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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.

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  • $\begingroup$ This is the dirac measure? $\endgroup$ – Marios Gretsas Sep 28 '19 at 19:08
  • $\begingroup$ 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. $\endgroup$ – Zzz Sep 28 '19 at 19:12
  • $\begingroup$ If you mean convolution of sequences then simply apply the definition $a \ast b(n) = \sum_m a(m) b(n-m)$ $\endgroup$ – reuns Sep 28 '19 at 19:19
  • $\begingroup$ 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. $\endgroup$ – Zzz Sep 28 '19 at 19:27
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$$\delta (t + 1) \ast \delta (n+1) = (\delta \ast \delta)(t+2)=\int_{-\infty}^{\infty}\delta(\tau)\delta(t+2-\tau)\mathrm{d}\tau=\delta(t+2)$$

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    $\begingroup$ Thanks for the clarification. $\endgroup$ – Zzz Oct 1 '19 at 14:05
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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.

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

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