Detuner question: How much is a cent in the detuner?
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PaulRiedel ·
Hi!
Another question:
As I understand correctly, doubling the value in the detuner raises the octave by one, so 1200 cent.
But how much is one cent then?
According to my leftover common sense, a cent would be 2 / 1200 = 0.00166666...7, but in the semitone detuner module posted by someone here, the value for one semitone are:
+1: 1.05946
+2: 1.12246
+3: 1.18921
+4: 1.25992
+5: 1.33484
+6: 1.41421
+7: 1.49831
+8: 1.5874
+9: 1.68179
+10: 1.7818
+11: 1.88775
Shouldn't they be this though?:
+1: 0.1666...7
+2: 0.333...3
+3: 0.5
+4: 0.666...4
+5: 0.8333...3
+6: 0.999...6
+7: 1.1666...2
+8: 1.333...28
+9: 1.4999...4
+10: 1.666...6
+11: 1.8333...26
And so how much does the "Fine" know detune the Signal? I believe it's not by any "real" cent value, correct?
Thanks a lot for all your guys' help!!
- Paul
doug4350 ·
Hi Paul,
If memory serves, think of a cent as centi- applies to the metric system. A cent is 1/100 of the frequency of the note being played (i.e. a semitone). So, an A4 (at 440 Hz) would have a cent of 4.40 Hz. An A5 (octave higher at 880 Hz) would have a cent of 8.80 Hz. I (still) haven't dove deep enough into dsp to realize the relationship between Hz and -1 - +1, but I'd figure there's some proportionality there... As there are 12 notes (semitones) in an octave explains why 1200 cents is an octave higher. I think you might be misunderstanding that an octave up is twice the frequency in Hz or something like that.
Hope that makes cents 🙂
Aloha,
Doug
stefan ·
The toolbox detuner works on frequency and not pitch. If you want to convert from frequency to pitch you do:
freq = baseFrequency * (2 ^ (semitones / 12))
which means that if you want to calculate what to multiply a frequency with to for instance raise it one semitone you'd do
multiplier = 1 * (2 ^ (1/12));
or two semitones:
multiplier = 1 * (2 ^ (2/12));
and so forth
hopefully this will get you the values seen in that list of values from the semitone detuner 🙂
PaulRiedel ·
Hmm, still doesn't make too much sense to me.
Don't you agree that one cent must be 0.001666...7 if an Octave change is NewFreq = OldFreq * (2^Octavechange)?
(so for 440 Hz: 880 = 440 * (2^1))
Because 2 / 1200 = 0.001666...7?
Maybe if thought about it too much now and ran head into a wall that I'm stuck in now 😉
stefan ·
PaulRiedel wrote: Hmm, still doesn't make too much sense to me.
Don't you agree that one cent must be 0.001666...7 if an Octave change is NewFreq = OldFreq * (2^Octavechange)?
(so for 440 Hz: 880 = 440 * (2^1))
Because 2 / 1200 = 0.001666...7?
Maybe if thought about it too much now and ran head into a wall that I'm stuck in now 😉
No, I do not agree 🙂
A cent change would be a multiplier of 2^(1/12), which is approx 1.059. If you multiply a frequency with this, you'll move up one cent.
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