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3. (30 pts) A dynamic system is represented by the following third-order transfer funct transfer function: 96 G(8) 75 +22)(52 +8s + 32) (a) Explain why a second-order approximation is valid for this system. For a step input, R(s) = . and an output, C(s), determine the following parameters for the non-normalized second-order approximation: (b) settling time (c) rise time (d) the steady-state value of c(t)

Answer 1

Give transfer function 96 GCS)= (S+22) (3 + 85 +32) The poles of the system are obtained by equating denomenator to zero (S+22) (87785+32) = 0 - P is very far from origin, so P2, P3 are dominant poles since they are close to origin.

compared to P2R P3, P, is located Very far from orizin; the behaviour system mainly located depends on the Poles located to close to origin, son it can be appronimated as 96 acs) = 32+85+32. 6 comparing 5 +85+32 with 57zę was twa. weget Win= 32 ) Wn=832 = 5.65 rad/ sec. 9 - 2x5.65 2 & wh=8 = &= 2x16s -201707 ② settering time for 5% error tsa awon agb2x 5705 = 0.25sec Sefteling time for 24 error. ts- en won a on 67x565 = 1sec Ics Scanned with CamScanner

- o T-o to Wd - wnri-&2 = tant (1 ) bana Vimo ?") - us 0.707 o in radius. ugxt = 0.785 - 3.14-01785 2,354 tra h - 0.7072 4 tr=01589 sec @ Steady State value Steady state Lt sx G (s). - 96x05 = 1.5 sto 96 0'5 .- so pr 8²85 +32 M - 0²+880 +32 -