Assume I have the following two sequences : $$ x[n]=\begin{cases} \alpha&\text{if $a\leq n\leq b$}\\ \\ \tag{1} 0&\text{if otherwise} \end{cases} \qquad \text{and} \qquad h[n]=\begin{cases} \beta&\text{if $c\leq n\leq d$}\\ \\ 0&\text{if otherwise} \end{cases} $$ I was wondering in this case if the convolution : $$ (x*h)[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k] \tag{2} $$ has a specific form applicable for these two piecewise sequences.

I have seen on the internet something like : $$ (x * h)[n]=\sum_{k=\max (a, n-d)}^{\min (b, n-c)} x[k] h[n-k] . \tag{3} $$ However, I can not confirm if this is true. I hope someone can provide a proper proof and thank you very much.

  • $\begingroup$ This is really straightforward. $k$ just needs to be inside $[a,b]$ and $n-k$ needs to be in $[c,d]$, that's all. $\endgroup$ – Matt L. Mar 1 at 22:06
  • $\begingroup$ I see, but why does the boundaries of the summation are given as $\max(a,n-d)$ and $\min(b,n-c)$? @MattL. $\endgroup$ – Mykael Yuday Mar 1 at 22:07
  • $\begingroup$ intersection of sets... $\endgroup$ – Fat32 Mar 1 at 22:08
  • $\begingroup$ Because $k$ needs to satisfy two sets of inequalities, so you need to make sure it does by using $max()$ and $min()$. $\endgroup$ – Matt L. Mar 1 at 22:09
  • $\begingroup$ Ah I see, one last question is there a closed form for the output of the convolution in terms of $\alpha$ and $\beta$? @MattL. $\endgroup$ – Mykael Yuday Mar 1 at 22:10

Given your Eq-3

$$ (x * h)[n]=\sum_{k=\max (a, n-d)}^{\min (b, n-c)} x[k] h[n-k] . \tag{1} $$

where $x[n]$ and $h[n]$ are as in Eq-1, then you will have

$$ \sum_{k=\max (a, n-d)}^{\min (b, n-c)} \alpha \beta = \alpha \beta (\min (b, n-c)- \max (a, n-d) + 1). \tag{2} $$

Where you should evaluate the closed form expression Eq.2 for the non-zero range of output given by the range of $n$ : $a+c \leq n \leq b+d $

  • $\begingroup$ Thank you very much! $\endgroup$ – Mykael Yuday Mar 1 at 22:20
  • $\begingroup$ have you verified it? :-)) $\endgroup$ – Fat32 Mar 1 at 22:20
  • $\begingroup$ Your answer has a beautiful closed form so now it makes sense :) $\endgroup$ – Mykael Yuday Mar 1 at 22:21

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