Question

Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is

\(\mathop \to \limits^{U\left( s \right)} \boxed{\ \frac{1}{s}\ }\mathop \to \limits^{Y\left( s \right)}\)

1. u(t)
2. tu(t)
3. \(\frac{{{t^2}}}{2}u\left( t \right)\)
4. e^-t u(t)

Answer 1

Correct Answer - Option 2 : tu(t)

L( u(t) )  = U(s) = 1/s

\(\frac{o}{p},\ Y\left( s \right) = \left( {\frac{1}{s}} \right)\left( {\frac{1}{s}} \right) = \frac{1}{{{s^2}}}\)

For zero initial condition, we check

\(u\left( t \right) = \frac{{dy}}{{dt}}\)

U (s) = sY(s) – y(0)

\(U\left( s \right) = s\left( {\frac{1}{{{s^2}}}} \right) - y\left( 0 \right)\)

For y(0) = 0

\(U\left( s \right) = \frac{1}{s}\)

Hence, the o/p is correct which is

\(Y\left( s \right) = \frac{1}{{{s^2}}}\)

Taking inverse Laplace, we get

Y(t) = tu (t)