The Laplace transform of a continuous - time signal x(t) is $$X\left( s \right) = {{5 - s} \over {{s^2} - s - 2}}$$. If the Fourier transform of tyhis signal exists, then x(t) is

The inverse Laplace transform of the function $${{s + 5} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ is

If $$F\left( s \right) = L\left[ {f\left( t \right)} \right] = {K \over {\left( {s + 1} \right)\,\left( {{s^2} + 4} \right)}}$$ then $$\matrix{ {Lim\,f\,\left( t \right)} \cr {t \to \infty } \cr } $$ is given by

The Laplace transform of the periodioc function f(t) describe4d by the curve below, i.e., $$f\left( t \right) = \left\{ {\matrix{ {\sin \,t\,\,\,if\,\left( {2n - 1} \right)\pi \le t \le 2n\pi } \cr {0\,\,\,\,\,\,\,\,otherwise} \cr } } \right.$$
is _________. (fill in the blank), n is an integer.