19 Oct
2008
19 Oct
'08
3:31 p.m.
Thanks Fons,
It's clear now.
On Sun, Oct 19, 2008 at 8:21 PM, Fons Adriaensen <fons(a)kokkinizita.net>wrote;wrote:
On Sun, Oct 19, 2008 at 08:00:26PM +0300, Arda Eden
wrote:
--
Arda EDEN
Cumhuriyet University
Faculty of Fine Arts
Department of Music Technology
Sivas/TURKEY
note_frqs[i] = (2.0 * 440.0 / 32.0) * pow(2,
(((jack_default_audio_sample_t)i - 9.0) / 12.0)) / srate;
I couldn't catch the relationship in the calculation above.
Midi note numbers are just a count in semitones.
There are twelve semitones in an octave, which
is a frequency ratio of 2:1.
So given two note numbers that are N semitones apart,
the ratio of their frequencies is
2 ^ (N / 12), or in C: pow (2.0, N / 12.0)
Now MIDI note number 9 (an 'A', note number 0 is the 'C'
below that) has the frequency 880 / 32 Hz, five octaves
below the 880 Hz 'A'.
So to get the scale right we use
F = 880.0 / 32 * pow (2.0, (i - 9) / 12.0)
For i = 9 this becomes
F = 880.0 / 32 * pow (2.0, 0)
= 880.0 / 32 * 1
= 880.0 / 32
as required.
Ciao,
--
FA
Laboratorio di Acustica ed Elettroacustica
Parma, Italia
Lascia la spina, cogli la rosa.
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