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me352_hw7_soln - ME 352 System Dynamics and Control...

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1 ME 352 System Dynamics and Control Homework #7 Solutions [1]. (a) = + + + + = + 0 2 3 5 1 ) ( 1 2 3 s s s s K s KG ( ) 2 ( ) 5 ( ) 3 2 ( 1)( 2) Q s s P s s s s s s s = + = + + = + + Poles: 0, -1, -2 (n=3) Zero: -5 (m=1) Asymptotes: K > 0: 90 270 o o , , Intersect of Asymptotes: 1 1 3 ) 5 ( 2 1 = - - - - - = α Breakaway-point Equation: s s s 3 2 9 15 5 0 + + + = , Breakaway Points: - 0.4475, - 1.609, - 6.9434 For j ω -crossing, construct Routh array: K s K s K s K s 5 : 3 ) 2 6 ( : 5 3 : 2 1 : 0 1 2 3 - + Stable if K <3. At K =3, characteristic equation becomes: 0 15 5 3 2 3 = + + + s s s , which has roots s = ±√ 5j, s =-3.
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2 MATLAB Plot: Root Locus Real Axis Imaginary Axis -6 -5 -4 -3 -2 -1 0 1 -20 -15 -10 -5 0 5 10 15 20 0.03 0.065 0.1 0.15 0.2 0.28 0.42 0.7 0.03 0.065 0.1 0.15 0.2 0.28 0.42 0.7 2.5 5 7.5 10 12.5 15 17.5 2.5 5 7.5 10 12.5 15 17.5 20 (b) ( ) ( ) ( ) 2 2 ( ) 1 2 ( ) 2 2 Q s s s P s s s s = - + = + + Poles: 0, -1 ± j (n=3) Zeros: ± 1, -2 (m=3), Since Q s ( ) and P s ( ) are of the same order, there are no asymptotes ( n - m =0) Breakaway-point Equation: 6 12 8 4 0 3 2 s s s + + + = , Breakaway Points: - 1.3848 For j ω -crossing, construct Routh array: K s s K K s K K s 2 : 2 : 2 ) 1 ( 2 : 2 1 : 0 1 2 3 - - + - + For K >0, there will always be one unstable root (real-axis branch)
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3 MATLAB Plot: -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 0.22 0.42 0.6 0.74 0.84 0.92 0.965 0.99 0.22
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