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A practical example: I perform a DCT on a time series of discrete values that are spaced in time by 1/30 of a second. What frequencies each bin of this DCT represents? What is the formula to find that?

If I want to filter the DCT to remove all signal bins that correspond to frequencies below 0.4 Hz and above 4 Hz, and keep the values between these intact, how should the filter impulse be constructed?

I understand that in order to filter that DCT I have to to build a filter array that is a series of zeros in the bins I want to remove and ones in the bins I want to keep and multiply elements in each array by its correspondent in the other, zeroing the bins I don't want.

But I need to know the frequency each bin on the original DCT represents.

Please help.

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  • $\begingroup$ you probably mean the DCT-II of the DCT. The wikipedia article has the formula. You'll find the cosine in there; the cosine argument depends on the bin number $k$. The derivative of the cosine's argument w.r.t. $n$ is the frequency. $\endgroup$ – Marcus Müller Feb 21 '19 at 7:58
  • $\begingroup$ Yes, I mean DCT-II. Sorry for my ignorance, but I do not have a clue on how to extract the frequency in Hertz from that formula. That is Klingon to me. wrtn???? $\endgroup$ – SpaceDog Feb 21 '19 at 17:42
  • $\begingroup$ So, the frequency of a cosine tone $\cos(2\pi f n)$ is $f$. You just need to set the argument to the $\cos$ in that formula $=2\pi f n$ and solve for $f$. $\endgroup$ – Marcus Müller Feb 21 '19 at 18:10
  • $\begingroup$ I appreciate your effort, thanks, but It is not that easy to find the frequency, because each bin is a sum from 1 to n. Can you please post an answer giving a numeric example? Suppose the 5 first bins of the DCT are: 3, 2, -1, 4. What frequencies in Hz these bins represent? $\endgroup$ – SpaceDog Feb 21 '19 at 18:44
  • $\begingroup$ Seriously, I can't simplify it any further. you literally just have to take the argument of the $\cos$ from that formula, in that sum, and set it $=2\pi fn$, and solve for $f$. $\endgroup$ – Marcus Müller Feb 21 '19 at 19:09

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