Question

If laplace transform of$x\left(t\right)isX\left(s\right)andX\left(s\right)={\int }_{-\infty }^s\frac{2}{s^2+1}ds$, then x(0)is ________

• A. -1

Answer 1

The correct option is A -1
$X\left(s\right)={\int }_{-\infty }^s\frac{2}{s^2+1}ds$

$\frac{dX\left(s\right)}{ds}=\frac{2}{s^2+1}$

$x\left(t\right)⟷X\left(s\right)$

$tx\left(t\right)⟷-\frac{dX\left(s\right)}{ds}$

$-tx\left(t\right)⟷\frac{dX\left(s\right)}{ds}$

$L^{-1}\left(\frac{2}{s^2+1}\right)=2sintu\left(t\right)$

$-tx\left(t\right)=2sintu\left(t\right)$

$x\left(t\right)=-\frac{2sint}{t}u\left(t\right)$

$x\left(0\right)=-\underset{t\to 0}{lim}\frac{2sint}{t}u\left(t\right)$

$=-2\underset{t\to 0}{lim}u\left(t\right)$

$=-2\left(\frac{1}{2}\right)=-1$