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

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 0.349 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 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

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),

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 are orthogonal

• Let

To transmit efficiently to mobiles with

we design a special precoding matrix

Transmission to

Transmission to

are orthogonal

are orthogonaland

Decoding

Q(3), |Q(3)|=1080 Random Code C, |C|=1080

• Complex

multiplications 0 8*1080

• Complex

summations 1500 7*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.