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\begin{align*} &(\delta[n]+\delta[n-1])*(\delta[n]+\delta[n-1])\\ \\ =\;&\delta[n]*\delta[n]+2(\delta[n]*\delta[n-1])+\delta[n-1]*\delta[n-1]\\ \\ =\;&(\delta[n]+2(\delta[n]*\delta[n-1])+\delta[n-1]) \end{align*}

I am not sure if its true that $\delta[n]*\delta[n]=\delta[n]$ and $\delta[n-1]*\delta[n-1]=\delta[n-1]$ for discrete signals but I proceeded with this assumption. Furthermore, how do we compute the convolution $\delta[n]*\delta[n-1]$?

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Convolution in time domain means product in frequency domain. DTFT of $\delta[n]$ is just $1$, therefore any arbitrary $x[n]$ when convolved with $\delta[n]$ remains same, as in frequency domain you are just multiplying $X(e^{j\omega})$ with $1$.

So, $Y(e^{j\omega}) = 1 \ . \ X(e^{j\omega}) = X(e^{j\omega})$. Hence, $y[n] = x[n]$.

Now, apply same principle for $\delta[n-1]*\delta[n-1]$ to see that the result will be $\delta[n-2]$.

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  • $\begingroup$ Hello, is there a method that doesn't involve Fourier because I have not reached this concept yet :( $\endgroup$
    – SPARSE
    Mar 3 '21 at 20:08
  • $\begingroup$ @deerclaysup Another way is to understand that convolution with $\delta[n-1]$ will bring 1 sample of delay to any arbitrary sequence $x[n]$. So, when $x[n] = \delta[n]$ gets convolved with $\delta[n-1]$, then it gets delayed by 1 sample and we get $y[n] = x[n-1] = \delta[n-1]$. And when $x[n] = \delta[n-1]$ gets convolved with $\delta[n-1]$, then it gets delayed by 1 sample and you get $y[n] = x[n-1] = \delta[n-2]$. $\endgroup$
    – DSP Rookie
    Mar 3 '21 at 20:28
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    $\begingroup$ This makes sense now, thank you very much! $\endgroup$
    – SPARSE
    Mar 3 '21 at 20:30
  • $\begingroup$ @deerclaysup No, you won't get $\delta[n]*\delta[n-1] = 2\delta[n-1]$, but you will get $\delta[n-1]$. $\delta[n]$ is your input signal and you are convolving it with $\delta[n-1]$, so the result is 1 sample delay in your input signal that is $\delta[n-1]$. $\endgroup$
    – DSP Rookie
    Mar 3 '21 at 20:33
  • $\begingroup$ Yep you are right, I meant to my original expression it had the term $2(\delta[n]*\delta[n-1])$. Thank you once more :) $\endgroup$
    – SPARSE
    Mar 3 '21 at 20:36

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