Question

We would like to approximately find the time response of a system with the following transfer function: T(s) - 25s + 45 (0.5s + 30)(10s + 19)(s2 + 0.45 + 1) a) (35p) Find the 2nd order transfer function which behaves closest to the above system. Justify your approach. b) (35p) Determine the type of the system response and find all of the transient response parameters. c) (30p) Draw the transient response of the system indicating all of the transient response parameter values on the graph.

Answer 1

Solutions :- 25s tu5 =-60 Given Toonsfer function TOS) = 6.55+30) (105 +19) (s?+0.45 +1) poles of the T(s) are at $?to. Usti = 0 -0.4+ Jeu)?_4 S1,2= 105+19=0 (S=1.9 51,2=-0.4 1.96j pole location of following on the S1,2 = -0.2 +0.983 Splomes jo Sa -30 05 2: 2 Х * -60 X -1.9 o Х By approximation, (0.5s+20) (105 +19) of S=lig, -60 poles are very far apost from the jw quis. Hence these poles are not dominant poles. These are not describe / affect the main charactersticas. only secouma orden S+0.45 = o equabon are dominating. Hence comsider it only and regled then other poles.

Hence, approximate Troms ben function 25stus T(S) S2+0.us+ b) Type of the system = mo.oy geroes at the origin Type=o charactesslic egy sto.us+1 = 0 Tomsient response posenmeters ase delined og : Rise 1) time :- Tda 7- tant Wasirea Here won=); P=0.2 0.2 Td = π- tam T-0.78 1,-0-2² 11-(0-2)2 0.98 pa ITd=2.40 sec 2) Settlingtime: Tsa A = A = 20 sec Econ 0.241 peak 3) time : ㅈ 3.14 Top- = 3. 204 seci wall-qz 0.98 4) maximum overshoof ६ = 0.5266= 72.66% mp= ୧ ہے -ول

f(t) 81 overshoot 72 63 54 45 36 settling time 27 18 0 2 6 00 10 12 14 16 18 t[s] ) peak time