Correct Answer - Option 3 : 4 s
System is given as
\(H\left( s \right) = \frac{2}{{s + 1}}\)
Step input R(s) = 1/s
o/p , Y(s) = H(s) R(s)
\(\begin{array}{l}
= \frac{2}{{\left( {s + 1} \right)}}\frac{{1}}{s}\\
= \frac{2}{s} - \frac{2}{{s + 1}}
\end{array}\)
Taking inverse laplace transform,
y(t) = (2-2e-t) u(t)
Final value of y(t),
\(y\left( {ss} \right)\left( t \right) = \begin{array}{*{20}{c}}
{lt}\\
{t \to \infty }
\end{array}\ y\left( t \right) = 2\)
Let time taken for step response to reach 98% of its final value is ts
So,
2-2e-ts = 2×0.98
0.02 = e-ts
ts = ln 50 = 3.91 sec.
