Cognitive Radio for Cognitive Radio for Dynamic Spectrum Allocation Dynamic Spectrum Allocation

SystemsSystemsXiaohua (Edward) Li and Juite HwuXiaohua (Edward) Li and Juite Hwu

Department of Electrical and Computer EngineeringDepartment of Electrical and Computer EngineeringState University of New York at BinghamtonState University of New York at Binghamton

{xli,jhuw1}@binghamton.edu{xli,jhuw1}@binghamton.edu

1. Introduction1. 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’)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’)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’)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 Model2. System Model

r 0

r1

r2

T0

T1

T2

Structure:

T0: primary transmitter

T1,T2: secondary transmitters

Circles: the coverage range of different antennas

Tx0Tx1Tx2

Tx0Tx1Tx2

Tx0Tx1Tx2

ACK

ACK

t

Channel access protocol:

2. System Model (cont’)2. System Model (cont’)

0 00 0

KP r

N

For successful transmissions, the power of transmitter has to satisfy the equation

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’)2. System Model (cont’)

T0

T1d

R1

R0

R2

00 x r d 0 0r d x r d Fig. 1 Fig. 2

Tx0

Txi

Rx3Rx0

Rx1

Rx2

d ∆x

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’)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…

2log 1S

CN

Shannon Hartley capacity equation

3. Capacity of a single 3. Capacity of a single secondary transmittersecondary transmitter

Tx0

Txid

One secondary transmitter only (Tx1), and primary receivers are distributed uniformly in a circle made by Tx0

Receivers with density β

x

00 x r d

With a Poisson distribution:

The probability of “no primary receivers in the small circle with radius x”

2 2

02

0!x x

xf x e e

r0

3. Capacity of a single 3. Capacity of a single secondary transmitter (cont’)secondary transmitter (cont’)

Tx0

Txid

0 0r d x r d

x

220

022 2

rxA x g g r g d g x

2 2 2 2 2 21 10 0 0

0

, cos , cos2 2 2

r d x d x r r d xg

xd r d

A xf x e

The probability of “no primary receivers in the slash area”

3. Capacity of a single secondary t3. Capacity of a single secondary transmitter (cont’)ransmitter (cont’)

00

10

0 00 0

1

, 0

,

KP d xx r d

KP x x Nx

KP rr d x r d

KP x x N

P0: power of primary transmitter

P1(x): power of secondary transmitter

N: very small noise power

00

0

1

0 00 0

0

, 0

,

P x d Nx x r d

KP x

P r Nx r d x r d

K

3. Capacity of a single 3. Capacity of a single secondary transmitter (cont’)secondary transmitter (cont’) Take the expect

20 0

1 1 1 10 0( )

r d r d xP E P x P x f x dx P x e dx

Why do we need this?

We have to consider all locations that primary receivers may be placed.

3. Capacity of a single 3. Capacity of a single secondary transmitter (cont’)secondary transmitter (cont’)

T0

T1d x

rRx

y

θ

2 2 2 cosy r d rd

11

0

, ,KP x y

x rKP r N

SINR of Rx is

Rx is secondary receiver thus the power from T0 becomes noise

For convenience, we use polar coordinate system instead of Cartesian

3. Capacity of a single 3. Capacity of a single secondary transmitter (cont’)secondary transmitter (cont’) The capacity of the transmission is

2 1, , log 1 , ,C x r x r

So the average capacity is written as

0

0, , , , ,

r dC r E C x r 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 3. Capacity of a single secondary transmitter (cont’)secondary transmitter (cont’)

Best case:

00 x r d

Tx0

Txi

Rx3Rx0

Rx1

Rx2

d ∆x

T0

T1d

R1

R0

R2

0 0r d x r d 11

0

KP yy

KP z N

0min ,d y r

3. Capacity of a single 3. Capacity of a single secondary transmitter (cont’)secondary transmitter (cont’)

Worst case:

00 x r d Fig. 1

Tx0

Txi

Rx3Rx0

Rx1

Rx2

d ∆x

T0

T1d

R1

R0

R2

0 0r d x r d Fig. 2

11

0

KP yy

KP d y N

1 11

00

KP y KP yy

KP z NKP d y N

3. Capacity of a single 3. Capacity of a single secondary transmitter (cont’)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

1 12 1 2

00

log 1 log 1KP y KP y

C yKP z NKP d y N

Compare it with the loss of primary transmitter capacity

Secondary access protocol provide large capacity when y is small

4. Capacity of multiple 4. Capacity of multiple secondary transmitterssecondary transmitters The scheme

The primary spectrum access is keeping stable

No interference between any two secondary receiver

0 00

1

M

i ii

KPb

K Pb N

01

0

M

i i bi

Pb NPb Q

K

4. Capacity of multiple 4. Capacity of multiple secondary transmitterssecondary transmitters Consider a special case

T2

y

T1

T0

R0

r0r1

x

12

1

,, , 0

L

s

A y xF y x y r x

r

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

4. Capacity of multiple 4. Capacity of multiple secondary transmitterssecondary transmitters

0

1 0

,L

ii

P NP y x y x

K

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

According to this bound, we can find the maximum capacity

5. Simulation5. Simulation

One secondary transmitter

Multiple primary receivers

Two approach methods 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

100 150 200 250 300 350 40010

-1

100

101

102

103

d (the distance between Tx0 and Tx1)

Cap

acity

of

Rx1

, un

it: b

ps/H

z, a

nd b

eta

Capacity of Rx1, relative to beta and d(distance between Tx1 and Tx0)

beta=1e-7

beta=1.1e-6

beta=2.1e-6(ana)beta=1e-7

(ana)beta=1.1e-6

(ana)beta=2.1e-6

6. Conclusion6. 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 youand

Any question?