Question

Question 3 a) Develop the transfer function X (s)/F(s) of the mechanical system shown in Figure 3(a). Give and explain one example the real application where you can relate with this system. (5 marks) b) Routh's stability criterion is of limited usefulness in linear control systems analysis mainly because it does not suggest how to stabilize an unstable system. Thus, we should evaluate the stability range of a parameter value. Consider the servo system with tachometer feedback as shown in Figure 3(b). Evaluate the ranges of stability for K and Kn. (Note that Kn must be positive). (5 marks) (10 MARKS] f X1 X2 ki kz k2 m m2 1 D2 D Figure 3(a) R(s) 20 (5+1) ($+4) H KA 8 / 10 Figure 3(b)

question 1

Answer 1

RI XL. ma mo mu D im R Fama - m dy dt F=force Di LDz = damping cond916- m=mats --displacemento Va velocity az acceleruhion External force.a) F= Resisting forces IF = md² At Node ① Family +6,21 + k₃ (81-ki) +DI 1.521 de LO X2 X! M DI AD₂ F At Nodec);- k3(2,-X2) - M dan + K2 (XL) +2 dl d6 Apply Laplace for 0.0 F82 = m, 8? Xi 6)+ Ke X,3)+K3 (XIS) – Xa8 ) & Disxis) FC) = ( 874 818 +(1+43)*18) - kg Mus) Kz (X16) – X269).- M 5?x28) + K2X28)+23/6) Ks X7(6): Masa+D28 +ka+K3]] x2(8) К3. ..X23) = M25 *D St Ketky) 210)

k3 1 m.8&destkett FB) = (mis-e Dis +($i+ky)] 88) – K3X$ya = { mus++25+kusky) - King (8) My SPD254*+ K3 FO= (0,8 +0,5 +K1+13)[_musfD23+ka+ks) k? m25² +225 +keiks X18) Me 8² +223 +K 2 tks FO) (Qurulisem co wy) 1925 + D]s? +(Os CK44kg)+Dickz+kg)3 +(ki+K3)C ke +18) hectric equivalent Busbem? - forsce - oroltage dobegy FEⓇ) KYR my o D C Re Riau MMM CL WI -65 Iz practical example R-L-C series circuit,

D RSS kt (3+0) (344) 2oK pyobebet Juins of 26 hes (1-0) 313+1)(8+4) 14 ho krkh 3+1%C9+) fosaadpaths 1-4) 1-0 na S A = 200P sain Aiz Loop gain of non-turecluing Loops Non-touching Loop meciurg. 2008 which is not touching to forward SQ = Lopi Jain product of fax non leucing Leur's Puth. c) 20K 20k $(8+1)(8+) 8(S+1) (s+4)+ 20KXRh8*** $(+1)(344) (8) S'(8+DC8+4) +20 KXkhd'+K chassasiberristic equation =o. 8C8+i) (3+4)+a0kk43 +K 20 3(8758+4) +20k kg 3 +K=0 33 75824 48 + 20% K4 $ +Kzo 33 +59* Hee 480XXX)3 +K TO

31 4+20k km к s! 15(420kKa)-K K for stability anti there is no sign change so Co-Efficients in first colomm must be positive kool 5(4+20k kh)-k 20+100% Kb -Kno 5 5C 4+20kKb) = k 20+ 200k kh= 100 kn -K) =-20 K-lookh=20 K-20 - Kb 100 Ky must be positive sokno K (k>20 -20 so for Stability in So, K320