Convolution of two discrete sequence

[nacho-man]

Hi,

New to this topic, and need some help.

My task is to find the convolution between
$ y= x ∗ h$

where $x = u_n - u_{n-N}$ and $h_n = u_n - u_{n-M}$ and $M\ge N$ are positive integers

My understanding is that

in general, $ y= x ∗ h = \sum\limits_{m=-\infty}^\infty x_m h_{n-m} $

so for my question i get

$\sum\limits_{m=-\infty}^\infty (u_m-u_{m-N})(u_{m-n}-u_{m-M})$
is there anything further i can do here? It doesn't feel complete, and to be honest, the idea of convolution still seems vague to me.

$u_n$ is the step function

 

[I like Serena]

Hi,

New to this topic, and need some help.

My task is to find the convolution between
$ y= x ∗ h$

where $x = u_n - u_{n-N}$ and $h_n = u_n - u_{n-M}$ and $M\ge N$ are positive integers

My understanding is that

in general, $ y= x ∗ h = \sum\limits_{m=-\infty}^\infty x_m h_{n-m} $

so for my question i get

$\sum\limits_{m=-\infty}^\infty (u_m-u_{m-N})(u_{m-n}-u_{m-M})$
is there anything further i can do here? It doesn't feel complete, and to be honest, the idea of convolution still seems vague to me.

$u_n$ is the step function

Hi nacho!

First off, it should be:
$$y_n = (x*h)_n = \sum\limits_{m=-\infty}^\infty (u_m-u_{m-N})(u_{n-m}-u_{n-m-M})$$

Can you eliminate the parentheses?From the definition of convolution, we have:
$$(u*u)_n = \sum\limits_{m=-\infty}^\infty u_m u_{n-m}$$

Then, with $k=m-N \Rightarrow m=k+N$, this implies for instance:
$$\sum_{m=-\infty}^\infty u_{m-N} u_{n-m}
=\sum_{k=-\infty}^\infty u_{k} u_{n-(k+N)}
=\sum_{k=-\infty}^\infty u_{k} u_{(n+N)-k}
= (u*u)_{n+N}$$

Can you write $y_n$ as a combination of such self-convolutions?