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The time taken for the output to settle within ± 2% of step input for the control system represented by \(\frac{25}{(s^2+5s+25)}\) is given by
1. 1.2 s
2. 1.6 s
3. 2.0 s
4. 0.4 s

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Best answer
Correct Answer - Option 2 : 1.6 s

Concept:

The characteristic equation is given by:

1 + G(s) H(s) = 0

Also, the standard second-order characteristic equation is:

\({s^2} + 2\zeta {\omega _n}s + \omega _n^2 = 0\)

Settling Time is the time taken by the response to reach ± 2%, tolerance band.

\({e^{ - \xi {\omega _n}{t_s}}} = \pm 5\% \;\left( {or} \right) \pm 2\% \)

 \({t_s} \simeq \frac{3}{{\xi {\omega _n}}}\) for a 5% tolerance band.

\({t_s} \simeq \frac{4}{{\xi {\omega _n}}}\) for 2% tolerance band

Calculation:

Characteristic equation:

s2 + 5s + 25 = 0

By comparing this with the standard second-order equation, we get:

\(\omega _n^2 = 25 \Rightarrow {\omega _n} = 5\)

\(2\zeta {\omega _n} = 5\)

\(\Rightarrow \zeta = \frac{5}{{2 \times 5}} = 0.5\)

Settling time is, therefore:

\({t_s} = \frac{4}{{0.5 \times 5}} = 1.6\;sec\)

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