# Tag Info

Because, the region of convergence in the Laplace transform $$X(s) = \int_{-\infty}^{\infty} x(t) e^{-st} dt$$ is related to the weighting provided by the real part of the complex $s = \sigma + j \omega$; as this will yield the weight $|e^{-st}| = e^{-\sigma t}$ applied on the input signal $x(t)$, and is a function of $\sigma$ alone and is a ...