The second-order system defined by \(\frac{{25}}{{{s^2} + 5s + 25

$The second-order system defined by \(\frac{{25}}{{{s^2} + 5s + 25$

| The second-order system defined by $\frac{{25}}{{{s^2} + 5s + 25}}$ is given a step input. The time taken for the output to settle within ± 2% is

A. 1.2 s

B. 2 s

C. 1.6 s 

D. 0.4 s

Please scroll down to see the correct answer and solution guide.

Right Answer is: C

SOLUTION

Concept:

The transfer function of the standard second-order system is:

$TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2\zeta {ω _n}s + ω _n^2}}$

ζ is the damping ratio

ωn is the natural frequency

Characteristic equation:

${s^2} + 2\zeta {ω _n} + ω _n^2 = 0$

Settling time (Ts): It is the time taken by the response to reach ± 2% tolerance band as shown in the fig above.

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

${t_s} \simeq \frac{3}{{\xi {ω _n}}}$ for a 5% tolerance band.

${t_s} \simeq \frac{4}{{\xi {ω _n}}}$ for 2% tolerance band

Calculation:

$G\left( s \right) = \frac{{25}}{{{s^2} + 5s + 25}}.$

By comparing the above transfer function with the standard second-order system,

ωn² = 25 ⇒ ω_n = 5

2ζωn = 5 ⇒ ζ = 0.5

As the damping ratio is less than unity, the system is underdamped.

Settling time,

${t_s} = \frac{4}{{\zeta {ω _n}}} = \frac{4}{{0.5 \times 5}} = 1.6\;sec$

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

For a second order system;

$1 + G\left( s \right)H\left( s \right) = {s^2} + \;2ζ {ω _n}s\; + {ω ^2}_n = 0\; \ldots \ldots \;\;\left( 1 \right)$

where ζ is damping factor and ω_n is natural frequency of oscillations.

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For 2% tolerance band:

$ts\; = \;\frac{4}{{\zeta {ω _n}}}$

$ts\; = \;\frac{3}{{\zeta {ω _n}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$