[PD] Relationship between constant-Q and FFT - can you do it in vanilla?

[Thomas Grill]

Btw., just committed the solution to github, additionally implementing the mel frequency scale.
https://github.com/grrrr/spect_scaled

Looking forward to contributions.
best, Thomas

--
Thomas Grill
http://grrrr.org



> Am 13.07.2017 um 02:00 schrieb Thomas Grill <gr at grrrr.org>:
> 
> Hi,
> apart from true CQ Gabor transformations there is a method for estimating the power spectra of arbitrary frequency scales. It's probably what you are looking for.
> It is based on summing up power spectrum bins by use of triangular windows.
> I made an implementation which is fairly general. It employs some dynamic patching to generate the individual filter bands.
> spect_scaled.pd is the main abstraction, it comes with a demo file spect_scaled-help.pd
> 
> have fun, Thomas
> 
> --
> Thomas Grill
> http://grrrr.org
> 
> <hz-log.pd><list-map.pd><log-hz.pd><scl2bin.pd><spect_part.pd><spect_parts.pd><spect_scaled-help.pd><spect_scaled.pd>
> 
> 
> 
>> Am 10.07.2017 um 22:01 schrieb Alexandre Torres Porres <porres at gmail.com>:
>> 
>> Howdy, trying to figure the relationship between constant-Q and FFT, and if you can make a constant-Q patch in vanilla based on [fft~]. This would be for an android app, so it'd be good if I didn't need an external.
>> 
>> I was checking the [constantq~] external, which is found here and in 'deken'. it's based on a paper by Puckette and Brown. I need help cause I can't read math or figure out code reasonably well, so I don't really know how it works. It seems it could be done just by rescaling the output of an FFT with logarithmic values, but I could be terribly wrong.
>> 
>> Thing is that I already have a pd patch with a logarithmic output here - this is not vanilla, but made on extended, but easily adaptable for vanilla (I'm also attaching it).
>> 
>> Also, how about just having a bank of filters with center frequencies around the tempered scale, could you have in practice the same thing as from [constantq~]?
>> 
>> thanks
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