Question

R(S) + + 1 C(s) 10 s(s+2) KS

For the control system in the figure, a) Draw the change of closed loop poles relative to K while K changes from zero to infinity. b) If available, determine the K values that make the system damped, critical damped and excessively damped c) Is it possible for any value of K to be Mp = 5% and ts = 2sec?

Answer 1

Sol- Given ISLS+2) Sista) SiS+2] itt yks SEST s+25+ks st 25 ks Sista RCS) (15) 10 3² +25ths C. (5) +25+ks lo RIS) S +5+kg+ Staseks co lo RUS) R(S) st (2+kisto compared with standard second oder Tronsler function, (s) Šta deinst high w = 10 who vio sud/ser 2. Wn= 1.tk 2 X VTO = 2tk 2 tk To for critical damping, 2-+k 2 tk = 2510 - 6.324 k = 4.324 cortical damping, x>! 2 tk

2 tk > 2010 ky 4.32 ht, k=5. for under daming 2 tk 2510 2 tk C2510 2+k 2432 K L 4.32 tet, k = 4.

characteristic equation, st25tks tlo=0 It ks Š+25 to The change of closed koop poles relative to kuhen k changes from o to infinity is called soot locus j 3 j3 angle of asymptotes -ji Polecat S = -2 2 1 (2) 2 - 4x10 -2 + 14-40 s=-27536 = -21 jt tj3 2 Here, poles =2, Zeros = 1 centroid = e poles-Ezeros p-z angle of asymptote, o = 1180(22+) p-2 ,01 (P-2) 180

re zeto on hare seen that only one the real anie cet s=0, So all Points on the left of 5=0 must lie on the root holly one a symptore that of on angle of 180 and intercepts at centroid -2. Breakaway points :- ITGES)H(s) = 0 Št2stks +10=0 ks - - (5+25+10) - (5+2stlo) S dk as (5 725+! +101 - (5 (2512)-15 S2 25 429 mg 25-10-05 10 20 15 10 S = 155o = + 3.162, Ass=-30182 lie, on root socy it is break away Point. 33 -3-16 O 53

to = i wna Vlorad/sec - e -0.000 X 100 Given, Settling time, ts=2 sec overshoot Mp=5% for 2% tolerance, 4 pun = 2+k 2010 2 k etk XVTO RUTO 8 2 s #2+k=4 2 tk X=4 -0.6325 aro Vio Ymp = ē V-22 100 - 0-6325x 3-14 k=2, 1-98for Atoo 07745 -2.564 xloo = 0.676X100=708% But given Mp = 5% K value not possible to matching both Mp and to