# Convolution of Dirac with itself

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.

• This is the dirac measure? – Marios Gretsas Sep 28 '19 at 19:08
• Well, the is a dirac delta function that is shifted to the left from 0 of 1 units. At n = -1, its value is 1. – Zzz Sep 28 '19 at 19:12
• If you mean convolution of sequences then simply apply the definition $a \ast b(n) = \sum_m a(m) b(n-m)$ – reuns Sep 28 '19 at 19:19
• I did but I am not sure about the answer. Convolution tells us to reverse the signal and shift it to the right. However, in this case, I cant get any answer to the multiplication unless i shift the reversed signal to the left. – Zzz Sep 28 '19 at 19:27

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

• Thanks for the clarification. – Zzz Oct 1 '19 at 14:05

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.

• Thanks for the clarification. – Zzz Oct 1 '19 at 14:05