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If there is a transmitter with three transmitted antennas,and a receiver with five antennas,and these five antennas are very close to each others.

Now transmitter use beamforming with these three transmitted antennas to transmit the same signal to the receiver,these five antennas will also receive the signal,but will their received signal wave be the same?i mean same magnitude,same phase.

Because if their phase are not the same,we should multiply different $e^{j\theta}$ to received signal that every antenna receive,then sum them,then decoding it.

But if their phase are totally the same,we can just sum them,them decoding it

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    $\begingroup$ What's "very close" in terms of wavelengths and/or near-field? Kind of intuitively, it all depends on that. $\endgroup$ – Marcus Müller Nov 9 at 15:56
  • $\begingroup$ @MarcusMüller their position $\endgroup$ – Sylvindrunner Nov 9 at 23:35
  • $\begingroup$ not what I meant; define how close they are in terms of wavelength. 1/2 wavelength? 1/1000 wavelength? $\endgroup$ – Marcus Müller Nov 10 at 10:03
  • $\begingroup$ very close i said means their position are very close,just like if i sit nearby you,so we are very close $\endgroup$ – Sylvindrunner Nov 11 at 15:22
  • $\begingroup$ OK, what is "very close", compared to 1 Wavelength? Is that question unclear somehow? I'm voting to close this as unclear, because it really all hinges on the question how the distance is, measured in wavelengths. And I've asked this exact question three times now. Your "very close" doesn't mean anything. You will have to put a number to it – the number of wavelengths that fit between to antennas. $\endgroup$ – Marcus Müller Nov 11 at 16:35
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Say you have the receive antennas very close together, like with spacing $\lambda/10$ or similar and they are uniformly spaced in a linear array. As you pointed out the the phase shifts between them will be fairly small, so what does it mean? It means that the spatial resolution of your beamformer will be poor. In the extreme case, consider something crazy where the spacing is $\lambda/1000$, all the receive antennas essentially receive the same copy of the signal. So why can't you just add them all up if they have no phase difference? You can, you just won't have any spatial discrimination and it is only really amplifying your signal by the number of receive elements.

I will leave a plot here that I generated in MATLAB which is the take away. You see as the spacing (specified in wavelengths) becomes smaller and smaller the beam response becomes flatter and flatter indicating that the receive array is unable to spatially discriminate because of the reasons outlined earlier. Hope this helps!

enter image description here

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