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For digital beamforming, we can form as many beams as we want. No matter where I read about this subject I cannot find any discussion of the interference that one would expect between beams (not within a single beam).

For example, in my picture, if the transmitter forms three simultaneous beams with receivers 1, 2, and 3, then communication with receiver 3 would be fine but wouldn't the tight spacing of receiver 1 and 2 introduce a lot of interference between the beam the transmitter forms with Rx 1 and the beam it forms with Rx 2? Or, are there ways in which beams can be set so that they don't interfere?

EDIT: To make the question more clear, say for example all nodes operate in half-duplex. Say during some time period, the transmitting node decides to send an identical packet to each of the receivers at the same time. There is no frequency, time, or code division multiplexing. In this case, would receiver 1 receive some of sidelobe power from the beam meant for receiver 2? And is it really considered interference if the transmitter is sending the same information to all receivers?

And vice versa, if each receiver is to receive independent, potentially different information, then would receiver 1 get some interference from the sidelobe power from the beam meant for receiver 2?

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  • $\begingroup$ I am not sure I get the problem. How are all of these terminals multiplexed? Is this a broadcasting situation? Is it supposed to be full duplex? Is there time division, code division multiplexing taking place? What I am trying to say is, it depends on what you are trying to achieve but in general, while in one mode of communication anything out of its operation would be considered noise even if it is some form of useful signal at a different time. See for example here. Do you think you could clarify the question a bit? $\endgroup$ – A_A Jan 10 at 9:35
  • $\begingroup$ Ok I have updated my question with some more details. $\endgroup$ – Engineer Jan 10 at 17:17
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    $\begingroup$ I think unless the beamforming vectors are perfectly orthogonal to each others, there will always be interference from other receivers streams at each receiver. $\endgroup$ – BlackMath Jan 10 at 17:42

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