Q2Signals and Systems

Statement: If $x(t)$ and its derivative are Laplace transformable, then $x(0^{+})={\lim }_{s\to \infty }sX(s)$.

Proof: Consider the transform of the derivative: $\mathcal{L}\{x^{\prime }(t)\}={\int }_0^{\infty }{x}^{\prime }(t)e^{-st}dt=sX(s)-x(0^{+})$. Take the limit as $s\to \infty$ on both sides. On the left side, as $s\to \infty$, $e^{-st}\to 0$, making the integral zero. $0={\lim }_{s\to \infty }[sX(s)-x(0^{+})]⟹x(0^{+})={\lim }_{s\to \infty }sX(s)$.

Statement: If $x(t)$ and its derivative are transformable, and all poles of $sX(s)$ lie in the left half of the s-plane, then $x(\infty )={\lim }_{s\to 0}sX(s)$.

Proof: Using the derivative property again: ${\int }_0^{\infty }{x}^{\prime }(t)e^{-st}dt=sX(s)-x(0^{+})$. Take the limit as $s\to 0$ on both sides. On the left side, $e^{-st}\to 1$. ${\int }_0^{\infty }{x}^{\prime }(t)dt={\lim }_{s\to 0}[sX(s)-x(0^{+})]$ $[x(\infty )-x(0^{+})]={\lim }_{s\to 0}sX(s)-x(0^{+})$ $x(\infty )={\lim }_{s\to 0}sX(s)$.