On Sun, Oct 19, 2008 at 08:00:26PM +0300, Arda Eden wrote:

> 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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