Lecture 5: Introduction to Multiuser MIMO

This is the video for Lecture 5 in the course Multiple Antenna Communications at Linköping University and KTH. The lecture covers: Limitations of point-to-point MIMO, the motivation behind multi-user MIMO, non-orthogonal and orthogonal multiple access, the capacity region and how to compute it in the uplink.
The slides are available here:
github.com/emilbjornson/multi...
0:00 Introduction
0:43 Recall: Point-to-Point MIMO Capacity . Compute SVD of channel matrix
4:17 Problems with point-to-point MIMO • Multiplexing gain: S = rank(G)
7:17 Multiuser MIMO Communication
8:32 Orthogonal multiple access . Two users want to communicate with base station
15:02 Non-orthogonal multiple access: Rate region Four operating points (R.R)
19:15 Uplink Multiuser MIMO: System model
21:44 What is the difference from point-to-point MIMO?
24:26 Motivating example
26:40 Shape of capacity region • One can pick two points and use them fractions of the time
27:49 Points in the capacity region • Combinations (RR) of rates that can be simultaneously achieved
30:46 Sum Capacity of Uplink Multiuser MIMO • Recall: Received signal
36:35 Summary Point-to-point MIMO channels - Large multiplexing gains are hard to achieve in practice

[@maryammohseni4507]

such a great lecture. thanks

[@Musiclover5258]

@32:47 (page 16), initially considering the individual rates, R_1 ... R_K as a vector and comparing that to the flow diagram for the equivalent P2P MIMO shown at 18:39 (page 10) of Lecture 4, am I correct in making the following inferences? 1. The precoder V would now be absent in multipoint MIMO, since there is no cooperation. 2. There is still the post-processing operation U^H for recovery at the receiver with U^H now being replaced with a normalized version of G^H (G Having been obtained via channel estimation using pilots the usual way, and repeated for each coherence interval)? Next, the expression (with the matrix determinants) shown for sum rate R is obtained by adding up the component rates, R_1,...,R_K, in the equivalent P2P MIMO, as stated. While I am able to make this connection sort of intuitively, by referring back to the singular values, is there a place where I can find the precise matrix derivation? In the third step, the argument is made that the sum capacity of the multipoint MIMO which uses successive interference cancellation is identical to the above expression for the sum rate of the corresponding P2P MIMO. Is this some sort of fundamental result in information theory? Is there a place where I can find the derivation? Thank you so much for your advice. I am thoroughly enjoying this series, though I have to absorb it slowly😊.

[@HarinduJayarathne]

Thank you for the video on Introduction to Multi user MIMO. Can you suggest a good reference to learn more about MU-MIMO?

[@WirelessFuture]

Many of the lectures are based on Fundamentals of Massive MIMO, but as a first step, I think you should read Chapter 1 in the book Massive MIMO Networks: massivemimobook.com

[@erikbertram6019]

4:50 In the case you are using dual polarization antennas at the receiver and the transmitter, you would be able to get a multiplexing gain of 2 in the LOS scenario right? Is this something you see often or is it too hard to properly seperate the two polarizations? PS your content is great!

[@WirelessFuture]

You are right! This is indeed used a lot. There are at least dual polarization at the base station and many devices also have dual antennas which might have different polarization. There might be leakage between the polarizations due to hardware effects, which means that the 2x2 MIMO matrix is not diagonal. One can then use the point-to-point MIMO theory to find the right precoding and receiver computation.

[@stellatauer761]

I have just a short understanding question regarding MU MIMO. In the DL we a a bit another situation than in the UL. The DL gives us several beams to different user, and we can use the same resources (frequency) for every user. We get spatial multiplexing. That seems to rather clear. In the UL it is a bit different, because the UEs send in an isotropic matter. So at the receiver (base station) we must handle the interference. So it is a bit more difficult, but possible to decode. We use SIC. Nevertheless, in the UL the base station uses the same beams like in the DL. Is my understanding correct? Many thanks.

[@WirelessFuture]

The downlink beams are not razor sharp, so there will be some remaining interference also in the downlink. The total interference is rather similar between the uplink and downlink, but divided differently between the UEs. While each signal is transmitted isotropically in the uplink, they reach the base station with different directivity. The base station can process the received signals over its antennas to filter out signals from undesired directions, in a way that is often called receive beamforming since the directivity that one can achieve is the same as when transmitting beams. The reason that we talk about SIC in the uplink, but didn’t cover the downlink counterpart, isn’t because the interference is worse in the uplink. It is just easier to implement advanced interference cancellation methods in the uplink than in the downlink. The downlink counterpart is called dirty paper coding.

[@mohammadsafa6113]

Thanks for the great lecture. I just have simple question, from the derivation of channel capacity with power constrain we know the optimum probability distribution is the Gaussian and its assumed for the transmitted signal x. however also the signal is modulated as BPSK for example, or any other higher order modulation, and it is just a points on the constellation diagram not Gaussian. so what is the philosophy or the idea behind this thinking and analysis in driving the capacity of a digital communication system.

[@WirelessFuture]

This is a good point that we discuss in Section 2.4 of the book "Introduction to Multiple Antenna Communications and Reconfigurable Surfaces" (Download here: www.nowpublishers.com/Article/BookDetails/9781638283140) The short answer is that the channel capacity is achieved by encoding the data using codewords whose length goes to infinity. The proof builds on generating those codewords with Gaussian distributed entries. In practice, we don't do that, but instead, we create the codewords using a combination of a modulation format (e.g., BPSK or 16-QAM) and a channel code. This makes the implementation much easier, and with the right design, we can come close to the capacity.

[@mohammadsafa6113]

@WirelessFuture Thank you very much for your reply. The book is great and contains everything I want to know. I have started reading it. Thank you for this contribution.

[@Julia-hu4xe]

Many thanks for all of your videos. A short question. On slide 6 you have this well known capacity equation. I refer so the SNR term (P*Beta/Alpha*B*N0). The whole noise term increases with higher bandwidth, that is clear. But with higher B, I assume the nominator term must increase too. Everybody wants more B - so does the term P*Beta increases too with higher B? Is that P something like P=B*P0?

[@WirelessFuture]

I thought that I had responded to your question, but it seems that my answer didn’t appear. Here is a new attempt: Since the total noise variance is proportional to the bandwidth, it is indeed preferable to also increase the transmit power P proportionally to it. This is indeed done at cellular base stations in the traditional sub-6 GHz bands, which might use something at order of 10 W per 10 MHz. The maximum transmit power of a mobile phone is, however, limited due to battery limitations and regulations on the electromagnetic exposure so it cannot increase with the bandwidth. When we reach for even higher bandwidths at mmWave frequencies, it becomes harder to build radio transceivers that uses high power. A mmWave transmitter for 5G typically use 1-10 W, even if it has 5 times more bandwidth than a 5G base station in the 3 GHz band that uses 100 W. So in these cases, P isn’t increasing with the bandwidth - rather the opposite.

[@tuanphamanh2849]

Thanks Prof for your great lecture. I have a couple of questions: 1) Why SNR is likely to be high in LOS and low in NLOS, could you explain more about it? 2) In the slide at 32:53, you said that the sum is "the sum rate of P2P MIMO but the sum capacity of uplink MU-MIMO". Does it mean that the capacity of P2P MIMO is theoretically larger than uplink MU-MIMO's one? And that is because it's unable to optimally allocate power to users in Uplink MU-MIMO as in P2P MIMO. Do I understand correctly?

[@WirelessFuture]

1) In NLOS scenarios, the signal has to be reflected on various objects and penetrate various materials to reach the receiver. This creates a lot of extra attenuation. 2) Yes, for the same channel matrix G, P2P MIMO gives a higher capacity than MU-MIMO, since the "user" antennas can cooperate in the P2P by sharing their power and apply precoding/combining.

[@tuanphamanh2849]

@@WirelessFuture Thanks very much. I'm still confused that if P2P MIMO gives higher capacity, so what is the actual motivation to use MU-MIMO? It is just to serve multiple UE simultaneously?

[@jasminnadic2103]

Thanks for your videos. The term "orthogonal" does not really mean a different of 90 degrees? Am I right?

[@erikbertram6019]

If two things are orthogonal to each other it means that the one thing does not influence the other. For example in a constellation diagram, you will have two dimensions that correspond to a signal that is modulated on a sine and one that is modulated on a cosine. Because a sine and a cosine are "orthogonal" to each other we can have independent modulation on both these components. The difference between a sine and a cosine is 90 degrees, but you can have orthogonal things that have nothing to do with angles. For example two independent spatial transmission paths are also orthogonal, or you can have two different frequency bands that are orthogonal. You can send data over these paths, without influencing the data that is communicated over the other path.

[@WirelessFuture]

When considering two-dimensional real valued vectors, then “orthogonal” means different by 90 or 270 degrees. When considering M dimensional complex values vectors, there are many more dimensions where orthogonality can appear, but it is essentially the same underlying geometry. Within an M-dimensional world, you can find a plane where the two vectors exist and are different by 90 degrees.

[@user-xw7ug1gq2m]

Good Video, sir! And I have a question. What is the difference between Non-orthogonal access and MU-MISO? In the 13-th page, you mentioned that MU-MISO's region is lager than Non-Orthogonal access, why? Thx!

[@WirelessFuture]

What we refer to as non-orthogonal access is a MU-SISO scenario, so one serves multiple users but has only one antenna at the base station.

[@user-xw7ug1gq2m]

@@WirelessFuture Clear explanation,thank you! I noticed that there is a technology named 'NOMA (Non-orthogonal Multiple Access)'. Does multiple means base station has multiple antennas? If so, Does NOMA equals MU-MISO?

[@knutlohmann8205]

A short question, please. I assume that the channel matrix H is measured very precisely, perhaps we get 4 digits after the point. With that granularity we get always the max. rank (columns will be always independent). Is that right? I think you must round it a bit to get perhaps just 2 digits. Then it could happen lower ranks. What are your thoughts about that? Thank you.

[@WirelessFuture]

The measurement noise will always lead to full rank, from a strict mathematical perspective. What matters in practice is how many of the singular values are “large” and caused by the channel. This is the useful rank of the channel from a MIMO communication perspective, because these are the ones that the waterfilling power allocation wants to put power on.

[@knutlohmann8205]

@@WirelessFuture thank you, the eigenvalue decomposition is also used. Is waterfilling also done on the basis of the eigenvalues (with quadratic matrix)?

[@WirelessFuture]

@@knutlohmann8205 Yes, the eigenvalues of the quadratic matrix (called the Gram matrix) are the same as the squared singular values. So you get the same result in both cases