This question was previously asked in

ESE Electronics 2016 Paper 2: Official Paper

Option 1 : 0.8 s

CT 3: Building Materials

3014

10 Questions
20 Marks
12 Mins

__Concept:__

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

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

ζ is the damping ratio

ωn is the natural frequency

The characteristic equation is given as,

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

Roots of the characteristic equation are,

\( - \zeta {\omega _n} + j{\omega _n}\sqrt {1 - {\zeta ^2}} = - \alpha \pm j{\omega _d}\)

α is the damping factor

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

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

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

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

Calculation:

\(\frac{C(s)}{R(s)}=\frac{100}{(s^2+10s+100)}\)

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

\(\omega _n^2 = 100\)

\( {\omega _n} = 10\)

2ζωn = 10 ⇒ ζ = 0.5

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

∴Settling time is

\({t_s} = \frac{4}{{\zeta {\omega _n}}} = \frac{4}{{0.5 \times 10}} = 0.8\;sec\)