What is the convolution of 2 dirac functions? I ran into a problem, which wants to find the convolution $\delta (3-t) * \delta (t-2)$ and I am stuck. How can I approach it? 
 A: It helps to look at the definition of a convolution: Given two functions $f$ and $g$ (assumed from $\mathbb R$ to $\mathbb R$), the convolution $f*g$ is defined as
$$(f*g) (t) = \int\limits_{-\infty}^\infty f(t-y) g(y) \,\text d y\,.$$
In your case, $f(t)=\delta(3-t)$ and $g(t)=\delta(t-2)$, so we have
$$(f*g) (t) =\int \delta(3-(t-y)) \,\delta(y-2) \,\text{d} y\,.$$
(You can switch which function you call $f$ and which one $g$ without changing the result, since the convolution is symmetric.) , Now the essential property of the $\delta$ funcion is that $\int f(y) \delta(y-a)\,\text d y=f(a)$, so the result is
$$(f*g) (t) =\int \delta(3-(t-y)) \,\delta(y-2) \,\text{d} y =\delta(3-t+y)\big|_{y=2}=\delta(5-t)\,.$$
A: You could do it using the Laplace transform and the convolution theorem for Laplace transforms. The Laplace transform of a Dirac delta is
$$\mathcal{L}(\delta(t-a)) = e^{-as}$$
and the convolution theorem states that $\mathcal{L} ((f*g)(t)) = \mathcal{L}(f(t))\mathcal{L}(g(t))$, so you can multiply the Laplace transforms of your deltas and then take the inverse. There is likely a more direct method though.
