Question

Question 2 a) Consider the control system in Figure 2(a). Determine the transient response characteristics (rise time, peak time, maximum overshoot and settling time) and the steady state error for the system. (2 marks) b) To improve the relative stability, the tachometer feedback are employed as shown in Figure 2(b). i Determine the value Kso that the damping ratio of the system is 0.5. (1 % marks) i. From the result obtained in (), evaluate the transient response characteristics (rise time, peak time, maximum overshoot and settling time) and steady state error. (1 % marks) i. Sketch the output response for the both systems in Figure (a) and (b). (2 marks) iv. State what type of the controller that can replace the tachometer feedback which can give the similar output response. Discuss your answer. (3 marks) [10 MARKS R(s) 10 $(3+1) R(3) 10 s+ Ky (b) Figure 2 (a) Control system; (b) Control system with tachometer feedback

Answer 1

Solution & 10. Given GI(S) = S(SH) Transfer function cul RCS) GLS) I+Gls) (3) R(S) To SCSE 1 + To SestD ((s) RUS) characteristic egh, By comparing S²+5+10 ²+26wns twn²=0 s²+ Stlo Wh o rad/ See 2 & 4n = 1 + {= = E= 0.158. 2510 Transient response characteristic Rise time (tr) = TT-0 WhJ1 - 82 where = fent

0= tant tant ST-0.15872 0.158 0 = lill - tx= T-1.41 T-d wnl-62 = Sio Ji-0.1582 tr 0.55 sec Peak Timelte) = T OnJ1-62 ㅠ Slo fi-(0.158) 2 1.006 See T 3.122 tp = 11006 see رتن Maximum oven shoot (६॥ JI-82 mp = &= 0.158 - (0.15817 mp e ме? 6049

mp = 0.6049 mp% 60.49 qw Setteling time (78) an 2% basis ts= 4) & Wn ts- 4 (0.158) Blo ts= d.oo5 sec on 5% Basis ts= 3 & wn ts= 3 (0.158)J lo ts= 6.004 See steady state enror (ess) for unit step input ess = 1+кр whore kp = Lim Gis)

kp= Lim so G(S) Kp = Lim sto lo sest) Kp= ess O Tike Ito esso for unit ramp input essa ku= Lim SGCO Sao kv = Lim so 10 SISH) kv= 10 > essa lo Therefor 0 Rise time (tr) w Peak time (t) = 0.55 sec 160o6 sec بين Maximum Oveel hoot (mp) = 60.49% (1) settling time cits) 8.005, 6.004sea! Steady state error (ess) for Unitstep=0 unit damp = 1 To Auswer 22

(Ь for figu. 2(b) To S+/ lis - Kh GO) (10) 16001) 5 Kh - Gis 10 (St) + lokh → characteristic eqh I+G6S2 = 0 | + =0 10 s{ (5+1) + lokn 3 32+ S citlokn) + 10 = 0 By comparing s²2eWnst wn²=0 Wn = to rad/ See cil Given 2ɛwn (It lokn) 2x0,54 Jo = It lokh

3.16 Itlokh ☆ Kh = 0.216. Transient respons charcutenstic Rise time (tr) T-6 Gon 11-६२ dan Shah & -0,982 toa TT-0.982 737 to = 0.91 See - 1.325 2.37 Peak time (tp) = I won JF- 62 t- 1.325 see Maximum Peak overshoot (mp) = xlu e Troish) x100 16.3%

Selling time te) = 4 & Wn 3 Ewn 3 0.5XTO 4 0.5+To 4 1.581 ナ B 1.50) = 252 see, 1.898 see Steady state error (ess) ess The V = Lim سجلماهره Ку = 2 stess for unit step ramp 1 [ Esto tlokh sto Kr= lo T+loxkh (SH) tlokh 10 7+10(0.2162 Kr= 3.16 ess 0.316 kr 3.16 Kr Ang