Question

By using the signal flow graph and the Mason Rule, derive the system transfer C(s) function G(s)= R(S) [4 marks) Analyze the values of Ki and Ky to yield a peak time, Tp of 1.0 seconds and a settling time, T. of 2.0 seconds. [4 marks) (ii) S (BENE 3223) R(s) C(s) K 10 s(s+2) Kas Figure Q2(b)

How to solve this question?

Answer 1

Solution 7 RS) ki 10 S(5+2) Yes KS ki S[s+2) lle R9 - kas There is only one forward path and have gain pi loki Sist 3) There are two loops 4= - K28.10 -10k2 stz $(st2) Lqz 10 ki Slst 2) and A = 1+1+12) = 1+ lok2 + lok) s(sty sta CS Scanned with CamScanner

& Mason's Gain Rule & W (( RLG = Pk DK ki n = no of forward paths, Pc = forward path gain determinant Ak= is the o value for those loops which are not touching Ith forward path 1. PIDI loki 1 OO RO s(s+2) 1+ toka slit) (342) 10k, t. 10 cº RS 8+12+1062)3 +10% Sis+2) + 105K2 tloky 3 10 oc Toansfer function, co RES s²7 (2110kg)stlok A i ciez s?+(2+1012) 3+ 1014 20 CS Scanned with CamScanner

Given Tp = 1 sec and To = 28ec t I sec and y = 2 Wd e un Wd : 3.14 zewn=2 ós pole position at S = & wn I j ad S 25 j 3.14 in CoE = (8+2 +3 3.14) (5+2 -;3.14) $?+48 + 4 + 9.8596 =0 2 > 32+ 457 13.8596 = 0 By comparing ☺ and ① 10k 13.8596 =) Kl = 1.38596 -0.2 and 2+ 10 kq = 4 = K2= / and K2=0.2 hence Ky=1.38596 CS Scanned with CamScanner