Power Allocation for Full-Duplex Relay Selection in Underlay Cognitive Radio Networks: Coherent versus Non-Coherent Scenarios
aa r X i v : . [ c s . I T ] M a r Power Allocation for Full-Duplex Relay Selectionin Underlay Cognitive Radio Networks: Coherentversus Non-Coherent Scenarios
Le Thanh Tan,
Member, IEEE,
Lei Ying,
Member, IEEE, and Daniel W. Bliss,
Fellow, IEEE
Abstract —This paper investigates power control and relayselection in Full Duplex Cognitive Relay Networks (FDCRNs),where the secondary-user (SU) relays can simultaneously receivedata from the SU source and forward them to the SU destination.We study both non-coherent and coherent scenarios. In thenon-coherent case, the SU relay forwards the signal from theSU source without regulating the phase; while in the coherentscenario, the SU relay regulates the phase when forwardingthe signal to minimize the interference at the primary-user(PU) receiver. We consider the problem of maximizing thetransmission rate from the SU source to the SU destinationsubject to the interference constraint at the PU receiver andpower constraints at both the SU source and SU relay. Wethen develop a mathematical model to analyze the data rateperformance of the FDCRN considering the self-interferenceeffects at the FD relay. We develop low-complexity and high-performance joint power control and relay selection algorithms.Extensive numerical results are presented to illustrate the impactsof power level parameters and the self-interference cancellationquality on the rate performance. Moreover, we demonstrate thesignificant gain of phase regulation at the SU relay.
Index Terms —Full-duplex cooperative communications, opti-mal transmit power levels, rate maximization, self-interferencecontrol, full-duplex cognitive radios, relay selection scheme,coherent, non-coherent.
I. I
NTRODUCTION
Cognitive radio is one of the most promising technologiesfor addressing today’s spectrum shortage [1]–[3]. This paperconsiders underlay cognitive radio networks where primaryand secondary networks transmit simultaneously over the samespectrum under the constraint that the interference caused bythe secondary network to the primary network is below a pre-specified threshold [4], [5]. Because one critical requirementfor the cognitive access design is that transmissions on thelicensed frequency band from PUs should be satisfactorilyprotected from the SUs’ spectrum access. Therefore, powerallocation for SUs should be carefully performed to meetstringent interference requirements in this spectrum sharingmodel.In particular, we consider a cognitive relay network wherethe use of SU relay can significantly increase the transmission
This paper will be presented in part at IEEE CISS 2017.The authors are with the School of Electrical, Computer and EnergyEngineering, Arizona State University, Tempe, AZ 85287, USA. Emails: { tlethanh,lei.ying.2,d.w.bliss } @asu.edu. L. T. Tan is the correspondingauthor. rate because of path loss reduction. Most existing research onunderlay cognitive radio networks has focused on the designand analysis of cognitive relay networks with half-duplex (HD)relays (e.g., see [1]–[3] and the references therein). Due tothe HD constraint, SUs typically require additional resourceswhere the SU relays receive and transmit data on differenttime slots or on orthogonal channels. Different from theseexisting work, this paper considers full-duplex relays, whichcan transmit and receive simultaneously on the same frequencyband [6]–[14]. Comparing with HD relays, FD relays canachieve both higher throughput and lower latency with thesame amount of spectrum.Design and analysis of FDCRNs, however, are very dif-ferent from HDCRNs due to the presence of self-interference ,resulted from the power leakage from the transmitter to the re-ceiver of a FD transceiver [7], [8], [12]. This self-interferencemay significantly degrade the communication performance ofFDCRNs. Hence this paper focuses on power control andrelay selection in FDCRNs with explicit consideration ofself-interference. We assume SU relays use the amplify-and-forward (AF) protocol, and further assume full channel stateinformation in both the non-coherent and coherent scenariosand the transmit phase information in the coherent scenario.This can be done by using the conventional channel estimationtechniques [15] and implicit/explicit feedback techniques [16]which are beyond the scope of our work. The contributions ofthis paper are summarized below.1) We first consider the power control problem in thenon-coherent scenario. We formulate the rate maximizationproblem where the objective is the transmission rate fromthe SU source to the SU destination, the constraints includethe power constraints at the SU source and SU relay and theinterference constraint at the PU receiver, and the optimizationvariables are the transmit power at the SU source and SU relay.The rate maximization problem is a non-convex optimizationproblem. However, it becomes convex if we fix one of twooptimization variables (i.e., fixing the transmit power at theSU source or the transmit power at the SU relay). Therefore,we propose an alternative optimization algorithm to solve thepower control problem. After calculating the achievable ratefor each FD relay using the alternative optimization algorithm,the algorithm selects the one with the maximum rate.2) We then consider the coherent scenario, where inaddition to control the transmit power, a SU relay further regulates the phase of the transmitted signal to minimize theinterference at the PU receiver. We also formulate a ratemaximization problem, which again is nonconvex. For thiscoherent scenario, we first calculate the phase to minimizethe interference at the PU receiver. Then we prove that thepower-control problem becomes convex when we fix either thetransmit power of the SU source then optimize the transmitpower of the SU relay or vice versa. We then propose analternative optimization method for power control. Based onthe achievable rate calculated from the alternative optimizationalgorithm for each relay, the relay with the maximum rate isselected.3) Extensive numerical results are presented to investigatethe impacts of different parameters on the SU network rateperformance and the performance of the proposed powercontrol and relay selection algorithms. From the numericalstudy, we observe significant rate improvement of FDCRNscompared with HDCRNs. Furthermore, the coherent mecha-nism yields significantly higher throughput than that under thenon-coherent mechanism.
A. Related Work
The FD technology can improve spectrum access efficiencyin cognitive radio networks [17]–[20] where SUs can senseand transmit simultaneously. [17] developed an FD MACprotocol that allows simultaneous spectrum access of the SUand PU networks where both PUs and SUs are assumed toemploy the MAC protocol for channel contention resolutionand access. This design is, therefore, not applicable to thehierarchical spectrum access in the CRNs where PUs shouldhave higher spectrum access priority compared to SUs. In[18], the authors propose the FD MAC protocol by using thestandard backoff mechanism as in the 802.11 MAC protocolwhere the system allows concurrent FD sensing and accessduring data transmission as well as frame fragmentation. Thisdesign has also been studied in [19], [20] for the single-and multi-channel scenarios, respectively. However all ofthese existing results assume the interweave spectrum sharingparadigm under which SUs only transmit when PUs are nottransmitting.Moreover, engineering of a cognitive FD relaying networkhas been considered in [21], [22], where various resourceallocation algorithms to improve the outage probability havebeen proposed. In [23], the authors developed a mathematicalmodel to analyze the outage probability for the proposedFD relay-selection scheme over both independent Nakagami- m and Rayleigh fading channels. The authors also extendedtheir analysis to two-way FD-based AF relays in underlaycognitive networks [24] where they analyzed various per-formance metrics such as outage probability, symbol errorprobability, etc. In addition, [25] proposed a joint routingand distributed resource allocation for FD wireless networks.[26] investigated distributed power allocation for a hybridFD/HD system where all network nodes operate in the HDmode except the access point (AP). These existing results SR h ! K SR h SP h R D h K R D h R R h K K
R R h SD h k R P h k R D h k SR h k k R R h Fig. 1. System model of power allocation with relay selection for thecognitive full-duplex relay network. focus on either minimizing the outage probability or analyzingperformance for existing algorithms. This paper considers bothnon-coherent and coherent FU relay nodes and focuses onmaximizing SU throughput given interference constraint at thePU and power constraints.The remaining of this paper is organized as follows. Sec-tion II describes the system model. Section III studies thedynamic power allocation policies for full-duplex cognitiverelay selection to achieve the maximum SU network rate.Sections IV and V consider the non-coherent and coherentscenarios, respectively. Section VI demonstrates numericalresults followed by concluding remarks in Section VII. Apart of this work will be presented at CISS 2017 [27].II. S
YSTEM M ODELS
A. System Model
We consider a cognitive relay network which consists ofone SU source S , K SU relays R k ( k = 1 , . . . , K ), one SUdestination D , and one PU receiver P . The system model forthe full-duplex cognitive relay network is illustrated in Fig. 1.The SU relays are equipped with FD transceivers to work inthe FD mode while all the other terminals work in the HDmode. Therefore the receiver performance of each SU relay isaffected by the self-interference from its transmitter since thetransmit power is leaked into the received signal.Each SU relay R k uses the AF protocol, and amplifies thereceived signal from S with a variable gain G k and forwardsthe resulting signal to SU destination, D . We denote h SR k , h R k D , h SD , h R k P and h R k R k by the corresponding channelcoefficients of links S → R k , R k → D , S → D , R k → P and R k → R k which follows the independent and identicallydistributed (i.i.d.) Gaussian with the powers of σ SR k , σ R k D , σ SD , σ R k P and σ R k R k . Let P S denote the transmit power ofSU source S . We also denote by x S ( t ) , y R k ( t ) and y D ( t ) thegenerated signal by the SU source, the transmitted signals atthe SU relay and the received signals at the SU destination,respectively.Let us consider a specific SU relay (say relay R k ). Fig. 2illustrates the signal processing at the relay. At time t , thereceived signals at SU relay R k and SU destination D are asfollows: y ( t ) = h SR k p P S x S ( t )+ h R k R k ( y ( t )+∆ y ( t ))+ z R k ( t ) (1) y D ( t ) = h R k D y R k ( t ) + h SD p P S x S ( t ) + z D ( t ) , (2)where z R k ( t ) and z D ( t ) are the additive white Gaussiannoises (AWGN) with zero mean and variances σ R k and σ D ,respectively; y D ( t ) and y ( t ) are the received signals at SUrelay R k and SU destination D ; and y ( t ) is the receivedsignal after the amplification. In the following, we ignore thedirect signal from the SU source to the SU destination (i.e.,the second part in equation (2)). Note that this assumption ishas been used in the literature [10], [22]–[24] when there isattenuation on the direct transmission channel.The transmitted signals at SU relay R k is y R k ( t ) = y ( t ) + ∆ y ( t ) , where y ( t ) = f (ˆ y ) = G k ˆ y ( t − ∆) . We should note that the SU relay amplifies the signal by afactor of G k and delays with duration of ∆ . In the noncoherentscenario, ∆ is fixed. In the coherence scenario, the delay ∆ will be optimized to minimize the interference at the PUreceiver. Furthermore, ∆ y ( t ) is the noise and follows thei.i.d. Gaussian distribution with zero mean and variance of P ∆ = ζP R k [9]–[11]. G k can be expressed as G k = h P S | h SR k | + ζP R k | h R k R k | + σ R k i − / . We assume that the channel h R k R k is perfectly estimated andhence the received signal after self-interference cancellation is ˆ y ( t ) = p P R k ( y ( t ) − h R k R k y ( t ))= p P R k h h SR k p P S x S ( t )+ h R k R k ∆ y ( t )+ z R k ( t ) i . (3)In the equation above, y ( t ) is known at SU relay R k andtherefore is used to cancel the interference. However, theremaining h R k R k ∆ y ( t ) is still present at the received signalsand is called the residual interference. So we can write thetransmitted signals at SU relay R k as follows: y R k ( t ) = G k h SR k p P R k p P S x S ( t − ∆) + ∆ y ( t )+ G k h R k R k p P R k ∆ y ( t − ∆)+ G k p P R k z R k ( t − ∆) . (4) !" !" $ % k SR h SP h k R D h !" K R P h k k R R h k R z ! k k R R h y y y ! f y y y k R y Fig. 2. The process at FD relay k . III. P
OWER C ONTROL AND R ELAY S ELECTION
In this section, we study the problem of maximizing the ratebetween SU source and SU destination while protecting thePU via power control and relay selection. Here the best relaywill be selected to help the transmission from the SU sourceto the SU destination.
A. Problem Formulation
Let C k ( P S , P R k ) denote the achieved rate of the FDCRNwith relay R k , which is the function of transmit power of SUsource S and transmit power of SU relay R k . Assume theinterference caused by the SU source and relay, I k is requiredto be at most I P to protect the PU.Now, the rate maximization problem for the selected relay k can be stated as follows: Problem 1: max P S ,P Rk C k ( P S , P R k ) s.t. I k ( P S , P R k ) ≤ I P , ≤ P S ≤ P max S , ≤ P R k ≤ P max R k , (5)where P max S and P max R k are the maximum power levels forthe SU source and SU relay, respectively. The first constrainton I k ( P S , P R k ) requires that the interference caused by theSU transmission is limited. Moreover, the SU relay’s transmitpower P R k must be appropriately set to achieve good tradeoffbetween the rate of the SU network and self-interferencemitigation.Then the relay selection is determined by k ∗ = arg max k ∈{ ,...,K } C ∗ k (6)where C ∗ k is the solution of (5). In the following, we showhow to calculate the achieved rate, C k ( P S , P R k ) and theinterference imposed by SU transmissions, I k ( P S , P R k ) . B. The Achievable Rate
When SU relay R k is selected, the achievable rate of thelink S → R k → D based on (1) and (2) is as follows: C k = log P Rk | h RkD | σ D P S | h SRk | ˆ ζP Rk + σ Rk A (7)where A = 1 + P R k | h R k D | σ D + P S | h SR k | ˆ ζP R k + σ R k (8) ˆ ζ = | h R k R k | ζ (9)Recall that we assume the direct signal from the SU sourceto the SU destination is negligible. C. The Imposed Interference at PU
We now determine the interference at the PU caused by theCRN. The interference is the signals from the SU source S and the selected relay R k : y PU I ( t ) = h SP p P S x S ( t )+ h R k P y R k ( t )+ z P ( t ) (10)where z P ( t ) is the AWGN with zero mean and variance σ P , and y R k ( t ) is defined in (4).We next derive and analyze the interference in twocases: coherent and non-coherent. In particular, we focus oncoherent/non-coherent transmissions from the SU source andthe SU relay to the PU receiver. All other transmissions areassumed to be non-coherent for simplicity. In the coherentscenario, the phase information is needed for a further regula-tion. This information can be obtained by using methods suchas the implicit feedback (using reciprocity between forwardand reverse channels in a time-division-duplex system), andexplicit feedback (using feedback in a frequency-division-duplex system) [16] or the channel estimation [15].
1) Non-coherent Scenario:
From (10) and (4), the receivedinterference at the PU caused by the SU source and theselected relay can be written as follows: I non k ( P S , P R k ) = | h SP | P S + | h R k P | ζP R k + G k | h R k P | P R k h | h SR k | P S + | h R k R k | ζP R k + σ R k i (11)After using some simple manipulations, we obtain I non k ( P S , P R k ) = | h SP | P S + | h R k P | P R k (1 + ζ ) (12)
2) Coherent Scenario:
Combining (10) with (4), the re-ceived interference at the PU caused by the SU source andthe selected SU relay is ¯ I coh k ( P S , P R k , φ ) = (cid:12)(cid:12) A + Be − jφ (cid:12)(cid:12) (13)where A = h SP p P S + h R k P p ζP R k = | A | ∠ φ A (14) B = (cid:18) h SR k p P S + h R k R k p ζP R k + σ R k √ j ) (cid:19) × G k h R k P p P R k = | B | ∠ φ B (15)and φ = 2 πf s ∆ , f s is the sampling frequency.Before using ¯ I coh k ( P S , P R k , φ ) in the constraint of the op-timization problem, we can minimize ¯ I coh k ( P S , P R k , φ ) overthe variable φ at given ( P S , P R k ) , i.e., min φ ¯ I coh k ( P S , P R k , φ ) (16) Theorem 1:
The optimal solution to (16) is φ opt = π + φ B − φ A I coh k ( P S , P R k ) = ¯ I coh k ( P S , P R k , φ opt ) = ( | A | − | B | ) . (17) Proof:
The proof is given in Appendix A.IV. P
OWER C ONTROL AND R ELAY S ECTION IN THE N ON - COHERENT S CENARIO
At the SU relay, we assume the self-interference is muchhigher than the noise, i.e., ˆ ζP R k >> σ R k . Therefore, we omitthe term σ R k in the object function. Moreover log (1 + x ) isa strictly increase function in x , so we rewrite Problem 1 as Problem 2: max P S ,P Rk ¯ C k ( P S , P R k ) s.t. I non k ( P S , P R k ) ≤ I P , ≤ P S ≤ P max S , ≤ P R k ≤ P max R k , (18)where ¯ C k ( P S , P R k ) = P Rk | h RkD | σ D P S | h SRk | ˆ ζP Rk ¯ A (19) ¯ A is given as ¯ A = 1 + P R k | h R k D | σ D + P S | h SR k | ˆ ζP R k (20)and ˆ ζ is calculated in (9).We characterize the optimal solutions for Problem 2 by thefollowing lemmas.
Lemma 1:
Problem 2 is a nonconvex optimization problemfor variables ( P S , P R k ) . Proof:
The proof is provided in Appendix B.
Lemma 2:
Given P S ∈ [0 , P max S ] , Problem 2 is a convexoptimization problem in terms of P R k . Similarly, given P R k ∈ (cid:2) , P max R k (cid:3) , Problem 2 is also a convex optimization problemin terms of P S . Proof:
The proof can be found in Appendix D.Since
Problem 2 is non-convex, we exploit alternating-optimization problem (according to Lemma 2, the problemis convex when we fix one variable and optimize the other)to solve
Problem 2 , where each step is a convex optimizationproblem and can be solved using standard approaches [29].Finally, we determine the best relay by solving (6).
We now consider the special case of ideal self-interferencecancellation, i.e., ˆ ζ = 0 . We characterize the optimal solutionsfor Problem 1 in the special case by the following lemma.
Lemma 3:
Problem 1 is a convex optimization problem forvariables ( P S , P R k ) when ˆ ζ = 0 . Proof:
The proof is provided in Appendix E.Based on Lemma 3, we can solve
Problem 1 when ˆ ζ = 0 by using fundamental methods [29].V. P OWER C ONTROL AND R ELAY S ELECTION IN THE C OHERENT S CENARIO
Again, we assume that the self-interference is much higherthan the noise at the selected relay, i.e., ˆ ζP R k >> σ R k . Problem 1 can thus be reformulated as
Problem 3: max P S ,P Rk ¯ C coh k ( P S , P R k ) s.t. I coh k ( P S , P R k ) ≤ I P , ≤ P S ≤ P max S , ≤ P R k ≤ P max R k , (21)where ¯ C coh k ( P S , P R k ) = P Rk | h RkD | σ D P S | h SRk | ˆ ζP Rk P Rk | h RkD | σ D + P S | h SRk | ˆ ζP Rk (22)and ˆ ζ is calculated in (9).To solve Problem 3 , the new variables are introduced as p S = √ P S and p R k = p P R k . Hence Problem 3 can beequivalently formulated as
Problem 4: max p S ,p Rk ˘ C coh k ( p S , p R k ) s.t. I coh k ( P S , P R k ) ≤ I P , ≤ p S ≤ p P max S , ≤ p R k ≤ p P max R k , (23)where the objective function is written as ˘ C non k ( p S , p R k ) = p Rk | h RkD | σ D p S | h SRk | ˆ ζp Rk p Rk | h RkD | σ D + p S | h SRk | ˆ ζp Rk (24)We give a characterization of optimal solutions for Problem4 by the following lemmas.
Lemma 4:
Problem 4 is not a convex optimization problemfor variable ( p S , p R k ) . Proof:
The proof is in Appendix F.
Lemma 5:
Given p S ∈ (cid:2) , p P max S (cid:3) , Problem 4 is a convexoptimization problem for variable p R k . Similarly, given p R k ∈ (cid:2) , p P max R k (cid:3) , Problem 4 is also a convex optimization problemfor variable p S . Proof:
The proof is provided in Appendix H.Based on Lemma 5, we again develop the alternating-optimization strategy to solve
Problem 4 , where each stepis a convex optimization problem and can be solved using basic approaches [29]. The relay selection is then determinedby solving (6).We now investigate the special case of ideal self-interferencecancellation, i.e., ˆ ζ = 0 . We then characterize the optimalsolutions for Problem 4 in the special case by the followinglemma.
Lemma 6:
Problem 4 is a convex optimization problem forvariables ( P S , P R k ) when ˆ ζ = 0 . Proof:
The proof can be found in Appendix J.According to Lemma 6, we can solve
Problem 4 in thisspecial case by using standard approaches [29].VI. N
UMERICAL R ESULTS
In the numerical evaluation, we set the key parametersfor the FDCRN as follows. We assume that each link isa Rayleigh fading channel with variance one (i.e., σ SR k = σ R k D = 1), except the negligible σ SD = 0.1. The noisepower at every node is also set to be one. The channelgains for the links of the SU relay-PU receiver and SUsource-PU receiver are assumed to be Rayleigh-distributedwith variances { σ SP , σ R k P } ∈ [0 . , . We also assume thatthe impact of imperfect channel estimation is included in onlyone parameter, i.e., ζ . Due to the space constraint, we onlyreport some essential results, more detailed results can befound in the online technical report [28].We first demonstrate the efficacy of the proposed algorithmsby comparing their achievable rate performances with thoseobtained by the optimal brute-force search algorithms. Numer-ical results are presented for both coherent and non-coherentscenarios where we set P max S = P max R k = P max for simplicity. InTable I, we consider the scenario with ζ = 0.001, 8 SU relaysand P max = 20 dB. We compare the achievable rate of theproposed and optimal algorithms for ¯ I P = { , , , , , } dB. These results confirm that our proposed algorithms achieverate very close to that attained by the optimal solutions for bothcoherent and non-coherent scenarios (i.e., the errors are lowerthan 1%).We then consider a FDCRN SU relays with ζ = 0, 0.001,0.01, and 0.4, which represent ideal, high, medium and lowQuality of Self-Interference Cancellation (QSIC), respectively.The tradeoffs between the achievable rate of the FDCRN andthe interference constraint are shown in Figs. 3, 4, 5 and 6under different values of ζ. In these numerical results, wechose P max S = P max R k = P max for simplicity.We have the following observations from these numericalresults. • The achievable rates of the coherent mechanism arealways significantly higher than those of the non-coherentmechanism. This is because the phase is carefully reg-ulated to reduce the interference at the PU receiverimposed by the SU transmissions, which allows highertransmit power both at the SU source and the SU relay. • As expected, the achievable rate decreases as the QSICincreases due to the increase of self-interference at theFD relay.
TABLE IA
CHIEVABLE RATE VS ¯ I P ( P max = 20 dB , ζ = 0 . ) ¯ I P (dB) 0 2 4 6 8 10 ζ = 0 . , Optimal 4.3646 5.1933 5.5533 5.6944 5.8162 5.9155Coherent Greedy 4.3513 5.1807 5.5496 5.6811 5.8131 5.8826scenario ∆ C (%) ζ = 0 . , Optimal 1.2390 1.6946 2.2118 2.7753 3.3718 3.9902Non-coherent Greedy 1.2309 1.6856 2.2018 2.7650 3.3610 3.9791scenario ∆ C (%) Achievable Rate vs P max and ¯ I P ¯ I P (dB) A c h i e v a b l e R a t e ( b i t s / s / H z ) Gap
Coherent P max = 10, 15, 20, 25 dB Noncoherent P max = 10, 15, 20, 25 dB Fig. 3. Achievable rate versus the interference constraint ¯ I P for K = 8 , ζ = 0 , P max = { , , , } dB, and both coherent and non-coherentscenarios. Achievable Rate vs P max and ¯ I P ¯ I P (dB) A c h i e v a b l e R a t e ( b i t s / s / H z ) NoncoherentCoherent
Gap P max = 10, 15, 20, 25 dB P max = 10, 15, 20, 25 dB Fig. 4. Achievable rate versus the interference constraint ¯ I P for K = 8 , ζ = 0 . , P max = { , , , } dB, and both coherent and non-coherentscenarios. • In all cases, if we increase ¯ I P , the performance in termsof data rate increases. Because the feasible range of { P S , P R k } is much larger. However there is the thresholdfor ¯ I P where we cannot obtain the higher data ratewhen we increase ¯ I P (see Fig. 6). Because to obtain thehigher performance of data rate with higher ¯ I P , we shallincrease P S and P R k . However the self-interference isalso higher due to the increase of P R k which results todecrease the data rate performance.We now show the achievable rates of the FDCRN under Achievable Rate vs P max and ¯ I P ¯ I P (dB) A c h i e v a b l e R a t e ( b i t s / s / H z ) NoncoherentCoherent
Gap P max = 10, 15, 20, 25 dB P max = 10, 15, 20, 25 dB Fig. 5. Achievable rate versus the interference constraint ¯ I P for K = 8 , ζ = 0 . , P max = { , , , } dB, and both coherent and non-coherentscenarios. Achievable Rate vs P max and ¯ I P ¯ I P (dB) A c h i e v a b l e R a t e ( b i t s / s / H z ) Gap
NoncoherentCoherent P max = 10, 15, 20, 25 dB P max = 10, 15, 20, 25 dB Fig. 6. Achievable rate versus the interference constraint ¯ I P for K = 8 , ζ = 0 . , P max = { , , , } dB, and both coherent and non-coherentscenarios. different values of P R k when fixing P S = 5 dB in Fig. 7. Thechannel gains of the links of the SU relay-PU receiver and SUsource-PU receiver were assumed to be Rayleigh-distributedwith variances { σ R k P , σ SP } ∈ [0 . , . Fig. 7 evaluates thenon-coherent scenario, K = 10 SU relays and ¯ I P = 8 dB.For each value of ζ ( ζ = { . , . , . , } ), there exists thecorresponding optimal SU relay transmit power P ∗ R k ( P ∗ R k = { . , . , . , . } dB) where the SUs achievesthe largest rate. Furthermore, the achieved rate significantlydecreases when the P R k deviates from the optimal value P ∗ R k .It is also easily observed that all four cases (low, medium, high -10 -5 0 500.20.40.60.811.21.41.6 P R k (dB) A c h i e v a b l e r a t e ( b i t s / s / H z ) Achievable Rate vs P R k Full-duplexHalf-duplex ζ = 0.4, 0.01, 0.001, 0 Fig. 7. Achievable rate versus the transmitted powers of SU relay P R k for fixed P S = 5 dB, K = 10 , ¯ I P = 8 dB, P max = 25 dB, and thenon-coherent scenario. P S (dB) A c h i e v a b l e r a t e ( b i t s / s / H z ) Achievable Rate vs P S Full-duplexHalf-duplex ζ = 0.4, 0.01, 0.001, 0 Fig. 8. Achievable rate versus the transmitted powers of SU source P S for fixed P R k = 5 dB, K = 10 , ¯ I P = 8 dB, P max = 25 dB, and thenon-coherent scenario. -5 0 5 10 15 20 2500.511.52 P R k (dB) A c h i e v a b l e r a t e ( b i t s / s / H z ) Achievable Rate vs P R k Full-duplexHalf-duplex ζ = 0.4, 0.01, 0.001, 0 Fig. 9. Achievable rate versus the transmitted powers of SU relay P R k forfixed P S = 5 dB, K = 10 , ¯ I P = 8 dB, P max = 25 dB, and the coherentscenario. and ideal QSIC) have similar behaviors and achieve higher ratethan the half-duplex case. For low QSIC (i.e., ζ = 0 . ), therate first increases then decreases as P R k increases where the P S (dB) A c h i e v a b l e r a t e ( b i t s / s / H z ) Achievable Rate vs P S Full-duplexHalf-duplex ζ = 0.4, 0.01, 0.001, 0 Fig. 10. Achievable rate versus the transmitted powers of SU source P S forfixed P R k = 5 dB, K = 10 , ¯ I P = 8 dB, P max = 25 dB, and the coherentscenario. rate decrease is due to the strong self-interference.Fig. 8 illustrates the achievable rates of the cognitive systemagainst P S for a fixed P R k = 5 dB in the non-coherentscenario. Here we also consider the cognitive radio networkwith the non-coherent scenario, K = 10 SU relays and theparameter setting of ¯ I P = 8 dB. Moreover, we compare theachievable rate of our proposed power allocation for the FD-CRN where SU relays can perform simultaneously receptionand transmission and the half-duplex cognitive radios whichuses different time slots for transmission and reception. Theresults from both Figs. 7 and 8 confirm that the proposedpower allocation for the FDCRN outperforms the HDCRN atthe corresponding optimal value of P ∗ S (or P ∗ R k ) required byour proposed scheme.We consider the cognitive radio network with the coherentscenario, K = 10 SU relays and the parameter setting of ¯ I P = 8 dB. Fig. 9 demonstrates the achievable rates of theproposed FDCRN under different values of P R k when fixing P S = 5 dB; while Fig. 10 demonstrates the achievable rates ofthe proposed FDCRN under different values of P S when fixing P R k = 5 dB. In the coherent scenario, we also have the sameobservations as those in the non-coherent scenario. However,the feasible range of P R k (or P S ) in the coherent scenario ismuch larger than that in the non-coherent scenario. That isthe result of coherent phase regulation which then improvesthe achievable rates. Recall that more numerical results can befound in the online technical report [28].VII. C ONCLUSION
This paper studied power control and relay selection inFDCRNs. We formulated the rate maximization problem,analyzed the achievable rate under the interference constraint,and proposed joint power control and relay selection algo-rithms based on alternative optimization. The design andanalysis have taken into account the self-interference of theFD transceiver, and included the both coherent and non-coherent scenarios. Numerical results have been presented to demonstrate the impacts of the levels of self-interference andthe significant gains of the coherent mechanism. Moreover, wehave shown that the proposed FDCRN achieves significantlyhigher data rate than the conventional HD schemes, whichconfirms that the FDCRN can efficiently exploit the FDcommunication capability.A
CKNOWLEDGMENT
This work was supported in part by the NSF under GrantCNS-1262329, ECCS-1547294, ECCS-1609202, and the U.S.Office of Naval Research (ONR) under Grant N00014-15-1-2169. A
PPENDIX AP ROOF OF T HEOREM ˜ B = Be − jφ = | B | ∠ φ ˜ B where (cid:12)(cid:12)(cid:12) ˜ B (cid:12)(cid:12)(cid:12) = | B | and φ ˜ B = φ B − φ . Let the subscripts R and I (at a R and a I ) bethe real and image of complex a . Note that we can obtain thefollowing equations | A | = (cid:0) A R (cid:1) + (cid:0) A I (cid:1) (25) (cid:12)(cid:12)(cid:12) ˜ B (cid:12)(cid:12)(cid:12) = (cid:16) ˜ B R (cid:17) + (cid:16) ˜ B I (cid:17) (26) A R = | A | Cos ( φ A ) (27) A I = | A | Sin ( φ A ) (28) ˜ B R = (cid:12)(cid:12)(cid:12) ˜ B (cid:12)(cid:12)(cid:12) Cos ( φ ˜ B ) = | B | Cos ( φ ˜ B ) (29) ˜ B I = (cid:12)(cid:12)(cid:12) ˜ B (cid:12)(cid:12)(cid:12) Sin ( φ ˜ B ) = | B | Sin ( φ ˜ B ) (30)From (13), we can rewrite the ¯ I coh k ( P S , P R k , φ ) as follows: ¯ I coh k ( P S , P R k , φ ) = (cid:12)(cid:12)(cid:12) A R + jA I + ˜ B R + j ˜ B I (cid:12)(cid:12)(cid:12) (31) = (cid:12)(cid:12)(cid:12) A R + ˜ B R + j (cid:16) A I + ˜ B I (cid:17)(cid:12)(cid:12)(cid:12) (32) = (cid:20)(cid:0) A R (cid:1) + (cid:16) ˜ B R (cid:17) + (cid:0) A I (cid:1) + (cid:16) ˜ B I (cid:17) (33) +2 (cid:16) A R ˜ B R + A I ˜ B I (cid:17)i (34)We substitute the parameters from (25)–(30) to the aboveresult. After using some simple manipulations, we get ¯ I coh k ( P S , P R k , φ ) = h | A | + | B | +2 | A | | B | Cos ( φ A − φ ˜ B ) i (35)From (35), we can obtain the minimum ¯ I coh k ( P S , P R k , φ ) when Cos ( φ A − φ ˜ B ) = − . Therefore, φ A − φ ˜ B = φ A − ( φ B − φ ) = π . Finally, we get φ opt = π + φ B − φ A (36) ¯ I coh k ( P S , P R k , φ opt ) = ( | A | − | B | ) (37)We now complete the proof of Theorem 1. A PPENDIX BP ROOF OF L EMMA
Problem2 is not the strictly convex optimization problem. We caneasily confirm that the constraints in
Problem 2 are convexsets due to the fact that all of them are linear functions. Toprove
Problem 2 is convex optimization problem, we mustprove that the function ¯ C k ( P S , P R k ) is concave. We rewrite ¯ C k ( P S , P R k ) as follows: ¯ C k ( P S , P R k ) = | h RkD | σ D | h SRk | ˆ ζ f ( P S , P R k ) (38)where f ( P S , P R k ) = 1 P S + P R k | h R k D | P S σ D + | h SR k | ˆ ζP R k (39)We can see that if both the terms in (38) are concave,then ¯ C k ( P S , P R k ) is concave. Here the first term, P S | h SD | σ D is concave. To prove that ¯ C k ( P S , P R k ) is concave, we caninstead prove that f ( P S , P R k ) = 1 /f ( P S , P R k ) is a concavefunction. To do so, we first determine the Hessian matrix of f ( P S , P R k ) which can be expressed as H f = " H H H H (40)Here {H ij } , i, j ∈ { , } are given as H = ∂ f∂P S = − f ∂ f∂P S − (cid:16) ∂f∂PS (cid:17) f H = ∂ f∂P S ∂P Rk = − f ∂ f∂PS∂PRk − ∂f∂PS ∂f∂PRk f H = ∂ f∂P Rk ∂P S = − f ∂ f∂PRk ∂PS − ∂f∂PS ∂f∂PRk f H = ∂ f∂P Rk = − f ∂ f∂P Rk − (cid:18) ∂f∂PRk (cid:19) f (41)where ∂ f∂P S = P S + P Rk | h RkD | σ D P S ∂ f∂P S ∂P Rk = − | h RkD | σ D P S ∂ f∂P Rk ∂P S = − | h RkD | σ D P S ∂ f∂P Rk = | h SRk | ˆ ζP Rk ∂f∂P Rk = | h RkD | P S σ D − | h SRk | ˆ ζP Rk ∂f∂P S = − P S − | h RkD | P Rk P S σ D (42)We can easily observe that H f is the symmetric matrix, i.e., H = H . Because ∂ f∂P S ∂P Rk = ∂ f∂P Rk ∂P S = − | h RkD | σ D P S .According to the Sylvester’s criterion [31], the Hessianmatrix H f is negative definite iff H < and H H −H H > . However we can choose the values of ( P S , P R k ) ( P S ∈ h , ˜ P S (cid:17) ) such that H H − H H < . Here theproof and the quantity of ˜ P S are given in Appendix C. In thiscase, the Hessian matrix H f is indefinite. Hence the function f ( P S , P R k ) is not the strictly convex function. Thus Problem2 is not the strictly convex optimization problem. So Lemma1 is completely done. A
PPENDIX CP ROOF OF H H − H H < WHEN P S ∈ h , ˜ P S (cid:17) In this section, we prove that for any given P R k we canchoose the value of P S ( P S ∈ h , ˜ P S (cid:17) ) such that H H −H H < , where ˜ P S = ˜ P S is from (52). We define SC as follows: SC = H H − H H = H H − ( H ) (43)We substitute all parameters in (42) to H , H , and H at (41). After using some manipulations, we obtain H = − | h SR k | f ˆ ζP R k P S σ D (cid:16) σ D + | h R k D | P R k (cid:17) (44) H = − f P S σ D " − | h R k D | σ D + | h SR k | P S (cid:16) σ D + 3 | h R k D | P R k (cid:17) ˆ ζP R k (45) H = − f P S σ D " − | h R k D | | h SR k | ˆ ζP R k + | h R k D | P S − | h SR k | σ D ˆ ζP R k ! | h R k D | P R k σ D ! (46)From (44), (45), (46) and (47), we can get SC = | h R k D | P S σ D (cid:0) aP S + bP S + c (cid:1) (47)where a = | h SR k | ˆ ζ P R k
12 + 11 | h R k D | P R k σ D ! > (48) b = 2 | h SR k | ˆ ζP R k | h R k D | P R k σ D ! (49) c = − | h R k D | σ D | h R k D | P R k σ D ! < (50)We can easily see that the quadratic function m ( P S ) = aP S + bP S + c has two roots, namely ˜ P S and ˜ P S . Because c < and a > , Ω = b − ac > . These quantities can bewritten as ˜ P S = − b − √ b − ac a (51) ˜ P S = − b + √ b − ac a (52) Note that ˜ P S < and ˜ P S > because − b − √ b − ac a < − b − √ b a = 0 (53) − b + √ b − ac a > − b + √ b a = 0 (54)Hence SC can be rewritten as SC = | h R k D | aP S σ D (cid:16) P S − ˜ P S (cid:17) (cid:16) P S − ˜ P S (cid:17) (55)We now can find that if < P S < ˜ P S , where ˜ P S = ˜ P S then SC < . So we complete the proof.A PPENDIX DP ROOF OF L EMMA
Problem 2 areconvex sets due to the fact that all of them are linear functions.To prove
Problem 2 is convex optimization problem, wemust prove that the function ¯ C k ( P S , P R k ) is concave. Notethat ¯ C k ( P S , P R k ) and f ( P S , P R k ) are from (38) and (69),respectively.We can see that if both the terms in (38) are concave,then ¯ C k ( P S , P R k ) is concave. Here the first term, P S | h SD | σ D is concave. So to prove that ¯ C k ( P S , P R k ) is concave, we caninstead prove that f ( P S , P R k ) is a convex function [29]. Forgiven P S ∈ [0 , P max S ] , we take the first-order partial derivativeof f ( P S , P R k ) with respect to P R k as ∂f∂P R k = | h R k D | P S σ D − | h SR k | ˆ ζP R k (56)Then the second-order partial derivative of f ( P S , P R k ) withrespect to P R k can be determined as ∂ f∂P R k = 2 | h SR k | ˆ ζP R k (57)We can see that ∂ f∂P Rk > , hence f ( P S , P R k ) is a convexfunction for P R k .Similarly, for given P R k ∈ (cid:2) , P max R k (cid:3) , we take the firstderivative of f ( P S , P R k ) with respect to P S as ∂f∂P S = − P S − | h R k D | P R k P S σ D (58)Then the second derivative of f ( P S , P R k ) with respect to P S can be calculated as ∂ f∂P S = 2 P S + 2 | h R k D | P R k P S σ D (59)Since ∂ f∂P S > , we can conclude that f ( P S , P R k ) is a convexfunction for P S . The proof of Lemma 2 is complete. A PPENDIX EP ROOF OF L EMMA ˆ ζ = 0 . Moreover the function log (1 + x ) isthe strictly increase function of variable x , so we can rewritethe objective function of Problem 1 as ˜ C k ( P S , P R k ) = P Rk | h RkD | σ D P S | h SRk | σ Rk P Rk | h RkD | σ D + P S | h SRk | σ Rk (60)Here we approximate the ˜ C k ( P S , P R k ) in the high SNR regionwhich is usually used in wireless communications [30]. So ˜ C k ( P S , P R k ) is rewritten as ˜ C k ( P S , P R k ) = P Rk | h RkD | σ D P S | h SRk | σ Rk P Rk | h RkD | σ D + P S | h SRk | σ Rk (61)We can easily confirm that the constraints in Problem 1 areconvex sets due to the fact that all of them are linear functions.To prove
Problem 1 is the convex optimization problem, wemust prove that the function ˜ C k ( P S , P R k ) is concave. Werewrite ˜ C k ( P S , P R k ) as follows: ˜ C k ( P S , P R k ) = | h RkD | σ D | h SRk | σ Rk ˜ f ( P S , P R k ) (62)where ˜ f ( P S , P R k ) = | h R k D | P S σ D + | h SR k | σ R k P R k (63)To prove that ˜ C k ( P S , P R k ) is concave, we can instead provethat f ( P S , P R k ) = 1 / ˜ f ( P S , P R k ) is a concave function. Todo so, we first determine the Hessian matrix of f ( P S , P R k ) which can be expressed as H ˜ f = " ˜ H ˜ H ˜ H ˜ H (64)Here {H ij } , i, j ∈ { , } are given as ˜ H = ∂ f∂P S = − ˜ f ∂ f∂P S − (cid:16) ∂ ˜ f∂PS (cid:17) ˜ f ˜ H = ∂ f∂P S ∂P Rk = − ˜ f ∂ f∂PS∂PRk − ∂ ˜ f∂PS ∂ ˜ f∂PRk ˜ f ˜ H = ∂ f∂P Rk ∂P S = − ˜ f ∂ f∂PRk ∂PS − ∂ ˜ f∂PS ∂ ˜ f∂PRk ˜ f ˜ H = ∂ ˜ f∂P Rk = − ˜ f ∂ f∂P Rk − (cid:18) ∂ ˜ f∂PRk (cid:19) f (65) where ∂ ˜ f∂P S = | h RkD | σ D P S ∂ ˜ f∂P S ∂P Rk = 0 ∂ ˜ f∂P Rk ∂P S = 0 ∂ ˜ f∂P Rk = | h SRk | σ Rk P Rk ∂ ˜ f∂P Rk = − | h SRk | σ Rk P Rk ∂ ˜ f∂P S = − | h RkD | σ D P S (66)Substitute (66) to (65), we obtain ˜ H = − f | h RkD | | h SRk | σ D σ Rk P S P Rk < H = f | h RkD | | h SRk | σ D σ Rk P S P Rk = ˜ H ˜ H = − f | h RkD | | h SRk | σ D σ Rk P S P Rk (67)We can see that H < and H H − H H = 0 , i.e., theHessian matrix H ˜ f is negative semi-definite according to theSylvester’s criterion [31]. Hence the function f ( P S , P R k ) isthe convex function. Thus Problem 1 is the convex optimiza-tion problem for the case of ˆ ζ = 0 . So the proof of Lemma3 is completely done. A PPENDIX FP ROOF OF L EMMA
Problem 4 is not the strictly convexoptimization problem by using contradiction. To prove
Prob-lem 4 is convex optimization problem, we must prove that thefunction ˘ C coh k ( p S , p R k ) is concave. We rewrite ˘ C coh k ( p S , p R k ) as follows: ˘ C coh k ( p S , p R k ) = | h RkD | σ D | h SRk | ˆ ζ g ( p S , p R k ) (68)where g ( p S , p R k ) = 1 p S + p R k | h R k D | p S σ D + | h SR k | ˆ ζp R k (69)To prove that ˘ C coh k ( p S , p R k ) is concave, we can insteadprove that g ( p S , p R k ) = 1 /g ( p S , p R k ) is a concave function.To do so, we first determine the Hessian matrix of g ( P S , P R k ) which can be expressed as G g = " G G G G (70) Here {G ij } , i, j ∈ { , } are expressed as follows: G = ∂ g∂p S = − g ∂ g∂p S − (cid:16) ∂g∂pS (cid:17) g G = ∂ g∂p S ∂p Rk = − g ∂ g∂pS∂pRk − ∂g∂pS ∂g∂pRk g G = ∂ g∂p Rk ∂p S = − g ∂ g∂pRk ∂pS − ∂g∂pS ∂g∂pRk g G = ∂ g∂p Rk = − g ∂ g∂p Rk − (cid:18) ∂g∂pRk (cid:19) g (71)where ∂ g∂p S = p S + | h RkD | p Rk σ D p S ∂ g∂p S ∂p Rk = − p Rk | h RkD | σ D p S ∂ g∂p Rk ∂p S = − p Rk | h RkD | σ D p S ∂ g∂p Rk = | h RkD | p S σ D + | h SRk | ˆ ζp Rk ∂g∂p Rk = h RkD p Rk p S σ D − | h SRk | ˆ ζp Rk ∂g∂p S = − p S − | h RkD | p Rk p S σ D (72)We can also observe that G g is the symmetric matrix, i.e., G = G . Because ∂ g∂p S ∂p Rk = ∂ g∂p Rk ∂p S = − p Rk | h RkD | σ D p S .According to the Sylvester’s criterion [31], the Hessian ma-trix G g is negative definite iff G < and G G − G G > . However, we can observe that G G − G G < in thecase that p R k ∈ [0 , ˜ p R k ) and p S ∈ [0 , ˜ p S ) . The proof and thevalues of (˜ p R k , ˜ p S ) are given in Appendix G. Therefore thefunction g ( p S , p R k ) is not the strictly convex function. Thus Problem 4 is not the strictly convex optimization problem. Sowe complete the proof of Lemma 4.A
PPENDIX GP ROOF OF G G − G G < WHEN p R k ∈ [0 , ˜ p R k ) AND p S ∈ [0 , ˜ p S ) In this section, we prove that for any given p R k ( p R k ∈ (cid:0) ˜ p R k , p P max R k (cid:3) ) we can choose the value of p S ( p S ∈ [0 , ˜ p S ) )such that G G − G G < , where (˜ p R k , ˜ p S ) is from ( 80)and (86)).We define SC as follows: SC = G G − G G = G G − ( G ) (73)Hence we must prove that SC < . Moreover, from (73), wecan instead prove that G G < which will be done in thefollowing.We substitute all parameters in (72) to G and G at (71).After using some manipulations, we obtain G =2 (cid:16) σ D + | h R k D | p R k (cid:17) g ˆ ζp R k p S σ D (cid:16) ˆ ζp R k − | h SR k | p S (cid:17) (74) G = 1 g p S (cid:0) a p S + 2 a p S + a (cid:1) (75) where a = 2 | h SR k | ˆ ζ p R k > (76) a = − | h SR k | ˆ ζp R k σ D (cid:16) σ D + 4 | h R k D | p R k (cid:17) < (77) a = − | h R k D | σ D (cid:16) σ D − | h R k D | p R k (cid:17) (78)From (74), if we choose p S < ˜ p S, , then G > . Here ˜ p S, is written as ˜ p S, = s ˆ ζ p R k | h SR k | (79)To complete the proof, we will find the ranges ( p S , p R k ) (where p S < ˜ p S, ) such that G < . Note that there are manysuch ranges ( p S , p R k ) . However we just give one exampleas follows. Let consider the quadratic function η ( p S ) = a p S + 2 a p S + a . From (78), if p R k < ˜ p R k , then a < .Here ˜ p R k is defined as ˜ p R k = σ D √ | h R k D | (80)In this case, let us check the Ω as Ω = a − a a (81)We can observe that Ω > a > (because a > and a < ).So the quadratic function η ( p S ) has two real roots as follows: p S, = 1 a (cid:16) − a − √ Ω (cid:17) < (82) ˜ p S, = 1 a (cid:16) − a + √ Ω (cid:17) > (83)We should note that ˜ p S, > because a (cid:16) − a + √ Ω (cid:17) > a (cid:18) − a + q a (cid:19) = 0 (84)Hence η ( p S ) can be written as follows: η ( p S ) = a ( p S − p S, ) ( p S − ˜ p S, ) (85)It is clearly seen that if ≤ p S < ˜ p S, , then η ( p S ) < andhence G < .In summary, let us define ˜ p S as ˜ p S = min { ˜ p S, , ˜ p S, } (86)where ˜ p S, and ˜ p S, are from (79) and (83), respectively. Wecan conclude that if ≤ p S < ˜ p S and p R k < ˜ p R k ( ˜ p R k isfrom (80)) then G > and G < ; and hence G G < ,i.e., SC < . So the proof is completely done. A PPENDIX HP ROOF OF L EMMA
Problem 4 are convex sets. Because the constraint ¯ I coh k ( p S , p R k ) ≤ ¯ I P is the convex set for variable p S and p R k (see the proof inAppendix I); and the all of remaining are linear functions. Toprove Problem 4 is the convex optimization problem, we mustprove that the function ˘ C coh k ( p S , p R k ) is concave.To prove that ˘ C coh k ( p S , p R k ) is concave, we can insteadprove that g ( p S , p R k ) is a convex function [29]. Here ˘ C coh k ( p S , p R k ) and g ( p S , p R k ) are given from (68) and (69),respectively. For given p S ∈ (cid:2) , p P max S (cid:3) , we take the first-order partial derivative of g ( p S , p R k ) with respect to p R k as ∂g∂p R k = 2 | h R k D | p R k p S σ D − | h SR k | ˆ ζp R k (87)Then the second-order partial derivative of g ( p S , p R k ) withrespect to p R k can be determined as ∂ g∂p R k = 2 | h R k D | p S σ D + 6 | h SR k | ˆ ζp R k (88)We can see that ∂ g∂p Rk > , hence g ( p S , p R k ) is a convexfunction for p R k .Similarly, for given p R k ∈ (cid:2) , p P max R k (cid:3) , we take the firstderivative of g ( p S , p R k ) with respect to p S as ∂g∂p S = − p S − | h R k D | p R k p S σ D (89)Then the second derivative of g ( p S , p R k ) with respect to p S can be calculated as ∂ g∂p S = 6 p S + 6 | h R k D | p R k p S σ D (90)Since ∂ g∂p S > , we can conclude that g ( p S , p R k ) is a convexfunction for p S . So we complete the proof of Lemma 5.A PPENDIX IP ROOF THAT ¯ I coh k IS A CONVEX FUNCTION FOR p S AND p R k In this section, we will prove that ¯ I coh k is a convex functionfor p S and p R k . ¯ I coh k ( p S , p R k , φ opt ) = (cid:12)(cid:12) A + Be − jφ (cid:12)(cid:12) (91)where A = h SP p S + h R k P p ζp R k = | A | ∠ φ A (92) B = (cid:20) h SR k p S + h R k R k p ζp R k + σ R k √ j ) (cid:21) × G k h R k P p P R k = | B | ∠ φ B (93)Let us define D as follows: D = h SR k p S + h R k R k p ζp R k + σ R k √ j ) (94) We calculate | D | = DD ∗ . After some manipulations, weobtain | D | = G − k + L (95)where L/ p S h RSR k p ζp R k h RR k R k + p S h RSR k σ R k / √ p ζp R k h RR k R k σ R k / √ p S h ISR k p ζp R k h IR k R k + p S h ISR k σ R k / √ p ζp R k h IR k R k σ R k / √ Here the subscripts R and I (at a R and a I ) are the real andimage of complex a . Using the CauchySchwarz Inequality[32], we have L/ ≤ p S h(cid:0) h RSR k (cid:1) + (cid:0) h ISR k (cid:1) i + ζp R k h(cid:0) h RR k R k (cid:1) + (cid:0) h IR k R k (cid:1) i + σ R k (96)So L/ ≤ p S | h SR k | + ζp R k | h R k R k | + σ R k = G − k (97)We use the approximation of L ≈ G − k for the remainingproof. So we have | B | = 3 p R k | h R k P | . Then we can rewrite B as follows: B = | B | ∠ φ B = √ p R k | h R k P | ∠ φ B (98)The interference is now written as ¯ I coh k ( p S , p R k ) = (cid:12)(cid:12)(cid:12) h SP p S + h R k P p ζp R k + √ p R k | h R k P | ∠ φ B − φ opt (cid:12)(cid:12)(cid:12) (99)We now prove that the constraint ¯ I coh k ( p S , p R k ) ≤ ¯ I P isthe convex set for variable p S . We can rewrite ¯ I coh k ( p S , p R k ) as follows: ¯ I coh k ( p S , p R k ) = ( h RSP p S + F p R k ) + ( h ISP p S + F p R k ) (100)where F = h RR k P p ζ + √ | h R k P | Cos ( φ B − φ opt ) (101) F = h IR k P p ζ + √ | h R k P | Sin ( φ B − φ opt ) (102)We take the first-order and second-order partial derivativesof ¯ I coh k ( p S , p R k ) with respect to p S as ∂ ¯ I coh k ∂p S = 2 h RSP (cid:0) h RSP p S + F p R k (cid:1) +2 h ISP (cid:0) h ISP p S + F p R k (cid:1) (103) ∂ ¯ I coh k ∂p S = 2 (cid:0) h RSP (cid:1) + 2 (cid:0) h ISP (cid:1) = 2 | h SP | (104)Since ∂ ¯ I coh k ∂p S > , we can conclude that ¯ I coh k is a convexfunction for p S . Similarly, we prove that the constraint ¯ I coh k ( p S , p R k ) ≤ ¯ I P is the convex set for variable p R k . To do so, we take the first-order and second-order partial derivatives of ¯ I coh k ( p S , p R k ) with respect to p R k as ∂ ¯ I coh k ∂p R k = 2 F ( h RSP p S + F p R k )+2 F ( h ISP p S + F p R k ) (105) ∂ ¯ I coh k ∂p R k = 2 F + 2 F (106)Since ∂ ¯ I coh k ∂p Rk > , we can conclude that ¯ I coh k is a convexfunction for p R k . Hence we complete the proof.A PPENDIX JP ROOF OF L EMMA
Problem 4 are convex sets. Because the constraint ¯ I coh k ( p S , p R k ) ≤ ¯ I P is the convex set for variable p S and p R k (see the proof inAppendix I); and the all of remaining are linear functions. Toprove Problem 4 is the convex optimization problem, we mustprove that the function ˘ C coh k ( p S , p R k ) is a increased functionfor p S > and p R k > .When ˆ ζ = 0 , we can rewrite ˘ C coh k ( p S , p R k ) as ˘ C coh k ( p S , p R k ) = | h R k D | p R k σ D . (107)We can easily observe that the quadratic function ˘ C coh k ( p S , p R k ) depends only on p R k with the criticalpoint of p ∗ R k = 0 . Therefore ˘ C coh k ( p S , p R k ) is monotonicallyincreasing in the range of p R k > . Hence we can find theglobally optimal p R k in its feasible range. So we completethe proof of Lemma 6.R EFERENCES[1] L. T. Tan and L. B. Le, “Distributed MAC protocol for cognitiveradio networks: Design, analysis, and optimization,”
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