Hi guys, s’been a while.
So, I’m making an old school equalizer using the parameter eq dsp. I have a few questions.
[b:1l7c6fgl]1- [/b:1l7c6fgl]first, I need help with the octave parameter.
Given that I want to specify a slider to affect a range from freq1 to freq2 here is my base frequency and octave calculation
fr1 = 0;
fr2 = 400;
basef = (fr2+fr1)/2;
octaverange = (basef-fr1)/(basef);
I just need confirmation that the calculation will affect the range I want.
[b:1l7c6fgl]2- [/b:1l7c6fgl]Given that there are many slider affecting various ranges, I also would like to know if the cut of is abrupt or not and if it is not, then what sort of overlap should I give if any
[b:1l7c6fgl]3-[/b:1l7c6fgl] Finaly, what is the standard frequency range for each slider; where can I get info on the subject. Right now I have 7 sliders, affecting frequencies 0-400, 400-800, 800-1600, 1600-3200, 3200-6400, 6400-9600 and 9600-22000. the last 2 I tweaked a little
- icuurd12b42 asked 6 years ago
[quote:13y84izc][i:13y84izc]the interval between one musical pitch and another with half or double its frequency[/i:13y84izc][/quote:13y84izc]
Fristly frequency zero is nonsensical, it is just DC. If you were to go from say 10Hz to 160Hz it would be doubling four times (10, 20, 40, 80, 160) making it four octaves. All of your ranges are a single octave except the first one and the last two.
The forumla would be something like this:
2^x * fr1 = fr2
where fr1 and fr2 are the two boundary frequency, solve for x (the number of octaves)
2^x = fr2 / fr1
ln(2^x) = ln(fr2/fr1)
x * ln(2) = ln(fr2/fr1)
x = ln(fr2/fr1) / ln(2)
- Guest answered 6 years ago
I got a document specifying the ranges, So that is better in my latest version.
But your sln is a drastic improvement. works almost like the real thing now
It does not take much to overdrive the data resulting is distortion (clipping) though. Reminds me of the first equalizer booster I bought for my car in the early 80’s…. Not sure how to address that
- icuurd12b42 answered 6 years ago
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