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Quantization Codes Comprising Multiple Orthonormal Bases

Quantization Codes Comprising Multiple Orthonormal Bases. MIMO Broadcast Transmission Quantizers Q(m) for MIMO Broadcast Systems transmission to mobiles with orthogonal channel vectors transmission to mobiles with almost orthogonal channel vectors Simulation Results

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Quantization Codes Comprising Multiple Orthonormal Bases

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  1. Quantization Codes Comprising Multiple Orthonormal Bases • MIMO Broadcast Transmission • Quantizers Q(m) for MIMO Broadcast Systems • transmission to mobiles with orthogonal channel vectors • transmission to mobiles with almost orthogonal • channel vectors • Simulation Results • Algebraic Constructions of Q(m) Alexei Ashikhmin Bell Labs

  2. Base Station • The Base Station (BS): • chooses some mobiles, for example mobiles 1,2,3 • forms and using computes a precoding matrix • transmits to mobiles 1,2,3 using the precoding matrix MIMO Broadcast Transmission is a quantization code

  3. Requirements for a quantization code • should provide good quantization (for given size ) • should afford a simple decoding • should have many sets of M orthogonal codewords (bases of ) is the channel vector of BS is the channel vector of is the channel vector of If are pairwise orthogonal then signals sent to do not interfere with each other

  4. Base Station • Mobiles quantize: • Base Station strategy – among find orthogonal codewords, say , and transmit to the corresponding mobiles 1,3,5 • The channel vectors of these mobiles will be almost orthogonal

  5. orthogonal codewords • If the number of mobiles (channel vectors) is large, e.g. , then • with a high probability all codewords will be occupied Let us have a quantization code If a channel vector is quantized into we say that is occupied and mark by • In this case even if we have only a few sets of orthogonal codewords, we easily find a set of occupied orthogonal codewords

  6. Example • Let and the number of mobiles is small, say • Let • If are many sets of orthogonal code vectors there is a chance to find occupied orthogonal codewords • For example, if • are sets of orthogonal codewords. Then

  7. Example: The number of antennas The first code in the family: (for practical applications we add four vectors to the code to make the code size 64) (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1) (1, 0, 1, 0), (0, 1, 0, 1), (1, 0, -1, 0), (0, -1, 0, 1) (1, 0, -i, 0), (0, 1, 0, -i), (1, 0, i, 0), (0, 1, 0, i) (1, 1, 0, 0), (0, 0, 1, 1), (1, -1, 0, 0), (0, 0, -1, 1) (1, -i, 0, 0), (0, 0, 1, -i), (1, i, 0, 0), (0, 0, 1, i) (1, 0, 0, 1), (0, 1, 1, 0), (1, 0, 0, -1), (0, 1, -1, 0) (1, 0, 0, -i), (0, 1, i, 0), (1, 0, 0, i), (0, 1, -i, 0) (1, 1, 1, 1), (1, -1, 1, -1), (1, 1, -1, -1), (1, -1, -1, 1) (1, 1, -i, -i), (1, -1, -i, i), (1, 1, i, i), (1, -1, i, -i) (1, -i, 1, -i), (1, i, 1, i), (1, -i, -1, i), (1, i, -1, -i) (1, -i, -i, -1), (1, i, -i, 1), (1, -i, i, 1), (1, i, i, -1) (1, -i, -i, 1), (1, i, -i, -1), (1, -i, i, -1), (1, i, i, 1) (1, -i, 1, i), (1, i, 1, -i), (1, -i, -1, -i), (1, i, -1, i) (1, 1, 1, -1), (1, -1, 1, 1), (1, 1, -1, 1), (1,-1,-1,-1) (1, 1,-i, i), (1, -1, -i, -i), (1, 1, i, -i), (1, -1, i, i) 105 orthogonal bases

  8. The number of mobiles • The bases form the constant weight code (n=60, |C|=105, w=4). • With probability 0.65 will find four orthogonal occupied codewords • With probability 0.349 will find three orthogonal occupied codewords

  9. Examples (continued) 1. The number of orthogonal bases is 105. Each codeword belongs to 7 bases. The bases form the constant weight code (n=60, |C|=105, w=4). 2. The number of orthogonal bases is 1076625. Each codeword belongs to 7975 bases. The bases form the constant weight code (n=1080, |C|=1076625, w=8) If K is small that the probability to find M occupied orthogonal codewords is also small What to do? - Use almost orthogonal codewords

  10. K=1000 Q(3) Yoo and Goldsmith greedy alg. with RVQ RVQ with Reg. ZF RVQ with ZF Simulation Results All results for M=8, i.e. the number of Base Station antennas is 8 Q(3)

  11. Q(3), Q(3), If K=50 typically we can find 5 or 6 occupied codewords

  12. Q(3) greedy alg.

  13. Bases form a full size MUB set Mutually Unbiased Bases (MUB) Def. Orthonormal bases of are mutually unbiased if for any we have Theorem The number of MUBs Def. (i.e. ) is a full size MUB set.

  14. MUB sets form a constant weight code C (n=15, |C|=6, w=5) • If K is small the chance that M occupied codewords are covered by • an MUB set is significantly higher than that they are covered by a basis

  15. There are 840 full size MUB sets , each belongs to 56 full size MUB sets • Let are orthogonal • Let are orthogonal • Let • Totransmit efficiently to mobiles with • we design a special precoding matrix

  16. Transmission to Transmission to are orthogonal and are orthogonal

  17. Q(3), |Q(3)|=1080 Random Code C, |C|=1080 • Complex • multiplications0 8*1080 • Complex • summations1500 7*1080 Decoding Example M=8

  18. Construction of Q(m) • Q(m) is a code in • There are two equivalent methods for construction of Q(m): • Group theoretic approach • Coding theory approach

  19. Orthogonal Projectors • A subspace of can be defined by its orthogonal • projector , i.e. • a • is an orthogonal projector iff

  20. Group Theoretic Construction of Q(m) Pauli matrices: where

  21. It is easy to check that Theorem is an orthogonal projector and

  22. Def. Vectors and are orthogonal (with respect • to the symplectic inner product) if • is a set of orthogonal independent vectors • . • Lemma 2 The operator is an orthogonal projector on a subspace , • and

  23. It is easy to check that and Thus defines a subspace . So is a line. therefore

  24. Construction of Q(m) • Take all sets of orthogonal independent vectors • Take all choices of • For each set and set compute • defines a line, in other words defines a code vector of Q(m).

  25. Coding Theory approach for construction of Q(m) • Q(m) is obtained by merging of • Binary Reed-Muller codes RM(r,m); • is the order or RM(r,m), • the code length is • 2. Codes B(m) over the alphabet {1,-1,i,-i} • the code length is

  26. 3. r=m-2=0: take the only minimum weight codeword of RM(r,m)=RM(0,m): (1,1,1,1) and substitute into its nonzero positions codewords of Merging RM(r,m) and B(m) into Q(m) r changes from m=2 to 0: • r=m=2: take the all minimum weight codewords of RM(2,2): • r=m-1=1: substitute codewords of • into the minimum weight codewords of RM(1,2) (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1) Minimum weight codeword of RM(1,2): Codewords of Q(2): (1,i) (1,-i) (1,1) (1,-1) (1,i,0,0) (0,1,i,0) (0,1,-i,0) (1,-i,0,0) (1,1,0,0) (0,1,1,0) (1,1,0,0) (0,1,1,0) (1,-1,0,0) (0,1, -1,0)

  27. r=0, minimum weights v codewords of RM(2,2) r=1, minimum weights v codewords of RM(1,2) v +codewords of B(1) r=2, minimum weights v codewords of RM(0,2) v +codewords of B(2) (1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1) (1,1,0,0),(1,i,0,0),(1,-1,0,0),(1,-i,0,0) (1,0,1,0),(1,0,i,0),(1,0,-1,0),(0,1,0,-i) (1,0,0,1),(1,0,0,i),(1,0,0,-1),(1,0,0,-i) (0,1,1,0),(0,1,i,0),(0,1,-1,0),(0,1,-i,0) (0,1,0,1),(0,1,0,i),(0,1,0,-1),(1,0,-i,0) (0,0,1,1),(0,0,1,i),(0,0,-1,1),(0,0,1,-i) (1,1,1,1), (1,-1,1,-1), (1,1,-1,-1), (1,-1,-1,1), (1,1,-i,-i), (1,-1,-i,i), (1,1,i,i), (1,-1,i,-i), (1,-i,1,-i), (1,i,1,i), (1,-i,-1,i), (1,i,-1,-i), (1,-i,-i,-1), (1,i,-i,1), (1,-i,i,1), (1,i,i,-1), (1,-i,-i,1), (1,i,-i,-1), (1,-i,i,-1),(1,i,i,1), (1,-i,1,i), (1,i,1,-i), (1,-i,-1,-i), (1,i,-1,i), (1,1,1,-1), (1,-1,1,1), (1,1,-1,1), (1,-1,-1,-1), (1,1,-i,i), (1,-1,-i,-i), (1,1,i,-i), (1,-1,i,i)

  28. Theorem Example: Theorem (Inner product distribution of Q(m)). For any we have and the number of such that is Example: in Q(2) there are 15 vectors such that in Q(3) there are 315 vectors such that

  29. Theorem For any basis there exist bases such that is an MUB set. TheoremThe maximum root-mean-square (RMS) inner product is

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