[time-nuts] 32768 Hz from 10 MHz
azelio.boriani at screen.it
Thu Feb 2 21:10:41 UTC 2012
Now I'm thinking that starting with a first run of 8 cycles at 500nS + 2
cycles at 400nS to be repeated for 10 times and then inserting 2 cycles of
400nS, a first approximation of my 2.048MHz can be done. Maybe with a
deltaF/F of 10 at -4 for tau 1 second but it can be done. In the very long
run the count will be correct and the accuracy gets better tau after tau.
Of course there is the source oscillator's limit.
On Thu, Feb 2, 2012 at 9:21 PM, Tom Van Baak <tvb at leapsecond.com> wrote:
> Hi Roberto:
>> By changing the timer count dynamically it's possible to lower the jitter
>> to one timer count. See:
>> Have Fun,
>> Brooke Clarke
> Hi Brooke,
> You're a fellow PIC guy; let me explain.
> Correct, that method works with a modest interrupt rate to count
> integer seconds without long-term rounding error; but to generate
> a total of 32,768 as-consistent-as-possible pulses *per* second
> is quite different.
> It's possible to use Bresenham with two integers 10,000,000 and
> 32,768 but I found no way to perform all the 24-bit calculations
> on an 8-bit PIC quick enough. Removing the GCD often helps
> but in this case the accumulator remains 3-bytes wide.
> To generate 32 kHz you have to toggle a pin and calculate if
> the next toggle must be 38 or 39 instructions in the future; all
> the math must occur within 37 instructions. That's why I came
> up with the binary leap year kind of algorithm; it's as close to
> math-less as you can get.
> By comparison, all the decimal dividers (1 Hz, 10 Hz, etc.) that
> you and I do are trivial because of the common factors with the
> 10 MHz clock. It's just that 32,768 has no factors of 5. Read the
> comments in the file 10m32k.c for more details.
> I'm curious how a 10 MHz-driven high-end DDS would generate
> 32 kHz with the lowest possible jitter?
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