Ramp Input MCQ Quiz in বাংলা - Objective Question with Answer for Ramp Input - বিনামূল্যে ডাউনলোড করুন [PDF]

Given,

\({\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{{\rm{s}}\left( {{\rm{s}} + 4} \right)\left( {{\rm{s}} + 5} \right)}}\) H(s) = 1

For ramp input \({{\rm{e}}_{{\rm{ss}}}} = \frac{1}{{{{\rm{K}}_{\rm{v}}}}}\)

\(\begin{array}{l} {{\rm{K}}_{\rm{v}}} ={\rm{\lim\limits_{s\to0}}}\ {\rm{s}}.{\rm{G}}\left( {\rm{s}} \right)\\ = {\rm{\lim\limits_{s\to0}}}\ \frac{{\rm{K}}}{{\left( {{\rm{s}} + 4} \right)\left( {{\rm{s}} + 5} \right)}} = \frac{{\rm{K}}}{{20}}\\ \therefore {{\rm{e}}_{{\rm{ss}}}} = \frac{{20}}{{\rm{K}}} = 0.4 \Rightarrow {\rm{K}} = 50 \end{array}\)

\(15{\rm{\;rpm}} = 15 \times \frac{{2{\rm{\pi }}}}{{60}} = \frac{{{{\rm{\pi }}^{\rm{c}}}}}{2} = 90^\circ\)

Given \({{\rm{e}}_{{\rm{ss}}}}{\rm{\;}} < 4^\circ {\rm{\;}},{\rm{\;let\;input}} = {\rm{Ar}}\left( {\rm{t}} \right)\)(\(\because\) type of system is 1)

But \({{\rm{e}}_{{\rm{SS}}}} = \frac{{\rm{A}}}{{{{\rm{K}}_{\rm{V}}}}} = \frac{{\rm{A}}}{{15{\rm{K}}}}\)  But \({\rm{A}} = 15{\rm{rpm}} =\frac{15\times{2\pi}^c}{60}= \frac{{{{\rm{\pi }}^{\rm{c}}}}}{2}\)

\(\therefore {{\rm{e}}_{{\rm{ss}}}} < 4 \times \frac{{\rm{\pi }^c}}{{180^\circ }}\)

\(\frac{{\rm{\pi }^c}}{2} \times \frac{1}{{15{\rm{K}}}} < \frac{{4{\rm{\pi }^c}}}{{180}}\)

\(\rm\frac {1}{K}<\frac {4}{6\ } \Rightarrow\ K>1.5\)

Velocity error constant \({k_v} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

\( = \mathop {\lim }\limits_{s \to 0} \frac{{s.500}}{{s\left( {s + 1} \right)\left( {s + 100} \right)}} = 5\)

The system is type – 1, since the initial slope is - 20 dB/Dec.

The Transfer function is given as , \(G(S)H(S) =\ \frac{K(1+\frac {S}{2})}{S(1+\frac {S}{10})}\)

At ω = 2   ,   M = 0 dB hence

\(- 20{\log _{10}}2 + 20{\log _{10}}K = 0\)

K = 2

Steady state error for ramp input \(= \frac{1}{{{K_V}}} = \frac{1}{2} = 0.5\)

\(\rm \begin{array}{l} TF = \frac{{C\left( s\right)}}{{R\left( s \right)}} = \frac{{\frac{1}{{sC}}}}{{R + sL + \frac{1}{{sC}}}}\\ \rm TF = \frac{1}{{{s^2}LC + sCR + 1}} = \frac{{\frac{1}{{LC}}}}{{{s^2} + \left( {\frac{R}{L}} \right)s + \frac{1}{{LC}}}} = \frac{{G\left( s \right)H\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}} \end{array}\)

∴ open loop transfer function \(\rm G\left( s \right)H\left( s \right) = \frac{{\frac{1}{{LC}}}}{{{s^2} + \frac{R}{L}.s}}\)

\(\rm G\left( s \right)H\left( s \right) = \frac{{\frac{1}{{LC}}}}{{s\left( {s + \frac{R}{L}} \right)}}\)

Velocity error constant \(\rm {K_V} = \begin{array}{*{20}{c}} {lt}\\ {s \to 0} \end{array}s.G\left( s \right)H\left( s \right)\)

\(\rm \begin{array}{l} {K_V} = \begin{array}{*{20}{c}} {lt}\\ {s \to 0} \end{array}\frac{{\frac{1}{{LC}}}}{{\left( {s + \frac{R}{L}} \right)}}\\ \rm = \frac{1}{{LC}} \times \frac{L}{R} = \frac{1}{{RC}} = \frac{1}{{100 \times {{10}^3} \times {{10}^{ - 6}}}} = 10 \end{array}\)