Information Theory for Wireless Communications, Part II:

Transcription

1 Information Theory for Wireless Communications, Part II: Lecture 5: Multiuser Gaussian MIMO Multiple-Access Channel Instructor: Dr Saif K Mohammed Scribe: Johannes Lindblom In this lecture, we give the capacity region of the Gaussian multiuser multiple-input multiple-output MIMO multiple-access channel MAC We also give an algorithm for computing the maximum sumcapacity of the Gaussian MIMO MAC for the -user scenario I GENERAL SYSTEM MODEL The system under study is illustrated in Fig There are K multiple-antenna users that want to transmit data to a single receiver equipped with multiple antennas User k {,,K} has n k antennas and the receiver has n r antennas The channel between user k and the receiver is k C nr n k User k transmits a vector x k C n k The transmissions are concurrent and take place in the same band, so the signals will sum up at the receiver ence, the received signal is y = K k x k +z, k= where z C nr is the additive noise that we model as a zero-mean complex-symmetric Gaussian random vector with covariance I The transmitted vectors are subject to a power constraint Tr { E { }} { x k x k = E xk } P k II REVIEW OF TE CAPACITY REGION FOR TE SISO TWO-USER MAC We studied the two-user SISO MAC in [] and [] ere we revisit a few of these results Consider the input distribution px,x = px px This input distribution corresponds to independent encoding For a given input distribution, we can achieve the rate region R IX ;Y X, R IX ;Y X, R +R IX,X ;Y An example of this region is illustrated in Fig Note that IX,X ;Y = IX ;Y X +IX ;Y = IX ;Y X +IX ;Y The term IX ;Y is the mutual information we get when X is decoded treating X as noise interference The point A in Fig is the rate pair we achieve when we first decode user and then user This document is a property of Communication Systems Division, Department of Electrical Engineering, Linköping University, Sweden Copyright mus be obtained by writing to {saif,eriklarsson}@isyliuse prior to usage

2 User x n n r User k x k n k k y K User K x K n K Fig System model of the Gaussian multi-user MIMO MAC A R [bits/channel use] R [bits/channel use] Fig Achievable rate region of the SISO MAC for a given input distribution px,x = px px For the Gaussian case, we have the simplified signal model Y = X +X +Z with E{X k } P k and Z CN0, The best rate of user in the absence of user is R log + P In the same way, the best rate of user is R log + P Consider the case where the receiver first decodes user treating X as noise The new signal model is then Y = X + Z with Z X + Z If X CN0,P then Z CN0,P + and user can

3 3 achieve rate R log + P P + When X is decoded, the receiver subtracts it from the received signal and obtains the signal Now, user can achieve rate Note that R +R log +P +P / Y Y X = X +Z R IX ;Y = log + P III CAPACITY REGION FOR GAUSSIAN SIMO MAC ere, we consider the scenario of where n k =, k =,,K and n r > This setup is the single-input multiple-output SIMO MAC For the two-user case, the signal model for the SIMO MAC is Y = h X +h X +Z Let us assume that E{ X k } = Define = [h h ] The capacity region rate region of this system is given by R IX ;Y X = log + h P R IX ;Y X = log + h P R +R IX,X ;Y = max log 0 α k P k I + [ ] α 0 0 α 3 The rates and are the single-user rates To see how we get to 3, we consider a point-to-point single-user MIMO link Y = X +Z with X = [X,X ] T, K X E { XX }, and Z CN0, I The capacity of such channel is max log I + K X Tr{K X } P,K X 0 In our case, we restrict to the case where the encoding is independent, ie, the signals at the transmit antennas are independent This restriction gives us a diagonal covariance matrix K X Also, we have a per antenna power constraint Next we will show that in order to maximize 3, we must have α k = P k Due to the monotonicity of the logarithm, maximize 3 is equivalent to maximizing I + [ ] α 0 0 α = I + [ ] h { } [α h α h ] h = I +AB = I +BA = I + [ ] h h [α h α h ] = I + [ α h α h h ] α h h α h = log + α h + α h α α h N h 0 = log + α h + α h + α α h h h h 4

4 4 Since h h h h 0 equality only if h and h are colinear, we maximize 4 for αk = P k, k =, ence, we can write 3 as R +R log I + [ ] P 0 0 P It is clear that also this capacity region is a pentagon similar to that in Fig At the corner point A, we have the rate pair R = log +P h I +P h h h 5 R = log + P h 6 We achieve this point by first decoding user, treating user as noise The components of the noise vector h X +Z is spatially correlated with covariance P h h +I So, we prewhiten the received signal, ie, Ỹ = h h + I / Y = h h + I / h }{{} h X +h h + I / h X +Z }{{} Z Since Z CN0,I, the rate of user is log +P h = log +P h I+P h h h Once user is decoded, the term is subtracted from the received signal and user is decoded, and it can achieve the rate R = log +P h / A Degrees of Freedom We study the behavior of the rates 5 and 6 when P,P First, we note that R lim = P log P Second, we have R { } lim = Matrix inversion lemma P,P log P = lim P,P { } = P log log P = lim P + P log P log h + P h h /P + h h h h h = To conclude, for the two-user SIMO MAC, we achieve two degrees of freedom This is a gain over the SISO MAC, where we achieved only a single degree of freedom B Orthogonal Transmission Scheme Consider the scenario where the transmissions of users and are divided in time or frequency in such way that they do not interfere with each other at the receiver That is, user uses the channel for a fraction α of the time During that time it transmits using power P /α in order to use power P in average For the remaining fraction α of the time, user transmits using power P / α For this scheme, the achievable rates are R = αlog + P P R = αlog αn + 0 α

5 5 for the SISO MAC and R = αlog + P h α R = αlog + P h α for the SIMO MACBy varying α from 0 to, we get the boundaries of the achievable rate regions As illustrated in Fig 3, we see that the orthogonal transmission scheme is suboptimal owever, for the SISO MAC, the orthogonal scheme achieves the sum-capacity [3, Exercise 04] SISO MAC, P = P = = SIMO MAC, P = P = = h = h = R [bits/channel use] R [bits/channel use] R [bits/channel use] R [bits/channel use] Fig 3 Illustration of the optimality of orthogonal transmission schemes The solid line is the boundary of the capacity region whereas the dashed line is the boundary of the region that is achievable by orthogonal transmission IV TWO-USER GAUSSIAN MIMO-MAC Now, we extend the model to the scenario of multiple antennas at transmitters and receiver We focus on the two-user case For a given pair of transmit covariance matrices K,K, we have the achievable rate region R log I + K, 7 R log I + K, 8 R +R log I + [ ] [ ][ ] K K We also have the power constraints Tr{K k } P k, k =, The reason for why the joint covariance matrix in 9 is block-diagonal is that the encoding of the users messages is independent Note that the pair of covariance matrices K,K that maximizes 7 and 8 does not maximize 9 in general To find the capacity region, we have to take the union over all pair of feasible transmit covariance matrices This also implies that the capacity region of the Gaussian MIMO MAC is not a pentagon

6 6 Next, we will focus on the sum-capacity and we will introduce the so-called iterative water-filling algorithm to solve max log Tr{K } P k,k k 0 I + K N + K 0 In the remainder of this section, we assume that = First, we start with a feasible pair of transmit covariance matrices K 0,K 0 for which we achieve the rate CK 0,K 0 = log K 0 +I + K 0 0 Compare this to the point-to-point scenarioy = X +W, where the noise vectorw has the covariance K 0 W I+ K 0 The mutual information is then log K 0 W + K K 0 log W ence, we can write 0 as CK 0,K0 = IK = K 0 +log K 0 IK = K +log K 0 where K = argmax Tr{K } P,K 0 W I 0 K W X ;Y K 0 We solve this using the same water-filling algorithm as we use for the point-to-point MIMO channel Now, for the pair K,K 0, we have the capacity CK,K 0 = log K 0 +I + K Next, we maximize with respect to K, and so on until convergence We see that we have an improvement of the sum-rate in each iteration Since the sum-capacity is bounded, the iterative water-filling algorithm A pair K,K is a optimal solution if K = argmax log K +I + K Tr{K } P,K 0 and vice versa for K In Fig 4, we illustrate the capacity region for the two-user Gaussian MIMO MAC with [ ] [ i i i 00+58i =, i i = i i and P = P = = As we can see, the region does not have the shape of a pentagon W ], REFERENCES [] Q Ngo, Information Theory for Wireless Communications: Part I, Lecture : Single-antenna multi-user uplink channels achievable rate region [] Information Theory for Wireless Communications: Part I, Lecture 3: Single-antenna multi-user uplink channels capacity region and converse [3] D Tse and P Viswanath, Fundamentals of Wireless Communication, Cambridge University Press, 008

7 7 3 5 R [bits/channel use] R [bits/channel use] 3 35 Fig 4 Illustration of the two-user Gaussian MIMO MAC capacity region The square marks the sum-capacity point