Cognitive Radio for
Dynamic Spectrum Allocation Systems
Xiaohua (Edward) Li and Juite Hwu
Department of Electrical and Computer Engineering
State University of New York at Binghamton
{xli,jhuw1}@binghamton.edu
1. Introduction
• DSA (Dynamic Spectrum Allocation):
– What is this?
New spectrum management rules.
– Why do we need it?
For efficiently utilizing spectrum
– How to apply it?
Two primary methods to approach
1. Introduction (cont’)
• Basic idea:
• Licensed and unlicensed users access the
spectrum at different time period.
• Licensed and unlicensed users access the
spectrum simultaneous with proper power.
1. Introduction (cont’)
• Our task
1. The 2nd method will be applied.
2. We use two different analysis approaches,
one is theoretical and the other is
simulation.
3. If allowed, find the maximum capacity that
unlicensed user can obtain.
1. Introduction (cont’)
• Access protocol
1. Give a very small power for the secondary
transmitter.
2. Check the SINR of primary receivers.
3. Adjust the power of secondary transmitter
according to the ACK.
4. Repeat step 2 and 3, find the maximum
power of secondary transmitter which is
allowed.
2. System Model
 Structure:


T0: primary transmitter

Circles: the coverage
range of different
antennas
r0
T1,T2: secondary
transmitters
T0
r2
T1
 Channel access protocol:
Tx0
Tx1
Tx2
T2
A Tx0 A Tx0
C Tx1 C Tx1
K Tx2 K Tx2
t
r1
2. System Model (cont’)
For successful transmissions, the power of transmitter has
to satisfy the equation
KP0 r0
N

  0  0
N : noise power (include AWGN and other
transmitters’ power)
α: path-loss exponent
K : constant
γ0: SINR
Γ0: minimum required SINR of T0
2. System Model (cont’)
T0
Tx0
d
Rx0
Rx3
∆x
Txi
x
R0
d
T1
Rx1
Rx2
R2
R1
Fig. 1
0  x  r0  d
Fig. 2
r0  d  x  r0  d
Here, we separate our scheme into two different
cases. If we can find a receiver that has the
minimum SINR, the threshold can be examined
2. System Model (cont’)
 At Fig.1 Rx0 has the smallest SINR., For Fig.2,
Rx0 and Rx1 have the minimum SINR.
 Why do we need the smallest SINR?
 It’s our threshold and the worst case!!
 Capacity calculation…
S

C  log 2 1  
 N
Shannon Hartley capacity equation
3. Capacity of a single secondary
transmitter
One secondary transmitter only (Tx1), and primary
receivers are distributed uniformly in a circle made
by Tx0
With a Poisson distribution:
The probability of “no primary
receivers in the small circle with
radius x”
f  x
x 


2
0!
Tx0
r0
d
x Txi
0
e
 x 2 
e
 x 2 
Receivers with
density β
0  x  r0  d
3. Capacity of a single secondary
transmitter (cont’)
 x 2 r0 2

A x  2 

 g  g  r0  g  d  g  x  
2
 2

Tx0
x
d
g
 d  x  r0 
r0  d  x
1  r0  d  x 
,   cos1 
,


cos



2
2
xd
2
r
d


0


2
2
2
2
f  x   e A x 
The probability of “no primary
receivers in the slash area”
2
2
Txi
r0  d  x  r0  d
3. Capacity of a single secondary
transmitter (cont’)
P0: power of primary
transmitter
P1(x): power of
secondary transmitter
N: very small noise
power
 KP0  d  x 
, 0  x  r0  d


 KP  x  x  N
 0  x   1
KP0 r0 

 KP  x  x   N , r0  d  x  r0  d
 1

 x

P1  x   
 
x

 P0  x  d  N 
  , 0  x  r0  d

0
K 

 P0 r0  N 
  , r0  d  x  r0  d

K
 0
3. Capacity of a single secondary
transmitter (cont’)
 Take the expect
P1  E  P1 ( x)  
r0  d
0
P1  x  f  x  dx  
r0 d
0
P1  x  e
 x2 
dx
 Why do we need this?
We have to consider all locations that primary receivers
may be placed.
3. Capacity of a single secondary
transmitter (cont’)
For convenience, we use polar coordinate system
instead of Cartesian
Rx
y
θ
2
r
y  r  d  2rd cos 
2
SINR of Rx is
 1  x, r ,   
T1
x
T0
d
KP1  x  y 
KP0 r   N
Rx is secondary receiver thus the
power from T0 becomes noise
3. Capacity of a single secondary
transmitter (cont’)
 The capacity of the transmission is
C  x, r ,    log 2 1   1  x, r ,   
 So the average capacity is written as
C  r ,   E C  x, r ,   
r0  d
0
C  x, r ,  f  x  dx
The discussion above is based on one fixed receiver, how
about the receiver is moving?
3. Capacity of a single secondary
transmitter (cont’)
Best case:
Tx0
Rx0
Rx3
∆x
Txi
d
T0
x
R0
d
T1
Rx1
R2
Rx2
R1
0  x  r0  d
KP1 y 
1  y  
KP0 z   N
r0  d  x  r0  d
min d  y, r0 
3. Capacity of a single secondary
transmitter (cont’)
Worst case:
Tx0
d
T0
Rx0
Rx3
∆x
Txi
x
R0
d
T1
Rx1
R2
Rx2
R1
Fig. 1
0  x  r0  d
KP1 y 
KP0 d  y
KP1 y 
KP0 d  y


N
 1  y 
Fig. 2
r0  d  x  r0  d
KP1 y 
 1  y  

KP
z
N
N
0
3. Capacity of a single secondary
transmitter (cont’)
 Why do we need the range?
This range gives us the best and worst case that we can calculate
the capacity gain



KP1 y 
KP1 y  
log 2 1 
  C1  y   log 2 1 



 KP0 d  y  N 
 KP0 z  N 
 Compare it with the loss of primary transmitter capacity
Secondary access protocol provide large capacity when y is small
4. Capacity of multiple secondary
transmitters
 The scheme
 The primary spectrum access is keeping stable
 No interference between any two secondary receiver
KP0b0 
K  i 1 Pb
i i
M

M

Pb
i 1 i i

N
P0b N


0
K
 0
Qb
4. Capacity of multiple secondary
transmitters
 Consider a special case
 A  y, x  
Fs  y, x   
 , 0  y  r1  x
2
  r1 
T1
L
A(y,x) is the area of the
cross section between the
circle of radius y and the
circle of radius r1
Fs(y,x) is the cumulative distribution
that all secondary transmitters are
inside a circle of radius x centered
around T0
r1
y
r0
T0
R0
x
T2
4. Capacity of multiple secondary
transmitters
Consider R0 with a distance x from T0, and to which all
secondary transmitters have distance at most y.
The upper bound is obtained with the equality sign
 P0  N 
Pi  y, x   y  x  

K
i 1
 0
L

According to this bound, we can find the maximum
capacity
5. Simulation
 One secondary transmitter
 Multiple primary receivers
3
Parameters:
PT0=100 watts
PAWGN: N=5*10-10
GT=GR=1 (transmitter
and receiver gain)
r0≈1000 (m)
α=3 (urban area)
Γ0=20dB
Capacity of Rx1, unit: bps/Hz, and beta
 Two approach methods
Capacity of Rx1, relative to beta and d(distance between Tx1 and Tx0)
10
2
10
1
10
0
10
-1
10
100
150
200
250
300
d (the distance between Tx0 and Tx1)
beta=1e-7
beta=1.1e-6
beta=2.1e-6
(ana)beta=1e-7
(ana)beta=1.1e-6
350(ana)beta=2.1e-6
400
6. Conclusion
• d increase, the power from Tx0 decrease,
P1 decrease, but the capacity raise still.
• Our scheme provide a large capacity
(GSM cellular provides 1.35bps/Hz)
Thank you
and
Any question?