6 Mar 2011
Doppler shift on 10m WSPR Transequatorial
This afternoon, before working some QRP DX on 10m and 12m SSB I had my WSPR system running and spotted FR1GZ (Reunion Is) several times at good levels. What was interesting was the Doppler shift on his signal which was up to -3Hz at times. I guess the propagation is trans-equatorial spread-F and what I'm observing are moving F-layers on the path between us.
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4 comments:
Hello Roger,
When the car passing the tone gets lower.
So I think that with -3Hz the layer is going up.
Bert, PA1B
Hi Roger and Bert, when can I say that a radio signal is coming in my direction or is passing by? How can I determine a Doppler effect on a radio signal? Sometimes I can't comprehend this matter. I tried hard, maybe you can explain to me? I am familiar with the Doppler effect, no doubt about it. But a radio signal with the speed of light seems to be omni directional present when propagation is good, because of relative short distance. In space where light travels with the speed of light and it travels for thousands of light years then it's easier to see the light is going in a certain direction or isn't? 73 Paul
Paul, on the WSPR screen Doppler shift can be see as a descending or ascending 2 minute line. This is very common on 6m and 2m WSPR signals and these are usually the result of moving aircraft. On 10m HF I was seeing a descending line on the WSPR screen and the reported drift was up to -3Hz. This suggests, as Bert said, that the effective layer is moving away (going higher in the ionosphere) I believe.
Hello Roger,
I did some calculation.
If the layer goes up, the path is getting longer and the frequency become lower. With a shift of -3 Hz the total path is getting 3 wavelengths longer in ONE second.
With the refective layer at 400 km, a drawing on scale shows me a double hop. I estimate, from the drawing on scale that the angle with the ionosperic layer is 20 degrees.
I calculated that, if the ionosphere is moving at one point, the layer is moving upward in that point with 5 m/s.
v=(3 * 10 * 1/2)/sin(20)= 5 m/s
1/2 --> if the layer is moving 1 m, the path will be 2 m longer
73,Bert PA1B
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