A stable linear time invariant (LTI) system has a transfer function H(s) = $${1 \over {{s^2} + s - 6}}$$. To make this system casual it needs to be cascaded with another LTI system having a transfer function H₁(s). A correct choice for H₁(s) among the following options is

A casual LTI system has zero initial conditions and impulse response h(t). Its input x(t) and output y(t) are related through the linear constant - coefficient differential equation $$${{{d^2}y\left( t \right)} \over {d{t^2}}} + \alpha {{dy\left( t \right)} \over {dt}} + {\alpha ^2}y\left( t \right) = x\left( t \right).$$$

Let another signal g(t) be defined as $$\left( t \right) = {\alpha ^2}\int_0^t {h\left( \tau \right)d\tau + {{dh\left( t \right)} \over {dt}} + \alpha h\left( t \right)} $$.

If G(s) is the Laplace transform of g(t), then the number of poles of G(s) is ______.

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