How does Parceval's theorem help with an efficient CQT?

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Parceval's theorem plays a crucial role in enhancing the efficiency of the constant Q transform (CQT) by allowing pre-calculation of a kernel that can be repeatedly multiplied with the input signal, thus reducing processing power requirements. The discussion highlights the algorithm by Judith Brown and Miller Puckette, which suggests that using FFTs for both the kernel and input signal can lead to significant speed improvements compared to time-domain multiplication. The efficiency gain arises from the fact that many terms in the summation of the FFT are close to zero, allowing for a reduction in computational load without significantly affecting pitch identification accuracy. The conversation also touches on potential issues with viewing the referenced article, indicating that some users may experience difficulties accessing the first page. Overall, understanding the implications of Parceval's theorem in this context is essential for optimizing real-time pitch detection applications.
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Hi,

First, apologies if this is the wrong forum or even the wrong website. The site popped up when searching for a math forum to help me implement an equation efficiently. I’m not sure if this is the place for discussing computational efficiency, but since the efficiency is achieved by way of mathematical theorems, I thought it might fit.

I’m trying to implement an efficient version of the constant Q transform in an attempt to determine pitch for a real-time application, but the math part of my brain has been sitting unused for some years and Google doesn’t seem to hold much information.

I’ve stumbled upon an algorithm derived by Judith Brown and Miller Puckette in an (easy to skim) article here:
http://academics.wellesley.edu/Physics/brown/pubs/effalgV92P2698-P2701.pdf

I’ve verified that the algorithm works (although not yet its speed), but I need to understand it better.
The gist of the article is (if I understand correctly, of course) that a kernel (equation 4) can be pre-calculated and then used repeatedly for multiplication with the input signal. That seems reasonable: less processing power required, much more memory needed.

The article seems to imply that using a form of Parceval’s relation (equation 3) is what provides the speed. I can’t understand the efficiency gain in multiplying an FFT’d kernel with an FFT’ed input, as opposed to just performing the multiplication in the time domain (first and second part of equation (5) in the article).

What’s the catch? I guess what I’d like to know is what the use of Parceval’s theorem really is (especially coupled to the CQT).

If this is such an elementary question that you don’t want to reward it with a lengthy answer, pointers to relevant articles would be sufficient.

Also, if the first thought that pops into your head is “that’s not at all efficient. I know a much better way to do this”, please write away.

Thanks in advance.
 
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When I look at that document,the first page is blank. Anyone else able to see the first page ?
 
When I click on the link in my post, I get the PDF document displayed as it should. I can't see anything different between page one and the others (apart from the information, of course), so I don't know why the first page would appear to be invisible for you.

I use Safari and its built-in PDF reader, btw, although it shouldn't matter.
 
I think the point is that in equation (5) the sum is not really over N-1 terms because most of the K* terms are close to zero, as shown in Fig 2. Fig 3 is presumably some measure of the error introduced by dropping the near-zero terms. If the objective is to identify discrete 12-tone or 24-tone pitches in a scale, errors of a fraction of a tone interval (e.g. less than half) probably don't matter.
 
Of course. It's funny how one's mind works. I've read the article several times, but somehow I've completely missed that (which some might argue is a central point of the article).

It makes a lot more sense now. Thanks.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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