# Is the expansion of this expression correct?

\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 $$\delta[n]*\delta[n-1]$$?

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 $$\delta[n-2]$$.
• @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 $y[n] = x[n-1] = \delta[n-2]$. Mar 3 '21 at 20:28
• @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 $\delta[n-1]$. Mar 3 '21 at 20:33
• Yep you are right, I meant to my original expression it had the term $2(\delta[n]*\delta[n-1])$. Thank you once more :) Mar 3 '21 at 20:36