Quantization Codes Comprising Multiple Orthonormal Bases Alexei Ashikhmin Bell Labs  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)

MIMO Broadcast Transmission Base Station is a quantization code 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

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 ) BS is the channel vector of If are pairwise orthogonal then signals sent to do not interfere with each other is the channel vector of

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 Base Station

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

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 Example

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) 105 orthogonal bases (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)

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 will find three orthogonal occupied codewords

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 Each codeword belongs to 7975 bases. The bases form the constant weight code (n=1080, |C|= , 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

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

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

Q(3) greedy alg.

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. Mutually Unbiased Bases (MUB) Bases form a full size MUB set

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

There are 840 full size MUB sets, each belongs to 56 full size MUB sets Let are orthogonal Let To transmit efficiently to mobiles with we design a special precoding matrix

Transmission to are orthogonal and

Decoding Q(3), |Q(3)|=1080 Random Code C, |C|=1080 Complex multiplications 0 8*1080 Complex summations *1080 Example M=8

Q(m) is a code in There are two equivalent methods for construction of Q(m): 1.Group theoretic approach 2.Coding theory approach Construction of Q(m)

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

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

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

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

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

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).

Q(m) is obtained by merging of 1.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 Coding Theory approach for construction of Q(m)

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

(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) 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)

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

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