Question

If Lapplace transform of x(t) is X(s) and $X\left(s\right)={\int }_{-\infty }^s\frac{2}{s^2+1}ds$ ,
Then x(0) is ________.

• A. -1

Answer 1

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

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

$x\left(t\right)⟷X\left(s\right)$

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

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

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

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

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

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

$x\left(0\right)=-{lim}_{t\to 0}u\left(t\right)$

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

$=-1$