Download M5-MIMO

MIMO Communication Systems
• So far in this course
We have considered wireless communications with
only one transmit and receive antenna- SISO
However there are a lot advantages to be had if we
extend that to multiple transmit and multiple
receive antennas
• MIMO
– MIMO is short for multiple input multiple output
systems
– The “multiple” refers to multiple transmit and
receiver antennas
– Allows huge increases in capacity and performance
– MIMO became a “hot” area in 1998 and remains
hot
1
Motivation
• Current wireless systems
– Cellular mobile phone systems, WLAN, Bluetooth,
Mobile LEO satellite systems, …
• Increasing demand
– Higher data rate ( > 100Mbps) IEEE802.11n
– Higher transmission reliability (comparable to wire
lines)
– 4G
• Physical limitations in wireless systems
–
–
–
–
Multipath fading
Limited spectrum resources
Limited battery life of mobile devices
…
2
Multiple-Antenna Wireless Systems
• Time and frequency processing hardly meet
new requirements
• Multiple antennas open a new signaling
dimension: space
Create a MIMO channel
• Higher transmission rate
• Higher link reliability
• Wider coverage
3
General Ideas
• Digital transmission over Multi-Input MultiOutput (MIMO) wireless channel
Tx
coding
info. bits
modulation
weighting/mapping
Rx
weighting/demapping
detected bits
demodulation
decoding
• Objective: develop Space-Time (ST) techniques
with low error probability, high spectral
efficiency, and low complexity (mutually
conflicting)
4
Possible Gains: Multiplexing
• Multiple antennas at both Tx and Rx
• Can create multiple parallel channels
• Multiplexing order = min(M, N), where M
=Tx, N =Rx
• Transmission rate increases linearly
Tx
Rx
Tx
Rx
Spatial Channel 1
Spatial Channel 2
5
Possible Gains: Diversity
• Multiple Tx or multiple Rx or both
• Can create multiple independently faded
branches
• Diversity order = MN
• Link reliability improved exponentially
Tx
Rx
Tx
Rx
Fading Channel 1
Fading Channel 2
Fading Channel 3
Fading Channel 4
6
Key Notation- Channels
• Assume flat fading for now
• Allows MIMO channel to be written as a matrix H
Tx
Rx
x1
x2
y1  h11x1  h12 x2
h11
h12
h21
h22
y1
y2
y 2  h21x1  h22 x2
y  Hx
 h11 h12 
H 

h
h
 21 22 
• Generalize to arbitrary number of inputs M and outputs N so
H becomes a NxM matrix of complex zero mean Gaussian
random variables of unity variance
• Can understand that each output is a mixture of all the
different inputs- interference
7
• We assume UNCORRELATED channel elements
Key Decomposition- SVD
• SVD- singular value decomposition
• Allows H of NxM to be decomposed into parallel channels as
follows
H  USV
H
• Where S is a NxM diagonal matrix with elements only along the
diagonal m=n that are real and non-negative
• U is a unitary N x N matrix and V is a unitary M x M matrix
• The superscript H denotes Hermitian and means complex
transpose
• A Matrix is Unitary if AH=A-1 so that AHA = I
• The rank k of H is the number of singular values
• The first k left singular vectors form an orthonormal basis for the
range space of H
• The last right N-k right singular vectors of V form an orthogonal
basis for the null space of H
• What does SVD mean?
8
Key Decomposition- SVD
• What does it mean?
• Implies that UHHV=S is a diagonal matrix
• Therefore if we pre-process the signals by V at the
transmitter and then post-process them with UH we
will produce an equivalent diagonal matrix
• This is a channel without any interference and
channel gains s11 and s22 for example
Tx
Rx
Tx
Rx
Spatial Channel 1
Spatial Channel 2
9
Key Decomposition- SVD
• What are the singular values?
• You can remember eigenvalues and eigenvectors Ae  e
• If A is any square matrix then it can diagonalized using
E-1AE = D where E is the matrix of eigenvectors
• Note we can generate a square M x M matrix as
HHH= (USVH)H(USVH)=V(SHS)VH
• Letting A=HHH so that E-1AE = D = VHHHHV= SHS
• Alternatively we can generate a square N x N matrix as
HHH= (USVH)(USVH)H= UH (SSH)U
• Therefore we can see that the square of the singular
values are the eigenvalues of HHH
• Also note that V is the matrix of Eigenvectors of HHH
• Similarly U is the matrix of eigenvectors of HHH
10
Capacity
• For a SISO channel capacity C is given by
C  log 2 (1   | h |2 ) b/s/Hz
where  is the SNR at a receiver antenna and h is the
normalized channel gain
• For a MIMO channel we can make use of SVD to
produce multiple parallel channels so that
M min

C   log 2 (1 
i ) b/s/Hz
M min  min( M , N )
M min
1
• Where i are the eigenvalues of W

 HH H
W  H

H H
NM
MN
11
Capacity
• We can also alternatively write the MIMO capacity



C  log 2 det(I N 
HH H ) b/s/Hz
M


• It can be demonstrated for Rayleigh fading channels that if
N<M then the average capacity grows linearly with N as
N

C  N log 2 1    b/s/Hz
M

• This is an impressive result because now we can arbitrarily
increase the capacity of the wireless channel just by adding
more antennas with no further power or spectrum required
• In these calculations it is also assumed the transmitter has no
knowledge about the channel
12
Note on SNR
• The definition of the SNR used previously is simply the
receiver SNR at each receiver antenna
• In this definition the channel must be normalized rather than
be the actual measured channel
• This approach is used since it is more usual to specify things
in terms of received SNR
• However in calculations it is perhaps easier to think of total
transmit power, un-normalized channel G and the received
noise power per receive antenna, a, so capacity becomes
P


C  log 2 det(I N  T GG H ) b/s/Hz
aM


13
Example
14
Example
15
Special Cases
• Take M=N and H =In and assume noise has cross-correlation
In then C  N log 2 1  P / N  b/s/Hz
• Let Hij = 1 so that there is only one singular value given by
NM and assume noise has cross-correlation In
• The first column of U and V is
1
1  
U
 

N  
1N 
1 
1  
V
 

M  
1M 
• Thus C  log 2 1  NP b/s/Hz
• Each transmitter is sending a power P/M and each is sending
the same signal Hx
• These M signals coherently add at each receiver to give
power P
• There are N receivers so the total power is NP
16
• Given the noise has correlation In SNR is also NP
Example
• Consider the following six wireless channels
G1  1
1 1
G4  

1
1


G 2  1 1
1 0
G5  

0
1


1
G3   
1
4 / 3 0 
G6  

0
2
/
3


• Determine the capacity of each of the six channels above,
assuming the transmit power is uniformly distributed over
the transmit antennas and the total transmit power is 1W
while the noise per receive antenna is 0.1W.
• Note which channels are SIMO and MISO
17
Example
18
Example
19
Example
20
Capacity
• These capacity results are however the
theoretical best that can be achieved
• The problem is how do we create receivers and
transmitters that can achieve close to these
capacities
• There are a number of methods that have been
suggested
–
–
–
–
Zero-forcing
MLD
BLAST
S-T coding
21
MIMO Dectection
• Consider a MIMO system with M transmit and N
receive antennas (M,N)
y  Hx  n
where
x is the Mx1 transmit vector with constellation Q
H is a NxM channel matrix
y is Nx1 received vector
n is a Nx1 white Gaussian complex noise vector
Energy per bit per transmit antenna is E
b
• Our basic requirement is to be able to detect or receive
our MIMO signals x
22
MIMO MLD
• Lets first consider optimum receivers in the sense of
maximum liklihood detection (MLD)
• In MLD we wish to maximize the probability of p(y|x)
• To calculate p(y|x) we observe that the distribution must
be jointly Gaussian and we can use previous results
from M-ary to write it as
p ( y | x) 
1
(2N 0 ) N
 1 | y  Hx |2 

exp  
 2

N0


23
MIMO MLD
• That is we need to find an x from the set of all
possible transmit vectors that minimizes
| y  Hx |
2
• If we have Q-ary modulation and M transmit
antennas then we will have to search through QM
combinations of transmitted signals for each
transmit vector and perform N QM multiplications
• Because of the exponent M the complexity can get
quite high and sub-optimal schemes with less
complexity are desired
24
MIMO Zero-Forcing
• In zero-forcing we use the idea of minimizing
| y  Hx |
2
• However instead of minimizing only over the
constellation points of x we minimize over all possible
complex numbers (this is why it is sub-optimum)
• We then quantize the complex number to the nearest
constellation point of x
• The solution then becomes a matrix inverse when N=M
and we force | y  Hx |2 to zero (zero-forcing)
• What about when M does not equal N?
25
Key Theorem- Psuedoinverse
• When H is square one way to find the transmitted symbols x
from Hx = y is by using inverse.
• What happens when H is not square? Need psuedo-inverse
• Note that HHH is a square matrix which has an inverse
• Therefore HHH x =HH y so that
(HHH)-1HHH x = (HHH)-1HH y
|
and the psuedo-inverse is defined as H+ = (HHH)-1HH
• The psuedo-inverse provides the least squares best fit solution
to the minimization of ||Hx-y||2 with respect to x
26
Example
• If we use a zero-forcing receiver in the previous example
what is the receiver processing matrix we need for each of
the 6 channels?
• G1- none needed
1
 1
 1
• G2-Inverse not possible- not needed  11 1 1
 
 
• G3- [1,1]
• G4- Inverse not possible- just MRC weights
• G51 0
In  

0 1
• G6-
 4 / 3 0
0 2 / 3


1
27
Performance analysis of ZF
• The zero-forcing estimate of the transmitted
signal ~
x can be written as:
~x  Gy
where G  (H H H) 1 H H (with elements g i , j ) is known as the
pseudo-inverse of the channel H and the superscript H
denotes conjugate transpose
• Substituting :
~
x  x  Gn
the ith row element of Gn is equal to a zero mean
Gaussian random variable with variance:
wi | g i1 | 2  | g i 2 | 2    | g iL | 2
28
Performance analysis of ZF
• The noise power is scaled by wi which is the
square 2-norm of the ith row of G
• The diagonal elements of GG’ however are the
square 2-norm of the rows of G
• In addition we can show that
GG '  (H' H) 1 H' ((H' H) 1 H' )'
 (H' H) 1 H' H((H' H) 1 )'
1
1
 ( H' H) I  ( H' H)
• Which is equal to the the diagonal element of (H' H) 1
29
Performance analysis of ZF
•
•
•
Since all wi are all identically distributed so we
drop its subscript
w follow the reciprocal of a Chi-Square random
variable with 2(N-M+1) degrees of freedom
The probability density function (PDF) of w
w ( D 2) e 1 / w
f W ( w) 
( D  1)! 2( D 1)
w0
where D=N-M
30
Why Chi-Square?
• Check out
• H. Winters, J. Salz and R. D. Gitlin, “The Impact of antenna Diversity on
the Capacity of Wireless Communication Systems”, IEEE Trans.
Commun., VolCOM-42, pp. 1740-1751, Feb./March/April.
• 24. J. H. Winters, J. Salz and R. D. Gitlin, “The capacity increase of
wireless systems with antenna diversity”, in Proc. 1992 Conf, Inform.
Science Syst., Princeton, NJ, Mar. 18-20, 1992
• For a N=M it is easy to show as
follows
|
• Matrix G, the inverse of H can then be written as

| A11 |
 | A21 |
(1) N 1 | AN 1 | 


1   | A12 |

G

|H |
(1) j i | A ji |


1

N
N

N
(1)
| A1N |
(1)
| ANN |
• Where Aij is the sub-matrix of H without row i and column j
31
Why Chi-Square?
• The square of the 2-norm for the i row of G is therefore
equal to
N
 (( 1) j i | A ji |) 2
wi 
• Noticing that
j 1
|H|
| H |
2
N
 hij (1) j i | A ji |
j 1
for i  1 to N
the equation above becomes
N
 (( 1) j i | A ji |) 2
wi 
j 1
N
(  hij (1) j i | A ji |) 2
for i  1 to N
j 1
• Since |Aji| is independent of hij we can condition on it so the
equation can be further simplified
• Remember hij are random variables (like noise so independent and add
up)
32
Why Chi-Square?
• The square of the 2-norm for the i row of G is
therefore equal to
N
 | (1) j i A ji |2
j 1
wi 
| h' | 2

2
for i  1 to N
N
 (1) j i | A ji |2
j 1
|
• Where h’ is a random variable following the same
distribution as hij
• Canceling common terms we get
wi 
1
Re( h' )
2

2
 Im( h' ) 2

for i  1 to N
33
Why Chi-Square?
• h’ is a random variable with the same distribution as
hij
• The weights, w are therefore distributed as the
reciprocal of the sum of the square of two Gaussian
random variables with zero mean and variance α/2
• That is the weights are distributed as the reciprocal
of a chi-squared random variable with 2 degrees of
freedom
• This turns out to be the reciprocal of a Rayleigh
fading variable for this special case
|
34
Performance analysis of ZF
• To obtain the error probabilities when w is
random, we must average the probability of error
over the probability density function ,

BER   Pb f w ( w)dw
0
where Pb is the probability of error in AWGN
channel with depend on the signal constellation.
35
Performance of BPSK and
QPSK
• For BPSK and QPSK
 2 Eb
Pb  Q
 wi N 0




Performing the integral and define  b   2 Eb / N 0 as the
SNR per bit per channel (see Proakis 4th ed, p825)
1

BER   (1   )
2

where

D 1 D
 D  k  1



(
1


)

 k  2



k 0 

k
b
1  b
36
Performance of BPSK and
QPSK using ZF
0
10
4-QAM (2,2)
4-QAM (3,4)
4-QAM (4,6)
4-QAM (1,4)
-1
10
-2
BER
10
-3
10
-4
10
-5
10
-6
10
2
4
6
8
10
12
14
SNR per bit per channel (dB)
16
18
20
Exact BER expression for QPSK compared with Monte
Carlo simulations
37
Performance of M-PSK
For M-PSK:
min( 2 , M / 4 ) 
2 Eb log 2 M
2
(2i  1)

Pb 
Q
sin


max(log 2 M ,2) i 1
wi N 0
M

BERPSK
min( 2 , M / 4 ) 
2
 1


(
1


)
 
i 
max(log 2 M ,2) i 1  2

D 1 D




 D  k  1



(
1


)

i 
 k  2


k 0 

k



(log 2 M ) sin 2(i  1) / M   b
i 
2
1  (log 2 M ) sin 2(i  1) / M   b
2
where
38
Performance of M-QAM
For M-QAM:
3Eb log 2 M
1 min( 2, M / 2) 

Pb 
1 
  Q (2i  1)
log 2 M 
( M  1) wi N 0
M  i 1

4
BERQAM
1 min( 2, M / 2)  1



1 
   (1   i )
log 2 M 
M  i 1  2

where
4
i 
D 1 D




 D  k  1



(
1


)

i 
 k  2

k 0 

k



3(log 2 M )( 2i  1) 2  b
2( M  1)  3(log 2 M )( 2i  1) 2  b
39
Comparison with simulation (ZF)
0
10
16-PSK (analysis)
16-PSK (simulation)
16-QAM (analysis)
16-QAM (simulation)
-1
10
-2
BER
10
(3,3)
-3
10
(4,6)
-4
10
(8,12)
-5
10
-6
10
2
4
6
8
10
12
14
SNR per bit per channel (dB)
16
18
20
BER approximations for 16-PSK and 16-QAM compared
with Monte Carlo simulations for (3,3), (4,6) and (8,12)
antenna configurations.
40
Comparison with simulation
0
10
64-PSK (analysis)
64-PSK (simulation)
64-QAM (analysis)
64-QAM (simulation)
-1
10
-2
BER
10
-3
10
-4
10
-5
10
-6
10
2
4
6
8
10
12
14
SNR per bit per channel (dB)
16
18
20
• BER approximations for 64-PSK and 64-QAM compared
with Monte Carlo simulations for (8,12) antenna
configurations.
41
Performance of MLD
-1
10
zero-forcing
SB-MLD (P=16)
SB-MLD (P=24)
ESS-NMLD (P=16)
ESS-NMLD (P=24)
MLD
-2
10
-3
BER
10
-4
10
-5
10
-6
10
7
9
11
13
SNR per receive antenna (dB)
15
17
BER of zero-forcing and MLD for a (4,6) system using 4QAM.
42
Performance of MLD
0
10
8-QAM zero-forcing
8-QAM ESS-NMLD (P=40)
8-QAM MLD
16-QAM zero-forcing
16-QAM ESS-NMLD (P=300)
16-QAM MLD
-1
10
-2
BER
10
-3
10
-4
10
-5
10
14
16
18
20
22
24
SNR per receive antenna (dB)
26
28
30
BER of zero-forcing and MLD for a (3,3) system using 8QAM and 16-QAM.
43
Performance of MLD
0
10
8-QAM zero-forcing
8-QAM ESV-NMLD (P=40)
8-QAM MLD
16-QAM zero-forcing
16-QAM ESV-NMLD (P=300)
16-QAM MLD
-1
10
-2
BER
10
-3
10
-4
10
-5
10
14
16
18
20
22
24
SNR per receive antenna (dB)
26
28
30
BER of zero-forcing and MLD for a (3,3) system using 8QAM and 16-QAM.
44
MIMO V-BLAST
• It turns out the performance of ZF is not good enough
while the complexity of MLD is too large
• Motivate different sub-optimum approaches
• BLAST is one well known on (Bell Laboratories
Layered Space Time)
• Based on interference cancellation
• A key idea is that when we perform ZF we detect all the
transmitted bit streams at once
45
MIMO V-BLAST
• Generally we would expect some of these bit streams to
be of better quality than the others
• We select the best bit stream and output its result using
ZF
• We then also use it to remove its interference from the
other received signals
• We then detect the best of the remaining signals and
continue until all signals are detected
• It is a non-linear process because the best signal is
always selected from the current group of signals
46
MIMO V-BLAST
• Basically layers of interference cancellation
Stage 1
Stage (M-1)
Stage M
Linear
Detector
Linear
Detector
Linear
Detector
Interference
Cancellation
Interference
Cancellation
47
Performance of V-BLAST
-1
10
zero-forcing
V-BLAST
ESV-NMLD (P=7)
ESV-NMLD (P=24)
MLD
-2
10
-3
BER
10
-4
10
-5
10
-6
10
7
9
11
13
SNR per receive antenna (dB)
15
17
BER of zero-forcing, V-BLAST and MLD for a (4,6)
system using 4-QAM.
48