Convolution of Dirac with itself

Question

How can we calculate the convolution $\delta[n+1] \ast \delta[n+1] $ ? Is it $\delta[n+2]$ ? We know already that the convolution of $\delta[n-1] \ast \delta[n-1] $ is $\delta[n-2] $ , but I am not sure for the former case.

Answer 1

$$\delta (t + 1) \ast \delta (n+1) = (\delta \ast \delta)(t+2)=\int_{-\infty}^{\infty}\delta(\tau)\delta(t+2-\tau)\mathrm{d}\tau=\delta(t+2)$$

Answer 2

I'll write $\delta_x[n]$ for $\delta[n-x]$, so that $\delta_x[n] = 0$ if $n \neq x$ and $\delta_x[x] = 1$. In other words, $\delta_x$ is the Dirac mass at $x$.

Then $(\delta_x * \delta_y)[n] = \sum_m \delta_x[m]\delta_y[n-m]$. The terms in this sum are all zero unless $m = x$ and $n-m=y$, in which case, the sum is 1. Solving for $n$ with these restrictions, we get that $n = x+y$.

Therefore, we get that $(\delta_x*\delta_y)[n] = 0$ if $n \neq x+y$, and $(\delta_x*\delta_y)[x+y] = 1$. So $\delta_x*\delta_y = \delta_{x+y}$.

Convolving Dirac masses at positions $x$ and $y$ will create a Dirac mass at $x+y$. So you're right that convolving two Dirac masses at -1 will produce a Dirac mass at -2.