Lecture 8 Multi-User MIMO I-Hsiang Wang ihwang@ntu.edu.tw 5/27, - - PowerPoint PPT Presentation

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

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Outline

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

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Uplink ¡with ¡ Multiple ¡Rx ¡Antennas

Spatial ¡Division ¡Multiple ¡Access

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= h1x1 + h2x2 + w x1 x2 h1 h2

User 1 User 2

y

Rx: decodes both users’ data

• 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 h2]

is well-conditioned ⟹ 2 spatial DoF for the 2 users to share

Capacity ¡Bounds

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• Individual rates: each user is faced with a SIMO channel
• Sum rate: viewed as a MIMO channel with V-BLAST and

identity precoding matrix: (

• )

= ⇒ Rk ≤ log

• 1 + Pk

σ2 ||hk||2

, k = 1, 2 H = ⇥h1 h2 ⇤ , Λ = diag (P1, P2) = ⇒ R1 + R2 ≤ log det ⇣ Inr + HΛH∗

σ2

⌘ = h1x1 + h2x2 + w x1 x2 h1 h2

User 1 User 2

y

Rx: decodes both users’ data

Capacity ¡Region ¡of ¡the ¡UL ¡Channel

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R1 R2

CUplink

CUplink = [ 8 > < > : (R1, R2) ≥ 0 : 8 > < > : R1 ≤ log

• 1 + P1

σ2 ||h1||2

R2 ≤ log

• 1 + P2

σ2 ||h2||2

R1 + R2 ≤ log det

• Inr +

1 σ2 HΛH∗

9 > = > ; H = ⇥h1 h2 ⇤ , Λ = diag (P1, P2)

How to achieve the corner points? From the study of V-BLAST we know the answer:

MMSE-SIC!

Decoding order: User 2 → User 1 Decoding order: User 2 → User 1

K-­‑user ¡Uplink ¡Capacity ¡Region

• The idea can be easily extended to the K-user case
• Again, can be achieved using MMSE-SIC architectures

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CUplink = [ 8 > < > : (R1, . . . , RK) ≥ 0 : ∀ S ⊆ [1 : K], P

k∈S

Rk ≤ log det

• Inr +

1 σ2 HSΛSH∗ S

• 9

> = > ; HS := ⇥hl1 hl2 · · · hl|S| ⇤ , l1, . . . , l|S| ∈ S ΛS := diag

• Pl1, Pl2, . . . , Pl|S|
• ,

l1, . . . , l|S| ∈ S

Comparison ¡with ¡Orthogonal ¡Access

• Orthogonal multiple access can achieve
• Unlike the single-antenna case, it’s cannot achieve the

sum capacity

• In total only 1 spatial DoF

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8 < : R1 = α log ⇣ 1 + P1||h1||2

ασ2

⌘ R2 = (1 − α) log ⇣ 1 + P2||h2||2

(1−α)σ2

⌘ α ∈ [0, 1]

A B

2

R2 R1

Because the rate expressions are the same as those in the single-antenna case!

Total ¡Available ¡Spatial ¡DoF

• With K single-antenna mobiles and nr antennas at the

base station, the total # of spatial DoF is min{K, nr} .

• When K ≤ nr , the multi-antenna base station is able to

distinguish all K users with SDMA

• When K > nr , the multi-antenna base station cannot

distinguish all K users

• Instead, divide the users into nr groups: in each group,

users share the single DoF by orthogonalization

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Downlink ¡with ¡ Multiple ¡Tx ¡Antennas

Downlink ¡with ¡Multiple ¡Tx ¡Antennas

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• Superposition of two data streams: x = u1x1+u2x2
• 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 2 spatial DoF with u1⟂h2 & u2⟂h1 .
• Similar to zero forcing (decorrelator) in point-to-point and uplink

y2 = h2*x + w2 y1 = h1*x + w1 h1 h2

User 1 User 2

x

Tx: encodes both users’ data

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

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Uplink-­‑Downlink ¡Duality ¡(1)

• Primal downlink:
• Superposition of data streams:
• Received signals and SINR:
• Vector channel:
• Vector SINR: let
• Let the matrix A have entry
• Then we have
• For given {uk}, we can compute the power vector p:

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xdl = PK

k=1 ukxk

SINRdl,k =

Pk|h⇤

kuk|2

σ2+P

j6=k Pj|h⇤ kuj|2 , k = 1, . . . , K

ydl = H∗xdl + wdl ydl,k = (h⇤

kuk) xk + P j6=k (h⇤ kuj) xj + wdl,k, k = 1, . . . , K

Ak,j = |h∗

kuj|2

(IK − diag (a) A) p = σ2a ak :=

1 |h∗

kuk|2

SINRdl,k 1+SINRdl,k , k = 1, . . . , K

p = σ2 (IK − diag (a) A)−1 a = σ2 (Da − A)−1 1 Da := diag (1/a1, . . . , 1/aK)

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User K ydl, K x dl uK H* User 1 ydl,1 wdl u1 ~ x1 ~ xK User K User 1 ^ xK ^ x1 yul wul uK u1 H xul,1 xul, K

Uplink-­‑Downlink ¡Duality ¡(2)

• Dual uplink:
• Vector channel:
• Filtered output SINR:
• Vector SINR: let
• Let the matrix A have entry
• Then we have
• For given {uk}, we can compute the power vector q:

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Ak,j = |h∗

kuj|2

yul = Hxul + wul SINRul,k =

Qk|h⇤

kuk|2

σ2+P

j6=k Qj|h⇤ kuj|2 , k = 1, . . . , K

bk :=

1 |h∗

kuk|2

SINRul,k 1+SINRul,k , k = 1, . . . , K

• IK − diag (b) AT

q = σ2b Db := diag (1/b1, . . . , 1/bK) q = σ2 IK − diag (b) AT −1 b = σ2 Db − AT −1 1

Uplink-­‑Downlink ¡Duality ¡(3)

• For the same {uk}, to achieve the same set of SINR

(a=b), the total Tx power of the UL and DL are the same:

• 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

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K

P

k=1

Pk = σ21T (Da − A)−1 1 = σ21T Da − AT −1 1 =

K

P

k=1

Qk

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?

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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
• f 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

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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

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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

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u s x

• Replicate the PAM constellation to tile the whole real line
• Represent information u by an equivalent class of

constellation points instead of a single point

Tomlinson-­‑Harashima ¡Precoding ¡(1)

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–3a 2 – a 2 a 2 3a 2

– 5a 2 – 7a 2 – 9a 2 – 11a 2 3a 2 – a 2 – 3a 2 11a 2 9a 2 7a 2 5a 2 a 2

Tomlinson-­‑Harashima ¡Precoding ¡(2)

• Given u and s, find the point in its equivalent class

closest to s and transmit the difference

24 transmitted signal x s – 11a 2 – 9a 2 – 7a 2 – 5a 2 – 3a 2 – a 2 a 2 3a 2 5a 2 7a 2 9a 2 11a 2

p

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
• ne can get to the capacity of the AWGN channel with

the interference completely removed

• Applying this to the downlink, can perform SIC at the

transmitter

• The pre-cancellation order in the downlink is the reverse
• rder of the SIC in the dual uplink

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