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venkyelectrical
Asked: 2022-04-10T14:47:12+00:00 In:

Design Project #2 : Compensation Design Using Root Locus Method Consider the control system in the following figure. Design a lead-lag compensator such that the closed-loop step response has a rise time of less than 0.7sec and an overshoot of less than 10% and the static velocity error constant kb, =50sec-1. —Qs 10 Gc(8) I 2(3+2)(8+5

Design Project #2 : Compensation Design Using Root Locus Method Consider the control system in the following figure. Design a lead-lag compensator such that the closed-loop step response has a rise time of less than 0.7sec and an overshoot of less than 10% and the static velocity error constant kb, =50sec-1. —Qs 10 Gc(8) I 2(3+2)(8+5Design Project #2 : Compensation Design Using Root Locus Method Consider the control system in the following figure. Design a lead-lag compensator such that the closed-loop step response has a rise time of less than 0.7sec and an overshoot of less than 10% and the static velocity error constant kb, =50sec-1. —Qs 10 Gc(8) I 2(3+2)(8+5Design Project #2 : Compensation Design Using Root Locus Method Consider the control system in the following figure. Design a lead-lag compensator such that the closed-loop step response has a rise time of less than 0.7sec and an overshoot of less than 10% and the static velocity error constant kb, =50sec-1. —Qs 10 Gc(8) I 2(3+2)(8+5

compensator design using root locus
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