If the Laplace transform of a signal y(t) is $$Y\left( s \right) = {1 \over {s\left( {s – 1} \right)}},$$ then its final value is

The correct answer is $\boxed{\text{C}}$.

The final value theorem states that the final value of a system is equal to the limit of its Laplace transform as $s$ approaches infinity, provided that the limit exists. In this case, we have

$$\lim_{s \to \infty} Y\left( s \right) = \lim_{s \to \infty} {1 \over {s\left( {s – 1} \right)}} = \lim_{s \to \infty} {1 \over s} \cdot \lim_{s \to \infty} {1 \over {s – 1}} = 0 \cdot 1 = 0.$$

Therefore, the final value of the signal is $0$.

Option A is incorrect because the final value of the signal is not $-1$.

Option B is incorrect because the final value of the signal is not $0$.

Option D is incorrect because the final value of the signal is bounded.