Question

Question 3 a) Develop the transfer function X2%)/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 K (Note that K, must be positive). (5 marks) [10 MARKS) X1 ki w m2 D Figure 3(a) R(s) C's) 20 (s) (+4) KA Figure 3(b)

Answer 1

given mechanical system X ki MA M21 2 파 P2 for mass (m) ; on Im Evet = Md²x dta 41 for Spring Constant; F= kx L for damper constant Bdx df dash pot 8 oor ret Ha & ret B drawing free body diagram for min f dki foi M, fk3 L) By D'Alembert's principle :sum of applied forces and herbal forces is zero for any virtual displacement of the system f. fmi + tk, tfort fra f= M, d²x, + ks 8 t Dvd + K3(x,- *2) b converting into s donais F(s): M, S x1(s) + ki XICs) + DSX.(s) + kz CX, C5)- X9(S) F(S) = Xics)[mis?+ +58,+ k,+R3] – x26s)[43] df 2

drawing free body diagro orom for M2 k fma M2 fk3 for fme +fts +ftet forro M2 d ²x₂ + K3(x2 + x) +k2X2 + B2d42 ett dt2 4 converting into s-domain My S² xqts) + SD₂ + K3 tk 2]*g(s) = Kg +, (s) 4 [M2 S + SD2 +63 tkz] X215) = K3 x1(5) in order to find *I'S) FCS) Substituting 0 in 0 FCs) = xi Cmis?+ Dis + k, +63)– K3 x165) k3 M2 S ² SD2+ kg 4kg f(s) = x1(s) mis tais +ki+tz) (M25+502 ++z+13) – (kal? M25² +502 + kg tkz 3,0 must XICS! ACS) M₂ S²+ SD at K₂ +€3 (mis'+Dis + kit*3](mas +502 +Ka+kz) – (kz)?

20 +1 (514) CCS) 092 4 finding on stare function for above block diagram Ris). G1s >() G (3) CCS) Res) It Gishes) HOS) RSS) 2ok 5+1)(S+y)+206-F CC) Then CCS) 20k SCS+)5+4) + (20%*) S + 206 20k 3 +55²+45 + (204.&n strok 2ok +55²+(4+20kkals 120k 4. The system will stable if roots of characteristic ecucion ore positive 83 + 55²+(4+20kkalst 20k 4 By Routh-huswita stability criterions L it The first column of coeffecients are positive in Routh-herwitz Table Then system cis positive and stable

! 4 + zokka S S2 5 Eܩܐ 504+2017) - 20k 20k 5 (4+20k kn)-2ok ১০ / 20k so 5 k১০ 5(4+20kka) - rok > 4+ zokka > 4K 2ok kn 46-4 KA > K-1 10 5k stable is Lirthe ronges of k and in for the system to be k zor kn K-1 5K