0
$\begingroup$

I was reading circular convolution topic from Digital Signal Processing Using Matlab 3rd Ed,by Proakis. I came across a strange term/symbol; I have highlighted it in attached snapshot

I have also attached snap of eq 5.24,in 2nd snap, where this RN(n) symbol appears also:

enter image description here

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Can you give a little more context on the topic being explained here? By looking at just this I can only imagine that this term represents $n^{th}$ component of a Vector in N dimension, usually represented as $\mathcal{R}^N$ $\endgroup$ – DSP Rookie Jun 20 at 11:44
  • $\begingroup$ Proakis / edited multiple books, in multiple revisions. Which one is this, which page? $\endgroup$ – Marcus Müller Jun 20 at 11:53
2
$\begingroup$

This symbol denotes a rectangular pulse of length $N$:

$$\mathcal{R}_N(n)=\begin{cases}1,&0\le n\le N-1\\0,&\textrm{otherwise}\end{cases}$$

I'm not sure where it is defined for the first time, but this definition is clear from the equation above Eq. $(5.24)$ on page $130$ of the 3rd edition of Digital Signal Processing Using Matlab by V.K. Ingle and J.G. Proakis.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I have attached snap of eq 5.24 ,but sorry for that equation is bit different from your $\endgroup$ – engr Jun 20 at 14:09
  • $\begingroup$ @engr: I don't see how it's different. If you interpret $\mathcal{R}_N(n)$ as in my answer, it matches exactly the equation given in the book. $\endgroup$ – Matt L. Jun 20 at 14:49
  • $\begingroup$ Because apparently the equation involves product of RN(n) & X(k) . Beacuse RN(n) is not alone $\endgroup$ – engr Jun 20 at 17:06
  • $\begingroup$ @engr: That's right, but from the equation you can infer what is meant by that symbol. $\endgroup$ – Matt L. Jun 20 at 17:11
  • $\begingroup$ Yes, i understood, that is why, i clicked accept answer $\endgroup$ – engr Jun 20 at 17:31

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.