Inverse laplace transform of the expression

Question

I am struggling to find the inverse Laplace transform of this expression $$X(s) = \frac{e^{-st_0}}{\omega^2 + s^2}$$ The inverse Laplace transform of the numerator is $\delta(t-t_0)$. However, I don't know how to tackle the denominator, or shall I need to tackle the whole expression. Any suggestions would be invaluable. Thanks.

Answer 1

I would use the time shifting property: $$\mathcal{L}^{-1}\!\left(e^{-as}F(s)\right)=f(t-a)\,u(t-a),$$ where $u$ is the Heaviside step function. $t_0$ plays the role of $a$ in this equation, and the inverse LT of the denominator is $\sin,$ so that you get \begin{align*} \mathcal{L}^{-1}\!\left(\frac{e^{-st_0}}{\omega^2+s^2}\right) &=\frac{1}{\omega}\,\mathcal{L}^{-1}\!\left(\frac{\omega\,e^{-st_0}}{\omega^2+s^2}\right)\\ &=\frac{\sin(\omega(t-t_0))\,u(t-t_0)}{\omega}. \end{align*}