math question -Sound Forge - PC - MAGIX Community Forums Archive

Subject:math question Posted by: tcrabb Date:7/30/2002 7:31:18 PM Octaves are non linear and the frequency doubles when you go up one octave. An example would be if I used a center frequency of 400Hz then one octave up would be 800Hz and one octave down would be 200Hz. The center frequency is not the frequency in the middle between 200Hz and 800Hz. Now my question comes from using the multi-band dynamics, paragraphic EQ and other band-notch type applications that use octaves as one of the settings. If I know the frequency range I want to work with, how can I determine the proper center frequency and octave settings so that I only get the the frequency range I am looking for? There has to be a way that I can plug it all into a calculator and get a center frequency and octave setting. I may be way off base and making things harder than they need to be, but I would sure like some insight. Thanks for your time.
Subject:RE: math question Reply by: Chienworks Date:7/30/2002 10:07:47 PM Very easy, especially if you have a calculator. Multiply the lower frequency by the higher one, then take the square root.
Subject:RE: math question Reply by: tcrabb Date:7/30/2002 11:32:18 PM Great, Thanks..... Now lets say I want to set up a band-notch between 315Hz and 3.15Khz. I come up with a center frequency of 996Hz. How do I figure out the octave setting?
Subject:RE: math question Reply by: Chienworks Date:7/31/2002 7:11:46 AM This is slightly more complicated; you need a calculator with a LOG key. Calculate the Log of 2. It doesn't matter if you use common log or natural log (or if you even understand what these are!) as long as you use the same one throughout. Most calculators use "Log" for common log and "Ln" for natural log. Pick one or the other and stick with it. For this example i'll use common log. The common log of 2 is 0.301029995. Now divide the high frequency by the center frequency 3150 / 996 = 3.162650602, take the log of this number to get 0.500051215 and divide this by the log of 2 to get 1.661134181. The higher frequency is 1.66 octaves above the center frequency. If you want to calculate the range from high to low, then 3150 / 315 = 10, log 10 = 1 (yes, exactly one), and 1 / log 2 = 3.32, so the span from 315 to 3.15K is 3.32 octaves. If you really want to know what common logs and natural logs are i can bore you with the information, but this formula works fine even if you have no clue. In general: octaves = ( log (high frequency / low frequency) ) / log 2
Subject:RE: math question Reply by: tcrabb Date:7/31/2002 10:01:08 PM Perfect......That is exactly what I was looking for......Thanks......Tim

Subject:math question

Posted by: tcrabb

Date:7/30/2002 7:31:18 PM

Octaves are non linear and the frequency doubles when you go up one octave. An example would be if I used a center frequency of 400Hz then one octave up would be 800Hz and one octave down would be 200Hz. The center frequency is not the frequency in the middle between 200Hz and 800Hz. Now my question comes from using the multi-band dynamics, paragraphic EQ and other band-notch type applications that use octaves as one of the settings. If I know the frequency range I want to work with, how can I determine the proper center frequency and octave settings so that I only get the the frequency range I am looking for? There has to be a way that I can plug it all into a calculator and get a center frequency and octave setting. I may be way off base and making things harder than they need to be, but I would sure like some insight.
Thanks for your time.

Subject:RE: math question

Reply by: Chienworks

Date:7/30/2002 10:07:47 PM

Very easy, especially if you have a calculator. Multiply the lower frequency by the higher one, then take the square root.

Subject:RE: math question

Reply by: tcrabb

Date:7/30/2002 11:32:18 PM

Great, Thanks.....
Now lets say I want to set up a band-notch between 315Hz and 3.15Khz. I come up with a center frequency of 996Hz. How do I figure out the octave setting?

Subject:RE: math question

Reply by: Chienworks

Date:7/31/2002 7:11:46 AM

This is slightly more complicated; you need a calculator with a LOG key. Calculate the Log of 2. It doesn't matter if you use common log or natural log (or if you even understand what these are!) as long as you use the same one throughout. Most calculators use "Log" for common log and "Ln" for natural log. Pick one or the other and stick with it. For this example i'll use common log. The common log of 2 is 0.301029995. Now divide the high frequency by the center frequency 3150 / 996 = 3.162650602, take the log of this number to get 0.500051215 and divide this by the log of 2 to get 1.661134181. The higher frequency is 1.66 octaves above the center frequency. If you want to calculate the range from high to low, then 3150 / 315 = 10, log 10 = 1 (yes, exactly one), and 1 / log 2 = 3.32, so the span from 315 to 3.15K is 3.32 octaves.

If you really want to know what common logs and natural logs are i can bore you with the information, but this formula works fine even if you have no clue. In general:

octaves = ( log (high frequency / low frequency) ) / log 2

Subject:RE: math question

Reply by: tcrabb

Date:7/31/2002 10:01:08 PM

Perfect......That is exactly what I was looking for......Thanks......Tim