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My question pertains to the reduction in SNR achieved by arraying parabolic antennas in deep space communications. My engineering education is a bit out of date, so my thanks for your patience.

I understand that when N similar antennas are arrayed, each with its own amplifier, signal power increases as N^2 while noise power increases only as as N, so that SNR increases as N (coherent integration gain).

This makes sense to me for internal receiver noise (e.g., thermal), which is uncorrelated between the N receivers, but I'm wondering about the cosmic microwave background (CMB) radiation.

It appears to me that since the N arrayed antennas are receiving the CMB simultaneously from the same small bit of the sky (all are aimed at the spacecraft along a common line of sight), the CMB does not behave like the uncorrelated noise sources within the N receivers. Rather, it should behave like point-source interference, an unwanted signal added to the spacecraft signal that appears identically at all N antennas. (Here I am looking at the treatment of point-source interference vs. internal noise in Sec. A of Lee et al., Large-Array Signal Processing for Deep-Space Applications -- which does not, alas, mention the CMB. I have been unable to find a source that explicitly addresses how arraying for improved SNR affects the CMB noise contribution.)

If my naïve reasoning is correct, arraying improves SNR with respect to internal noise but not with respect to the CMB.

Is this right, or am I missing something?

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  • \$\begingroup\$ The CMB is not a point source. \$\endgroup\$
    – John Doty
    Commented Jul 3 at 16:27

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You're right. However, remember that coherent addition maximimizing the received signal power means that you would correct the phase of the individual received signals before adding up. In that, you build an antenna array that has additional directivity – you get an additional array gain; which means that you only collect background radiation coming from the same direction of your deep space craft. Thus, depending on the size of your array and the accuracy of your addition, you also drastically reduce the amount CMB power you "collect".

Couple caveats:

  • antenna array directivity patterns have side lobes – you get power from these as well. This is especially problematic for terrestrial antennas in populated areas, as there might be man-made interference from there.
    • As in filter design (literally the same problem), you can suppress sidelobes at the cost of precision of the main lobe. So, trade-offs.
  • atmospheric "flimmer" might distort your signal's phase sufficiently that you don't want to overdo it with directional selectivity.
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  • \$\begingroup\$ Thank you, Marcus, that is very helpful. Apologies, but I have some lingering confusion. It's clear to me that the signals from the N antennas are phase-shifted for coherent addition. But doesn't this coherent addition affect the spacecraft signal and the CMB equally? If so, I don't understand how arraying could increase signal power while decreasing CMB power. \$\endgroup\$ Commented Jul 3 at 16:34
  • \$\begingroup\$ CMB is a distributed phenomenon: If you look only at a very narrow angle of the sky, you get less than when you look at the whole sky. Array processing means you make that angle narrower! \$\endgroup\$ Commented Jul 3 at 16:52
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It appears to me that since the N arrayed antennas are receiving the CMB simultaneously from the same small bit of the sky (all are aimed at the spacecraft along a common line of sight)

Two physically displaced antennas that are aimed at a wanted source i.e. a distant space craft/probe will pick-up CMB radiation along two slightly different trajectories. This means that the CMB radiation picked up on both, have to be different.

With two displaced antennas pointing at the same wanted source, any unwanted CMB radiation beyond that wanted source will become gradually less coherent as distance extends beyond that source. The rest of space beyond that point will naturally contribute the vast majority of the received CMB.

enter image description here

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  • \$\begingroup\$ Thanks for the reply, Andy (including original figure!) which does help my comprehension. \$\endgroup\$ Commented Jul 5 at 0:39

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