1
$\begingroup$

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.

$\endgroup$
11
  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ Mar 1 at 22:10
0
$\begingroup$

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 $

$\endgroup$
3
  • $\begingroup$ Thank you very much! $\endgroup$ 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 :) $\endgroup$ Mar 1 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.