# [time-nuts] First success with very simple, very low cost GPSDO

Dennis Ferguson dennis.c.ferguson at gmail.com
Fri Apr 11 14:19:26 EDT 2014

```On 10 Apr, 2014, at 22:06 , Chris Albertson <albertson.chris at gmail.com> wrote:
>> You originally described a system that counts to 5M every second.  Tom and
>> others pointed out that you do not need the complete 5M count, all you need
>> is the remainder of a modulo count.  The question then is, how much of a
>> remainder do you need to be sure that it spans all anticipated errors in
>> both the PPS and the oscillator?
>
>
> Yes this right.  I'm sure that under limited conditions I can get by
> looking only at the remainder.  It is harder than was said because the
> overflows per second rate is a non-interger but there are till only two
> "flavors" of seconds:  Those with N overflows and those with N+1 overflows.

I think you are seeing a complexity that isn't there.  5,000,000 is
19531 * 256 + 64, so if you get an 8-bit counter sample t0 in one
second and t1 in the next then the difference

t1 - t0

will be equal to 64 if the oscillator is on frequency, and

t1 - t0 - 64

will be equal to 0, when computed with 8 bit arithmetic.  If the result of
the latter, interpreted as a signed value, is less than zero your oscillator
is going too slow, if greater it is going too fast.  This will be true whether
there have been 19531 or 19532 counter overflows in the second; the (t1-t0)
subtraction will set the borrow bit in the latter case if you want to know
that but there is no reason to care about it.  8 bits is enough to measure a
+/- 25 ppm error in this case, which seems sufficient for any oscillator
you are likely to want to discipline with this.

All keeping the full count seems to do is require you to subtract two
much larger numbers to find the same small difference.

Dennis Ferguson
```