convolution of $\delta(t+4) \ast \delta(t-1)$?

Question

How does one solve convolution of $\delta(t+4) \ast \delta(t-1)$ where $\delta$ is dirac delta function?

In ordinary function convolution, tricks are obvious but does dirac delta function share the same property as ordinary function when convolving?

Answer 1

Using the property that $\delta(t+t_1)*\delta(t+t_2)=\delta(t+t_1+t_2)$ the answer is $\delta(t+3)$

Answer 2

Use the distributional definition of convolution:

Let $\delta_1 = \delta(t + 4), \delta_2 = \delta(t-1)$. Then

$$ \langle \delta_1 * \delta_2,\phi\rangle = \langle\delta_1(s),\langle\delta_2(t),\phi(t + s)\rangle\rangle\\ = \langle\delta_1(s),\phi(1 + s)\rangle\\ = \phi(1 + (-4)) = \phi(-3) $$

So $\delta_1*\delta_2 = \delta(t + 3)$. (and if you do this with arbitrary shifts $t_1,t_2$, you can get the formula stated in Sina's answer).