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@xxxxxxxxxxxxxxxxxxxx http://lists.linuxaudio.org/mailman/listinfo/linux-audio-user
