Question

The peak percentage overshoot of the closed loop system is :

(a) 5.0%

(b) 10.0%

(c) 16.3%

(d) 1.63%

This question was addressed to me during an online exam.

My enquiry is from Time Response of Second Order Systems topic in portion Time Response Analysis, Design Specifications and Performance Indices of Control Systems

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 %