Usually it's as Kassen said - looks like an exponential curve to me,
but the theory behind it is that the frequency bands should be double
with each step, like Kassen said. So, if we're running at 44100 then
the frequency interval btw bands is 22100/256 (highest freq we can
detect is always 1/2 the sample rate) = 86.328125
;-------
(define num-bins 256)
(define freqs (list 1 2 3 4 6 8 11 15 21 29 41 55 75 102 138 188 256))
(define half-sr 22100)
(define bin-width (/ half-sr num-bins))
(define bin-freqs (map (lambda (f) (round (* f bin-width))) freqs))
(display bin-freqs)(newline)
;-------
Gives:
(86 173 259 345 518 691 950 1295 1813 2504 3539 4748 6475 8805 11913
16230 22100)
Which isn't ideal - we could easily throw out the 86 at the start and
the 16230 and 22100 at the end and increase the range in the middle.
Of course, this means that the range won't be perceptually accurate,
but who cares - that won't help its usefulness.
So, taking bins (starting numbering at bin 1, not 0):
2 3 4 5 6 7 8 11 15 21 29 41 55 75 102 138
(freqs: 173 259 345 432 518 604 691 950 1295 1813 2504 3539 4748 6475
8805 11913)
Looks more useful, musically, to me. Up for debate, though. Its a
pretty arbitrary thing.
Also, how come you were using 17 bins in the original? Shouldn't that
have been 16?
Cheers
Evan
On Jan 17, 2010, at 9:33 PM, Matt Jadud wrote:
Hi all,
On Sun, Jan 17, 2010 at 12:40 PM, Kassen
<[email protected]> wrote:
of this was obvious. I didn't do the math, I'll trust Matt that it
checks
out. I would BTW expect these numbers to depend on the samplerate
that
All I did was ask Excel fit an exponential to those numbers (leaving
out the zero). As the saying goes, "It looks like an exponential, and
quacks like an exponential..." Whether those are meaningful
coefficients, that's another thing entirely...
Cheers,
Matt
