Question

Consider a casual second-order system will be transfer function
$G\left(s\right)=\frac{1}{1+2s+s^2}$
with a unit-step $R\left(s\right)=\frac{1}{s}$ as an input. Let $C\left(s\right)$ be the corresponding output. The time taken by the system output $c\left(t\right)$ to reach $94\%$ of its steady-state value $\underset{t\to \infty }{lim}c\left(t\right)$, rounded off to two decimal places, is

• A. $5.25$
• B. $4.50$
• C. $2.81$
• D. $3.89$

Answer 1

The correct option is B $4.50$

$G\left(s\right)=\frac{1}{s^2+2s+1}=\frac{1}{(s+1{)}^2}$
$R\left(s\right)=\frac{1}{s}$
$C\left(s\right)=G\left(s\right)R\left(s\right)=\frac{1}{(s+1{)}^2}$
Using partial fraction expansion, we get,
$C\left(s\right)=\frac{A}{s}+\frac{B}{(s+1)}+\frac{c}{(s+1{)}^2}$ $A(s^2+2s+1)+B(s^2+s)+Cs=1$
$A=1$
$A+B=0\Rightarrow B=-1$
$2A+B+C=0\Rightarrow C=-1$
$\therefore C\left(s\right)=\frac{1}{s}-\frac{1}{(s+1)}-\frac{1}{(s+1{)}^2}$
and $c\left(t\right)=(1-e^{-t}-t{e}^{-t})u\left(t\right)$
$\underset{t\to \infty }{lim}c\left(t\right)=1$
In order to reach 94% of its steady-state value, $(1-e^{-t}-t{e}^{-t})=0.94$
By trial and error ,we get,
$t=\approx 4.50sec$