Alphabeta Math
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A numerical asymmetric Gaussian channel in the strong regime

Example

Use the complex Gaussian channel with P1=P2=1, unit noise variance, h11=1, h22=3/2, h21=2, and h12=5/2 in the convention of Real and complex two-user Gaussian interference channels. Then SNR1=1,SNR2=94,INR1=4,INR2=254, and the exact capacity region is R1≥0,R2≥0,R1≤1,R2≤log⁡2134,R1+R2≤log⁡26 bits per use.

Facts & Assumptions

Given: The complex Gaussian interference channel and parameters in the example.

[A1]

Assume AC (The Axiom of Choice), as required by the Gaussian information and coding suppliers used in [F2].

[F1]

Real and complex two-user Gaussian interference channels: INR1=∣h21∣2P2 and INR2=∣h12∣2P1; complex scalar rates use κ=1.

[F2]

Exact strong-interference region for the two-user Gaussian interference channel: if ∣h12∣≥∣h11∣ and ∣h21∣≥∣h22∣, capacity is the stated intersection of the two Gaussian MAC regions.

Proof

1.1A1F1F2givenalgebra

Assume [A1] and use the complex channel model [F1] with the stated unit power budgets and unit-variance noises. The gains satisfy ∣h12∣=2.5≥1=∣h11∣ and ∣h21∣=2≥1.5=∣h22∣, so the strong-regime hypothesis of [F2] holds. The displayed SNR and INR values follow by squaring the gains and multiplying by unit powers.

2.1F1F2step 1.1algebra∎

The individual constraints in [F2] become R1≤log⁡2(2)=1 and R2≤log⁡2(1+9/4)=log⁡2(13/4). The two sum constraints are log⁡2(1+1+4)=log⁡26 and log⁡2(1+9/4+25/4)=log⁡2(19/2); since 6<19/2, the first is active. This gives exactly the stated region.

Depends on

Used by

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