Let h(t) denote the impulse response of a casual system with transfer function $${1 \over {s + 1}}$$. Consider the following three statements.

S1: The system is stable.
S2: $${{h\left( {t + 1} \right)} \over {h\left( t \right)}}$$ is independent of t for t > 0.
S3: A non-casual system with the same transfer function is stable.

For the above system,

A system is described by the following differential equation, where u(t) is the input to the system and y(t) is output of the system $$\mathop y\limits^ \bullet \left( t \right) + 5y\left( t \right) = u\left( t \right)$$

When y(0) = 1 and u(t) is a unit step function, y(t) is

A system is described by the differential equation $$${{{d^2}y} \over {d{t^2}}} + 5{{dy} \over {dt}} + 6y\left( t \right) = x\left( t \right)$$$
Let x(t) be a rectangular pulse given by $$$x\left( t \right) = \left\{ {\matrix{ {1\,\,\,\,\,\,\,\,\,0 \le \,t\, \le 2} \cr {0\,\,\,\,\,otherwise} \cr } } \right.$$$

Assuming that y(0) = 0 $${{dy} \over {dt}} = 0$$ at t = 0, the Laplace transform of y(t) is

If $$F\left( s \right) = L\left[ {f\left( t \right)} \right] = {{2\left( {s + 1} \right)} \over {{s^2} + 4s + 7}}$$ then the initial and final values of f(t) are respectively