télécom paristech · introduction next frontier for wireless networks cells in cellular wireless...

L’interférence dans les réseaux non filairesDu contrôle de puissance au codage et alignement

Jean-Claude BelfioreTélécom ParisTech

7 mars 2013

Séminaire Comelec

Part 1 Interference in Wireless NetworksPart 2 Resource AllocationPart 3 Han & KobayashiPart 4 Interference AlignmentPart 5 The Compute-and-Forward tool

2 / 54J.-C. Belfiore - Interference in Wireless Networks

Part I

Interference in Wireless Networks

Introduction

Outline of current Part

1 Introduction

2 In cellular systems

3 In ad hoc networks

Introduction

Next Frontier for Wireless Networks

Cells in cellular wireless networks are becoming smaller and smaller as the density ofusers per space unit is becoming higher and higher. In wireless sensor networks, thedensity of sensors is becoming higher and higher as well.

Unwanted signalsEach node receives a combination of its own signal and many unwanted ones. Wirelessnetworks become more and more Interference Limited.

Introduction

Next Frontier for Wireless Networks

Cells in cellular wireless networks are becoming smaller and smaller as the density ofusers per space unit is becoming higher and higher. In wireless sensor networks, thedensity of sensors is becoming higher and higher as well.

Unwanted signalsEach node receives a combination of its own signal and many unwanted ones. Wirelessnetworks become more and more Interference Limited.

In cellular systems

1 Introduction

In cellular systems

Cellular network

Base StationR

Figure: One Cell : Many Users

In cellular systems

Cellular network

Base StationR

The Relay Channel

Figure: Accessing to hidden Terminals

In cellular systems

Cellular network

Base Station

The Multiple Access Channel

Figure: Uplink

In cellular systems

Cellular network

Base Station

The Broadcast Channel

Figure: Downlink

In cellular systems

Cellular network

Other system

Base Station 2

Base Station 1

The Interference Channel

Figure: Many cells sharing the same Physical Resources

In ad hoc networks

1 Introduction

In ad hoc networks

Ad Hoc and Wireless Sensor Networks

InterferencesAd Hoc and Wireless sensor networks can experience a high level of interference betweennodes when the number of nodes per area unit is high and the physical resource is scarce.

Properties of the Wireless MediumMain properties are

Braodcast property.

Superposition property.

In ad hoc networks

Ad Hoc and Wireless Sensor Networks

InterferencesAd Hoc and Wireless sensor networks can experience a high level of interference betweennodes when the number of nodes per area unit is high and the physical resource is scarce.

Properties of the Wireless MediumMain properties are

Braodcast property.

Superposition property.

Part II

Resource Allocation

Solve the problem at the RRM level

4 Solve the problem at the RRM level

5 Power Control

6 Subchannel allocation

At the Physical Layer

Orthogonal MultiplexingTransmit signals from different cells/users at different

Subbands (FDMA)

Time Slots (TDMA)

Interference problem is solved by avoiding Interference. But it is not enough ...

Interference as noiseAt receivers, the sum of all interfering signals is considered as noise. Definition of newparameters as

SINRi =Pi ·

∣∣hii∣∣2

N +∑j 6=i Pj ·

∣∣∣hji

∣∣∣2

for user i, where Pj is the transmit power of user j, hji is the attenuation from transmitcell j to receive cell i and N is the power of noise.

Interference has to be mitigated at the R adio R esource M anagement level.

Subbands (FDMA)

Time Slots (TDMA)

SINRi =Pi ·

∣∣hii∣∣2

N +∑j 6=i Pj ·

∣∣∣hji

∣∣∣2

Subbands (FDMA)

Time Slots (TDMA)

SINRi =Pi ·

∣∣hii∣∣2

N +∑j 6=i Pj ·

∣∣∣hji

∣∣∣2

Power Control

5 Power Control

Power Control

An Optimization Problem (Power Minimization)

Consider many interfering cells (or pairs of users) sharing the same physical resource (timeor frequency),

Fixed RateTarget Data rates Rk are given =⇒ Target SINR, γk are given. Optimization problem:

min~P∑

subject to SINRk ≥ γk

and Pi ≤ Pmax

Problem with solution[Pischella & B., 08] This problem has solutions whenever

∣∣hkk∣∣2

∑j 6=i

∣∣∣hjk

∣∣∣2> γk . (1)

When (1) is not satisfied for some cell, then we say that the network is interference-limited. No more degree of freedom is available.

Power Control

min~P∑

and Pi ≤ Pmax

∣∣hkk∣∣2

∑j 6=i

∣∣∣hjk

∣∣∣2> γk . (1)

Power Control

min~P∑

and Pi ≤ Pmax

∣∣hkk∣∣2

∑j 6=i

∣∣∣hjk

∣∣∣2> γk . (1)

Power Control

Rate Maximization

Same assumptions as before. Now we want to maximize a function ϕ (R1, . . . ,RK ) of the userrates.

Which functionϕ?A natural function can be the the weighted sum rate

where weights wk are proportional to users’ queue length.

Resolution[Pischella & B., 10] It is a nonconvex optimization problem that can only be solved whenSINR is high enough.

Power Control

Rate Maximization

Power Control

Rate Maximization

Subchannel allocation

5 Power Control

Use OFDMA ccess for the “orthogonality” option.

Rate-Constrained UsersFrequency allocation can be done optimizing a criterion fostering carriers withbetter SINR. Solution is “water filling”-type.

Rate MaximizationSubcarriers are not chosen when SINR falls below some threshold. Graph coloring algo-rithms may also be used for subcarrier allocation.

Part III

Han & Kobayashi

Beyond Interference as Noise

7 Beyond Interference as Noise

8 Han and Kobayashi [Han & Kobayashi, 81]

9 The W curve

What is possible in Interference-Limited Networks?

Figure: Channel Model

Model of the 2-user Interference Channel. Addi-tive Gaussian noise is present at each receiver.

Is it better to decode interference?When the level of interference becomes high enough, then it is better, for the receiver,to decode both the legitimate user and the interferers. For low level interference: stillconsider interference as noise.

Han and Kobayashi [Han & Kobayashi, 81]

9 The W curve

Han and Kobayashi Coding Scheme

TransmittersEach transmitter splits the data into

Private data.

Common data.

It transmits both flows using superposition cod-ing.

ReceiversRi decodes all data coming from Ti. It only de-codes common data from Tj , j 6= i. Private datafrom Tj are treated as noise.

Achievable RatesBy optimizing the ratio between Rc (common) and Rp (private), significantly higher ratesare achievable.

Private data.

Common data.

Private data.

Common data.

A toy example

PAM Constellation

8−PAM Transmitted Constellation

1 bit private and 2 bits common

Non legitimate receiver decodes

Figure: Example of an 8−PAM constellation

Hierarchical Modulation.

A toy example

PAM Constellation

8−PAM Transmitted Constellation

1 bit private and 2 bits common

Non legitimate receiver decodes

Figure: Example of an 8−PAM constellation

Hierarchical Modulation.

The W curve

9 The W curve

The W curve

Generalized Degrees of Freedom

Generalized D.O.F.Define

α= log INR

logSNR.

Use Han and Kobayashi.

Then the generalized DOF are defined as

D(α) = limSNR→∞

C(SNR,α)

logSNR

C is the capacity of the channel. C ≈ logSNR inabsence of interference.

No more degree of freedom

Interference as noise

1/2 2/3

Generalized Degrees of freedom (W curve)[Etkin, Tse & Wang, 08]

D(α) = 1−α 0 ≤α≤ 1/2 priv.

D(α) =α 1/2 ≤α≤ 2/3 priv.+com.

D(α) = 1−α/2 2/3 ≤α≤ 1 priv.+com.

D(α) = α/2 1 ≤α≤ 2 com.

D(α) = 1 α≥ 2 com.

The W curve

α= log INR

logSNR.

C(SNR,α)

logSNR

1/2 2/3

D(α) = 1−α 0 ≤α≤ 1/2 priv.

D(α) =α 1/2 ≤α≤ 2/3 priv.+com.

D(α) = 1−α/2 2/3 ≤α≤ 1 priv.+com.

D(α) = α/2 1 ≤α≤ 2 com.

D(α) = 1 α≥ 2 com.

The W curve

α= log INR

logSNR.

C(SNR,α)

logSNR

1/2 2/3

D(α) = 1−α 0 ≤α≤ 1/2 priv.

D(α) =α 1/2 ≤α≤ 2/3 priv.+com.

D(α) = 1−α/2 2/3 ≤α≤ 1 priv.+com.

D(α) = α/2 1 ≤α≤ 2 com.

D(α) = 1 α≥ 2 com.

Part IV

Interference Alignment

Alignment

PrinciplesInterference Alignment consists in reserving space (not linear in general) for all inter-fering signals. The remaining space will be used by the wanted signal and will be free ofinterference.

Which spacesSeveral types of alignment have been studied among which,

1 Linear over R or C. Needs space, time or frequency diversity.

2 Linear over Q. Does not need diversity.

3 Arithmetic (coding over residue rings). Does not need diversity.

Main ConstraintNeeds channels knowledge at all transmitting sides.

Alignment

Linear Interference Alignment [Cadambe & Jafar, 09]

10 Linear Interference Alignment [Cadambe & Jafar, 09]

11 Integer Interference Alignment [Jafarian & Vishwanath, 11]

12 What is really alignment?

Principle

An exampleSuppose we have 3 receivers for 5 un-knowns,

y1 = 3x1 +2x2 +3x3 +x4 +5x5

y2 = 2x1 +4x2 +x3 −3x4 +5x5

y3 = 4x1 +3x2 +5x3 +2x4 +8x5

where receiver 1 only wants x1.

Remark 2Vector u = [

17 −1 −10]>

is or-thogonal to all interfering vectors.

Remark 1In fact, the interfering beams span a 2−D space leavingone dimension free from interference for the wanted sig-nal,

H∗4 = H∗3 −H∗2 and H∗5 = H∗3 +H∗2.

Recovering x1

Projecting the received vector along u gives,

9x1 = 17y1 −y2 −10y3

recovering x1 from the 3 observed values yi.

Linear Interference alignment allows many interfering users to communicate simulta-neously over a small number of signalling dimensions by putting the interfering signalsin a space of small dimension so that the desired signal can be projected into the nullspace of the interference.

Principle

y1 = 3x1 +2x2 +3x3 +x4 +5x5

y2 = 2x1 +4x2 +x3 −3x4 +5x5

y3 = 4x1 +3x2 +5x3 +2x4 +8x5

17 −1 −10]>

H∗4 = H∗3 −H∗2 and H∗5 = H∗3 +H∗2.

Recovering x1

9x1 = 17y1 −y2 −10y3

Principle

y1 = 3x1 +2x2 +3x3 +x4 +5x5

y2 = 2x1 +4x2 +x3 −3x4 +5x5

y3 = 4x1 +3x2 +5x3 +2x4 +8x5

17 −1 −10]>

H∗4 = H∗3 −H∗2 and H∗5 = H∗3 +H∗2.

Recovering x1

9x1 = 17y1 −y2 −10y3

Principle

y1 = 3x1 +2x2 +3x3 +x4 +5x5

y2 = 2x1 +4x2 +x3 −3x4 +5x5

y3 = 4x1 +3x2 +5x3 +2x4 +8x5

17 −1 −10]>

H∗4 = H∗3 −H∗2 and H∗5 = H∗3 +H∗2.

Recovering x1

9x1 = 17y1 −y2 −10y3

Principle

y1 = 3x1 +2x2 +3x3 +x4 +5x5

y2 = 2x1 +4x2 +x3 −3x4 +5x5

y3 = 4x1 +3x2 +5x3 +2x4 +8x5

17 −1 −10]>

H∗4 = H∗3 −H∗2 and H∗5 = H∗3 +H∗2.

Recovering x1

9x1 = 17y1 −y2 −10y3

Feasibility

Feasibility of interference alignment depends on the structure and probabilistic behav-ior of the channel matrix.

MIMO caseFor instance, for the symmetric K−user inter-ference channel where all transmitters have ntantennas and all receivers, nr antennas, inter-ference alignment is feasible with probability 1when

d ≤ nt +nr

where d is the number of flows of each pair.

Symbol extensionBeamforming across multiple channel uses toincrease space dimension.

Feasibility

d ≤ nt +nr

Feasibility

d ≤ nt +nr

Integer Interference Alignment [Jafarian & Vishwanath, 11]

Alignment on Ideals

This type of alignment will be studied on an example. We assume 3 pairs of transmit-ters/receivers. PAM constellations are used (transmission of integer symbols), i.e. xi ∈Z.

Decoded signals are supposed to be:

y1 = x1 +4x2 +3x3

y2 = 2x1 +x2 +3x3

y3 = 6x1 +2x2 +x3

AlignmentInterference is aligned on ideals (6Z, 2Z and 3Z).Residue Symbols are received free of interfer-ence.

Now, precoding is done at the transmitters such that

x1 = u1 ; x2 = 3u2 ; x3 = 2u3

where ui carries information.

Equation (2) becomes

r1 = y1 = u1 +12u2 +6u3

r2 = y2 = 2u1 +3u2 +6u3

r3 = 12 y3 = 3u1 +3u2 +u3

If u1 ∈Z6, u2 ∈Z2 and u3 ∈Z3, then, u1 = r1 mod 6, u2 = r2 mod 2 and u3 = r3 mod 3.

Alignment on Ideals

y1 = x1 +4x2 +3x3

y2 = 2x1 +x2 +3x3

y3 = 6x1 +2x2 +x3

x1 = u1 ; x2 = 3u2 ; x3 = 2u3

r1 = y1 = u1 +12u2 +6u3

r2 = y2 = 2u1 +3u2 +6u3

r3 = 12 y3 = 3u1 +3u2 +u3

Alignment on Ideals

y1 = x1 +4x2 +3x3

y2 = 2x1 +x2 +3x3

y3 = 6x1 +2x2 +x3

x1 = u1 ; x2 = 3u2 ; x3 = 2u3

r1 = y1 = u1 +12u2 +6u3

r2 = y2 = 2u1 +3u2 +6u3

r3 = 12 y3 = 3u1 +3u2 +u3

Alignment on Ideals

y1 = x1 +4x2 +3x3

y2 = 2x1 +x2 +3x3

y3 = 6x1 +2x2 +x3

x1 = u1 ; x2 = 3u2 ; x3 = 2u3

r1 = y1 = u1 +12u2 +6u3

r2 = y2 = 2u1 +3u2 +6u3

r3 = 12 y3 = 3u1 +3u2 +u3

Alignment on Ideals

y1 = x1 +4x2 +3x3

y2 = 2x1 +x2 +3x3

y3 = 6x1 +2x2 +x3

x1 = u1 ; x2 = 3u2 ; x3 = 2u3

r1 = y1 = u1 +12u2 +6u3

r2 = y2 = 2u1 +3u2 +6u3

r3 = 12 y3 = 3u1 +3u2 +u3

Alignment on Ideals

y1 = x1 +4x2 +3x3

y2 = 2x1 +x2 +3x3

y3 = 6x1 +2x2 +x3

x1 = u1 ; x2 = 3u2 ; x3 = 2u3

r1 = y1 = u1 +12u2 +6u3

r2 = y2 = 2u1 +3u2 +6u3

r3 = 12 y3 = 3u1 +3u2 +u3

What is really alignment?

Framework

Linear or integerTechniques of alignment all rely in fact on the same idea.

Linear: Received signal (noiseless) belongs to a ring of multivariate polynomials (variables arechannel gains). Interferences are put in an ideal.

Integer: After approximation, channel gains are integer-valued. Received signal (noiseless)belongs to the ring of integers. Interferences are put in an ideal as well.

CodingLet R denote the ring of all possible (noiseless) received signal. Put all interferers in anideal J of R and encode the desired signal in the residue ring R/J .

Framework

Linear or integerTechniques of alignment all rely in fact on the same idea.

Linear: Received signal (noiseless) belongs to a ring of multivariate polynomials (variables arechannel gains). Interferences are put in an ideal.

Integer: After approximation, channel gains are integer-valued. Received signal (noiseless)belongs to the ring of integers. Interferences are put in an ideal as well.

CodingLet R denote the ring of all possible (noiseless) received signal. Put all interferers in anideal J of R and encode the desired signal in the residue ring R/J .

Part V

The Compute-and-Forward tool

Lattices

13 Lattices

14 Compute-and-Forward [Nazer & Gastpar 09]

15 1 interferer: (Almost) Achieving Capacity? [Nazer, Ordentlich & Erez, 12]

16 Is practical alignment feasible?

Lattices

From Modulation and Coding ...

What a lattice element could be

Binary Encoderin the signal space

Lattice element?

Modulator

Labeling

Figure: Encoder and Modulator

RequirementsEncoder must be linear

Modulation should be PAM, QAM or HEX

Labeling (modulator) between binary codewords and modulated symbols has to respectsome criteria

Lattices

Lattice element?

Modulator

Labeling

Lattices

Lattice element?

Modulator

Labeling

Lattices

An example: the D4 lattice (partition)

QAM Partition à la Ungerboeck

B subsetA subset

+3+1−1−3

Figure: Labeling of subsets A and B

Lattices

An example: the D4 lattice (coding)

Encoder

1 11Labeling

Binary data (QAM1,QAM2)

(A,A) ∪ (B,B)

Binary data (uncoded)

Figure: D4 encoder

The binary code is the (2,1) repetition code (linear)

Modulation is QAM, labeling is the Ungerboeck labeling

One of the simplest examples of “Construction A”

D4 = (1+ ı)Z[ı]2 + (2,1)F2

Lattices

Encoder

1 11Labeling

(A,A) ∪ (B,B)

Figure: D4 encoder

D4 = (1+ ı)Z[ı]2 + (2,1)F2

Lattices

Encoder

1 11Labeling

(A,A) ∪ (B,B)

Figure: D4 encoder

D4 = (1+ ı)Z[ı]2 + (2,1)F2

Lattices

Definition

Lattice pointsAn element v ofΛ can be written as :

v = a1v1 +a2v2 + . . .+anvn , a1 ,a2 , . . . ,an ∈Zwhere (v1 ,v2 , . . . ,vn) is a basis of Rn.

The lattice Λ can be defined as :

i=1aivi | ai ∈Z

Lattices

Lattices : Generator matrix

The set of vectors v1 ,v2 , . . . ,vn is a lattice basis.

DefinitionMatrix M whose columns are vectors v1,v2, . . . ,vn is a generator matrix of the latticedenotedΛM .

Each vector x = (x1 ,x2 , . . . ,xn)> in ΛM , can be written as,

x = M ·z

where z = (z1 ,z2 , . . . ,zn)> ∈Zn.

Lattice ΛM may be seen as the result of a linear transform applied to lattice Zn (cubic lattice).

Lattices

Lattices : Generator matrix

The set of vectors v1 ,v2 , . . . ,vn is a lattice basis.

DefinitionMatrix M whose columns are vectors v1,v2, . . . ,vn is a generator matrix of the latticedenotedΛM .

Each vector x = (x1 ,x2 , . . . ,xn)> in ΛM , can be written as,

x = M ·z

where z = (z1 ,z2 , . . . ,zn)> ∈Zn.

Lattice ΛM may be seen as the result of a linear transform applied to lattice Zn (cubic lattice).

Lattices

Lattices : Geometric properties

The generator matrix M describes the lattice ΛM , but it is not unique. All matrices M ·T where T hasinteger entries and detT =±1 are generator matrices of ΛM . T is called a unimodular matrix.

G = M> ·M is the Gram matrix of the lattice .

DefinitionsThe fundamental parallelotope of ΛM is the region,

P = {x ∈Rn p x = a1v1 +a2v2 + . . .+anvn , 0 ≤ ai < 1, i = 1. . .n

The fundamental volume is the volume of the fundamental parallelotope. It is denotedVol

The fundamental volume of the lattice is vol(ΛM

)= |det(M)|, which isp

det(G) either.

A bad basis is a basis with long vectors (large orthogonality defect).

A good basis (or reduced basis) is a basis with short vectors (small orthogonality defect).

Lattices

det(G) either.

Lattices

det(G) either.

Lattices

Z2 lattice

Lattice Basis

(v1, v2)

Lattice Point

Voronoi region

Fundamental Parallelotope

Z2 lattice

PropertiesGenerator matrix is

1 00 1

A QAM constellation is a finite part of Z2.

Lattices

Z2 lattice

Lattice Basis

(v1, v2)

Lattice Point

Voronoi region

Fundamental Parallelotope

Z2 lattice

1 00 1

A QAM constellation is a finite part of Z2.

Lattices

A2 lattice

Lattice basis

(v1, v2)

Lattice point

Voronoi region

The A2 lattice

Fundamental parallelotope

An HEX constellation is a finite part of A2, thehexagonal lattice.

Lattices

A2 lattice

Lattice basis

(v1, v2)

Lattice point

Voronoi region

The A2 lattice

Fundamental parallelotope

An HEX constellation is a finite part of A2, thehexagonal lattice.

Compute-and-Forward [Nazer & Gastpar 09]

13 Lattices

Principles

RelayThe Relay wants to reliably decode the result of computation λ = ∑k

j=1 ajxj . If xj are

lattice points of some integer lattice, thenλ is also a lattice point for some lattice.

h j x j

h kx k

∑kj=1 a j x j

∑Relay

Received signalReceived signal is y =∑k

j=1 hjxj +z where xj ∈Λ are lattice points,

hj ∈R and z iid Gaussian noise. Note that aj ∈Z.

GoalCompute λ reliably.

Principles

h j x j

h kx k

∑kj=1 a j x j

∑Relay

Principles

h j x j

h kx k

∑kj=1 a j x j

∑Relay

Principles

h j x j

h kx k

∑kj=1 a j x j

∑Relay

Computation Rate

Computation RateThe computation rate defined in is

Rcomp(h,a) = log2

(‖a‖2 − SNR

1+SNR‖h‖2

and is achievable by using lattice codes for xi.

MaximizationMaximization of Rcomp requires to choose aopt ∈Zk] as [Feng et al. 11],

aopt = argmina 6=0

I − SNR

1+SNR‖h‖2H

)a = argmin

a 6=0a>.Q.a

where H = [hihj]. Minimization of the symmetric form Q (Lattice Shortest Vector prob-lem. Λ has Gram matrix Q).

Successive MinimaFind the k successive minima ofΛ. ReduceΛ.

Computation Rate

Rcomp(h,a) = log2

(‖a‖2 − SNR

1+SNR‖h‖2

aopt = argmina 6=0

I − SNR

1+SNR‖h‖2H

)a = argmin

a 6=0a>.Q.a

Computation Rate

Rcomp(h,a) = log2

(‖a‖2 − SNR

1+SNR‖h‖2

aopt = argmina 6=0

I − SNR

1+SNR‖h‖2H

)a = argmin

a 6=0a>.Q.a

1 interferer: (Almost) Achieving Capacity? [Nazer, Ordentlich & Erez, 12]

13 Lattices

Use Compute-and-Forward

Each user transmits 2 data flows: private + common.

Each receiver sees k = 2 or k = 3 data flows to decode.

Rates with common data

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

tion R

Computation Rate for SNR = 40db

First Equation

Second Equation

Figure: Sum Rate with Gauss Reduction

Find the k successive minima related to Gram matrixQ.

2 or 3 computation rates R1, R2 and maybe R3 withR1 > R2 > R3 .

R2 is always achievable.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

tion R

First Equation

Second Equation

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

tion R

First Equation

Second Equation

Achievable Rates

Symmetric interference channel: Received Signals are y1 = x1 +gx2 +z1 and y2 = gx1 +x2 +z2;

Medium SNR

10−1

[Bits/C

hannel U

Upper Bound And Achievable Rate On The Capacity (SNR = 20db)

The Upper Bound

Achievable Rate

Figure: SNR= 20 dB ; Achievable Sum-Rate

High SNR

10−1

[Bits/C

hannel U

The Upper Bound

Achievable Rate

Achievable Rates

Medium SNR

10−1

[Bits/C

hannel U

The Upper Bound

Achievable Rate

High SNR

10−1

[Bits/C

hannel U

The Upper Bound

Achievable Rate

Achievable Rates

Medium SNR

10−1

[Bits/C

hannel U

The Upper Bound

Achievable Rate

High SNR

10−1

[Bits/C

hannel U

The Upper Bound

Achievable Rate

Is practical alignment feasible?

13 Lattices

From theory to practice

Linear alignmentProblems to overcome:

Perfect knowledge of all channel gains at all transmitters.

Based on Zero Forcing (becomes efficient when SNR→∞).

No idea of its behavior at finite SNR.

Integer alignment (or more generally lattice alignment)Problems to overcome:

Approximation of channel gains by integers generates additional noise.

No design criterion for lattice codes right now.

Fractal behaviorSome values of channel gains lead to performances much worse than very close otherones.

Conclusion

Open Problems

On the Coding+Alignment sideFind Lattice Codes adapted to the interference channel and find a practical way to aligninterferers.

Other pointsAsynchronous Codes?

Conclusion

Open Problems

On the Coding+Alignment sideFind Lattice Codes adapted to the interference channel and find a practical way to aligninterferers.

Other pointsAsynchronous Codes?

Merci !!

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