Comments on: FFT Averages

By: Discrete Fourier Transform (eeks)…

Discrete Fourier Transform (eeks)… — Fri, 12 Feb 2010 01:28:54 +0000

[...] (creating sub bands by concatenating and averaging the amplitudes for beat detection as described here), but it would still be nice to know what exactly i can expect. From what i can tell it is [...]

By: Beat It!

Beat It! — Mon, 08 Feb 2010 23:47:39 +0000

[...] way to do is is to use octaves as explained in this fine blog entry. This way of sub-band layout makes a lot more sense from a musical point of view. The article comes [...]

By: Ben Stucki » The Math Behind Flash’s FFT Results

Ben Stucki » The Math Behind Flash’s FFT Results — Mon, 16 Nov 2009 18:32:24 +0000

[...] of searching on my magic number and a little bit of luck, I came across this amazing blog post: http://code.compartmental.net/2007/03/21/fft-averages/ – and Eureka! Each point i in the FFT represents a frequency band centered on the frequency [...]

By: Jan

Jan — Sun, 29 Mar 2009 15:29:51 +0000

What I’ve noticed is that the number of bands equals the log2 of N, N being the number of samples in your FFT.
For an FFT with size 1024 this makes log2(1024) = 10.

Now from the second band to the last this is no problem.
You just use the ranges from 2^(bandno-2) to twice this number. So band 2 gives 2^(2-2) = 1 to 2 and band 10 gives 2^(10-2) = 256 to 512. Exactly like in the example given.

But the first band is somewhat of a difficult example. If u use a frequency of N/2 with amplitude 1 you get a nice 1 in your result (at the appropriate frequency sample), but if you use a constant DC with the value 1, you get 2 as a result in the FFT.

So my question is how to use the first band exactly. The page at http://www.dspguide.com/ch8/5.htm shows the first band runs from 0Hz to f0/N Hz where f0 is the sampling rate and N the size of the FFT. When we use the example of 44100 and 1024, we get band 0 running from 0 to 22Hz. Now how do we combine this band with band 1 (running from 22 to 66 with a center of 44Hz) to get our center frequency of 33Hz?

My guess is (0.5*band0 + band1)/2. This so we take into account the double size of band0.

Another idea if you want a nicer output, we hear things not only with a log frequency, but with a log amplitude as well. So when you have the results, take the log10 of those. (but beware log10(0) is undefined).

These are just my 2 cents, any comments would be really appreciated.

By: balu

balu — Mon, 23 Mar 2009 17:57:22 +0000

hello

I am using minim: FFT and I am testing Logarithmic Averages.
How can I get the spectrum[j]? please help me.

By: dekoffie

dekoffie — Tue, 17 Mar 2009 00:57:03 +0000

Please erase my last comment, I was being too fast and not taking enough time to think it through. It’s impossible to get more than 10 ranges, because the bandwidth for each frequency bin is 43.0664062Hz (in the files I am using here, with a samplerate of 44100 and an fft on 1024 samples). Range 1 & 2, as in the example, will always return the index 0 because the requested frequency interval is smaller than each bin contains.

By: dekoffie

dekoffie — Thu, 26 Feb 2009 20:17:15 +0000

About the low and hibound being zero;
when I implement your code, I get the following:

range 1=[0; 10], indexes 0-0
range 2=[10; 21], indexes 0-0
range 3=[21; 43], indexes 0-1
range 4=[43; 86], indexes 1-2
range 5=[86; 172], indexes 2-4
range 6=[172; 344], indexes 4-8
range 7=[344; 689], indexes 8-16
range 8=[689; 1378], indexes 16-32
range 9=[1378; 2756], indexes 32-64
range 10=[2756; 5512], indexes 64-128
range 11=[5512; 11025], indexes 128-256
range 12=[11025; 22050], indexes 256-512

So,should I just discard the two first ranges? How should I interpret these two ranges? Thanks for any tips

By: SoundRoom » Blog Archive » Day in GCCIS NML

SoundRoom » Blog Archive » Day in GCCIS NML — Tue, 20 Jan 2009 01:43:55 +0000

[...] http://code.compartmental.net/2007/03/21/fft-averages/ [...]

By: catchit2000

catchit2000 — Thu, 17 Jul 2008 00:00:09 +0000

what about geometric averages or arithmetic squared averages? not sure if it even makes sense.
thanks

By: ddf

ddf — Sun, 09 Sep 2007 19:50:13 +0000

Ah, you are correct, I was misremembering how Math.pow() works. I just checked out the code for doing this in Minim and the potential divide by zero line looks like this:

avg /= (hiBound – lowBound + 1);

That probably means I ran into the divide by zero problem while developing the library. Sorry about that. I’m going to update this post.