[time-nuts] Sawtooth correction with a slower TIC

Dr Bruce Griffiths bruce.griffiths at xtra.co.nz
Fri Jul 20 21:07:16 EDT 2007

Richard H McCorkle wrote:
> Gentlemen,
> I would like to pose a theory question and see if my thinking is
> faulty. My original design used a 50 MHz TIC clock and the TMR0
> prescaler as the phase counter giving 20ns single sample resolution
> in the prototype system. A synchronizer is used to insure only
> whole clock pulses are counted. If a 1-second sawtooth correction
> having a 1ns sample resolution is divided by 20 with the low bits
> retained and added to the 20ns single sample phase count, does this
> result in a resolution of 1ns per sample when the corrected samples
> are averaged? Does a faster TIC clock with an external high-speed
> counter really result in better sample resolution as long as the
> TIC clock granularity is within the sawtooth correction limits with
> the TIC giving the the MSB's of the sample and the sawtooth
> correction providing the LSB's of the sample?
> Just Wondering?
> Richard

You are confusing resolution with noise, they are not the same.

Since the quantisation error of the TIC measurements are statistically 
independent of the sawtooth correction, one cannot achieve a measurement 
noise of 1ns using this technique. The TIC noise contribution will be 
20/SQRT(12) ns rms (assuming a uniform error distribution) per 
measurement whereas the sawtooth quantisation error will be around 
1/SQRT(12) ns rms, and the receiver noise will also contribute a noise 
of several ns rms. The resultant noise in the measurement is calculated 
by squaring all the noise terms, adding them together and taking the 
square root (assuming these errors are uncorrelated ie statistically 
independent). Averaging reduces the noise by a factor equal of 1/SQRT(N) 
where N is the number of samples averaged.


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