Transcript Extrimes of Information Combining

Grassmannian Packings for Efficient
Quantization in MIMO Broadcast
Transmission
Alexei Ashikhmin
and RaviKiran Gopalan
Bell Labs
Texas Instrument
 MIMO Broadcast Transmission
 Examples GRM(m) for MIMO Broadcast Systems
• transmission to mobiles with orthogonal channel vectors
• transmission to mobiles with almost orthogonal
channel vectors
 Simulation Results
 Algebraic Construction of GRM(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
)
)
is the channel vector of
BS
is the channel vector of
is the channel vector of
If
are pairwise orthogonal then signals sent to
not interfere with each other
do
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
Let us have a quantization code
orthogonal codewords
If a channel vector
and mark
is quantized into
we say that
is occupied
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
• The number of mobiles is small, say
• Still if there are many sets of orthogonal codewords, there is a chance
to find occupied orthogonal codewords
• For example, let
be sets of orthogonal codewords. Then
Example:
The number of antennas
(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)
The first code in the family:
(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)
(for practical applications we
(1, 0, 0, -i), (0, 1, i, 0), (1, 0, 0, i), (0, 1, -i, 0)
add four vectors to the code
(1, 1, 1, 1), (1, -1, 1, -1), (1, 1, -1, -1), (1, -1, -1, 1)
to make the code size 64)
(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)
105 orthogonal bases
(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
Mutually Unbiased Bases (MUB)
Def. Orthonormal bases
if for any
of
are mutually unbiased
we have
Theorem The number of MUBs
Def.
(i.e.
Bases
) is a full size MUB set.
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
full size MUB sets
, each
belongs to 56
Simulation Results
All results for M=8, i.e. the number of Base Station antennas is 8
K=1000
GRM(3)
GRM(3)
Yoo and Goldsmith greedy
alg. with RVQ
RVQ with Reg. ZF
RVQ with ZF
If K=50 typically
we can find 5 or 6
GRM(3),
occupied codewords
GRM(3),
GRM(3)
greedy alg.
are orthogonal
and
are orthogonal
Transmission to
Transmission to
Construction of GRM(m)
GRM(m) is a code in
There are two methods for construction of GRM(m):
1. Group theoretic approach – a particular case of the Operator ReedMuller codes (A.Ashikhmin and A.R.Calderbank, ISIT 2005)
2. Coding theory approach
Group Theoretic Construction of GRM(m)
Pauli matrices:
where
Def. Vectors
and
are orthogonal (with respect
to the symplectic inner product) if
Construction of GRM(m)
•
is a set of orthogonal independent vectors
• .
Lemma 2 The operator
is an orthogonal projector on a subspace
,
Coding Theory approach for construction of GRM(m)
GRM(m) is obtained by merging of
1. Binary Reed-Muller codes RM(r,m)
2. Reed-Muller codes ZRM(2,m) codes over
ZRM(2,m) is generated by the Boolean functions:
ZRM(2,m) is generated by the Boolean functions
Let us construct the code
For example
from ZRM(2,m) by mapping
Merging RM(r,2) and CRM(2,2) into GRM(2)
r changes from m=2 to 0:
1. r=m=2: take the all minimum weight codewords of RM(r,m)=RM(2,2):
(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)
2. r=m-1=1: substitute codewords of
into the minimum weight codewords of RM(r,m)=RM(1,2)
Minimum weight codeword
of RM(1,2):
(1,i)
(1,-i)
Codewords of GRM(2):
(1,i,0,0)
(0,1,i,0)
(1,1,0,0)
(0,1,1,0)
(0,1,-i,0)
(1,-i,0,0)
(1,1)
(1,1,0,0)
(0,1,1,0)
(1,-1)
(1,-1,0,0)
(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
r=0, minimum weights
v
codewords of RM(2,2)
(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)
r=1, minimum weights
v
codewords of RM(1,2)
v +codewords of CRM(2,1)
(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),
r=2, minimum weights
v
codewords of RM(0,2)
v +codewords of CRM(2,2)
(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)
Theorem
Example:
Theorem (Inner product distribution of GRM(m)). For any
we have
and the number of
such that
Example: in GRM(2) there are 15 vectors
in GRM(3) there are 315 vectors
is
such that
such that
Theorem For any basis
such that
there exist bases
is an MUB set.
Theorem The maximum root-mean-square (RMS) inner product is
Decoding
Example M=8
GRM(3), |GRM(3)|=1080
Random Code C, |C|=1080
• Complex
multiplications
0
8*1080
1500
7*1080
• Complex
summations
Mobiles quantize:
If some channel vector
is quantized into
we say that
is occupied
The number of mobiles
is large, say
In this case, even if we
have only one set of
Example: The number of BS antennas M=4, hence
(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)
105 orthogonal
(1, 0, 0, 1), (0, 1, 1, 0), (1, 0, 0, -1), (0, 1, -1, 0)
bases
(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)
Example: The number of BS antennas M=4, hence
(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)
105 orthogonal
(1, 0, 0, 1), (0, 1, 1, 0), (1, 0, 0, -1), (0, 1, -1, 0)
bases
(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)
Merging of RM(r,m) and CRM(2,m)
changes from m to 0:
1. r=m=2: take the all minimum weight codewords of RM(r,2)=RM(2,2):
(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)
2. r=m-1=1: substitute codewords of
into minimum weight codewords of RM(r,2)=RM(1,2)
Minimum weight codeword
of RM(1,2):
(1,i)
(1,-i)
Codewords of G-ZRM(2):
(1,i,0,0)
(0,1,i,0,0)
(1,1,0,0)
(0,1,1,0)
(0,1,-i,0)
(1,-i,0,0)
(1,1)
(1,1,0,0)
(0,1,1,0)
(1,-1)
(1,-1,0,0)
(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
Lemma 1 The operator
Def. Vectors
is an orthogonal projector,
and
are orthogonal (with respect
to the symplectic inner product) if
•
is a set of orthogonal vectors
• .
Lemma 2 The operator
is an orthogonal projector on a subspace
Mutually Unbiased Bases (MUB)
Def. Orthonormal bases
if for any
of
are mutually unbiased
we have
Theorem The number of MUBs
Def.
(i.e.
Bases
) is full size MUB set.
form a full size MUB set