Lecture 8 Multi- User MIMO

Lecture 8 Multi- User MIMO I-Hsiang Wang ihwang@ntu.edu.tw 5/7, 014

Multi- User MIMO System So far we discussed how multiple antennas increase the capacity and reliability in point-to-point channels Question: how do multiple antennas help in multi-user uplink and downlink channels? Spatial-Division Multiple Access (SDMA): - Multiple antennas provide spatial resolvability for distinguishing different users signals - More spatial degrees of freedom for multiple users to share

Plot First study uplink/downlink scenarios with single-antenna mobiles and a multi-antenna base station Achieve uplink capacity with MMSE and successive interference cancellation Achieve downlink capacity with uplink-downlink duality and dirty paper precoding Finally extend the results to MIMO uplink and downlink 3

Outline Uplink with multiple Rx antennas - MMSE-SIC Downlink with multiple Tx antennas - Uplink-downlink duality - Dirty paper precoding MIMO uplink and downlink 4

Uplink with Multiple Rx Antennas 5

Spatial Division Multiple Access y = h1x1 + hx + w h1 h Rx: decodes both users data x1 x User 1 User Equivalent to the point-to-point MIMO using V-BLAST with identity precoding matrix Rx beamforming (linear filtering without SIC ) distinguishes two users spatially (and hence the name spatial division multiple access (SDMA)) - MMSE: the optimal filter that maximizes the Rx SINR - As long as the users are geographically far apart H := [h1 h] is well-conditioned spatial DoF for the users to share 6

Capacity Bounds y = h1x1 + hx + w h1 h Rx: decodes both users data x1 x Individual rates: each user is faced with a SIMO channel =) R k apple log 1+ P k h k, k =1, Sum rate: viewed as a MIMO channel with V-BLAST and H = h 1 h, = diag (P1,P ) identity precoding matrix: (!!!!!!!!!! ) =) R 1 + R apple log det User 1 User I nr + H H = log det (I nr + P 1 h 1 h 1 + P h h ) 7

Capacity Region of the UL Channel C Uplink = [ 8 >< >: (R 1,R ) 0: 8 >< R 1 apple log 1+ P 1 h 1 R apple log 1+ P h >: R 1 + R apple log det I nr + 1 H H 9 >= >; R Decoding order: User 1 User H = h 1 h, = diag (P1,P ) C Uplink How to achieve the corner points? From the study of V-BLAST we know the answer: MMSE-SIC! Decoding order: User User 1 R 1 8

K- user Uplink Capacity Region The idea can be easily extended to the K-user case C Uplink = [ 8 >< >: (R 1,...,R K ) 0: 8S P [1 : K], R k apple log det I nr + 1 H S S H S ks = log det I nr + 1 P H S := h l1 h l h l S, l1,...,l S S S := diag P l1,p l,...,p l S, l 1,...,l S S Again, can be achieved using MMSE-SIC architectures ks P k h k h k 9 >= >; 9

Comparison with Orthogonal Access Orthogonal multiple access can achieve Unlike the single-antenna case, it s cannot achieve the sum capacity 8 < R 1 = log 1+ P 1 h 1 [0, 1] : R =(1 ) log 1+ P h (1 ) In total only 1 spatial DoF R A Because the rate expressions are the same as those in the single-antenna case! B 10 h R 1

Total Available Spatial DoF With K single-antenna mobiles and n r antennas at the base station, the total # of spatial DoF is min{k, nr}. When K n r, the multi-antenna base station is able to distinguish all K users with SDMA When K > n r, the multi-antenna base station cannot distinguish all K users Instead, divide the users into n r groups: in each group, users share the single DoF by orthogonalization 11

Downlink with Multiple Tx Antennas 1

Downlink with Multiple Tx Antennas x h1 h Tx: encodes both users data y1 = h1 * x + w1 y = h * x + w User 1 User Superposition of two data streams: x = u 1x1+ux - uk: Tx beamforming signature for user k Downlink SDMA: - Design goal: given a set of SINR s, find the power allocation & the beamforming signatures s.t. the total Tx power is minimized Achieve spatial DoF with u 1 h & u h1. - Similar to zero forcing (decorrelator) in point-to-point and uplink 13

Downlink SDMA: Power Control Problem Finding the optimal Tx signatures & power allocation: - SINR of each user depends on all the Tx signatures (and the power allocation); in contrast to the uplink case - Hence maximizing all SINR is not a meaningful design goal Our design goal is to solve a power control problem: - Given a set of SINR s, find the power allocation & a set of Tx signatures such that the total amount of Tx power is minimized - It turns out that the power control problem is dual to a power control problem in a dual uplink channel Through the uplink-downlink duality, the downlink problem can be solved 14

Uplink- Downlink Duality (1) Primal downlink: - Superposition of data streams: - Received signals and SINR: x dl = P K k=1 u kx k y dl,k =(h k u k) x k + P j6=k (h k u j) x j + w dl,k, k =1,...,K SINR dl,k = P k h k u k + P j6=k P j h k u j, k =1,...,K - Vector channel: Vector SINR: let y dl = H x dl + w dl a k := 1 h k u k SINR dl,k 1+SINR dl,k, k =1,...,K - Let the matrix A have entry A k,j = h k u j - Then we have (I K diag (a) A) p = a For given {u k}, we can compute the power vector p: p = (I K diag (a) A) 1 a = (D a A) 1 1 D a := diag (1/a 1,...,1/a K ) 15

w dl ~ x1 u 1 y dl,1 User 1 x dl H * ~ xk u K y dl, K User K w ul User 1 x ul,1 u 1 ^x 1 H y ul User K x ul, K u K ^x K 16

Uplink- Downlink Duality () Dual uplink: - Vector channel: - Filtered output SINR: Vector SINR: let y ul = Hx ul + w ul SINR ul,k = 1 SINR b k := ul,k h k u k 1+SINR - ul,k, k =1,...,K Let the matrix B have entry B k,j = u k h j - Then we have! I!!!!!!!! since B = A T K diag (b) A T q = b For given {u k}, we can compute the power vector q: q = I K diag (b) A T 1 b = D b A T 1 1 D b := diag (1/b 1,...,1/b K ) Q k u k h k + P j6=k Q j u k h j, k =1,...,K 17

Uplink- Downlink Duality (3) For the same {u k}, to achieve the same set of SINR (a=b), the total Tx power of the UL and DL are the same: KP k=1 P k = 1 T (D a A) 1 1 = 1 T D a A T 1 1 = KP Hence, to solve the downlink power allocation and Tx signature design problem, we can solve the dual problem in the dual uplink channel Q k k=1 Tx signatures will be the MMSE filters in the virtual uplink 18

Beyond Linear Strategies Linear receive beamforming strategies for the uplink map to linear transmit beamforming strategies in the downlink But in the uplink we can improve performance by doing successive interference cancellation at the receiver Is there a dual to this strategy in the downlink? 19

Transmit Precoding In downlink Tx beamforming, signals for different users are superimposed and interfere with each other With a single Tx antenna, users can be ordered in terms of signal strength - A user can decode and cancel all the signals intended for the weaker user before decoding its own With multiple Tx antennas, no such ordering exists and no user may be able to decode information beamformed to other users However, the base station knows the information to be transmitted to every user and can precode to cancel at the transmitter 0

Symbol- by- Symbol Precoding A generic problem: y = x + s + w - x : desired signal - s : interference known to Tx but unknown to Rx - w : noise Applications: - Downlink channel: s is the signal for other users - ISI channel: s is the intersymbol interference 1

Naive Pre- Cancellation Strategy Want to send point u in a 4-PAM constellation But this is very power inefficient if s is large Transmit x = u s to pre-cancel the effect of s u s x

Tomlinson- Harashima Precoding (1) Replicate the PAM constellation to tile the whole real line 3a a a 3a 11a 9a 7a 5a 3a a a 3a 5a 7a 9a 11a Represent information u by an equivalent class of constellation points instead of a single point 3

Tomlinson- Harashima Precoding () Given u and s, find the point in its equivalent class closest to s and transmit the difference transmitted signal x s p 11a 9a 7a 5a 3a a a 3a 5a 7a 9a 11a 4

Writing on Dirty Paper Can extend this idea to block precoding - Problem is to design codes which are simultaneously good source codes (vector quantizers) as well as good channel codes Somewhat surprising, information theory guarantees that one can get to the capacity of the AWGN channel with the interference completely removed transmitter Applying this to the downlink, can perform SIC at the The pre-cancellation order in the downlink is the reverse order of the SIC in the dual uplink 5

MIMO Uplink and Downlink 6

MIMO Uplink Channel model:! y = H 1x1 + Hx + w Now the mobiles (Tx) have multiple antennas, and hence can form their own Tx covariance matrices For the two-user case, capacity bounds become - Individual rate bounds: R k apple log det I - nr + 1 H k K xk H, k =1, Sum rate bound: P R 1 + R apple log det I nr + 1 H k K xk H Note: in general there are no single K x 1 and K x that can k=1 simultaneously maximize the three rate constraints 7

Capacity Region (Two Users) C Uplink = conv [ k=1,, K xk 0 Tr(K xk )applep k 8 >< >: (R 1,R ) 0: R k apple log det I nr + 1 H k K xk H k, k =1, P R 1 + R apple log det I nr + 1 H k K xk H k k=1 9 >= >; R Two pentagon regions are achieved by different choices of K x1 and K x B 1 B A 1 Hence the capacity region is NOT a pentagon region anymore A R 1 MMSE-SIC can achieve all corner points in each pentagon region 8

MIMO Uplink with Fast Fading (1) Full CSI: need to solve a joint optimization problem regarding power allocation and precoding matrix design - Cannot use SVD because the two channel matrices may not have the same factoring left matrix U - The problem can be solved by iterative water-filling efficiently - Reference: W. Yu et al, Iterative Water-Filling for Gaussian Vector Multiple-Access Channels, IEEE Transactions on Information Theory, vol.50, no.1, pp.145 15, January 004 9

MIMO Uplink with Fast Fading () Receiver CSI: - Capacity region is the convex hull of the collection of rate pairs (R1,R) that satisfy the following inequalities for some covariance matrix K x1 and K x with Tr(K x1 ) < P1 and Tr(K x ) < P apple : R k apple E log det I nr + 1 H kk xk H k, k =1, "!# R 1 + R apple E log det I nr + 1 X H k K xk H k For i.i.d. Rayleigh {H k}, it is straightforward to see that uniform power allocation and identity precoding matrices k=1 maximize all bounds simultaneously the capacity region is a pentagon with K xk = P k n t,k I nt,k 30

Nature of Performance Gains For the uplink MIMO, regardless of CSIT, the total # of spatial DoF is min K P k=1 n t,k,n r CSIT is NOT crucial in obtaining multiplexing gain in the uplink, as long as receiver CSI is available Power gain is increased with CSIT Multi-user diversity gain is limited 31

MIMO Downlink Compared to the case with single-antenna users, one needs to further design the receive filters at the users Uplink-downlink duality can be naturally extended to the case with multiple Rx antennas: - The Rx linear filters are the Tx linear precoding filters in the dual uplink channel Hence in the case without fading, the sum capacity of the MIMO downlink channel is equal to that of the dual uplink channel (with total power constraint) 3

MIMO Downlink with Fast Fading With full CSI one can assort to the uplink-downlink duality to solve the joint optimization problem However, with receiver CSI: - Not possible for the base station to carry out Tx beamforming - In the symmetric downlink channel with CSIR, time-sharing is optimal in achieving the capacity region - DoF drops significantly from min{nt,k} to 1 Some partial CSI at Tx could recover the spatial DoF: - Channel quality of its own link rather than the entire channel - No phase information 33

Opportunistic Beamforming Opportunistic beamforming with multiple beams: - Form nt orthogonal beams - Whenever a user falls inside the beam, it feedback this piece of information back to the base station - As the number of users get large, one is able to find a user in each beam with high probability Hence the full DoF could be recovered Still, need instantaneous channel information user 1 user 34

Transmitter CSI Affects DoF Drastically Behaviors of power gain and multiuser diversity gain are similar to those in uplink For the downlink MIMO, CSIT is critical for obtaining spatial multiplexing gain (assuming i.i.d. Rayleigh below) n - PK o Full CSI: Total DoF = min k=1 n r,k,n - t CSIR: Total DoF = min max - k[1:k] n r,k,n t For the case of single-antenna users, DoF with CSIR is merely 1 Hence it might be beneficial to spend some resource for estimating the channel and predict the current channel from the past observations Under i.i.d. Rayleigh, prediction is not possible. Question: can past CSI at Tx still help? 35

Two- user MISO Downlink Delay h1[m], h[m] : h1[m] i.i.d. Rayleigh Delay h[m] Without transmitter CSI (CSIT), DoF = 1 With instantaneous CSIT H[m] at time m, DoF = With delayed CSIT H[1:m 1] at time m, DoF =? Prediction is useless delayed CSIT is useless? 36

DoF = 4/3 with Delayed CSIT Delay m = 1 m = m = 1 m = h1[m] L1(u1,v1) L3(u,v) u1 u v1 v h[m] L(u1,v1) L4(u,v) Delay m = 3 L(u1,v1)+L3(u,v) 0 h1[m] h[m] m = 3 h1,1(l+l3) h,1(l+l3) 37

DoF = 4/3 with Delayed CSIT Delay h1[m] L1(u1,v1), L3(u,v) L(u1,v1)+L3(u,v) Delay h[m] L(u1,v1), L4(u,v) L(u1,v1)+L3(u,v) L 1(u1,v1) L(u1,v1) & L3(u,v) L4(u,v) with prob. 1 Both decode their desired symbols over 3 time slots DoF = (+)/3 = 4/3 38

Exploiting Side Information m = 3 Delay h1[m] L1(u1,v1), L3(u,v) L(u1,v1)+L3(u,v) 0 h[m] L(u1,v1), L4(u,v) Delay L (u1,v1) is useful for user 1 but shipped to user L 3(u,v) is useful for user but shipped to user 1 With delayed CSIT, Tx forms u 1 := L(u1,v1)+L3(u,v) - User 1 can extract L because it has L3 as useful side info. - User can extract L3 because it has L as useful side info. 39

Hierarchy of Messages One can view u 1 := L(u1,v1)+L3(u,v) as a common message for both users Order-1 message: aimed at only 1 user - User 1: u1, v1 ; User : u, v Order- message: aimed at users - User {1,}: u1 Define DoF k := the DoF for sending all order-k messages Then we see that DoF 1 = 4 + 1 = 4 + DoF 1 1 = 4 3 40

Two Transmission Phases Transmission is divided into two phases Phase 1: time m = 1,: - Transmit two order-1 messages using two time slots End of Phase 1: - From the delayed CSI, Tx is able to form one order- message Phase : time m = 3: - Transmit this order- message using one time slot Only phase 1 sends fresh data; the rest is to refine the reception by exploiting delayed CSIT and Rx side info. The idea can be extended to scenarios with more users and more Tx antennas, where higher-order messages have to be formed to achieve optimality 41

Optimality of 4/3 It is remarkable that delayed CSIT is useful and we can achieve DoF = 4/3 > 1 Can we do better? The answer is no Rx antennas Stronger! Enhance user 1 by feeding user s signal to user 1 Now we have a natural ordering of the users and we find again time-sharing is DoF optimal Hence d1 + d Similarly d + d1 3DoF 4 4