Question

1. A system with unity feedback is shown below. The feed-forward transfer function is G(s). Sketch the root locus for the variations in the values of pi. R(9)+ 66) 69? Fig. 1: Unity-feedback closed-loop system G(s)= 100 s(s+ p) 2. The following closed-loop systems in Fig. 2 and Fig. 3 are operating with a damping ratio of 0.866 (S =0.866). The system in Fig. 2 doesn't have a PI controller, while the one in Fig. 3 does. Gain Plant R(S) RS) + Q EN CE(s) a < (s+1)(s +2)(s +10) R(s) Fig. 2: Closed-loop system without PI controller Compensator Plant R(S) + K(s +0.1) (s +1)(s +2)(s +10) Fig. 3: Closed-loop system with a practical PI controller, a=0.1 a. Please use Matlab to find the intersection point between & line and the root locus of the system in Fig. 2. Then find the K value and one complex closed-loop pole corresponding to the intersection point. Calculate the steady-state error. Show the Matlab codes in your answer sheet. b. Please use Matlab to find the intersection point between & line and the root locus of the system in Fig. 3. Then find the K value and one complex closed-loop pole associated with the intersection point. Compare the complex closed-loop pole with the one you just found in task a. Are they very different from each other? Calculate the steady-state error. Show the Matlab codes in your answer sheet. c. Use Matlab/Simulink to plot the step responses of two systems in Fig. 2 and Fig. 3.

Answer 1

>> G = tf([0 1 0], [1 0 100]) G = --------- s^2 + 100 Continuous-time transfer function. >> rlocus (G)

75=-10 is the break in point. ow. ila t b Root locus.

1. 413) = S(SHP) I characteristic epce?, => 1 + G(s)H(s) = 0 1+ 100 s(s+P) s s'+P,5+100 = 0 = 1 + P, s S+100 Now, he es Zero, s=o Poles, P = tj 10 · Ceufoorid, o · No. of asymptotes = 2-1 = 1 0, = (2x +!) X180° - 186, for kso · Room © P, = -(87100) et -[s(25) – (**)). =) S = -10,10 2-1