243 Control Systems RL_Tut

243 Control Systems RL_Tut - University of Wales Swansea...

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1 University of Wales Swansea SCHOOL OF ENGINEERING EG 243: Control Systems Root Locus 1 Plot and calibrate the root locus diagrams of the following systems as a function of the open-loop gain K: ( 29 ( 29 ( 29 ( 29 ( 29( 29 5 1 2 ) ; 1 2 ) ; 1 ) + + + + + + s s s s K c s s s K b s s K a 2 Use the root locus technique to determine the range of values of K for which the system with open-loop transfer function G o (s) is closed-loop stable. ( 29 ( 29( 29( 29 8 2 1 + + - = s s s K s G o 3 Plot the root locus of ( 29 ( 29 ( 29 9 1 2 + + = s s s K s G o as a function of K. 4 A position control system has plant transfer function ( 29 ( 29 4 4 + = s s s G . A cascade proportional plus derivative compensator with transfer function D(s) = K P + T D s is added to the system so that the closed-loop poles of the compensated system have a natural frequency of ω n = 10 rad s -1 and damping ratio ζ = 0.5. Determine the proportional gain, K P and derivative time, T D required to achieve this response. Calculate the steady-state velocity error of the compensated system. Hint: use the root locus angle criterion to find the location of the compensator zero when the closed-loop poles are positioned as specified.
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2 1 Plot and calibrate the root locus diagrams of the following systems as a function of the open-loop gain K: ( 29 ( 29 ( 29 ( 29 ( 29( 29 5 1 2 ) ; 1 2 ) ; 1 ) + + + + + + s s s s K c s s s K b s s K a See L17-20. Root locus sketched using 5 rules. L19, S7, S10, S12, S14. a) See L17, S14-17, which is a similar problem. ( 29 ( 29 ( 29 ( 29 ( 29 25 . 0 5 . 0 ; 1 2 2 - + + = + + = + = K s K K s s K s G s s K s G c o At K = 0.25 there are concurrent poles at σ = -0.5, which is the break-away point Rule 1,2: there are 2 branches; n=2; poles at s = 0 and –1. Symmetry Rule 3: the root locus is on the real axis for –1 < s < 0 Rule 4 there are no finite zeros; m = 0 2 infinite zeros The branch start and end points are: 1 at [0, - 0.5 + j ]; 2 at [-1, -0.5 - j ] Rule 5 using L19, S14/15 the asymptotes are at 90º and 270º with real-axis intercept at σ 0 = ( 29 ( 29 5 . 0 0 2 0 0 1 - = - - + - = - - m n zeros finite poles finite numKGH=1; denKGH=conv([1,0],[1,1]) denKGH = 1 1 0 rlocus(numKGH,denKGH)
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3 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 K s K s s K s G s s s K s G b c o 2 1 2 ; 1 2 ) 2 + + + + = + + = Rule 1,2: there are 2 branches; n=2; poles at s = 0 and –1. Symmetry
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This note was uploaded on 05/17/2010 for the course ENGINEERIN EG 243 taught by Professor Murphy during the Spring '10 term at Swansea UK.

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243 Control Systems RL_Tut - University of Wales Swansea...

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