# The open-loop transfer function of a unity-feedback control system is given by ${\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{{\rm{s}}\left( {{\ 0 votes 62 views closed The open-loop transfer function of a unity-feedback control system is given by \({\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{{\rm{s}}\left( {{\rm{s}} + 2} \right)}}$

For the peak overshoot of the closed-loop system to a unit step input to be 10%, the value of K is ____________.

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Concept:

Peak overshoot: It denotes the normalized difference between the steady-state output to the first peak of the time response. It is given as:

${{\rm{M}}_{\rm{p}}} = {\rm{\;}}{{\rm{e}}^{ - \frac{{{\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}}} \times 100$

where ξ : damping factor.

Calculation:

The open-loop transfer function is given as:

${\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{{\rm{s}}\left( {{\rm{s}} + 2} \right)}}$

Characteristics equation is ${{\rm{s}}^2}{\rm{\;}} + {\rm{\;}}2{\rm{s}} + {\rm{\;K\;}} = 0{\rm{\;}}$

Compare with the standard equation we get

$2{\rm{ξ \;}}{{\rm{\omega }}_{\rm{n}}} = 2{\rm{\;and\;\omega }}_{\rm{n}}^2{\rm{\;}} = {\rm{\;K\;}}$

${\rm{\;}}{{\rm{\omega }}_{\rm{n}}}{\rm{\;}} = {\rm{\;}}\sqrt {\rm{k}}$

$\therefore {\rm{\;\;ξ }} = \frac{1}{{\sqrt {\rm{k}} }}$

Given peak overshoot as:

${{\rm{M}}_{\rm{p}}} = {\rm{\;}}10{\rm{\% \;}} = {\rm{\;}}{{\rm{e}}^{ - \frac{{{\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}}} \times 100$

$\ln \left( {\frac{{10}}{{100}}} \right) = \frac{{ - {\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}$

$-2.3 = \frac{{ - {\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}$

$\frac{{{{\rm{ξ }}^2}}}{{1 - {{\rm{ξ }}^2}}} = 0.537$

${\rm{ξ }} = 0.6$

${\rm{K}} = \frac{1}{{{{\rm{ξ }}^2}}} = 2.86$