Question

Problem 1. (20pts) Consider the closed-loop system shown in the following figure. + NET 1 RO (s +0.25)2 (52 +0.01) s(s+1) (a) What is the condition on the gain, K, for the closed-loop system to be stable? (b) What is the system Type with respect to the reference input? (c) What is the system Type with respect to the disturbance input, W? (d) Prove that the system can track a sinusoidal input, r = sin(0.1t), with zero steady- state error.

Answer 1

solution :- RIS) * 15 +0.25) s2 tool 9 Lets wro, then RISO RIS 52+0.01 ci forward transfer function of the syetem - G15)= kls+0.25) 2 515+1) (52001) This is imity teedback system, i H182 = 1 now, Characteristics eaun of 878tem. 1+ GT 18) H15) = 0

1+ R (S+0.2592 - + s²+0.01 sist) s(s+1019270.01)+ KIS+0.2532 = s(+0.015 +52 +0.01) AR15?0,0625 +0.55)0 54 +0.0152 +53 +0.018 + Ks+0.0625K+0:5KS =0 St+33 +5270.014 K) +56.5K+0:01 +0,0625R now , by Routh criteria to find stabilith st 53 4 K +0.01 1 0:01+0.51 II R70.01)-110.01+0.5k) 0,0625 K 0.0625k 0 6 i en a osk al 0.53 (0:01+0.Sk) - 1 x 0.0625 K o o.sk 0.5k-olls = co (6.5k-115) X 0.0 625k- o O.SK -0.115

A & we know, for stability of element of firat same sign. ii 0.5k so the system all the column should have 0.5K -0.115 7o - K> 0.115 - = 0.23 Ik »23 Hence, K for the stability should be greater of the system than 0.23. b) type of 8ystem with wzo i G18) HIS) = R 15+0.25)? S15+1(52+0.01) type of system defined as the open toob GIS) HIS) has no. of pales at orgin. For this system. Pole & ad orgin P= 1 .. this is type 1 system.

c) system type with respect to disturbance make Ro R=0 KIS+0.257² s20.01 SIS+1) Rearrange this system Y52 SIS+1) R15+0.25121 S270.01 Sist IH G15) = 1 , Ho - K15+0.25) 2 s² tool G15) H18) = R15 +0.25)2 S18+1)(+0.01) .: +ype of Ye I no. of podes cet orgin i type=1 Hence, it is type-1 848tem.

d) prove that the system can track a sinu - soidal input, &=sinolt, with zero steady State error. As we known Els) = RIS) If GISHIS) ess = lim SRIS) so - IfGls)HIS) RIB) = L [sino.id] s2+0.12 s2+0.01 take reference from past a 6718) = R15+0.25)2 ,HIS) = 1 S15+) | SZA ess = lims. Oil So s 2+0.01 1 + K 18+0.25 )2 SIS+1)(S2+0.01) lim ools 15+1) S-o SIS+D (52+0.01)+R(S+035)2 i less = 0 Hence steady state eroor is zero for this sinusoidal input