slide11 icc2015

User Scheduling for Massive MIMO OFDMASystems with Hybrid Analog-Digital Beamforming

Tadilo Endeshaw Bogale

Institute National de la Recherche (INRS), Canada

June 9, 2015, (ICC 2015)

Presentation outline

1 IntroductionScenario and ObjectiveSystem Model and Problem Formulation

2 Proposed Solution

3 Performance Analysis

4 Simulation Results

5 Conclusions

Tadilo (ICC 2015, London, UK) User Scheduling June 9, 2015, (ICC 2015) 2 / 11

Introduction Scenario and Objective

System Scenario and Objective

a1 · · · aM

System Scenario

MS1, MS2, MSKt are decentralized in spaceDownlink Communications⇒ Downlink Multiuser systemMS1, MS2, MSKt have single antennas⇒ Downlink Multiuser MISO systemBS has N antennas but Na < N RF chains⇒ BS use hybrid analog-digital architecture

(To get better performance (Clear later!))Channel between Tx and Rx is Freq. Selective

a1 · · · aM

System Scenario

Objective

Schedule Kt MSs (Ki ,∀i)To maximize the Overall Data RateS.t. Per sub-carrier Power constraint

Introduction Scenario and Objective

System Scenario and Objective

a1 · · · aM

System Scenario

a1 · · · aM

System Scenario

Objective

Schedule Kt MSs (Ki ,∀i)To maximize the Overall Data RateS.t. Per sub-carrier Power constraint

Introduction System Model and Problem Formulation

System Model and Problem FormulationSource Tx (Digital part) RF Chain Tx (Analog part)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

1 Discard CP& take FFT

di1 Decoded11, · · · , dNf1

K Discard CP& take FFT

diK Decoded1K , · · · , dNfK

••

d1, · · · ,dNfB1, · · · ,BNf A

dik = hHikABidi + nik

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

log(1 + γik ) ,Nf∑

f (ABi)

s.t tr{BHi AHABi} ≤ Pi

Ki : Users served by sub-carrier iB = [B1,B2, · · · ,BNf ], Bi ∈ CNa×Si

γik : SINR of i th sub-carrier k th userA ∈ CNa×N : Realized using PSs only

(Expected to be more constrained!)

Source Tx (Digital part) RF Chain Tx (Analog part)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

For one Sub-carrier (Flat fading)

maxA,B

K∑k=1

log(1 + γk )

s.t tr{BHAHAB} ≤ Pmax

Kt : Total number of usersK : Scheduled usersAssume that we have Na = K RF chains

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

s.t tr{BHAHAB} ≤ Pmax , |Aij |2 = 1

Disadvantage:How far from digital scheduler which isan optimal approach?(Not clearly known for general channel)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

s.t tr{(Bd )HBd} ≤ Pmax

Exploit the solution of digital scheduler:Fact: rank(Bd ) ≤ K for any schedulerClever Method [1]: Bd = UD, U ∈ CN×2K ,D ∈ R2K×K , D = blkdiag matrix∴ 2K RF chain: No performance lossVIP: No need to constrain |Aij |2 = 1

REREFENCES1 X. Zhang, A. F. Molisch, and S-Y. Kung, ”Variable-phase-shift-based RF-Baseband codesign for MIMO antenna

selection” IEEE Trans. Signal Process., vol. 53, no. 11, pp. 4091 - 4103, Nov. 2005.

2 T. E. Bogale, L. Le, A. Haghighat, and L. Vandendorpe ”On the Number of RF Chains and Phase Shifters, andScheduling Design with Hybrid Analog-Digital Beamforming,” IEEE Trans. (Submitted),http://arxiv.org/abs/1410.2609.

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

U and D are not uniqueQuestion: Can we make U and D uniqueand reduce RF chain < 2KAnswer: Yes, by employing [2]x = ej cos−1( x

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

⇒ Bd = UD, with |Uij |2 = 1 andD = blkdiag two consecutive diag elementsare the same⇒ K RF chain is enough [2]∴ Scheduling without |Aij |2 = 1 (Simpler)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

System Model and Problem Formulation

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Freq. Dom.

data source

D(1,:)

D(2,:)

D(K,:)

••

Freq. Dom.

••

IFFT (row)

& add CP

••

analog

••

AnalogBF

••

d1, · · · ,dNfB1, · · · ,BNf A

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

maxA,B

K∑k=1

log(1 + γk )

First Possibility

maxA,B

K∑k=1

log(1 + γk )

Second Possibility

K∑k=1

log(1 + γk )

Observations

2 ) + e−j cos−1( x2 ), for −2 ≤ x ≤ 2

Problem Formulation

maxA,Bi

Nf∑i=1

Ki∑k=1

f (ABi)

Proposed Solution

Input: Hki ,Ki ,Pi ,Na

Solve Relaxed Problem

rank(B) ≤ Na?

Get A,Bi : From SVD(B)

FINISH

Determine A

Given A: Optimize Bi

FINISH

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Assume the system as if it is DB(i.e., implicitly uses Na = N and A = I)⇒ Scheduling independently for each SC

Ki∑k=1

log(1 + γik ) , f (Bi),

s.t tr{BHi Bi} ≤ Pi

Gready based SchedulerZero forcing (ZF) beamformingSet B = [B1, B2, · · · , BNf ]

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

S: Min No of SC with rank(B) ≥ Na

Compute SVD(B) = UΛVH

Λ arranged in decreasing order.Set A: First Na columns of U

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Optimize Bi for fixed A

For fixed A: Each sub-carrier canperform scheduling independently

Ki∑k=1

f (Bi)

Gready based SchedulerZero forcing (ZF) beamforming

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Ki∑k=1

f (Bi)

Conclusions

A user may be scheduledin one (more) sub-carriersProposed solution maynot be global optimal

Proposed Solution

rank(B) ≤ Na?

FINISH

Determine A

FINISH

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Ki∑k=1

f (Bi)

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Ki∑k=1

f (Bi)

Conclusions

Proposed Solution

rank(B) ≤ Na?

FINISH

Determine A

FINISH

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Ki∑k=1

f (Bi)

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Ki∑k=1

f (Bi)

Conclusions

Proposed Solution

rank(B) ≤ Na?

FINISH

Determine A

FINISH

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Ki∑k=1

f (Bi)

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Ki∑k=1

f (Bi)

Conclusions

Proposed Solution

rank(B) ≤ Na?

FINISH

Determine A

FINISH

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Ki∑k=1

f (Bi)

rank(B) ≤ Na?

FINISH

Determine A

FINISH

Determination of A

Sort f (B1) ≥ f (B2) ≥, · · · ,≥ f (BNf )

Get Bd = [B1, B2, · · · , BS]

Relaxed Problem

Ki∑k=1

log(1 + γik ) , f (Bi),

Ki∑k=1

f (Bi)

Conclusions

Performance Analysis

Performance AnalysisApproaches

Three approaches examined:Antenna Selection Beamforming (ASB):Implicitly assumes A = IN×Na

Proposed Hybrid Beamforming (HB)Digital Beamforming (DB):Assumes N RF chains and A = IN

Intuitive expectationP(ASB) ≤ P(HB) ≤ P(DB)

To what extent?In what scenario?

Approaches

Performance: Rayleigh Channel

Rayleigh Fading Channel:

Lemma 1: ZF beamforming, Rayleigh fading hHk and large Kt ,

we have

RHBi ≥ RASB

i when KHBi = KASB

i , and

RHBi = RDB

i when KHBi = KDB

KHBi = KDB

i may not hold, ⇒ Average performance makes senseTheorem 1: ZF beamforming with Pi = P,Ki = K and Rayleighfading channel hH

k , we have

E{RASB} ≤ KNf log2

E{χNa−K+1max (Kg)}

)E{RDB} ≤ KNf log2

E{χN−K+1max (Kg)}

)E{RHB} ≤ K S log2

E{χN−K+1max (Ks)}

)+ K (Nf − S) log2

)where E{χM

max (L)}: Max. of L independent χ2 each with M DOF,S ≥ 1, Kg = dKt

K e and Ks = dKt NfKNae

Simulation: Bound is tight

Approaches

we have

RHBi ≥ RASB

i when KHBi = KASB

i , and

RHBi = RDB

i when KHBi = KDB

KHBi = KDB

k , we have

)+ K (Nf − S) log2

)where E{χM

Performance: ULA Channel

Rayleigh:DB Scheduler: Likely chooses usersclose to orthogonal each otherProposed HB achieves lower sum rate

(Mainly due to rank(H) > Na)∴ if rank(H) ≤ Na then RHB

i ≈ RDBi

ULA: Condition rank(H) / Na existsLemma 2: When d = λ

2 and the AOD ofthe Kt users satisfy sin (θkm) ∈n sin (θ)[− 1

2N ], n = 1,2, · · · ,Na,where θ is an arbitrary angle,

KHBi = KDB

i and RHBi = RDB

i , ∀i

Approaches

we have

RHBi ≥ RASB

i when KHBi = KASB

i , and

RHBi = RDB

i when KHBi = KDB

KHBi = KDB

k , we have

)+ K (Nf − S) log2

)where E{χM

Approaches

we have

RHBi ≥ RASB

i when KHBi = KASB

i , and

RHBi = RDB

i when KHBi = KDB

KHBi = KDB

k , we have

)+ K (Nf − S) log2

)where E{χM

i ≈ RDBi

KHBi = KDB

i and RHBi = RDB

i , ∀i

Approaches

we have

RHBi ≥ RASB

i when KHBi = KASB

i , and

RHBi = RDB

i when KHBi = KDB

KHBi = KDB

k , we have

)+ K (Nf − S) log2

)where E{χM

Approaches

we have

RHBi ≥ RASB

i when KHBi = KASB

i , and

RHBi = RDB

i when KHBi = KDB

KHBi = KDB

k , we have

)+ K (Nf − S) log2

)where E{χM

i ≈ RDBi

KHBi = KDB

i and RHBi = RDB

i , ∀i

Simulation Results

Simulation ResultsParameter Settings

Number of antennas: 64Channel: Lk = 8 tap, Nf = 64,Scheduled users: Ki = Kmax = 8Power: Pi = P,∀iSINR: SNR = Nf P

Kmaxσ2

Total number of users: Kt = 8Policy: Equal power allocation

Parameter Settings

Kmaxσ2

Rayleigh: Theory Vs Simulation

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

SimulationTheory (Upper bound rate)

Existing ASB

Proposed HB

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

ULA: Selected phases, Ls = 8

0 5 10 15 200

SNR (dB)

Existing ASBProposed HBDB

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

0 5 10 15 200

SNR (dB)

Other Results

Effect of power allocation policyObservation: Adaptive superior to Equal powerEffect of Kt

Observation: Rate increases as Kt increasesEffect of Na

Observation: Rate increases as Na increases

Simulation Results

Parameter Settings

Kmaxσ2

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

0 5 10 15 200

SNR (dB)

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

0 5 10 15 200

SNR (dB)

Other Results

Simulation Results

Parameter Settings

Kmaxσ2

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

0 5 10 15 200

SNR (dB)

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

0 5 10 15 200

SNR (dB)

Other Results

Simulation Results

Parameter Settings

Kmaxσ2

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

0 5 10 15 200

SNR (dB)

Parameter Settings

Kmaxσ2

−1.5 1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.50

SNR (dB)

Existing ASB

Proposed HB

0 5 10 15 200

SNR (dB)

Other Results

Conclusions

In this work, we accomplish the following main tasks.We propose greedy like user scheduling algorithm with hybridanalog-digital beamformingThe proposed hybrid scheduling achieves better than antennaselection approachWe analyze the performance of the proposed hybrid schedulingfor Rayleigh and ULA channelsTheoretical results are confirmed via extensive simulations

Conclusions

References I

O. E. Ayach, S. Rajagopal, S. Abu-Surra, Z. Pi, and R. W. Heath,Spatially sparse precoding in millimeter wave MIMO systems,IEEE Trans. Wireless Commun. 13 (2014), no. 3, 1499 – 1513.

T. E. Bogale and L. B. Le, Beamforming for multiuser massiveMIMO systems: Digital versus hybrid analog-digital, Proc. IEEEGlobal Telecommun. Conf. (GLOBECOM) (Austin, Tx, USA), 10 –12 Dec. 2014.

T. E. Bogale, L. B. Le, A. Haghighat, and L. Vandendorpe, On thenumber of RF chains and phase shifters, and scheduling designwith hybrid analog-digital beamforming, IEEE Trans. WirelessCommun. (Submitted) (2014).

Conclusions

References II

S. Hur, T. Kim, D. J. Love, J. V. Krogmeier, T. A. Thomas, andA. Ghosh, Millimeter wave beamforming for wireless backhaul andaccess in small cell networks, IEEE Trans. Commun. 61 (2013),no. 10.

T. L. Marzetta, Noncooperative cellular wireless with unlimitednumbers of base station antennas, IEEE Trans. Wireless Commun.9 (2010), no. 11, 3590 – 3600.

S. Thoen, L. Van der Perre, B. Gyselinckx, and M. Engels,Performance analysis of combined Transmit-SC/Receive-MRC,IEEE Trans. Commun. 49 (2001), no. 1, 5 – 8.

Conclusions

References III

V. Venkateswaran and A-J. V. Veen, Analog beamforming in MIMOcommunications with phase shift networks and online channelestimation, IEEE Trans. Signal Process. 58 (2010), no. 8, 4131 –4143.

T. Yoo and A. Goldsmith, On the optimality of multiantennabroadcast scheduling using zero-forcing beamforming, IEEE Trans.Sel. Area. Commun. 24 (2006), no. 3, 528 – 541.

E. Zhang and C. Huang, On achieving optimal rate of digitalprecoder by RF-Baseband codesign for MIMO systems, Proc.IEEE Veh. Technol. Conf. (VTC Fall), 2014, pp. 1 – 5.

X. Zhang, A. F. Molisch, and S-Y. Kung, Variable-phase-shift-basedRF-Baseband codesign for MIMO antenna selection, IEEE Trans.Signal Process. 53 (2005), no. 11, 4091 – 4103.

slide11 icc2015

Engineering

system scenario ms1

objective system scenario

objective system model

n antennas

introduction system

rf chain tx analog

discard cp

multiuser miso system