Interference Alignment and Degrees of Freedom of the $K$ -User Interference Channel
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...Unfortunately, the regulatory bodies governing spectrum allocation have not shown much appetite for change since their inception in the early 1900s....
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...ference alignment, has been shown to achieve significantly improved multiplexing gain for the MIMO interference network, where both the transmitters and the receivers are equipped with multiple antennas [72]....
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..., [72])....
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References
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"Interference Alignment and Degrees ..." refers background in this paper
...beamforming [ 18 ]. While the pre-log term is the principal determinant of capacity as SNR approaches infinity, terms are quite significant at SNR values of practical interest....
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1,872 citations
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...Recently, a special case of the Han–Kobayashi scheme [3] is shown in [10] to achieve the capacity of the two-user interference channel within one bit....
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...A combination of Han–Kobayashi [3] type achievable schemes and structured codes is a promising avenue in the quest for the capacity of wireless networks....
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...926344 While less common in practice due to the complexity of multi-user detection, this approach is supported by the capacity results on the “very strong interference” [1], and “strong interference” [3], [4] scenarios in the context...
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...],i,j ∈ {1,2,3} have a full rank of M almost surely, the above equations can be equivalently represented as span(V[1]) = span(EV[1]) (50) V[2] = FV[1] (51) V[3] = GV[1] (52) where E = (H[31])−1H[32](H[12])−1H[13](H[23])−1H[21] F = (H[32])−1H[31] G = (H[23])−1H[21] Let e 1,e 2,...eM be the M eigenvectors of E. Then we set V 1 to be V[1] = [ e 1 ... (M/2)] Then V[2] and V[3] are found using equations (5...
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...of the interference has to be less than or equal to M/2. The following three interference alignment equations ensure that the dimension of the interference is equal to M/2 at all the receivers. span(H[12]V[2]) = span(H[13]V[3]) (47) H[21]V[1] = H[23]V[3] (48) H[31]V[1] = H[32]V[2] (49) where span(A) represents the vector space spanned by the column vectors of matrix A We now wish to choose V[i],i = 1,...
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...ced by equations (40), the interference vectors from transmitters 3,4...K are perfectly aligned with the interference from transmitter 2 and therefore, all interference arrives along the nN vectors H¯[12]V¯[2]. In order to prove that there are (n+1)N interference free dimensions it suffices to show that the col umns of the square, Mn ×Mn dimensional matrix h H¯[11]V¯[1] H¯[12]V¯[2] i (46) are linearly ...
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...corresponds to the term αmk = 0,∀(m,k)), the linear independence condition boils down to the condition that 1 = 0 with non-zero probability - an obvious contradiction. Thus the matrix h H¯[11]V¯[1] H¯[12]V¯[2] i can be shown to be non-singular with probability1. Similarly, the desired signal can be chosen to be linearly independent of the interference at all other receivers almost surely. Thus ( ( n+1...
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...mensional received signal vector Y¯[1], the dimension of the interference should be not more than nN. This can be ensured by perfectly aligning the interference from transmitters 2,3...K as follows H¯[12]V¯[2] = H¯[13]V¯[3] = H¯[14]V¯[4] = ... = H¯[1K]V¯[K] (37) At the same time, receiver 2 zero-forces the interference from X¯[i],i 6= 2 . To extract nN interference-free dimensions from a Mn = (n+1)N +...
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