[time-nuts] Basic question regarding comparing two frequencies

J. Forster jfor at quik.com
Fri Jul 30 14:08:24 UTC 2010


Partly.

There are hourly jogs in the WWVB signal and also diurnal shifts of the
order of a cycle at 60 KHz.

The Fluke receivers havs a counter for microseconds, but it's difficult to
intrerpret w/o the stripchart too.

Frankly, 60 KHz is a PITA IMO. Oh for LORAN!

-John

==========




> So on a 60 khz signal the long strip chart recorder is simply a super long
> low pass filter averaging out the doppler somewhat. It really doesn't do
> that well. The mark-1 eyeball does a better job. Right?
>
> On Tue, Jul 27, 2010 at 4:53 AM, Geoff <vk2tfg at ozemail.com.au> wrote:
>
>> On Tue, 27 Jul 2010 09:08:49 am Chuck Harris wrote:
>> > I suppose that you could always cheat?  Since you know where the
>> > transmitter is going to be, if you could get a timenut near to the
>> > transmitter to give you a beacon to measure 24hrs prior to the event,
>> > you could use the diurnal variations that you observed (observe?) on
>> > the beacon to predict the skywave offset due to Doppler at the time
>> > of the event.
>> >
>> > -Chuck Harris
>> >
>> > Murray Greenman wrote:
>> > > You guys are trying to crack a nut with a sledgehammer!
>> > >
>> > > For a start, as Didier says, you can't possibly read the frequency
>> of a
>> > > sky-wave signal to 0.01Hz in any short time frame since the Doppler
>> on
>> > > the signal can be as much as 1ppm (i.e. 10Hz at 10MHz). You can only
>> > > infer it closer than that by studying the frequency in the very long
>> > > term.
>> > >
>> > > In addition, you'll never know how much of the daily variation is
>> > > ionospheric, and how much is due to thermal changes at the source.
>> snipped
>>
>> There is one possible way of getting an accurate reading from a sky wave
>> signal over a short(ish) period. Plot a doppler shift curve with as fine
>> a
>> resolution as you can manage. Then look for a point of inflexion in the
>> curve, that is a point where the second derivative of the curve function
>> is
>> zero. The frequency at that time will be that transmitted as at that
>> instant
>> the path length is not changing. You may have to examine your data set
>> visually and mathematically examine a much smaller section. Of course if
>> you
>> don't get a point of inflexion you'll need much more data :-).
>>
>> Cheers, Geoff vk2tfg.
>>
>>
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