Voit lisätä twiitteihisi sijainnin, esimerkiksi kaupungin tai tarkemman paikan, verkosta ja kolmannen osapuolen sovellusten kautta. Halutessasi voit poistaa twiittisi sijaintihistorian myöhemmin. Lue lisää
Right, I've been trying to work this out for about 3 hours now. I feel like I'm on the cusp of understanding but just need a straightforward answer from someone knowledgeable in audio-related maths. I'll reply to myself with the problem since I've run out of space in this tweet.
I have an FFT of size S (thus S/2-1 bins), and a Nyquist frequency of F. I want to display a spectrum analyzer with B bars. Given these, how do I calculate the number of FFT bins to combine for each bar from 0 to B-1?
In simpler terms: I have a histogram with N values, but i want to show a histogram with M values. It's the same problem exactly :) One way would be to think of these as rectangles and literally integrate the old histogram over each new rectangle.
Like, if a new histogram bin goes from 0.5 to 0.6, what is the average height of the data in the old histogram from 0.5 to 0.6? That's the height to use for the new histogram bin.
You might want to consider grouping the Fourier coefficients by Barks, which correspond to perceptually correct bands. If the number of bars you want to display is less than 24, then group barks into a single bar.
If you have a real FFT, there should be S/2+1 bins (S/2-1 complex frequencies + 1 real DC + 1 real nyquist). I’ve done this multiple times and it never looks particularly great. My usual approach is to assume each edge has a bin index of A exp(B n) ...
Where n is the band index. A and B are derived to ensure that the final band index gives the last bin index and that the low bands use more than one bin.
I wish I would have seen this 2 days ago....
Why's that?
Twitter saattaa olla ruuhkautunut tai ongelma on muuten hetkellinen. Yritä uudelleen tai käy Twitterin tilasivulla saadaksesi lisätietoja.
