[LAU] Calculating the sine wave frequencies from the midi key

[Arda Eden]

Thanks Fons,
It's clear now.

On Sun, Oct 19, 2008 at 8:21 PM, Fons Adriaensen <fons at kokkinizita.net>wrote:

> 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.
>
> _______________________________________________
> Linux-audio-user mailing list
> Linux-audio-user at lists.linuxaudio.org
> http://lists.linuxaudio.org/mailman/listinfo/linux-audio-user
>



-- 
Arda EDEN
Cumhuriyet University
Faculty of Fine Arts
Department of Music Technology
Sivas/TURKEY
-------------- next part --------------
An HTML attachment was scrubbed...
URL: http://lists.linuxaudio.org/pipermail/linux-audio-user/attachments/20081019/b0531ad7/attachment.htm