HW12soln

HW12soln - ME 375 PROBLEM 1: (30%) GIVEN: HOMEWORK #12...

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ME 375 HOMEWORK #12 SOLUTION Spring 2009 PROBLEM 1 : (30%) GIVEN : characteristic equations: 2 1 10 ( 2)( 10 34) K ss s += ++ + FIND : y roots of this characteristic equation when K = 0 y K for which roots cross the imaginary axis y root locus sketch y range of gain for stability SOLUTION : open-loop poles (for K = 0): 2, 5 3 j =− =− ± ( ) 3 p N = imaginary axis crossing: 32 12 54 68 0 sss K + + = 12 54 68 0 jj K ωω ω −− + + + = real: 2 12 68 0 580 KK −+ + = = imag: 3 54 0 7.35 = ⇒= ± root locus sketch (see next page) range of gain for stability: 0 < K
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Sketch: Matlab plot:
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PROBLEM 2 : (35%) GIVEN : The following transfer function: () 2 22 1 2 Xs Fs ss = + with . ax = ±± FIND : (a) Transfer function relating A 2 ( s ) to F ( s ). (b) A block diagram that indicates how you would generate an acceleration feedback system. (c) If cp Gs K = , closed-loop transfer function between 2 desired a and a 2 . (d) Controller transfer function for closed-loop damping ratio to be nonzero. (e) If 1, a K = the controller gains for the controller in (d) that achieves a settling time for the response to be within 2% of final value of 2 seconds and 16% overshoot for a step change in desired acceleration. SOLUTION : (a) = 2 A X s = 2 2 11 (2 ) 2 sXs s =⇒ = ++ 2 2 1 2 As s = + (b) (c) GK = 2 2 2 2 2 1 2 desired c a ac Gs s Ts K KG s s + == + + 2 2 ap KK sK K = This closed loop system is undamped. (a) + Accelerometer Cartesian Robot K a Potentiometer Controller G c ( s ) 2 K a A 2 2 desired A (c)
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(d) () cp d Gs K Ks =+ Then, 2 2 2 2 ap d ad d KK K s KKs KK Ts sK K s K K K K s + + == ++ + + (e) 2 1; ( ) 2 dp a Ks K KT s s K + + 2 22 nd n
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HW12soln - ME 375 PROBLEM 1: (30%) GIVEN: HOMEWORK #12...

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