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UPRVUNL AE EC 2016 Official Paper

Option 2 : 0.726 sec, 1.6 sec

ME Subject Test 1: Strength of Materials

9166

20 Questions
20 Marks
18 Mins

Given:

**Concept:**

Time-domain specification (or) transient response parameters:

__Peak Time (tp):__

It is the time taken by the response to reach the maximum value.

\({\left. {\frac{{dc\left( t \right)}}{{dt}}} \right|_{t = {t_p}}} = 0,{\text{}}{t_p} = \frac{\pi }{{{\omega _d}}}\)

__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

If ξ increases, rise time, peak time increases and peak overshoot decreases.

**Calculation:**

\(G\left( s \right) = \frac{{25}}{{s\left( {s + 5} \right)}}\;\;\;\;\;\;\;\;H\left( s \right) = 1\;\)

Characteristic equation: 1 + G(s) H(s) = 0

s^{2} + 5s + 25 = 0

Compare with standard characteristic Equation, We’ll get:

\(\omega _n^2 = 25,\)

\(2\;\xi {\omega _n} = 5\)

ω_{n} = 5,

ξ = .5

\({\rm{Peak\;time\;}}\left( {{t_p}} \right) = \frac{\pi }{{{\omega _n}\sqrt {1 - {\xi ^2}} }}\)

\(= \frac{\pi }{{5\sqrt {1 - .25} }} = 0.726\;sec\)

\({\rm{Settling\;time}}\left( {{t_s}} \right) = \frac{4}{{\xi{\omega _n}}}\)

\(= \frac{4}{{2.5}} = 1.6\;sec\)