Question

A second order control system has a transfer function \(\frac{{16}}{{{s^2} + 4s + 16}}\) What is the time for the first overshoot?
1. \(\frac{{2\pi }}{{\sqrt 3 }}s\)
2. \(\frac{{\pi }}{{\sqrt 3 }}s\)
3. \(\frac{{\pi }}{{2\sqrt 3 }}s\)
4. \(\frac{{\pi }}{{4\sqrt 3 }}s\)

Answer 1

Correct Answer - Option 3 : \(\frac{{\pi }}{{2\sqrt 3 }}s\)

Concept:

Peak time:

It is the time required for the response to reach the peak time for the first time at t = tp, the first derivative of the response is zero.

\({t_p} = \frac{{n\pi }}{{\omega d}}\)

n = 1, 3, 5 …. For overshoot

2, 4, 6 …. For undershoot

Let n = 1

\({t_p} = \frac{\pi }{{\omega d}}\)

ωd = damping frequency (rad/sec)

\({\omega _d} = {\omega _n}\;\sqrt {1 - {\xi ^2}}\)

ωn = natural frequency.

ξ = damping ratio

Calculation:

The given transfer function is \(\frac{{16}}{{{s^2} + 4s + 16}}\)

ω_n = 4 r/s

2ζω_n = 4

⇒ ζ = 0.5

<!--[if gte msEquation 12]>ωd=41-14=23<![endif]--><!--[if !msEquation]--><!--[if gte vml 1]> <![endif]--><!--[if !vml]-->\({\omega _d} = 4\sqrt {1 - \frac{1}{4}} = 2\sqrt 3 \)<!--[endif]--><!--[endif]-->

Peak time, \({t_p} = \frac{\pi }{{2\sqrt 3 }}\)<!--[endif]--><!--[endif]-->