Lecture ¡8 Multi-‑User ¡MIMO I-Hsiang Wang ihwang@ntu.edu.tw 5/27, 2014
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 2
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 = h 1 x 1 + h 2 x 2 + w Rx: decodes both users’ data h 1 h 2 x 1 x 2 User 1 User 2 • 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 := [ h 1 h 2 ] is well-conditioned ⟹ 2 spatial DoF for the 2 users to share 6
Capacity ¡Bounds y = h 1 x 1 + h 2 x 2 + w Rx: decodes both users’ data h 1 h 2 x 1 x 2 User 1 User 2 • Individual rates: each user is faced with a SIMO channel 1 + P k � σ 2 || h k || 2 � = ⇒ R k ≤ log , k = 1 , 2 • Sum rate: viewed as a MIMO channel with V-BLAST and identity precoding matrix: ( �� � � � � � � � � ) ⇥ h 1 ⇤ H = , Λ = diag ( P 1 , P 2 ) h 2 ⇣ ⌘ I n r + H Λ H ∗ = ⇒ R 1 + R 2 ≤ log det σ 2 = log det ( I n r + P 1 h 1 h ∗ 1 + P 2 h 2 h ∗ 2 ) 7
Capacity ¡Region ¡of ¡the ¡UL ¡Channel 8 8 9 1 + P 1 � σ 2 || h 1 || 2 � R 1 ≤ log > > > < < = [ 1 + P 2 � σ 2 || h 2 || 2 � C Uplink = ( R 1 , R 2 ) ≥ 0 : R 2 ≤ log > > 1 > � σ 2 H Λ H ∗ � R 1 + R 2 ≤ log det I n r + : : ; ⇥ h 1 ⇤ Decoding order: H = , Λ = diag ( P 1 , P 2 ) h 2 User 1 → User 2 R 2 How to achieve the corner points? From the study of V-BLAST we know the answer: C Uplink MMSE-SIC! Decoding order: User 2 → User 1 R 1 8
K -‑user ¡Uplink ¡Capacity ¡Region • The idea can be easily extended to the K -user case 8 9 ( R 1 , . . . , R K ) ≥ 0 : > > > ∀ S ⊆ [1 : K ] , > > > > > > > 1 � � P < σ 2 H S Λ S H ∗ = R k ≤ log det I n r + [ C Uplink = S k ∈ S > > ✓ ◆ > > > > 1 P = log det I n r + P k h k h ∗ > > > > σ 2 k : ; k ∈ S ⇥ h l 1 ⇤ · · · h l 2 h l |S| l 1 , . . . , l |S| ∈ S H S := , � � Λ S := diag l 1 , . . . , l |S| ∈ S P l 1 , P l 2 , . . . , P l |S| , • Again, can be achieved using MMSE-SIC architectures 9
Comparison ¡with ¡Orthogonal ¡Access • Orthogonal multiple access can achieve 8 ⇣ ⌘ 1 + P 1 || h 1 || 2 R 1 = α log < ασ 2 α ∈ [0 , 1] ⇣ ⌘ 1 + P 2 || h 2 || 2 R 2 = (1 − α ) log : (1 − α ) σ 2 • Unlike the single-antenna case, it’s cannot achieve the sum capacity R 2 A • In total only 1 spatial DoF B Because the rate expressions are the same as those in the single-antenna case! R 1 2 10
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 , n r } . • 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 12
Downlink ¡with ¡Multiple ¡Tx ¡Antennas x Tx: encodes both users’ data h 1 h 2 y 1 = h 1* x + w 1 y 2 = h 2* x + w 2 User 1 User 2 • Superposition of two data streams: x = u 1 x 1 + u 2 x 2 - u k : 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 2 spatial DoF with u 1 ⟂ h 2 & u 2 ⟂ h 1 . - 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: - x dl = P K Superposition of data streams: k =1 u k x k - Received signals and SINR: y dl ,k = ( h ⇤ j 6 = k ( h ⇤ k u k ) x k + P k u j ) x j + w dl ,k , k = 1 , . . . , K P k | h ⇤ k u k | 2 SINR dl ,k = k u j | 2 , k = 1 , . . . , K j 6 = k P j | h ⇤ σ 2 + P - y dl = H ∗ x dl + w dl Vector channel: • Vector SINR: let SINR dl ,k 1 a k := 1+ SINR dl ,k , k = 1 , . . . , K | h ∗ k u k | 2 - Let the matrix A have entry k u j | 2 A k,j = | h ∗ - ( I K − diag ( a ) A ) p = σ 2 a Then we have • For given { u k } , we can compute the power vector p : p = σ 2 ( I K − diag ( a ) A ) − 1 a = σ 2 ( D a − A ) − 1 1 D a := diag (1 /a 1 , . . . , 1 /a K ) 15
w dl y dl,1 ~ u 1 User 1 x 1 x dl H * y dl, K ~ u K User K x K w ul ^ u 1 x 1 x ul,1 User 1 y ul H User K x ul, K ^ u K x K 16
Uplink-‑Downlink ¡Duality ¡(2) • Dual uplink: - y ul = Hx ul + w ul Vector channel: - Q k | u ⇤ k h k | 2 Filtered output SINR: SINR ul ,k = k h j | 2 , k = 1 , . . . , K j 6 = k Q j | u ⇤ σ 2 + P • Vector SINR: let SINR ul ,k 1 b k := 1+ SINR ul ,k , k = 1 , . . . , K | h ∗ k u k | 2 - Let the matrix B have entry k h j | 2 B k,j = | u ∗ - since B = A T � I K − diag ( b ) A T � q = σ 2 b Then we have � � � � � � � � � • For given { u k } , we can compute the power vector q : I K − diag ( b ) A T � − 1 b = σ 2 � D b − A T � − 1 1 q = σ 2 � D b := diag (1 /b 1 , . . . , 1 /b 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: K K D a − A T � − 1 1 = P k = σ 2 1 T ( D a − A ) − 1 1 = σ 2 1 T � P P Q k k =1 k =1 • Hence, to solve the downlink power allocation and Tx signature design problem, we can solve the dual problem in the dual uplink channel • 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 20
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 21
Naive ¡Pre-‑Cancellation ¡Strategy • Want to send point u in a 4-PAM constellation • Transmit x = u – s to pre-cancel the effect of s • But this is very power inefficient if s is large u s x 22
Tomlinson-‑Harashima ¡Precoding ¡(1) • Replicate the PAM constellation to tile the whole real line –3 a – a 3 a a 2 2 2 2 – 11 a – 9 a – 7 a – 5 a – 3 a – a 3 a 5 a 7 a 9 a 11 a a 2 2 2 2 2 2 2 2 2 2 2 2 • Represent information u by an equivalent class of constellation points instead of a single point 23
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