Correct Answer - Option 4 : zero
Transfer function: It is defined as the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable assuming all initial conditions to be zero.
Open-loop transfer function ⇒ \(G\left( s \right) = \frac{{2s}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\)
Nothing about feedback path, so H(s) = 1
Close-loop transfer function ⇒ \(T\left( s \right) = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\;\)
\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{2s}}{{{s^2} + 5s + 2}}\)
Where C(s) = Output function
R(s) = Input function
Given input function is a unit step ⇒ \(R\left( s \right) = \frac{1}{s}\)
\(C\left( s \right) = T\left( s \right)R\left( s \right) = \frac{1}{s} \times \frac{{2s}}{{{s^2} + 5s + 2}} = \frac{2}{{{s^2} + 5s + 2}}\)
\(C\left( s \right) = \frac{2}{{{s^2} + 5s + 2 + \frac{{25}}{4} - \frac{{25}}{4}}}\)
\(C\left( s \right) = \frac{2}{{{{\left( {s + \frac{5}{2}} \right)}^2} - {{\left( {\sqrt {\frac{{17}}{4}} } \right)}^2}}}\)
\(C\left( s \right) = \frac{1}{{\sqrt {\frac{{17}}{4}} }}\left( {\frac{{2\sqrt {\frac{{17}}{4}} }}{{{{\left( {s + \frac{5}{2}} \right)}^2} - {{\left( {\sqrt {\frac{{17}}{4}} } \right)}^2}}}} \right)\)
Taking Inverse Laplace
\(c\left( t \right) = \frac{2}{{\sqrt {\frac{{17}}{4}} }}{e^{ - \frac{{5t}}{2}}}\sinh \left( {\sqrt {\frac{{17}}{4}} t} \right)u\left( t \right)\)
\(c\left( t \right) = \frac{4}{{\sqrt {17} }}{e^{ - \frac{{5t}}{2}}}\sinh \left( {\frac{{\sqrt {17} }}{2}t} \right)u\left( t \right)\)
The steady-state value of c(t) for t → ∞ is zero as the value e(-2.5t) tends to zero.
Laplace Transform:
\({e^{ - at}}u\left( t \right) \Leftrightarrow \frac{1}{{s + a}}\)
\(\sin \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{s^2} + {\omega ^2}}}\)
\(\sinh \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{s^2} - {\omega ^2}}}\)
\({e^{ - at}}\sin \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{{\left( {s + a} \right)}^2} + {\omega ^2}}}\)
\({e^{ - at}}\sinh \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{{\left( {s + a} \right)}^2} - {\omega ^2}}}\)
