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