Cognitive Radio for Dynamic Spectrum Allocation Systems

1. 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

2. 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

3. 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.

4. 1. Introduction (cont’) • Our task • The 2nd 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.

5. Access protocol 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. 1. Introduction (cont’)

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

7. 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 γ0: SINR Γ0: minimum required SINR of T0

8. 2. System Model (cont’) Fig. 1 Fig. 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

9. Shannon Hartley capacity equation 2. System Model (cont’) • At Fig.1 Rx0 has the smallest SINR., For Fig.2, Rx0and Rx1have the minimum SINR. • Why do we need the smallest SINR? • It’s our threshold and the worst case!! • Capacity calculation…

10. With a Poisson distribution: The probability of “no primary receivers in the small circle with radius x” x 3. Capacity of a single secondary transmitter One secondary transmitter only (Tx1), and primary receivers are distributed uniformly in a circle made by Tx0 r0 Receivers with density β

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

12. 3. Capacity of a single secondary transmitter (cont’) P0: power of primary transmitter P1(x): power of secondary transmitter N: very small noise power

13. 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.

14. Rx is secondary receiver thus the power from T0 becomes noise 3. Capacity of a single secondary transmitter (cont’) For convenience, we use polar coordinate system instead of Cartesian SINR of Rx is

15. 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?

16. 3. Capacity of a single secondary transmitter (cont’) Best case:

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

18. 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

19. 4. Capacity of multiple secondary transmitters • The scheme • The primary spectrum access is keeping stable • No interference between any two secondary receiver

20. 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 r1 Fs(y,x) is the cumulative distribution that all secondary transmitters are inside a circle of radius x centered around T0

21. 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 According to this bound, we can find the maximum capacity

22. 5. 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

23. 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)

24. Thank you and Any question?