# Determining Stability of a continuous time system using Laplace Transform

I'm following Oppenheim's book. In exapmles, the laplace transforms of of the following signals

$$e^{-t}u(t)$$ and $$e^{-t-1}u(t+1)$$

is given as $$\frac{s}{(s+1)}$$ and $$\frac{e^{-s}}{(s+1)}$$ both having the same ROC $$Re(s)>-1$$.

however, the first signal is causal and the second one is non-causal. The system function is not rational for the second one.But why is that the problem when the ROC is essentially same?

The first signal, $$x(t) = e^{-t} u(t)$$ , has the Laplace transform $$X(s) = \frac{1}{s+1}$$ and its ROC is $$\mathcal{Re}\{s\} > -1$$ , since $$x(t)$$ was defined to be causal.
The second signal, $$y(t)$$, is obtained by advancing $$x(t)$$ by $$1$$ unit in time. Hence $$y(t) = x(t+1)$$. By Laplace transform property it's seen that $$Y(s) = \frac{e^{s}}{s+1}$$ , and it has the same ROC with $$X(s)$$, since they are both right sided sequences. However, $$y(t)$$ is not causal despite being right sided, due to the advance of $$1$$ unit to the left.