Discrete Convolution of two piecewise sequences having this specific form Assume I have the following two sequences :
$$
x[n]=\begin{cases}
\alpha&amp;\text{if $a\leq n\leq b$}\\ \\ \tag{1}
0&amp;\text{if otherwise}
\end{cases}
\qquad
\text{and}
\qquad
h[n]=\begin{cases}
\beta&amp;\text{if $c\leq n\leq d$}\\ \\
0&amp;\text{if otherwise}
\end{cases}
$$
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.
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 $