[sdiy] divider ratios

[Andre Majorel]

On 2001-11-25 19:20 -0500, harry wrote:

> > I have been thinking about my problem and if I am right I
> > don't need the division ratios I gave.  Lets consider that
> > the organ oscillator creates the highest note (say C) and
> > the decision chip divides this into the Cs below this than
> > the division ratios would be:
> > 1/8
> > 2/8
> > 3/8
> > 4/8
> > 5/8
> > 6/8
> > 7/8
> > 8/8
> > the input frquency.
> > 1/8,1/2, 1/4 are easy with a four bit counter but the ones
> > in the middle would need multiplying the bottom
> > octave(1/8)by 7,6,5 and 3.
> >
> > Well I have to measure the working chips and see if my
> > assumptions are right.
>
> I think the fractions are all wrong...
> 
> You want (for octaves)
> 1/2
> 1/4
> 1/8
> 1/16
> 1/32....
> 
> Each octave lower is a divide by two...
> (or I don't understand your application, still...)

Harry is right.

$ perl -e 'for ($n = 1; $n <= 8; $n++) {
    print "$n/8 = ", log ($n/8) / log (2) * 12, " semitones\n"
  }'
1/8 = -36 semitones
2/8 = -24 semitones
3/8 = -16.9804499913461 semitones
4/8 = -12 semitones
5/8 = -8.13686286135165 semitones
6/8 = -4.98044999134613 semitones
7/8 = -2.31174093530875 semitones
8/8 = 0 semitones

What these ratios produce are the fundamental and the seven
first harmonics of a note that's three octaves below. Are you
sure this is what you want ???

-- 
André Majorel <URL:http://www.teaser.fr/~amajorel/>
(Not speaking for my employer, etc.)