If I have block diagram with input $X(s)$ that goes to a block with $\frac{1}{s + 2}$ in it and then by way of $w(s)$ to a block with $s$ in it, and finally to the output $Y(s)$, how do I find the response $w(t)$ given that $x(t) = \delta(t)$.
Since $x(t)$ is defined as above, $X(s) = 1$. So can I write $w(s) = \frac{X(s)}{s + 2}=\frac{1}{s + 2}$? Therefore, $w(t) = e^{-2t}$.
Well this can't be entirely correct since it says to find $y(t)$ by differentiating $w(t)$. We know that $Y(s) = \frac{s}{s + 2} = 1 - \frac{2}{s + 2}$ So $$ y(t) = \frac{1}{2\pi i}\int_{-\infty}^{\infty}(1 - 2/(s + 2))e^{st}ds = \delta(t) -2e^{-2t} $$ If we differentiate $w(t)$ obtained above, we don't pick up the $\delta(t)$.
$$\mathcal{L}\{e^{-2t} u(t)\} = \int_{-\infty}^\infty e^{-2t} u(t) e^{-st} dt = \int_{0}^\infty e^{-2t} e^{-st} dt = \frac{1}{s+2}$$
$\mathcal{L}\{e^{-2t}\}$ doesn't converge for bilateral Laplace transformation.