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

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. 1.The 2 nd method will be applied We use two different analysis approaches, one is theoretical and the other is simulation If allowed, find the maximum capacity that unlicensed user can obtain.

1. Introduction (cont’) Access protocol 1. 1.Give a very small power for the secondary transmitter Check the SINR of primary receivers Adjust the power of secondary transmitter according to the ACK Repeat step 2 and 3, find the maximum power of secondary transmitter which is allowed.

2. System Model  Structure:  T0: primary transmitter  T1,T2: secondary transmitters  Circles: the coverage range of different antennas  Channel access protocol:

2. System Model (cont’) 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 γ 0 : SINR Γ 0 : minimum required SINR of T 0

2. System Model (cont’) Fig. 1Fig. 2 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 Rx 0 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… Shannon Hartley capacity equation

3. Capacity of a single secondary transmitter One secondary transmitter only (Tx 1 ), and primary receivers are distributed uniformly in a circle made by Tx0 Receivers with density β x With a Poisson distribution: The probability of “no primary receivers in the small circle with radius x” r0r0

3. Capacity of a single secondary transmitter (cont’) x The probability of “no primary receivers in the slash area”

3. Capacity of a single secondary transmitter (cont’) P 0 : power of primary transmitter P 1 ( x ): power of secondary transmitter N: very small noise power

3. Capacity of a single secondary transmitter (cont’) Take the expect 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’) SINR of R x is R x is secondary receiver thus the power from T 0 becomes noise For convenience, we use polar coordinate system instead of Cartesian

3. Capacity of a single secondary transmitter (cont’) The capacity of the transmission is So the average capacity is written as 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:

3. Capacity of a single secondary transmitter (cont’) Worst case: Fig. 1 Fig. 2

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 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

4. Capacity of multiple secondary transmitters Consider a special case A(y,x) is the area of the cross section between the circle of radius y and the circle of radius r 1 F s (y,x) is the cumulative distribution that all secondary transmitters are inside a circle of radius x centered around T 0

4. Capacity of multiple secondary transmitters Consider R 0 with a distance x from T 0, 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. Simulation One secondary transmitter Multiple primary receivers Two approach methods Parameters: P T0 =100 watts P AWGN : N=5* G T =G R =1 (transmitter and receiver gain) r 0 ≈1000 (m) α=3 (urban area) Γ 0 =20dB

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?