# A second order control system has a transfer function $\frac{{16}}{{{s^2} + 4s + 16}}$ What is the time for the first overshoot?

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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. $\frac{{\pi }}{{4\sqrt 3 }}s$

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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

${t_p} = \frac{\pi }{{\omega d}}$

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

ωn = natural frequency.

ξ = damping ratio

Calculation:

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

ωn = 4 r/s

2ζωn = 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]-->