This question was previously asked in

ESE Electronics 2010 Paper 2: Official Paper

Option 4 : 20 sec

CT 3: Building Materials

3174

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

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

The 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:

\(G(s) = \frac{4}{{{s^2} + 0.4s}}\)

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

\(\omega _n^2 = 4 \Rightarrow {\omega _n} = 2\)

⇒ 2ζωn = 0.4 ⇒ ζ = 0.1

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

Settling time, \({t_s} = \frac{4}{{\zeta {\omega _n}}} = \frac{4}{{0.1 \times 2}} = 20\;sec\)