MHB DFT for convolution like operation.

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The discussion focuses on using the Discrete Fourier Transform (DFT) to optimize convolution operations for finite real-valued sequences. It suggests redefining sequences to include only relevant terms in the convolution sum, allowing for the application of the convolution theorem. The proposed method involves taking the Fast Fourier Transforms (FFTs) of the redefined sequences, performing point-wise multiplication, and then applying the inverse FFT, achieving a complexity of O(n log n). A participant raises a concern about the dependency of the second sum on the index k, which complicates the factorization of the double sum into a product of DFTs. The conversation highlights the challenges in directly applying the convolution theorem under certain conditions.
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Let x,y be finite real valued sequences defined on 0...N-1 and let g be a non negative integer .

define
View attachment 2231
also on 0..N-1.

In addition, the DFT of y is known in closed form.
Is there a way to write z as some cyclic convolution, so that with the help of the convolution theorem z can be calculated in NLOG N istead of N^2?

Thank you
 

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It would seem you can do that. First note that the unusual upper limit of your sum, arbitrarily cut off at $g$, poses no issue for this algorithm. Simply re-define the $x$ and $y$ sequences to include only the terms that show up in the convolution sum. I'm not sure that knowing the FFT of $y$ in advance will help you, since you really need to truncate the sequence, at least in general.

1. Take the FFT's of the re-defined $x$ and $y$. Complexity is $\mathcal{O}(n \, \log(n))$.
2. Multiply the two transformed sequences point-wise. Complexity is $\mathcal{O}(n)$.
3. Take the inverse FFT of the result. Complexity is $\mathcal{O}(n \, \log(n))$.
 
Hi,

lets assume there is no g (g=inf).

notice the the sum end at the index n, not N.

I'm trying to see directly the convolution theorem but I get stuck:
View attachment 2233

The problem is that the second sum depends on k so the double sum doesn't factor to the product of DFTs.

what am I missing?

thank you
 

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