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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.

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  • $\begingroup$ You asked almost the same question there math.stackexchange.com/questions/2194187/… Do you read the answers ? $\endgroup$
    – reuns
    Commented Mar 21, 2017 at 5:50
  • $\begingroup$ $u[k] u[n-k] = 1_{k \in [0,n]}$ $\endgroup$
    – reuns
    Commented Mar 21, 2017 at 5:52
  • $\begingroup$ @user1952009 I don't see how they're the same. And there is no answer, just a comment with no explanation. $\endgroup$
    – Jonathan
    Commented Mar 21, 2017 at 6:07
  • $\begingroup$ 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. $\endgroup$
    – reuns
    Commented Mar 21, 2017 at 7:51

1 Answer 1

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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 $$ \sum_{k=-\infty}^{\infty}\alpha^nu[k]u[n-k]=\sum_{k=0}^n\alpha^n$$

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  • $\begingroup$ Do the exponents not have to change as the limits change? $\endgroup$
    – Jonathan
    Commented Mar 21, 2017 at 5:40
  • $\begingroup$ No, it's always $\alpha^n$. $\endgroup$ Commented Mar 21, 2017 at 5:42

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