Question

For the system shown in Figure 1, a step torque is applied at 0,(t). (a) Develop the transfer function, G(s) e2(S) T(s)

Answer 1

Sol a) Given Mechanical Rotational system, TIL 9,2) OLE) 1.67 kg-ig (70002 MS 1-92 ne-mprad 153 N-mistrad Tranejar tunction 2 TLS) Free-body diagram Ji TJ B 0 I TJ, = J, 028, de ² 1.07 doo, de ² TB = B (0, -02) 153 18,-2) at By D'Alemberty principle. T(A) = Tj, + To 1-07 aro, + 1.5] 16, -6,-) Taking laplace transform T(S) = 1.07 5²8, 6) + 1.535o, (o) - 1.5380, U)

Free body diagram for In TB O JA - TB=B d 10-o) 1-53d (0₂-0) dt at TK = KO2 1.9202 By D'Alembert's principle. TB + Tx = 0 1.53 a (02-0) + ke2 at 1953 (0₂-0) + 1.920 = 0 Takking taplace transform 1.5350₂ S) - 1.530, (s) + 1.92 '0₂) = FO (1.535 +1.92) 0₂ (1) - 1.550,(6) 50 1.5350,-) D.U) = (1.535 +1.92) O₂ W) (11535+1.92) 1.535 02 (8) Substituting O, (s) in eq T(S) = (1.0752 + 1.53) 0,01) 1535 0,41) (1.076² +1.538) (1.575+1.92)84) 1.5350₂ (6) 1.535

T(S) (10781153 83 + (1.67 x 1.92)5² + (1534155) + (1-53 x 1-92)s - (536)s) 1.535 1.631153 + 2.05445² +2.93765 TI) = 026) 3 1.535 T(s) $ (163715 + 0.05445 + +2.9376 1.58 O₂ (s) 1.53 020) T(S) 1.63715242-05445 +2.9376 (یارا 1.53 TO) 1.6371 [s²+ 1.25495 + 1-7943] O2 (s) TIS) 0.9345 s²+1-25495 + 1.7943 The characteristic equation of tunction is the cabove transfer s+1.25495 + 1.794370 Comparing the above transfer function with standard form S+ 2 Gunst wont to

on = 1.7943 2 gown 1.2549 n = 1.3395 rad see 12549 24 WD 1.2549 2x 1.3395 9 = 0.4684 Percent over-shoot % Mp = e e salsię 0.4684871 1-(6-4684)² 1.4715 / 0.8 0.8835 é - 1.6655 XO 6.189 xloo %.mp = 18.909/ Settling time ts 4 11 6.3752 0.4684 x 1.3395 to -6.3752 sec Peat time T tp = Wd 1-3395 1 -0.4684 tp: - 2.6547 1-1834 = 2.6547 Sec