12- OFDM with Multiple Antennas
Multiple Antenna Systems (MIMO)

TX


 RX
NT
NR
Transmit
Antennas
Receive
Antennas
NT  N R
Different paths
Two cases:
1. Array Gain: if all paths are strongly correlated to which other the SNR can be
increased by array processing;
2. Diversity Gain: if all paths are uncorrelated, the effect of channel fading can be
attenuated by diversity combining
Recall the Chi-Square distribution:
1. Real Case. Let
y  x  x  ... x
2
1
Then
with
y   n2
2
2
2
n
xi  N (0,1) real, i.i.d .
E{ y}  n
var{y}  2n
2. Complex Case. Let
y | x1 |2  | x2 |2 ... | xn |2
xi  ai  jbi  CN (0,1) complexgaussian, i.i.d.
Then
with
1 2
 2n
2
E{ y}  n
y
1
var{y}  n
2
Receive Diversity:
h1
TX
s
NT  1
Transmit
Antennas
y1

hN R
 RX
yNR
NR
Different paths
NR
Receive
Antennas
 y1   h1 
 w1 

  





E
s

N

0

   S

 y N  hN 
 wN 
 R  R
 R
Energy per
symbol
Noise PSD
Assume we know the channels at the receiver. Then we can decode the signal as
NR
NR
NR
i 1
i 1
i 1
y   hi* yi  ES | hi |2 s  N 0  hi*wi
signal
and the Signal to Nose Ratio
 NR
2  ES
SNR    | hi | 
 i 1
 N0
noise
NR
In the Wireless case the channels are random, therefore
| h |
i 1
i
2
is a random variable
Now there are two possibilities:
1. Channels strongly correlated. Assume they are all the same for simplicity
h1  h2  ...  hNR  h
Then
NR
2
2
2
|
h
|

N
|
h
|

N

 i
R
R 2
i 1
assuming
 
E | h |2  1
and


ES 
1 2  ES
SNR  N R | h |
  NR 2 
N0 
2  N0
2
From the properties of the Chi-Square distribution:
mSNR  ESNR  N R
 SNR
ES
N0
N R ES
 varSNR 
2 N0
Define the coefficient of variation
better on average …
… but with deep fades!
 var 
In this case we say that there is no diversity.
 SNR
mSNR
1

2
2. Channels Completely Uncorrelated.
 NR
2  ES
SNR    | hi | 
 i 1
 N0
NR
Since:
1 2
| hi |   2 N R

2
i 1
2
 1 2  ES
SNR    2 N R 
2
 N0
with
ESNR  N R
var SNR 
Diversity of order N R
ES
N0
N R ES
2 N0
 var 
 SNR
mSNR

1
2 NR
Example: overall receiver gain with receiver diversity.
15
N R  10
10
5
NR  2
0
-5
NR  1
-10
-15
-20
-25
0
20
40
60
80
100
120
140
160
180
200
Transmitter Diversity
h1

s
TX 
y
RX
hN R
NR  1
NT
NT
Transmit
Antennas
 ES
y  
 NT
Different paths

hi  s  N 0 w

i 1

Receive
Antennas
NT
Total energy equally distributed
on transmit antennas
Equivalent to one channel,
with no benefit.
However there is a gain if we use Space Time Coding (2x1 Alamouti)
Take the case of Transmitter diversity with two antennas
h1
x1[n]
y[n]
TX
RX
h2
x2 [ n ]
s1[n], s2 [n]
Given two sequences
code them within the two antennas as follows
antennas
x1
x2
s1
s
s2
s
*
2
*
1
2 n 2n  1
time
ES
h1s1  h2 s2   N0 w1
y[2n] 
2
ES
y[2n  1] 
 h1s2*  h2 s1*  N 0 w2
2


This can be written as:
 w1 
 y[2n] 
ES  h1 h2   s1 
 y*[2n  1]  2 h*  h*  s   N 0 w* 


1  2 
 2
 2
To decode, notice that
 z1   h1*
z    *
 2   h2
 s
h2   y[2n]   ES
2  1
 
|| h ||    
 *

h1   y [2n  1]  2
  s2 

 w1 
N 0 || h ||  
 w2 

Use a Wiener Filter to estimate “s”:


 K h y[2n]  h y [2n  1]
sˆ1  K h1* y[2n]  h2 y *[2n  1]
sˆ1
*
2
*
1
with
2 / ES
K
| h1 |2  | h2 |2 2 N 0 / ES
It is like having two independent channels
s1
ES
|| h ||2
2
N0 || h || w1
z1
s2
ES
|| h ||2
2
|| h ||2 ES
SNR 
2 N0
z2
N0 || h || w2
1 2
|| h || | h1 |  | h2 |   4
2
2
2
2
Apart from the factor ½, it has the same SNR as the receive diversity of order 2.
2x2 MIMO with Space Time Coding (2x2 Alamouti)
x1[n]
h11
y1[n]
h21
h12
TX
RX
x2 [ n ]
h22
y 2 [ n]
 y1[n]  h11 h12   x1[n]  w1[n]

 y [n]  h




h
x
[
n
]
w
[
n
]
 2   21 22   2   2 
Same transmitting sequence as in the 2x1 case:
antennas
x1
x2
s1
 s2*
s2
s1*
2 n 2n  1
time
Received sequences:
y1[2n] 
ES
h11s1  h12 s2   N 0 w1[2n]
2
y1[2n  1] 
y 2 [ 2n] 


ES
 h11s2*  h12 s1*  N 0 w1[2n  1]
2
ES
h21s1  h22 s2   N 0 w2 [2n]
2
y2 [2n  1] 


ES
 h21 s2*  h22 s1*  N 0 w2 [2n  1]
2
Write it in matrix form:
 y1[2n] 
 y *[2n  1]
 1

 y 2 [ 2n] 
 *

 y2 [2n  1]
 h11
 h*
ES  12
2  h21
 *
h22
h12 
 h11*   s1 
 N 0 w[n]


h22   s2 
* 
 h21 
Combined as
 z1   h11*
z    *
 2  h12
h12
 h11
*
h21
*
h22
 y1[2n] 
 *

h22   y1 [2n  1]

 h21  y2 [2n] 
 *

 y2 [2n  1]
to obtain
 z1   h11*
z    *
 2  h12
h12
 h11
*
h21
*
h22

 h11

 *
h22  ES  h12

 h21  2  h21
 *

h22


h12 

* 
 h11   s1 

 N 0 w[n] 


h22   s2 

* 

 h21 

After simple algebra:
 z1 
2

||
h
||
z 
 2
ES
2
 s1 
s   || h || N 0 w[n]
 2
with
2
1 2
|| h ||   | hij |  8
2
i , j 1
2
2
diversity 4
This yields an SNR
|| h ||2 ES
SNR 
2 N0
WiMax Implementation
h1
h2
Subscriber
Station
Base Station
Down Link (DL): BS -> SS Transmit Diversity
Uplink (UL):
SS->BS
Receive Diversity
Down Link: Transmit Diversity
Use Alamouti Space Time Coding:
Transmitter:
Data in
Error
Coding
X 2m
Xn
M-QAM
buffer
IFFT
TX
STC
IFFT
TX
X 2m1
Block to be
transmitted
Space Time Coding
X 2m
 X 2*m1
X 2m1
*
X 2m
2m
2m  1
time
Receiver:
Y2 m
X 2m
Data out
X n P/S
Error
Correction
M-QAM
S/P
2
STD
2
FFT
Y2 m1
X 2m1
Space Time Decoding:

For each subcarrier k
compute:

Xˆ 2 m [k ]  K H1*[k ]Y2 m [k ]  H 2 [k ]Y2*m 1[k ]
Xˆ
[k ]  K H *[k ]Y [k ]  H [k ]Y * [k ]
2 m 1
with

2
2m
1
2 / ES
K
| H1[k ] |2  | H 2 [k ] |2 2 N 0 / ES
2 m 1

Preamble, Synchronization and Channel Estimation with
Transmit Diversity (DL)
The two antennas transmit two preambles at the same time, using different sets of
subcarriers
p1[n]
EVEN subcarriers
CP
128
+
+
64
128
128

319
0
p2[n]
CP
64
128
128
time
+

n
0
 100
ODD subcarriers
128
frequency
k
 100
Both preambles have a symmetry:
p1[n]  p1[n  128]
p2 [n]   p2 [n  128]
p0 [ n ]
n  128,...,319
h0 [n]
received signal from
the two antennas
y[n]
p1[n]
h1[n]
Problems:
• time synchronization
• estimation of both channels
Symmetry is preserved even after the channel spreading:
h1[n] * p1[n]
CP
128
+
+
64
128
128
CP
128
+
64
128
128
h2 [n] * p2 [n]
One possibility: use symmetry of the preambles
y1[n]  2h1[n] * p1[n]


y[n]
64
n0
256
64 128
n0  128
z 128


y2 [n]  2h2 [n] * p2 [n]
64 128
n0  128
The two preambles can be easily separated
MIMO Channel Simulation
Take the general 2x2 channel
e j 3
e j 1
x1[n]
y1[n]
Rayleigh
T
T
T
T
Rayleigh
x2 [ n ]
e j 4
  [ 1   N ] sec
P  [ P1  PN ] dB
0  T  1 Correlation at the transmitter
0   R  1 Correlation at the receiver
y 2 [ n]
e j 2