Communcation over interference channels Dustin Cartwright 1 February - - PowerPoint PPT Presentation

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Communcation over interference channels Dustin Cartwright 1 February - - PowerPoint PPT Presentation

Communcation over interference channels Dustin Cartwright 1 February 24, 2011 1 work in progress with Guy Bresler and David Tse Multiple-input multiple-output channel transmitter receiver The transmitter sends a signal v C N by


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SLIDE 1

Communcation over interference channels

Dustin Cartwright1 February 24, 2011

1work in progress with Guy Bresler and David Tse

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SLIDE 2

Multiple-input multiple-output channel

transmitter receiver

◮ The transmitter sends a signal v ∈ CN by transmitting across

N = 3 antennas.

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SLIDE 3

Multiple-input multiple-output channel

transmitter receiver

◮ The transmitter sends a signal v ∈ CN by transmitting across

N = 3 antennas.

◮ The receiver detects Hv ∈ CN across its N antennas, where

each entry of the H ∈ CN×N depends on the signal path.

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SLIDE 4

Multiple-input multiple-output channel

transmitter receiver

◮ The transmitter sends a signal v ∈ CN by transmitting across

N = 3 antennas.

◮ The receiver detects Hv ∈ CN across its N antennas, where

each entry of the H ∈ CN×N depends on the signal path.

◮ If H is known and invertible, then the receiver can reconstruct

the message v.

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SLIDE 5

Multiple users of the same channel

H22 H11 H12 H21

transmitter 1 transmitter 2 receiver 1 receiver 2

◮ K transmitter-receiver

pairs using the same channel.

◮ Determined by K 2

channel matrices Hij of size N × N.

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SLIDE 6

Multiple users of the same channel

H22 H11 H12 H21

transmitter 1 transmitter 2 receiver 1 receiver 2

◮ K transmitter-receiver

pairs using the same channel.

◮ Determined by K 2

channel matrices Hij of size N × N.

◮ Reciever 1 only cares

about transmitter 1’s message, etc.

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SLIDE 7

Strategies for interference alignment

◮ Each transmitter has a subspace Vj ⊂ CN to transmit in. ◮ Each receiver has a subspace Ui ⊂ CN and only pays

attention to its signal modulo Ui.

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SLIDE 8

Strategies for interference alignment

◮ Each transmitter has a subspace Vj ⊂ CN to transmit in. ◮ Each receiver has a subspace Ui ⊂ CN and only pays

attention to its signal modulo Ui. In order for this to work, we need:

◮ For i = j, HijVj ⊂ Uj. ◮ (HiiVi) ∩ Ui = ∅.

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SLIDE 9

Strategies for interference alignment

◮ Each transmitter has a subspace Vj ⊂ CN to transmit in. ◮ Each receiver has a subspace Ui ⊂ CN and only pays

attention to its signal modulo Ui. In order for this to work, we need:

◮ For i = j, HijVj ⊂ Uj. ◮ (HiiVi) ∩ Ui = ∅.

If each Hii is generic, the second condition is satisfied automatically.

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SLIDE 10

Questions

◮ For which N, K, and (d1, . . . , dK) will generic channel

matrices have a solution?

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SLIDE 11

Questions

◮ For which N, K, and (d1, . . . , dK) will generic channel

matrices have a solution?

◮ What is the information capacity of this channel?

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SLIDE 12

Questions

◮ For which N, K, and (d1, . . . , dK) will generic channel

matrices have a solution?

◮ What is the information capacity of this channel? ◮ How to parametrize spaces of solution strategies?

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SLIDE 13

Incidence correspondence

  • CN×NK(K−1)

×

K

  • i=1

Gr(di, N) × Gr(N − di, N) Subvariety of those (H12, . . . , HK−1,K, V1, . . . , VK, U1, . . . , UK) such that HijVj ⊂ Ui for 1 ≤ i = j ≤ K

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SLIDE 14

Incidence correspondence

  • CN×NK(K−1)

×

K

  • i=1

Gr(di, N) × Gr(N − di, N) Subvariety of those (H12, . . . , HK−1,K, V1, . . . , VK, U1, . . . , UK) such that HijVj ⊂ Ui for 1 ≤ i = j ≤ K This is a vector bundle over a product of Grassmannians.

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SLIDE 15

Incidence correspondence

  • CN×NK(K−1)

×

K

  • i=1

Gr(di, N) × Gr(N − di, N) Subvariety of those (H12, . . . , HK−1,K, V1, . . . , VK, U1, . . . , UK) such that HijVj ⊂ Ui for 1 ≤ i = j ≤ K This is a vector bundle over a product of Grassmannians.

Question

Is the projection onto

  • CN×NK(K−1) surjective?
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Existence of solutions

Theorem

Assume that d = d1 = · · · = dK and K ≥ 3. Then a generic set of channel matrices has a solution if and only if 2N ≥ d(K + 1). If so, the dimension of the solution variety is dK

  • 2N − d(K + 1)
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Existence of solutions

Theorem

Assume that d = d1 = · · · = dK and K ≥ 3. Then a generic set of channel matrices has a solution if and only if 2N ≥ d(K + 1). If so, the dimension of the solution variety is dK

  • 2N − d(K + 1)
  • For non-constant di, we have the necessary conditions:

di + dj ≤ N for all i, j

  • i∈S

2di(N − di) ≥

  • i=j∈S

didj for all subsets S ⊂ {1, . . . , K}

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SLIDE 18

K = 3

The threshold case for feasibility is (d1, d2, d3) = (d, d, N − d), where d1 ≤ N/2.

=

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SLIDE 19

K = 3

The threshold case for feasibility is (d1, d2, d3) = (d, d, N − d), where d1 ≤ N/2.

=

◮ After change of coordinates, can assume that all but one

channel matrix is the identity.

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SLIDE 20

K = 3

The threshold case for feasibility is (d1, d2, d3) = (d, d, N − d), where d1 ≤ N/2.

=

◮ After change of coordinates, can assume that all but one

channel matrix is the identity.

◮ For dimension reasons, inclusions become equalities:

V1 = U3 = V2 ⊂ U1 = V3 = U2 ⊃ H21V1

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SLIDE 21

An eigenvector-like problem

Given generic N × N matrix H, find

◮ V ⊂ CN, subspace of dimension d ◮ U ⊂ CN, subspace of dimension e = N − d

such that V ⊂ U and HV ⊂ U

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SLIDE 22

An eigenvector-like problem

Given generic N × N matrix H, find

◮ V ⊂ CN, subspace of dimension d ◮ U ⊂ CN, subspace of dimension e = N − d

such that V ⊂ U and HV ⊂ U

◮ For d = e = 1, this is equivalent to V = U being the span of

an eigenvector.

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SLIDE 23

An eigenvector-like problem

Given generic N × N matrix H, find

◮ V ⊂ CN, subspace of dimension d ◮ U ⊂ CN, subspace of dimension e = N − d

such that V ⊂ U and HV ⊂ U

◮ For d = e = 1, this is equivalent to V = U being the span of

an eigenvector.

◮ More generally, for d = e > 1, take V = U to be spanned by

d eigenvectors. In particular, N

d

  • solutions.
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SLIDE 24

An eigenvector-like problem

Given generic N × N matrix H, find

◮ V ⊂ CN, subspace of dimension d ◮ U ⊂ CN, subspace of dimension e = N − d

such that V ⊂ U and HV ⊂ U

◮ For d = e = 1, this is equivalent to V = U being the span of

an eigenvector.

◮ More generally, for d = e > 1, take V = U to be spanned by

d eigenvectors. In particular, N

d

  • solutions.

◮ For e > d, variety of solutions of dimension

  • N − (e − d)
  • (e − d)
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SLIDE 25

Parametrizing the solution variety

Recall: Want to find V , U such that V ⊂ U and HV ⊂ U.

◮ Set ℓ :=

  • d

e − d

  • ◮ Choose S ⊂ CN of dimension d − ℓ(e − d).
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SLIDE 26

Parametrizing the solution variety

Recall: Want to find V , U such that V ⊂ U and HV ⊂ U.

◮ Set ℓ :=

  • d

e − d

  • ◮ Choose S ⊂ CN of dimension d − ℓ(e − d).

◮ Choose S + HS ⊂ T ⊂ CN of dimension e − ℓ(e − d).

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SLIDE 27

Parametrizing the solution variety

Recall: Want to find V , U such that V ⊂ U and HV ⊂ U.

◮ Set ℓ :=

  • d

e − d

  • ◮ Choose S ⊂ CN of dimension d − ℓ(e − d).

◮ Choose S + HS ⊂ T ⊂ CN of dimension e − ℓ(e − d). ◮ Set

U = S + T + . . . Hℓ−1T V = T + . . . + HℓT

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SLIDE 28

Parametrizing the solution variety

Recall: Want to find V , U such that V ⊂ U and HV ⊂ U.

◮ Set ℓ :=

  • d

e − d

  • ◮ Choose S ⊂ CN of dimension d − ℓ(e − d).

◮ Choose S + HS ⊂ T ⊂ CN of dimension e − ℓ(e − d). ◮ Set

U = S + T + . . . Hℓ−1T V = T + . . . + HℓT Structure of whole variety seems complicated: when e = d + 1, then it is the toric variety for the Minkowski sum of hypersimplices ∆e,N + ∆N,d.

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SLIDE 29

Numbers of solutions

Return to K ≥ 3 d = d1 = . . . = dk Zero-dimensional when N = d(K+1)

2

. The number of solutions is: K d 3 4 5 6 7 1 2

  • 216
  • 1,975,560

2 6 3700 388,407,960 3 20

  • 4

70 . . . . . . d 2d

d

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SLIDE 30

Number of solutions when d = 1

Assume d = d1 = · · · dK = 1 and 2N = K + 1. Degenerate each Hij to a rank 1 matrix: HijVj ⊂ Uj ⇐ ⇒ Vj ⊂ ker Hij or Uj ⊃ im Hij

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SLIDE 31

Number of solutions when d = 1

Assume d = d1 = · · · dK = 1 and 2N = K + 1. Degenerate each Hij to a rank 1 matrix: HijVj ⊂ Uj ⇐ ⇒ Vj ⊂ ker Hij or Uj ⊃ im Hij

Theorem

Number of solutions = number of balanced orientations of the graph G G has edges tj − si whenever i = j. Balanced orientation means that in degree(v) = out degree(v) = K − 1 2 for all vertices v of G.

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SLIDE 32

Further questions

◮ If the di are not necessarily all equal, when does a feasible

strategy exist?

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SLIDE 33

Further questions

◮ If the di are not necessarily all equal, when does a feasible

strategy exist?

◮ Can we parametrize the solution variety in more cases?

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SLIDE 34

Further questions

◮ If the di are not necessarily all equal, when does a feasible

strategy exist?

◮ Can we parametrize the solution variety in more cases? ◮ What if the receivers and transmitters have different numbers

  • f antennas?
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SLIDE 35

Further questions

◮ If the di are not necessarily all equal, when does a feasible

strategy exist?

◮ Can we parametrize the solution variety in more cases? ◮ What if the receivers and transmitters have different numbers

  • f antennas?

◮ What if the channel matrices have the form

Hij =      ˜ Hij · · · ˜ Hij . . . ... . . . · · · ˜ Hij     ?