With the advent of 5G Era, base station density increases, the cell radius decreases, and LOS propagation gets more dominant. So, why do we need for beamforming in 5G NR, although we have smaller cells and we have a small number of devices in a cell?

  • $\begingroup$ The question should be why beamforming is mandatory in 5G NR, because beamforming does exist in 4G LTE and WiMax. Also, your premise "small number of devices in a cell" is not true. Indeed, the purpose of small cells is to support a very high density of devices that leads to more or less the same cell capacity. For other things, I agree with the answer of Florian. $\endgroup$
    – AlexTP
    Oct 9, 2020 at 11:39

1 Answer 1


One reason is that higher frequencies are envisioned. With higher frequencies, the path loss grows (cf. Friis equation). Also, the wavelength is reduced and thus, $\lambda/2$ radiators start becoming quite small. The power you can radiate from a small aperture is limited as well so overall our power budget suffers. The only way out is directivity (cf. Friis equation again): we need to make sure our energy is directed towards the users of interest. This is possible by employing large antenna arrays (i.e., arrays comprised of a large number of elements). If each element radiates a certain power, $N$ elements can add up coherently to the $N$-fold power, provided we send it in the right direction. This is what beamforming does.

Another reason is that beamforming allows us to spatially separate users so that communication links can share the same time and frequency resources and be separated only spatially. In theory, an $N$-element antenna array can serve $N$ data streams at the same time and frequency, either to one user (that receives $N$ spatially multiplexed data streams, as it is common in WiFi) or to $N$ users (that each receive one). This increases network capacity even more.

Both factors play well together: higher frequencies and smaller wavelengths lead to a more rapid decorrelation of the channel in space so that even relatively closely spaced users can still be spatially multiplexed. Think of very narrow "pencil" beams, each pointed to a different receiver (if we are serving their lines of sight) or path (if we are using multipath).

While this sounds good, it also poses significant challenges. If we only slightly depoint the beam, the signal may be lost entirely. For moving users, this can be quite a challenge. Initial acquisition becomes harder too because signals transmitted/received omnidirectionally might be too faint to detect.

  • $\begingroup$ Yes we have significant pathloss in mmWaves I admit. The complexity and also cost of the overall system will be increased since we have a large number of cells and devices connected to each other in 5G. How do we overcome this computational complexity and cost issue? $\endgroup$
    – ozi2020
    Oct 9, 2020 at 11:27
  • $\begingroup$ by having more data capacity. You need multiple antennas anyways due to the spatially selectiveness of the paths, so there's no big extra cost. You're still caught in the idea that spectral resources are still infinite in 5G. They're not! $\endgroup$ Oct 9, 2020 at 11:37
  • $\begingroup$ I'm not caught anymore I just want to understand the concepts of the 5G correctly and its background. By data capacity do you mean data storage capacity of the devices (mobile or base stations)? $\endgroup$
    – ozi2020
    Oct 9, 2020 at 12:07
  • $\begingroup$ We overcome the complexity and cost issue by finding new ways of building RF receivers for high frequencies. A hot topic right now, especially in mmWave regime is hybrid beamforming, where we try to do part of the beamforming in specialized analog hardware so that fewer RF chains are needed and less processing in digital needs to be carried out. That's still ongoing work though and far from market-ready IMO. For the processing itself, check papers on "massive MIMO", they come up with various solutions to simplify the beamforming weight calculation. $\endgroup$
    – Florian
    Oct 9, 2020 at 13:36
  • $\begingroup$ Okay, thank you four your answers. I will check all these topics! $\endgroup$
    – ozi2020
    Oct 9, 2020 at 13:44

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