Question

The peak percentage overshoot of the closed loop system is :

(a) 5.0%

(b) 10.0%

(c) 16.3%

(d) 1.63%

Answer 1

The correct answer is (c) 16.3%

To explain I would say: C(s)/R(s) = 1/s^2+s+1

C(s)/R(s) = w/ws^2+2Gws+w^2

Compare both the equations,

w = 1 rad/sec

2Gw = 1

Mp = 16.3 %