# Summation with unit step function

I am confused on why the following summation simplifies as it does. $$\sum_{k =-\infty}^{\infty}\alpha^ku[k]u[n-k]\alpha^{n-k} = \sum_{k = 0}^{n}\alpha^n$$ Here, $u[n]$ is the unit step function. My understanding of this statement is the following. The lower bound was changed from $-\infty$ to $0$ because of $u[k]$, since the unit function is only $1$ for values $\geq 0$. Similarly, for $u[n-k]$, we have time shift and reversal. So $u[n-k]$ is defined for all values of $n$ from $[-\infty, n]$. Hence, the upper bound becomes $n$. That gets rid of the two step functions. However, how does the two alphas reduce to just $\alpha^n$? This is what I'm not able to understand.

• You asked almost the same question there math.stackexchange.com/questions/2194187/… Do you read the answers ? Commented Mar 21, 2017 at 5:50
• $u[k] u[n-k] = 1_{k \in [0,n]}$ Commented Mar 21, 2017 at 5:52
• @user1952009 I don't see how they're the same. And there is no answer, just a comment with no explanation. Commented Mar 21, 2017 at 6:07
• You know what I mean. If you didn't understand something then explain it. Asking exactly the same question reveals you probably only want the solution of your exercice. Commented Mar 21, 2017 at 7:51

First note that $\alpha^k\alpha^{n-k}=\alpha^{n-k+k}=\alpha^n$. Therefore $$\sum_{k=-\infty}^{\infty}\alpha^{k}u[k]u[n-k]\alpha^{n-k}=\sum_{k=-\infty}^{\infty}\alpha^nu[k]u[n-k]$$ Then as you noted, $u[k]=0$ unless $k\geq 0$, in which case it is equal to $1$. Similarly $u[n-k]=0$ unless $n-k\geq 0$, i.e. $k\leq n$, in which case it is $1$. Therefore $$\sum_{k=-\infty}^{\infty}\alpha^nu[k]u[n-k]=\sum_{k=0}^n\alpha^n$$
• No, it's always $\alpha^n$. Commented Mar 21, 2017 at 5:42