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2.) Design compensator for zero steady-state error with 10% overshoot and 0.4s of Peak time for the open loop transfer functi
these are useful formjlas to solve this problem
Percent overshoot = %OS = e 7/1-3? x 100 4. Settling time = Ts = 7cm Peak time = T, = - Roots of second order characteristics
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ANSWERS Given transfer function 18 G13) - (3+2) (914) (+90) Zero Steady error with 10%. Overshoot and peak time tp : 0:4 geeRoot laws of uncompensated system with K=/ Steady stale eroy for stap input Kp = It Go at k 390 840 (949)(974) (9420) Januar-180 + 186-1604115.770+ 102.49°+98 83° : 199.950 where 02 60° the design 4. lead compensator 0:48.94 Gread (8) - ($pe ne the

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