Question

A system having an open loop transfer function of G(S) = K10/(S+2)(3+1) has a root locus plot as shown below. The location of the roots for a system gain of K= 0.248 is show on the plot. At this location the system has a damping factor of 0.708 and a settling time of 4/1.5 = 2.67 seconds. A lead compensator is to be used to improve the transient response. (Note that nothing is plotted on the graph except for that one pole location at -1.5+1.49j). Please draw the RL yourself of the uncompensated system and the compensated system. • Considering the root locus for the uncompensated system, and that the system has 2 poles moving from K= 0 to infinity. Where is the other dominant pole location for the same system conditions (K = 0.248)? List two other controller configurations (other than lead) that could be used to improve the transient response of this system Choose pole and zero location of a lead compensator that can be used to reduce the settling time to about 1.5 seconds while maintaining a damping ratio of 0.7. Indicate the pole zero locations on the plot and show the effect of your choice of the lead network pole/zero combination (root locus plot change, approximate transfer function of new system) Indicate the desired closed loop pole position on the negative half of the plot. Root Locus 0.91 0.33 0.72 0.58 0.4 30.96 System:g Gain: 0.248 Pole: -1.5 +1.491 Damping: 0.708 Overshoot (*): 427 Frequency (rad/s): 2.12 10.99 Imaginary Axis (seconds) 30.96 0.91 0.83 0.72 0.58 0.4 -10 Real Axis (seconds)

Answer 1

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fenetren. lok ulampesated suten sol openloop trausd a aus = 517 (812). 0-700 at k=0.240 , &= 0.708, rond 5 2 on=1 un= ls - 2-11 Redes now lead Network have the tracy Patenction tes = 549 when qzb. Stb Root locus of uncompanyated syst en ay at no.ug poles - 2 S=-1,- 2 = 1100 (2p sample of asymptotes 92 =90, 278 Confroid ise Elle(R) - Rel2i) P-2 and break away point characteuntean 523512 tloka on - - 2-1-0 =1.5 Not Exist no og 2 en 25+3 +1o4k.o-dk.0 5 -15] ds

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