# HW2 Solution - HW 2 Solution 1 2 25.4 3 4 5 Poles 0-6-9 no...

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HW 2 Solution 1) 2) 3) 25.4% - - -

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4)
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 ± = -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.
6. Angles of Departure ! " z # ! " p = # 180˚ ! " z = \$ (S + 3) = tan # 1 2 0 % & ( ) * = 90˚ ! " p = \$ (s + 1) + \$ + 3 + 2 j) + \$ (s + 5) + " d = tan # 1 2 # 2 % & ( ) * + tan # 1 4 0 % & ( ) * + tan # 1 2 2 + , - . / 0 + " d = 135˚ + 90˚ + 45˚ + " d = 270˚ + " d 1 90˚ # (270˚ + " d) = # 180˚ # 180˚ # " d = # 180˚ 1 " d = jw-axis crossing start with char. Eq'n. s 4 + 12s 3 + 54s 2 + 108s + 65 + K(s + = 0 let s = j ! " 4 # 12 j 3 # 54 2 + (108 + k ) j + (65 + 3 k ) = 0 imag : # 12 3 + + K ) = 0 real : 4 # 54 2 + (65 + 3 K ) = \$ } can be solved for k & 7. 4 3 2 1 -1 -2 -3 -4 X X -9 -8 -7

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HW2 Solution - HW 2 Solution 1 2 25.4 3 4 5 Poles 0-6-9 no...

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