1
$\begingroup$

This will be maybe quite easy fore somebody but I am not sure how to solve it. If I have a signal which is equal to

$$ y(n)=x(n)\star g(n), \quad n\in[0,1,...,N] $$ where $\star$ is convolution operator, how do I get expression for taking every $K^{\textrm{th}}$ sample of $y(n)$, i.e., $y(Kn)$?

$\endgroup$
1
$\begingroup$

how do I get expression for taking every Kth sample of y(n), i.e., y(Kn)?

What I understand from this is you want a notation that represents $y[Kn]$ as a convolution operator. That doesn't exist. Or I cant remember.

For example the notation: $$y[Kn] = x[Kn]*g[n]$$ is wrong.

The following is wrong either: $$y[Kn] = x[Kn]*g[Kn]$$

Fundamentally $y[Kn]$ is defined as the samples of $y[n]$ as: $$y[Kn] = y[n]|_{n=Kn} = \sum{x[k]g[Kn-k]}$$

This last sum cannot be defined in a compact and simple manner by using the base signals $x[n]$,$g[n]$ and convolution operator alone. Rather the preferred way is to define $y[n] = x[n]*g[n]$ and indicate $y[Kn]$ to be used explicitly.

I strongly suggest you to look at the books and papers on the multirate signal processing subject in which such operations are abundant and some exclusive notation might have been inrtoduced so far.

$\endgroup$

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.