Question

Assume unit-step response of the closed-loop system with K = 2. Adding which of the following compensators improves the steady-state error without altering the transient response? Step Response 1 0.8 0.6 data 0.4 0.2 0 -0.2 0 0.5 1 1.5 2 2 Time (seconds) Gc(s) = s+0.3 8+3 O Gc(s) = s+0.03 s+0.3 O Ge(s) 8+3 s+0.3 O Gc(s) = S s+0.3

Answer 1

Q 닉 - 6 . 4 O .5 1.5 2 Bolation To Improve Steady state exror and Steady state response with out altering the transient response we use Lag Compensator" Genral form of hag compensator Ge(s) (+ ST) ( + 8 ST) for Lug compensator, BT 7T LB > 1 or Tw Pole zero plot of X - BT

From oltion A Gels) = [S+:3) $_+3) 3 3x (Ls +1) (5+1 from this: T = i 3 and BI = B = 0.) 3 XT IX:3 3 so 7 above_compensator is not so it can not be Ley componsator use. OP Hon_(A)_is_ wrongo From option B Ge(s) 03 St1 st St. 3 0.03 (tig. S + (tes +1) '3 from above, and BT= 03 TE +3 '3 Ba I I X:03 0 1 • BT .3 above Compensator_is_not bay. Compensatox so it can not use option (B) is wrong B<1 So 7 be

Option ! - Gels) = s +3 3 (1) )) S + .3 -3 (ES+1) from above T= 1 BT = to B 3. 3 10 • 3T B>L So above Compensator is "Lag compensator so it can be use to reduce Steady state error. Option (c) is Correct. Option (D) :- Gc(s) = S St13 Zero Since the at origin and and Pole Left Side is awowy from origin on ay it is not hag compensator. so option (P) is wrong. S Plane so
Note:- option (C) is correct answer . If you have any query regarding to solution please feel free to ask.