[time-nuts] TimeLab phase difference (slope sec/sec)

[Chris Burford]

Hi Bob,

I'm seeing 4.22E-12 as the slope value in the upper right of the TimeLab 
phase difference plot. Is that telling me that my DUT is within +4.22ps 
/ sec from my reference 1PPS for the 24 hour measurement duration?

I have attached a screen capture that will hopefully make its way 
through for viewing.

Thanks,

Chris

On 07/02/19 11:50:10, Bob kb8tq wrote:
> Hi
>
> The difference in seconds between the start phase and the end phase divided by the number
> of seconds duration gives you the parts in whatever of the error.
>
> If you see 1us ( = 1x10^-6 seconds)  of change in a second, you are off by 1 ppm (or 1x10^-6).
> If you see 1 us of change in 1,000 seconds you are off by 1 ppb (or 1x10^-9). At a bit over 10
> days (1,000,000 seconds) your 1 us change is 1 ppt (or 1x10^-12).
>
> Bob
>
>> On Jul 2, 2019, at 10:17 AM, Chris Burford<cburford1 at austin.rr.com>  wrote:
>>
>> Is the slope value for the phase difference shown in TimeLab an average of the overall data sample duration? The reason I ask is that my service manual for my RFS says:
>>
>> /"//A faster way to make the comparison between the reference frequency and the DUT is to use the time interval measurement mode of the counters. In this case, the time intervals between the 10MHz zero crossings of the reference frequency and the DUT are measured and averaged. If this time interval changes by less than 10ps per second, then the DUT is within 1 part in //10^11 of the frequency reference."/
>>
>> I'm just curious if the phase difference slope value can be plugged in to this equation.
>>
>> Regards,
>>
>> Chris
>>
>>
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