A unity feedback system has the open loop transfer function $G(s)=\frac{2s}{(s+1)(s+2)}$. The steady state response of the closed loop system to a unit step reference input is

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.

Important Points

Laplace Transform: 

$e^{-at}u\left(t\right)\iff \frac{1}{s+a}$

$\sin \omega tu\left(t\right)\iff \frac{\omega }{s^2+{\omega }^2}$

$\sinh \omega tu\left(t\right)\iff \frac{\omega }{s^2-{\omega }^2}$

$e^{-at}\sin \omega tu\left(t\right)\iff \frac{\omega }{{\left(s+a\right)}^2+{\omega }^2}$

$e^{-at}\sinh \omega tu\left(t\right)\iff \frac{\omega }{{\left(s+a\right)}^2-{\omega }^2}$