Question

A unity feedback system is shown in Fig. 1. The closed-loop transfer function ?(?) of this system is given as ?(?)=?1?4+2?3+(?2+1)?2+?2?+?1.

1. a) (20%) Using Routh-Hurwitz criteria, find expression (in terms of ?1 and ?2) and range of value of ?1 and ?2 such that the above system is stable.
2. b) (4%) It is desired to achieve steady-state error of less than 0.3 with a unit ramp input. Find an additional constrain in terms of ?1 and ?2 such that the requirement for steady-state error is satisfied.
3. c) (3%) One decided to make a guess on ?1 and fine tune ?2 until the steady-state error requirement in b) is satisfied. In the following cases (i and ii), use expressions of ?1 and ?2 to verify whether the requirement is satisfied and give the corresponding range of value of ?2. i. ?1 = 10.
4. ii. ?1 = 150.
5. d) (3%) Do you think the parameter tuning method in c) is efficient? Explain why or suggest another method.

A unity feedback system is shown in Fig. 1. The closed-loop transfer function T(s) of this K system is given as T(S) = 54+253+(K+1)52+K,s+K7. R(s) + Y(S) Ki 1 s(s+1)(s2 +s+K2) Figure 1: A unity feedback system a) (20%) Using Routh-Hurwitz criteria, find expression (in terms of Ky and K2) and range of value of Ky and Ky such that the above system is stable. b) (4%) It is desired to achieve steady-state error of less than 0.3 with a unit ramp input. Find an additional constrain in terms of Kį and K2 such that the requirement for steady- state error is satisfied. c) (3%) One decided to make a guess on K1 and fine tune K2 until the steady-state error requirement in b) is satisfied. In the following cases (i and ii), use expressions of K1 and K, to verify whether the requirement is satisfied and give the corresponding range of value of K2 i. K1 = 10. ii. K1 = 150. d) (3%) Do you think the parameter tuning method in c) is efficient? Explain why or suggest another method.

Answer 1

Solution Given that - the closed loop transfer function Tis) - Ki 5+233 + (2+1) 57 kgstki The unity feedback system. Pe t a Sito us Tes)= Y(S) : KI Res) $4263+(65+)87 Stk, The characteristic equation. s+as+ (x3 +00$+Kgs+kę = 0 Routh - Hurwitz criteria - s i (83+1 Kg be low • oo * Š 2(6+1)- kg K, 0 ar-2167 wo or

dky ta-kaca 2= 12th to be stable for the system K, 20; 2557aly yo ; kita so > (1572 ) 65 324 hºol> 13,54% :{}, 24 ... This is the condition for the system to be stable ♡ Steady state error ess 20-3 for unit ramp input B5 = 2 1 - 1 tv - velocity enor Constant. = lim SG(S) S)0 so s(s+1(527s+ky)

w ess <0.3 - 4 0 ) w - [kg 20:3k need to be satisfied to obtain a steady Constraint State enor 20.3 10) 11 K10 kg 20.3k, range of Kg = (-2,3) ply k, = 150 , us range of kq = (-2,45) for k=10; K = 6-2,3) The condition to taky > 44 is not satisfied for any value of Ky in the range, Hence System becomes unstable, so this is not a suitable value of a qkg.

por = 150 : 5 = 6-2,45) 1. The condition Kit2k >4k, is satisfied for some value of k in the range (+2, 45) 2+2kg > 600 ķ > 23.52 ; Ķ <-as-52. hot feasible. Range of g = (23-52; 45). (dit so; i we consider Parameter turing method in c! which takes into account only conditions obtained. in 'b! We may land in system which is unstable; Hence it is not an efficient method. The following and conditions should be satisfied for ker kg i K>0 - 0 Ka 20:3k, - Ky +2ķy >u2,-0 K = 0.3k - k = K2 0.3

0:3 K3 +2ky - uke 30 ka (ka-11:33)>0. Ky - the Ky > 11:33 So; the find. above system take k>11:33 ke from U