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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?

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ROC is a strip of plane which does not contain any poles. The region is determined by the location of poles.

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.

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