IPB

Welcome Guest ( Log In | Register )

 
Reply to this topicStart new topic
Frequency bands and octaves, about limited frequency ranges, and FFT depth
pkh
post Jan 20 2012, 13:44
Post #1





Group: Members
Posts: 13
Joined: 28-December 10
Member No.: 86869



Hi,

I'm trying to implement "Polyphonic Audio Key Finding Using the Spiral Array CEG Algorithm by Ching-Hua Chuan and Elaine Chew", but I'm not sure to understand a few things.

It is stated: "We limit the frequency range to be from 32 Hz (C1) to 1975 Hz (B6)". My input is sampled at 44100Hz and I have a FFT depth of 9 bits (→ windows size is 512), If I want to limit my analysis to these ~1900Hz, do I have to use an average of 22 bins only (512 * (1975-32) / 44100) and discard all the others?

And worse, these frequencies seem to be split in 12 octaves (isn't it supposed to be 8 btw?), which mean I would only have 2 bins per octave (assuming octaves are evenly distributed in the 32 to 1975 range which is afaik not the case...).

I'm certainly confused about a few things, so any explanation is welcome…

Thanks,
Go to the top of the page
+Quote Post
alexeysp
post Jan 21 2012, 13:55
Post #2





Group: Members
Posts: 130
Joined: 3-April 09
Member No.: 68627



QUOTE (pkh @ Jan 20 2012, 15:44) *
If I want to limit my analysis to these ~1900Hz, do I have to use an average of 22 bins only (512 * (1975-32) / 44100) and discard all the others?


Basically, yes. If you need better frequency resolution, you have to either increase the window length, or downsample the signal prior to analysis.

QUOTE
these frequencies seem to be split in 12 octaves (isn't it supposed to be 8 btw?)


An octave corresponds to doubling or halving the base frequency, so the number of octaves in a given range depends on what frequency is chosen as a base.

QUOTE
assuming octaves are evenly distributed in the 32 to 1975 range which is afaik not the case


The octaves divide the frequency range in a logarithmic manner, and FFT operates on a linear frequency scale. So yes, you will have varying number of bins per octave.
Go to the top of the page
+Quote Post
alexeysp
post Jan 22 2012, 09:42
Post #3





Group: Members
Posts: 130
Joined: 3-April 09
Member No.: 68627



Okay, correction to my previous post (I have now actually read the paper you linked).

What I said above about the number of octaves is, of course, incorrect. Regardless, the number 12 mentioned in the paper is not the number of octaves, it's a number of pitch classes, corresponding to chromatic scale. Every pitch class contains all frequency bands corresponding to the standard frequencies for a given pitch. For example, class C will contain the bands centered around the frequencies of 32.703, 65.406, 130.813, 261.626, 523.251 and 1046.502 Hz; class A will contain the bands centered around the frequencies of 55.000, 110.000, 220.000, 440.000, 880.000 and 1760.000 Hz etc. The width of each band is determined by the adjacent standard pitch frequencies within the same octave. You can find the table of standard pitch frequencies here: http://people.virginia.edu/~pdr4h/pitch-freq.html. Peak values found from FFT analysis are then summed within each class to determine the pitch strength.

Go to the top of the page
+Quote Post

Reply to this topicStart new topic
1 User(s) are reading this topic (1 Guests and 0 Anonymous Users)
0 Members:

 



RSS Lo-Fi Version Time is now: 25th July 2014 - 15:30