[time-nuts] signal transit time through WWV receiver

Tim Shoppa tshoppa at gmail.com
Mon Jan 6 21:52:53 UTC 2020


I can tell you, that with say a 10kHz bandwidth AM receive filter on an
analog receiver, the transit time through that filter will be ballpark 100
microseconds. Possibly more if it's a high-order filter.

I have used DSP-based (non-PC) SDR's for WWV reception and characterized
them as you suggest. I found that for a 6kHz SDR receive filter, 200
microseconds was the delay between RF in and audio out.

I suspect a PC-based SDR will be more delay than the pure-DSP solution.

For comparison, the propagation distance from Colorado to my home in
Maryland is about 1560 miles.  That right there is a 8.3 millisecond delay.
The few hundred microseconds delay in the SDR was negligible compared to
the 8.3 millisecond delay.

Dave Mills, the developer of NTP, tells me that he could see the night vs
day ionospheric path difference (circa.... 30 miles out of 1560?) by doing
extended analysis on the data. I looked for that in my data, and it might
have been there, but was kinda in the noise.

Tim N3QE



On Mon, Jan 6, 2020 at 3:24 PM GERRY ASHTON <ashtongj at comcast.net> wrote:

> I am aware that WWV signals are intended to reflect the correct time at
> the instant the signal leaves WWV's antenna. It's up to me to allow for
> propagation through air and space to my antenna, and transit through my
> receiver. With newer direct-sampling receivers, the processing time in the
> receiver's computer circuitry may greater than we are accustomed to.
>
> Are there any simple tricks to measuring the transit time. My thought is
> to create a pulsed 10 MHz source, with AM modulation when the signal is on,
> and compare a path directly into one channel of an oscilloscope with the
> audio output of the receiver feeding into the other channel. Any easier or
> better suggestions? My test equipment is much more limited than many other
> members of this list.
>
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