Question

Question No.3: (4 taries) For the system given in the following ligure, determine the value of which yields critically closed loop response for the unil step reference input. Compute the values of Wn....T, anu ketul thy stco response of the closed loop sysiten R(S) + Cisi 50+ ********************* GOOD LUCK *************

Answer 1

Given system is. R(5) 1 sisust TCC) It is a negative feedback system > c) = Goes) - 5STK form of and order control compare with standard System cc 7 Š+ 246nsten 7 WAFIK] + (28.955 given the system is critically damped system. for critically damped system [&=1] 2&wn=5 2(1) wn=s → Wn==> [on= 2.5 Dvadisec 7. WzK K=(2.5)> K= 6.25),

- - cls) R(S) 6.25 ŠISSIK S45576.25 input is a step input R(5)=1/5) given >> crs) = R65) * 6.25 _. $755+6.25 = 6.25_ 5 (5755+6.25) - 6.25 S(5125²

By using partid fractions → CCs) = 6.25 - 9 + 9+253 Stasit s(s2.5) = (AC stas)'+b(s) (s+2.5)7c5 = 6,25 Put são (A=D compare Š coefficients > A+B=0 ) B=-A = (B=-1) Put S=-2.5 = c(-25) = 6.25 => C= -2.5 CS) = $ - como erat = [(Ct) = [ (ers)] and following plot it is given by the ACCE) / 1 teo

→ delay time (td) is given by : ta = 1+0.76 wn 1=1+0.71) Ita=0.68 sec] we already know wn=2.5], [a=1] ₃ settling time is given by Ts= 5 un for 8% tollerance band | Ts = 4 con for at tollerence bend (2.5) → Tsaisessa 1s = a sce] => for o% tolles and bead t=4 Ts = 1,65€c) for 24. tollerance band. (2-3) Wd = on iTGE Wd=25V1-ci wdzo radisec + peak time (te) = L = L + tp = 0 → Rise time this given by Tr= TT-calle) TV = TP-casli) Tr=0, ما