• 31 Band Eq Frequencies

The standard ISO for a 31 band Eq is as followsHZ:20/25/31.5/40/50/63/80/100/125/160/200/250/315/400/500/630/800/1K/1.25K/1.6K/ 2K/ 2.5K/3.15K/4K/5K/6.3K/8K/10K/12.5K/16K/20K.But I think that this wasn’t the question, the question was how to calculate it, right?First, every octave doubles or divides per two a chosen frequency.Let’s take as a reference 400 Hz, the upper octave of this frequency is 800, and the lower 200. Let’s assume that you want a 1/3 octave Eq: 800Hz-400Hz=400Hz;Then 400Hz/3 = 133.33^HzThis means that the frequencies would be: 400/ 533.33^ (400+133.33^)/ 666.66^ (400+133.33^x2)/ 799.99^ (400+133.33^x3)And for the lower octave 400-200=200; 200/3=66.66^;So: 200/ 266.66/333.3266/399.99Notice that standard ISO has rounded those numbers 200/250/315/400If you want an 1/12 octave for the frequencies among 400 and 800, then 400/12=33.33^Then: 400/433.33^/466.66^/etc.

And then rounded.Notice also that the upper octave of 16K is 32K, so out of human range.I hope it helps.Regards. This is some code that I wrote for GNU Octave that calculates the lower cutoff, center frequency, and upper cutoff. From what I understand it should be close or identical to code that would run in MATLAB. Basically a notes perceived pitch is directly proportional to the logarithm of it's frequency.

The idea is to pick frequencies so that the difference of the logarithms of two adjacent frequencies are constant throughout the spectrum you want to equalize. It's pretty well commented so even if you don't know Octave you should be able to pick it up.Edit: I went ahead and made an account but can't comment in reply to n00dles because some kind of reputation thing.Anyway you're close, n00dles. The difference is my code wants the entire frequency range you want to equalize while the nominal frequency of 20Hz and 20kHz are the center frequencies of the lowest band and the highest band. So since each band is 1/3 of an octave wide, that means the -3dB cutoff of each band is 1/6 of an octave from the center frequency. So the low frequency cutoff for the 20Hz band is 2^(-1/6).

20Hz = 17.818Hz and the upper cutoff frequency is 2^(1/6).20kHz = 22.449kHz. So in order to reproduce the ISO 31 band frequencies, you should use logeqbands(17.818, 22449, 31).Yes this means a 31 band equalizer is equalizing sounds you can't hear. Putting the center frequencies at the edges of the audible range allows you to have a flat response over the audible range. A filter has -3db at the cutoff frequency so if that cutoff frequency was at 20Hz, you would have that much attenuation by the time you reached the bottom of the audible spectrum.I'll go ahead and list the frequencies. Gcc 4.1.2 14 el5 i386 rpm. Note There seems to be some floating point round off error occurring after the 4th significant figure, so I'll give 5. That's still more than what they put on equalizers.19.999, 25.170, 31.687, 39.893, 50.224, 63.229, 79.603, 100.22, 126.17 158.84, 1.9997, 251.76, 316.95, 399.03, 502.36, 632.46, 796.23, 1002.4, 1262.0, 1588.8, 2000.3, 2518.2, 3170.4, 3991.3, 5024.9, 6326.2, 7964.4, 10027, 12623, 15892, 20008.As you can see these numbers match up really well when you round them.

There's a.04% error at both the 20Hz, and 20kHz bands which I assume we know with absolute precision. So I would feel safe trusting these figures to be accurate within 0.1%. That's 0.1442% of an octave or 1.73% of a half step on the chromatic scale.I release this code under the WTFPL License.

If you want to credit someone call me bollocks. ##This function is used to design an audio equalizer.

31 Band Eq Frequencies

The user determines the##frequency range they want the equalizer to handle and how many bands they want##the equalizer to have. The function then calculates the upper and lower cut-off##frequency for each band as well as the middle, a.k.a. Nominal frequency.##The lower bound of the frequencies the user wants the equalizer to handle is##given in the parameter lowerFreq and the upper bound is similarly given in the##upperFreq parameter. The number of bands desired is given in the parameter##numBands. These must all be real positive scalars. NumBands will be cast as a 32-##bit integer if it is not already an integer because fractional bands don't make sense##The output is a matrix with three rows and a number of columns equal to numBands##bands(1,:) is the list of lower cut off frequencies.##bands(2,:) is the list of nominal(center) frequencies##bands(3,:) is the list of upper cut off frequencies##This function is intended to be used to design equalizers for audio use so the##equalizer bands are chosen so the center frequencies sound like the difference##in pitch is constant. This means##diff(log(bands(1,:))) diff(log(bands(2,:))) diff(log(bands(3,:))) K##where K is a constant the upper cut-off frequency of one band will equal the##lower cut-off frequency of the band above it.

For a 31-band equalizer, the formula for the nth band appears to be 10^(( n+12)/10) Hz. If you take these numbers, and round the 3rd digit from the left to 0 or 5, the first few frequencies you get are:. 1 - 20 Hz. 2 - 25 Hz. 3 - 31.5 Hz.

4 - 40 Hz. 5 - 50 Hz. 6 - 63 Hz. 7 - 79.5 Hz. 8 - 100 Hz. 9 - 125 Hz. 10 - 160 Hz.

11 - 200 Hz. 12 - 250 HzThese frequencies exactly match the ones in your question, except that the standard apparently uses 80 Hz instead of 79.5 Hz.

The leading digits repeat every 10 steps, so band 13 is 315 Hz, and band 23 is 3,150 Hz.It seems some of them are doubles. When does it change from double ranges to decade ranges?There is no change-over point. The scale is 10-based, not 2-based, throughout the entire range.

Each step is one decibel of frequency (so 10 steps make a factor of 10).It just so happens, though, that one decibel of frequency is 0.3322 octaves, or almost exactly one third of an octave. This is why the scale approximately doubles every 3 steps.