Deep space SDR enthusiast and amateur radio operator Edgar Kaiser @df2mz's 28-Dec-2018 tweet says:

Chang'e 4 in lunar orbit last night on 8479.77 MHz. The spacecraft seems to be locked to the Queqiao relay satellite at the moon-earth L2 point during most of the passes. Short unlocked intervals before LOS at 07:08 and 09:05.

I see a sinusoidal shape, but the observer must be seeing more here than I can.

How can the author know that Chang'e-4 was in "locked to the Queqiao" from this SDR spectrogram?

Chang'e 4 in lunar orbit last night on 8479.77 MHz.

This observation was also reported by Scott Tilley's 02-Jan-2019 tweet:

Amateur observations of the radio signal from #ChangE4 support a landing window at the Van Karman crater complex Jan 3, 00:30-01:00 UTC. CE4 has been busy in constant ground lock during observations made from EU from @DF2MZ https://twitter.com/AJ_FI/status/1080474805969797120

enter image description here

enter image description here

enter image description here

  • 1
    $\begingroup$ I was just about to write an answer to explain in more detail what you can read from the plot, but then noticed that the measured shift is off by a factor of 4 compared to the calculated one... :/ $\endgroup$ – asdfex Apr 23 '19 at 17:54
  • $\begingroup$ @asdfex tell me more! In the first image I can see an apparent oscillation with a range of about 87,000 Hz. But what do you mean "the calculated one"? Are you doing it yourself or is there something I'm missing? I think that anything you can add to help decode this will be very interesting, please feel free to post a partial answer toward that goal. Thanks! $\endgroup$ – uhoh Apr 23 '19 at 22:11

For example http://www.svengrahn.pp.se/trackind/Selene/Selene.html

The reason for the frequency jumps - a hypothesis Why does the frequency suddenly jump just before loss-of-signal? The velocity relative to earth when the probe disappears behind the moon is about 1.54 km/s. Of course, at the equator the circumferential speed is about 0.45 km/s, but at 59 N and with the Moon at 70 degrees elevation, let us disregard this effect for the moment. So, if the Selene transmitter were not locked to an uplink the carrier would be shifted downwards by 11.6 kHz. If the donwlink were locked to a fixed-frequency uplink the frequency would be shifted downwards by another 10.7 kHz (one has to take the turn-around ratio 240/221 into account), again neglecting the effect of the earth's rotation. So, just before LOS the uplink is turned off and the Selene transponder returns to its free-running transmit frequency and the uplink downward shift disappears. That is why the frequency jumps up by 10-13 kHz. I think the fact that it is not alwyas abou + 11 kHz may be related to which antenna is providing the uplink and where it s in relation to the Moon.

But why do we not see simular big jumps near AOS he observed frequency jumps are much smaller, often just a few kHz- I tink that the uplink transmit frequency is adjusted so that the downlink frequency does not jump very uch. This will simplify receiver lock-on on the ground.

A complete Doppler cycle during a pass across the Moon In the waterfall picture below I have edited out all the interference (while they are still visible in the spectrum display) and hand -edited every pixel where the carrier was visible by making it yellow. Whe running the display this slowly (12 seconds between updates) and the signla is varying very much close to the nise level, the display catches the carrier rather seldom. So, overview and signal clarity do not quite match when the signal is this weak. Anyway the 13 kHz frequency jump just before loss-of-signal is clearly seen.

  • $\begingroup$ Your quoted link is about Selene. Is it possible to add one or two sentences and explain how this applies to the image in the question? We're almost there, thanks! $\endgroup$ – uhoh Jan 24 '19 at 0:21
  • $\begingroup$ As far as can be understood from the text of this author, the satellite is almost always in the field of view of the receiving antenna, except for short periods of time. The satellite in orbit of the moon disappears from sight for a long time. For example Apollo-8: "Each time Astronauts Borman, Lovell, and Anders vanished behind the far side of the Moon they lost all contact with the Earth for 45 minutes on each of the 10 orbits. During the first long silence the black void crackled with tension until Mission Control in Houston reported, "We've got it! Apollo 8 is in lunar orbit."" $\endgroup$ – A. Rumlin Jan 26 '19 at 17:55

TL;DR: The plot shows that Chang'e 4 is not locked to any external source but uses its own on-board clock to send data.

To answer this, we can ignore the Queqiao relay satellite and assume that we have a direct connection between Earth Chang'e-4. We treat Queqiao as just some kind of passive mirror for the signal.

The sinusoidal shape is produced by the Doppler effect as the satellite is in orbit around the Moon. The orbital velocity is about 1.5 km/s. This is the velocity it approaches us with when it comes visible and the velocity it moves away from us shortly before it vanishes behind the moon.

The total change in velocity is therefore $\Delta v = $ 3 km/s which converts to a Doppler shift of $$\Delta f = \frac{\Delta v}{c} f = \frac{3}{3 \cdot 10^5} \cdot 8479.77~\rm{MHz} = 84.8~kHz$$ If we look to the frequency spectrum we notice that the peak-to-peak amplitude of the frequency shift is 85 kHz - quite precisely the calculated one.

Many space probes operate in a mode that they receive a signal from Earth and use exactly the frequency of the received signal to transmit their data back. When these don't receive a signal, they use their own, on-board clock to generate the signal. As two separate clocks are never precisely the same, the switch between both modes can be visible as a frequency jump in the spectrum.

In both modes the observed amplitude of the Doppler shift is different: We calculated the shift in the case of the "unlocked" transmitter - the spacecraft transmits always at the same frequency, but due to its motion we observe a changing frequency. In the "locked" mode, we would see twice the Doppler shift: It gets doubled because the spacecraft already receives a changing frequency signal from Earth, then uses this shifting frequency to transmit a signal that again is Doppler shifted on the way back to us. This is clearly not the case for Chang'e 4.

This "double Doppler" shift is visible for example in this recording of the Beresheet probe, again by Edgar Kaiser @df2mz. You can see how the received signal follows two separate sinusoidal curves: One with a larger amplitude for "locked" operation and a second one with half the amplitude during "unlocked" phases.

Now we assumed that Queqiao is just a "passive mirror" for the signal. There is a hint in the diagram that tells us that this is wrong: Two times, right when the signal vanishes, we see a sudden jump in the frequency (the two short, white lines). This seems to be the change between "Queqiao is locked to Chang'es signal" and "Queqiao uses its own clock" to send data.

Footnote: Thanks to @df2mz for the conversation and pointing out that I was using the wrong speed of light in my first attempt to write this.

  • $\begingroup$ Thank you for the excellent sleuthing! Aha! I see that for an object in lunar orbit the total line-of-sight velocity change will be about 3 km/s, and if the object were transmitting a stable beacon tone at ~8480 MHz that would produce an 84.8 kHz peak-peak frequency shift. So it is not locked to anything but instead using a fixed local frequency. In the linked tweet I see peak-peak ranges of about 42 kHz and 21 kHz, if that based off of ~2280 MHz then it all makes sense, the taller one is double the amplitude of a ~ 3 km/s swing. Very nice work! $\endgroup$ – uhoh Apr 24 '19 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.