Stability Analysis in a Cognitive Radio System with Cooperative Beamforming Mohammed Karmoose 1 Ahmed Sultan 1 Moustafa Youseff 2 1 Electrical Engineering Dept, Alexandria University 2 E-JUST
Agenda 1 Motivation and Related Work 2 System and Data Models 3 Problem Insights and Formulation 4 Queue Service Rates 5 Stability analysis using dominant systems
Motivation and Related Work Queuing analysis in CRNs to study the stability of the primary and secondary networks Shortcomings of such models: no time-frequency sharing between primary and secondary networks Cooperative beamforming to increase the available spectrum opportunities for the secondary networks No queuing analysis to characterize the stability regions of such systems Our goal: study a CR scenario in which cooperative beamforming is enabled for the secondary network, and obtain the stability region of the system for the primary and secondary network
System and Data Models PU - Tx H p PU - Rx Primary Network: H ps H sp1 H sp2 H spk Single link (Tx-Rx pair) R 1 H s1 Primary transmit power P p Infinite buffer Q p - Bernoulli arrival R.P with mean λ SU - Tx R 2 H s2 H sk SU - Rx Noise at the receivers CN (0, 1) R K K Relaying Nodes Figure: System Model
System and Data Models Secondary Network: Single link (Tx-Rx pair) Relay-assisted transmission (K relays) operating in a decode-and-forward fashion Relays are close to the secondary source Power control on total transmitted power by all relays (P s P max Infinite buffer Q s - Bernoulli arrival R.P with mean λ s Noise at the receivers CN (0, 1) PU - Tx SU - Tx H p H ps R 1 R 2 R K H sp1 H s1 PU - Rx H sp2 H spk H s2 H sk K Relaying Nodes Figure: System Model SU - Rx
System and Data Models Channel Conditions: PU - Tx H p PU - Rx Small pathloss between secondary source and relays (error free communications) Imperfect sensing of primary transmission SU - Tx H ps R 1 R 2 H sp1 H sp2 H spk H s1 H s2 H sk SU - Rx Channel between relays and receivers are CN (0, 1) Slow fading (channel constant over many time slots) R K K Relaying Nodes Figure: System Model
System and Data Models PU - Tx H p PU - Rx System Operation: Perfect channel estimation at relays based on transmitted ARQs H ps R 1 H sp1 H sp2 H spk H s1 Channel estimations are reported to secondary transmitter (negligible time) SU - Tx R 2 H s2 H sk SU - Rx PU packets are sensed by secondary transmitter - imperfect sensing (p md and p fa R K K Relaying Nodes Figure: System Model
System and Data Models PU - Tx H p PU - Rx Secondary transmitter computes required beamforming weights H ps R 1 H sp1 H sp2 H spk H s1 Secondary data packet is sent to relays which decode perfectly (beamforming vector included in the header) SU - Tx R 2 H s2 H sk SU - Rx Time required for relay reception and decoding is negligible R K K Relaying Nodes Figure: System Model
System and Data Models PU - Tx H p PU - Rx If PU is detected w p = (I Φ)H s P s H H s (I Φ) H s SU - Tx H ps R 1 R 2 H sp1 H sp2 H spk H s1 H s2 SU - Rx If PU is not detected H sk H s w a = P s H. s R K K Relaying Nodes Figure: System Model
Problem Insights and Formulation If sensing was pefect: PU will not be aware of the secondary transmitter. What would only limit its achievable service rate (µ p ) is the channel conditions between PU Tx-Rx If sensing is not perfect: SU can unintentionally interfere with PU transmission, reducing achievable service rate of primary user For the SU: Two factors govern its achievable service rate (µ p ): 1 Channel conditions between relays and secondary destination 2 The rate at which PU uses the channel (i.e., λ p and µ p )
Problem Insights and Formulation Problem Formulation Stability for a queue is achieved iff λ < µ We try to find the stability region of the system (the regions of λ p and λ s for which the two queues are stable), i.e.: λ p < µ p λ s < µ s
Queue Service Rates Primary Queue: If secondary queue is empty OR secondary queue is not empty and PU is detected: p out,p = Pr{P p H p 2 < β p } = 1 exp( β p P p ) If secondary queue is not empty and PU is misdetected: { pout,p md Pp H p 2 } = Pr Hspw H a 2 + 1 < β p Service rate: ( ) µ p = (1 p out,p ) Pr{Q s = 0} + (1 p md )Pr{Q s 0} + (1 p md out,p)p md Pr{Q s 0}
Queue Service Rates Secondary Queue: PU queue is empty AND not detected: p out,s = Pr{P s H s 2 < β s } PU queue is empty AND detected (false alarm): p fa out,s = Pr{ H H s w p 2 < β s } PU queue is not empty AND detected: PU queue is not empty AND misdetected: { p (d) H H out,s = Pr s w p 2 } { P p H ps 2 + 1 < β s pout,s md Ps H s 2 } = Pr P p H ps 2 + 1 < β s ( µ s = p out,s (1 p fa ) + pout,s fa p fa )Pr{Q p = 0} ( + p (d) out,s (1 p md) + pout,s md p md )Pr{Q p 0}
Stability analysis using dominant systems Continuing analysis is hard to interacting queues We use the concept of dominance : We assume two auxiliary systems 1 Primary queue is dominant (sends dummy packets when queue is empty) 2 Secondary queue is dominant (sends dummy packets when queue is empty) It is proven that the union of the stability regions of both systems is exactly the stability region of the original system
Stability analysis using dominant systems Primary dominant queue: Pr{Q p = 0} = 0 µ pd s = ( p (d) out,s (1 p md) + p md out,s p md ) Secondary queue does not depend on the state of primary queue (channel ( stationarity is guaranteed). ) We apply Little s theorem Pr{Q s = 0} = 1 λs µ pd s ) µ pd p = (1 p out,p ) λs p md (p out,p md p out,p µ pd s
Stability analysis using dominant systems Effect of changing P s : On secondary rate: Increasing P s increases µ pd s On primary rate: Increasing P s increases interference on the primary receiver when misdetection occurs Increasing P s helps SU empty its queue faster and evacuates the channel The impact of both effects depend on λ s and µ pd p Optimization problem over P s to maximize µ pd p max. P s λ p = µ pd p s.t. λ s < µ pd s, P s P max
Stability analysis using dominant systems Primary dominant queue: 1 1st Dominant System 0.9 0.8 0.7 0.6 pd µ s 0.5 0.4 0.3 0.2 0.1 P s range for Q s stability (λ s = 0.35) 0 0 0.5 1 1.5 2 P s Figure: P p = 1, K = 4, p md = 0.1, p fa = 0.01, β p = β s = 1 and P max = 2.
Stability analysis using dominant systems Primary dominant queue: 0.37 1st Dominant System 0.368 0.366 0.364 pd µ p 0.362 λ s = 0 λ s = 0.1 λ s = 0.35 0.36 0.358 0.356 0.354 0 0.5 1 1.5 2 P s Figure: P p = 1, K = 4, p md = 0.1, p fa = 0.01, β p = β s = 1 and P max = 2.
Stability analysis using dominant systems Secondary dominant queue: Pr{Q s = 0} = 0 µ sd p = (1 p out,p ) (1 p md ) + (1 pout,p)p md md Primary queue does not depend on the state of secondary queue (channel ( stationarity is guaranteed). ) We apply Little s theorem Pr{Q p = 0} = 1 λp µ sd p µ sd s = λ ( p µ sd p (d) out,s (1 p md) + p md p p fa out,s p fa ) + out,s p md p out,s (1 p fa ) ( p out,s (1 p fa ) + p fa out,s p fa )
Stability analysis using dominant systems Effect of changing P s : On primary rate: Increasing P s decreases µ sd p On secondary rate: Increasing P s enhances SINR of secondary transmission and increase µ sd s Increasing P s interferes with primary transmission - PU will occupy the channel for longer times which decreases µ sd s The impact of both effects depend on λ p and µ sd s Optimization problem over P s to maximize µ sd s max. P s λ s = µ pd s s.t. λ p < µ pd p, P s P max
Stability analysis using dominant systems Secondary dominant queue: 0.37 2nd Dominant System 0.365 0.36 sd µ p 0.355 0.35 0.345 P s range for Q p stability (λ p = 0.35) 0.34 0 0.5 1 1.5 2 P s Figure: P p = 1, K = 4, p md = 0.1, p fa = 0.01, β p = β s = 1 and P max = 2.
Stability analysis using dominant systems Secondary dominant queue: 1 2nd Dominant System 0.9 0.8 0.7 0.6 sd µ s 0.5 λ p = 0 λ = 0.1 p λ p = 0.35 λ p = 0.36 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 2 P s Figure: P p = 1, K = 4, p md = 0.1, p fa = 0.01, β p = β s = 1 and P max = 2.
Stability analysis using dominant systems Stability region of the original system: 1 Stability region of the 2nd dominant system Stability region of the 1st dominant system λ p 0.5 0 0 0.2 0.4 0.6 0.8 1 λ s 1 λ p 0.5 Stability region of the system with two interacting queues 0 0 0.2 0.4 0.6 0.8 1 λ s Figure: K = 4, p md = 0.1, p fa = 0.01, β p = β s = 1 and P max = 2.
Stability analysis using dominant systems Stability region of the original system: 1 0.9 0.8 Stability region of the system with two interacting queues (P p = 10) Stability region of the system with two interacting queues (P p = 1) 0.7 0.6 λ 0.5 p 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 λ s Figure: P p = 1, 10, K = 4, p md = 0.1, p fa = 0.01, β p = β s = 1 and P max = 2.
Thank you For any questions, feel free to contact the author at m h karmoose@alexu.edu.eg