Joint power control and beamforming for cognitive radio networks

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Joint power control and beamforming for cognitive radio networks
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  Joint Power and Channel Resource Allocation forF/TDMA Decode and Forward Relay Networks Yin Sun † , Yuanzhang Xiao † , Ming Zhao † , Xiaofeng Zhong † , Shidong Zhou † and Ness B. Shroff  ‡† State Key Laboratory on Microwave and Digital Communications † Tsinghua National Laboratory for Information Science and Technology † Department of Electronic Engineering, Tsinghua University, Beijing, China. ‡ Departments of ECE and CSE, The Ohio State University.  Abstract —In this paper, we study the joint power and channelresource allocation problem for a multiuser F/TDMA decode-and-forward (DF) relay network under per-node power con-straints and a total channel resource constraint. Our goal is tomaximize the total throughput achieved by the systems. To thatend, we formulate a joint power and channel resource allocationproblem. We develop an iterative optimization algorithm to solvethis problem, whose convergence and optimality are guaranteed.Due to the per-node power constraints, more than one relaynode may be needed for a single data stream. Our solution alsoprovides a way of finding the optimal relays among the assistingrelay nodes. I. I NTRODUCTION Cooperative relaying is a promising technique for providingcost effective enhancements of network coverage and through-put [1]. The relay nodes exploit the broadcast feature of wireless channels. They can “hear” the transmitted signals of the source nodes and assist forwarding the information [2].In wireless access networks, the transmission power of the nodes and the channel resources (time and frequancy)are limited. Hence, appropriate power and channel resourceallocation is needed to fully utilize the available radio resource.It has been shown that power and channel resource allocationcan result in significant performance gains for  single user relaynetworks  [3]-[5].The study of   multiuser relay networks  is more crucial forwireless access networks. When multiple relay nodes areinvolved in the network, the number of access links increasesgreatly. How to select proper access links and allocate powerand channel resource for them is very important for the systemperformance of wireless relay networks.In [6], the authors considered relaying strategy selectionand power allocation at the relay nodes for F/TDMA relaynetworks, where the power allocation at the source node andthe relay node selection are not jointly considered. The power The work of Yin Sun, Yuanzhang Xiao, Ming Zhao, Xiaofeng Zhong,Shidong Zhou is supported by MIIT Project of China (2008ZX03003-004),National Basic Research Program of China (2007CB310608), China’s 863Project (2007AA01Z2B6), National Science Foundation of China (60832008),and Program for New Century Excellent Talents in University (NCET).Email:  { sunyin02, xiaoyz02 } @mails.tsinghua.edu.cn,  { zhaoming, zhongxf,zhousd } @tsinghua.edu.cn.The work of Ness B. Shroff is supported by NSF Awards CNS-0626703, CNS- 0721236, ANI-0207728, and CCF-0635202, USA. Email:shroff@ece.osu.edu. and channel resource allocation for orthogonal multiple accessrelay networks was addressed in [7], where the data rate of one user is maximized subject to target rate requirements forthe other nodes.In this paper, we focus on the joint power and channelresource allocation problem of a multi-user F/TDMA decode-and-forward (DF) relay network. We adopt the assumptionthat each node is subject to separate power constraints [4].Further, we suppose that the total channel resource of thenetwork is limited 1 . We show that the joint power and channelresource allocation problem is a convex optimization problem.Therefore, we can develop a fast iterative algorithm for thisproblem based on the duality theory. The dual method isbeneficial since the dual problem not only has fewer variablesand simpler constraints but also is easily decomposable.However, this problem is hard because the objective func-tion of the problem is neither differentiable nor strictly concaveeven if only the power allocation subproblem is considered. Inthis paper, the non-differentiability of the objective function issolved by using auxiliary variables. This approach is equiv-alent to the max-min method [4]. The  proximal optimizationmethod   is used to handle the non-strict concavity of the primalobjective function [8], [9]. The channel resource allocationproblem is given by the root of the Karush-Kuhn-Tucker(KKT) condition [10, pp. 243].By performing power allocation and channel resource allo-cation iteratively, the joint optimal power and channel resourceallocation solution is derived. The convergence and global op-timality of this iterative optimization algorithm are guaranteedby using similar argument as in [4]. Due to the per-node powerconstraints, more than one relay nodes may be needed for asingle data stream. The optimal relay nodes selection is derivedsimultaneously in our algorithm.The outline of this paper is given as follows:In Section II, the system model is introduced. In Section III,we show that the joint power and channel resource allocationproblem is a convex optimization problem. In Section IV, wepresent the iterative optimization algorithm for this problem.The numerical results are shown in Section V. And finally, inSection VI, we give the conclusion. 1 For distributed controlled network, the channel resource is pre-assigned tothe source nodes. The source nodes can allocate the channel resource amongdifferent relay links. The distributed implentation of this problem is out of the scope of this paper. This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings. 978-1-4244-4148-8/09/$25.00 ©2009  II. S YSTEM  M ODEL We consider a multiuser F/TDMA DF relay network, whichconsists of   R  relay nodes and  N   user nodes. Here, the termuser nodes encompasses all possible source and destinationnodes. Let R be the set of relay nodes and  N   be the set of usernodes, i.e. R  =  { 1 , 2 ,...,R } and  N   =  { 1 , 2 ,...,N  } . In eachtime frame, certain data streams, each of which is between asource-destination pair, are scheduled. Let  m  = ( s,d ) ( s,d  ∈ N  )  be a source-destination pair, and let  M  be the set of data streams, which satisfies  M ⊆ { ( i,k ) | i,k  ∈ N  ,i   =  k } .Each data stream  m  = ( i,k )  ∈ M  can be assisted by somenearby relay nodes, which is the practical situation. Clearly,these relay nodes only compose a sub-set of   R . However, weassume all  R  relay nodes are potential relay nodes for eachdata stream, and let the joint optimization algorithm to selectthe best relay nodes for each data stream.Suppose the network operates in a slow fading environ-ment. Each node performs channel estimation and the channelstrength information is fed back to a central node (e.g., thebase station of the relay-assisted cellular network). The centralnode performs the power and channel resource allocation, andthen broadcasts the result to the other nodes.For practical consideration [3], all nodes are assumed to op-erate in half-duplex mode. In order to prevent inter-stream in-terference and facilitate simpler transmitter design, we requirethat all source and relay nodes transmit in orthogonal sub-channels [6] and [11]-[12]. When some data stream  m  = ( i,k ) is assisted by a relay node  j , the source node  i  transmits inone sub-channel with channel resource proportion  θ mj / 2  andthe relay node  j  transmits in another orthogonal sub-channelwith channel resource proportion  θ mj / 2 . The received signalof the destination  k  in the first sub-channel is y dm 1  =   2 P  smj /θ mj β  m x sm 1  +  n dm 1 ,  (1)where  x sm 1  is the transmitted signal of the source node  i ,  P  smj is total transmitted energy of the source node  i ,  β  m  denotes thenormalized channel gain of the source-destination pair  m  with n dm 1  as the zero mean AWGN with unit variance. Similarly,the received signal at relay node  j  is given by y rmj  =   2 P  smj /θ mj α mj x sm 1  +  n rmj ,  (2)where  α mj  denotes the normalized channel gain between thesource node  i  and the relay node  j , and  n rmj  is the zero meanAWGN with unit variance at the relay node  j . In the secondsub-channel, the received signal at the destination  k  is y dm 2  =   2 P  rmj /θ mj γ  mj x rmj  +  n dm 2 ,  (3)where  x rmj  is the transmitted signal of the relay node  j ,  P  rmj is the total transmitted energy of relay node  j ,  γ  mj  denotes thenormalized channel gain between relay node  j  and destination k  with  n dm 2  as the zero mean AWGN with unit variance.We also allow the user to transmit to the destination directly.Suppose the source node  i  transmits in one sub-channel withchannel resource proportion  θ m , the received signal of thedestination  k  is y dm  =   P  sm /θ m β  m x sm  +  n dm ,  (4)where  x sm  and  P  sm  are the transmitted signal and total trans-mitted energy of the source node  i  and  n dm  is the zero meanAWGN with unit variance.III. J OINT  P OWER AND  C HANNEL  R ESOURCE A LLOCATION  P ROBLEM The achievable data rate of decode-and-forward (DF) relay-ing strategy given in [13] is C  mj  =  θ mj / 2min  C   2( P  smj β  2 m  +  P  rmj γ  2 mj ) /θ mj  ,C   2 P  smj α 2 mj /θ mj  ,  (5)where  C  ( x ) = log 2 (1 +  x ) . The data rate of directiontransmission (DT) is simply the capacity of adaptive whiteGaussian noise channel C  m  =  θ m C   P  sm β  2 m /θ m  .  (6)If   α mj  ≤  β  m , using the property that the function  θ  → θC  ( a/θ )  is increasing, we can show that  C  mj  is no largerthan the  C  m  when  θ m  =  θ mj . Therefore, we only adopt theDF relaying strategy when  α mj  > β  m  as in [6]. If   α mj  ≤  β  m ,we simply let  P  smj  =  P  rmj  = 0 .One can show that both  C  m  and  C  mj  are strictly concavewith respect to the power allocation variables. Using theproperty of   perspective function  given in [10, pp. 89], itis easy to prove that  C  m  and  C  mj  are also concave withrespect to the power and channel resource variables. Ourobjective is to maximize the achievable sum rate of all thedata streams. The rate of a data stream is the sum rate of one DT link and  R  DF relay links. We note that DF and DTuse orthogonal channels. Their power and channel resourceallocation variables are independent variables, even for onesource-destination pair. Suppose that each node is subjectto separate average power constraints (or total transmissionenergy in a scheduling frame). The total channel resource of the F/TDMA relay network is limited to 1. Therefore, the jointpower and channel resource allocation problem is described bythe following convex optimization problem max P  sm ,θ m ,P  smj ,P  rmj ,θ mj  m ∈M ( C  m  +  j ∈R C  mj )  (7) s.t.  m ∈ S  l ( P  sm  +  j ∈R P  smj )  ≤  P  sl,max , ∀  l  ∈ N   (8)  m ∈M P  rmj  ≤  P  rj,max , ∀  j  ∈ R  (9)  m ∈M ( θ m  +  j ∈R θ mj )  ≤  1  (10) P  sm ,P  smj ,P  rmj ,θ m ,θ mj  ≥  0 , ∀  m,j  (11)where  C  mj  and  C  m  are given by (5) and (6),  S  l   { ( i,k )  ∈M  :  i  =  l }  is defined as the set of data streams with the samesource node  l ,  P  sl,max  and  P  rl,max  are the average transmittedpower constraints of source node and relay node.IV. I TERATIVE  O PTIMIZATION  A LGORITHM In this section, we develop an iterative algorithm to solvethe joint power and channel resource allocation problem. We This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings. 978-1-4244-4148-8/09/$25.00 ©2009  have mentioned that the data rate of DF relaying (5) is aconcave function, but it is neither differentiable nor strictlyconcave. Because the objective function is not strictly concave,the primal optimal solution is not unique and the dual functionis non-differentiable [8]. Using the standard dual solutionwill cause the primal variables to oscillate during the dualiterations. This is explained in detail in [9].To alleviate this difficulty, we use the proximal optimizationmethod to solve the power allocation subproblem. The basicidea is to make the primal objective function strictly concaveby subtracting a quadratic term from it. This ensures that thedual optimzation algorithm is stablized and converges quitefast, and the converged point is one of the optimal solutions of the srcinal problem [8, Section 3.4.3]. The non-differentiableproperty of (5) is handled by using auxiliary variables. Sucha method is equivalent to the max-min method given in [4].The channel resource allocation solution is given by theroot of the KKT condition. The famous rapidly convergentNewton’s method [14, Section 5.5.3] is used to solve theKKT condition. By performing power allocation and channelresource allocation iteratively, we can prove that the jointoptimal power and channel resource allocation solution isderived.  A. Proximal optimization method for power allocation In this subsection, we utilize the dual decomposition basedproximal optimization method [8], [9] to solve the powerallocation subproblem for fixed channel resource variables.The proximal optimization method considers the followingmodified problem max P  sm ,P  smj ,P  rmj ,Q smj ,Q rmj  m ∈M [ C  m  +  j ∈R C  mj −  c mj 2 ( P  smj  − Q smj ) 2 −  c mj 2 ( P  rmj  − Q rmj ) 2 ]  (12) s.t. P  sm ,P  smj ,P  rmj  ∈  W, Q smj ,Q rmj  ≥  0 , ∀  m,j, (13)where  c mj  is a positive number chosen for each  m,j , and W   =  { P  sm ,P  smj ,P  rmj |  m ∈M P  rmj  ≤  P  rj,max , ∀  j  ∈ R ,  m ∈ S  l ( P  sm  +  j ∈R P  smj )  ≤  P  sl,max , ∀  l  ∈ N  ,P  sm ,P  smj ,P  rmj  ≥  0 , ∀  m,j }  (14)One can show that the optimal power allocation variablesof the original problem (7)-(11) coincide with the optimalsolution of (12)-(13) [8, Section 3.4.3].The  proximal optimization algorithm  [8], [9] as applied inour problem is given by: Algorithm  A : At the  t -th iteration, (A1)  Fix  Q smj ( t ) ,Q rmj ( t )  and maximize the objective func-tion (12) with respect to  P  sm ,P  smj ,P  rmj . More precisely,this step solves the problem max P  sm ,P  smj ,P  rmj  m ∈M { C  m  +  j ∈R C  mj − c mj 2 [ P  smj  − Q smj ( t )] 2 −  c mj 2 [ P  rmj  − Q rmj ( t )] 2 }  (15) s.t. P  sm ,P  smj ,P  rmj  ∈  W   (16)Since the objective function (15) is strictly concave, thesolution to the problem exists and is unique. (A2)  Suppose the solution of (A1) is  P  sm ( t ) ,P  smj ( t ) ,P  rmj ( t ) .Let  Q smj ( t  + 1) =  P  smj ( t ) , Q rmj ( t  + 1) =  P  rmj ( t ) .Now, we use standard duality techniques to solve (15) inStep (A1). Let  µ l  ( l  ∈ N  )  and  ν  j  (  j  ∈ R )  be the Lagrangedual variables for constraints (8) and (9), respectively. TheLagrangian of (15) can be given in a dual decomposition form L ( P  sm ,P  smj ,P  rmj ,µ l ,ν  j )=  l ∈N   m ∈ S  l ( C  m  − µ l P  sm )+  l ∈N   m ∈ S  l  j ∈R  C  mj  −  c mj 2 [ P  smj  − Q smj ( t )] 2 − c mj 2 [ P  rmj  − Q rmj ( t )] 2 − µ l P  smj  − ν  j P  rmj  +  l ∈N  µ l P  sl,max  +  j ∈R ν  j P  rj,max .  (17)Therefore, the objective function of the dual problem is D ( µ l ,ν  j ) = max P  sm ,P  smj ,P  rmj ≥ 0 L ( P  sm ,P  smj ,P  rmj ,µ l ,ν  j )=  l ∈N   m ∈ S  l H  m ( µ l ) +  l ∈N   m ∈ S  l  j ∈R I  mj ( µ l ,ν  j )+  l ∈N  µ l P  sl,max  +  j ∈R ν  j P  rj,max ,  (18)where H  m ( µ l ) = max P  sm ≥ 0 ( C  m  − µ l P  sm ) ,  (19) I  mj ( µ l ,ν  j ) = max P  smj ,P  rmj ≥ 0  C  mj  −  c mj 2 [ P  smj  − Q smj ( t )] 2 − c mj 2 [ P  rmj  − Q rmj ( t )] 2 − µ l P  smj  − ν  j P  rmj  .  (20)The dual problem of (15) is given by min µ l ,ν  j ≥ 0 D ( µ l ,ν  j ) .  (21)Since the objective function of (15) is strictly concave, thedual function is differentiable on the whole region [8]. Thegradient of the dual function  D  is ∂D∂µ l =  P  sl,max  −  m ∈ S  l [ P  s⋆m  ( u ) +  j ∈R P  s⋆mj ( u )] ,  (22) ∂D∂ν  j =  P  rj,max  −  m ∈M P  r⋆mj ( u ) ,  (23)where  P  s⋆m  ( u ) ,P  s⋆mj ( u ) ,P  r⋆mj ( u )  solve (19) and (20) for  µ l  = µ l ( u )  and  ν  j  =  ν  j ( u ) . Therefore, the dual problem (21) canbe solve by the gradient project algorithm [8, Section 3.3.2] µ l ( u  + 1)=  µ l ( u ) +  ρ l [  m ∈ S  l ( P  s⋆m  +  j ∈R P  s⋆mj ) − P  sl,max ]  † ν  j ( u  + 1)=  ν  j ( u ) +  σ j (  m ∈M P  r⋆mj  − P  rj,max )  † (24)where  ( · ) † = max {· , 0 } . It can be shown that the dualiterations (24) converge to the optimal solution, if the stepsize  ρ l , σ j  are small enough [8, Section 3.3.2]. This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings. 978-1-4244-4148-8/09/$25.00 ©2009   B. Recovery of the power allocation variables In this subsection, we consider the power allocation vari-ables, which optimize the problems (19) and (20). The solutionof (19) is just the common “water-filling” result [10, pp. 245] P  sm  =  θ   1 µ l  ln2  −  1 β  2 m  † ,  if   m  ∈  S  l .  (25)For the problem of (20), we first define R 1  =  θ 2 C   2 θ ( P  s β  2 +  P  r γ  2 )  , R 2  =  θ 2 C   2 θP  s α 2  .  (26)We omit the subscripts of the variables in this subsectionfor brevity. substituting the data rate of DF relaying withauxiliary variable  t  in (20), we can change the problem (20)to a differentiable convex optimization problem with twonew inequality constraints (like [10, pp. 150-151]). Then, bycalculating the Lagrangian of the two new constraints and thederivative on the auxiliary variable  t , it is easy to verify that(20) is equivalent to the following problem max P  s ,P  r ,τ  τR 1  + (1 − τ  ) R 2  − µ l P  s − ν  j P  r − c 2[ P  s − Q s ( t )] 2 −  c 2[ P  r − Q r ( t )] 2 s.t.  0  ≤  τ   ≤  1 ,P  s ,P  r ≥  0 ,  (27)with the relationship of   R 1  and  R 2  determined by  τ  ⋆  if   τ  ⋆ = 0 , R 1  ≥  R 2 ; if   τ  ⋆ = 1 , R 1  ≤  R 2 ; if   0  < τ  ⋆ <  1 ,R 1  =  R 2 ; (28)where  τ  ⋆ is the optimal value of   τ   in (27). We note that thismethod is equivalent with the technique presented in [4, Sec.III.A], which has a geometric interpretation. When  c mj  = 0 ,the problem (27) reduces to standard Lagrange problem, andthen the solution is expected to be similar to the water-fillingsolution, but have some difficulty for convergence.Using the equivalent result of (20), which given in (27) and(28), we can deduce the solution of (20) in a case by casebasis. First, when  α 2 ≤  β  2 , we just let  P  s =  P  r = 0 . If  α 2 > β  2 , the solution of (20) is divided into 3 cases: Case 1 : if   τ  ⋆ = 0 , we have  R 1  ≥  R 2 , the KKT conditions of (27) are given by α 2 ln2(1 + 2 α 2 P  s /θ )  ≤  µ l  +  c [ P  s − Q s ( t )] , with equality if   P  s >  0 ,  (29) 0  ≤  ν  j  +  c [ P  r − Q r ( t )] , with equality if   P  r >  0 .  (30)1) If   − ν  j /c  +  Q r ( t )  >  0 , (30) yields  P  r >  0 , thus  P  r = − ν  j /c  +  Q r ( t ) .  Solving (29),  P  s is given by P  s =  f  ( θ,c,µ l ,Q s ( t ) ,α 2 , 1) ,  (31)where f  ( θ,c,µ,Q,h,v )  12   θµ ln2 − θh +   x 2 + y − x  † (32) x  =  µcv  −  Qv  +  θµ ln2  −  θ 2 h,  (33) y  = 2 Qθµv ln2 +  θ 2 hµ ln2  −  θ 2 µ 2 ln2 2 .  (34)It is interesting to look at the case  c  →  0  in (31). Onecan show that   x 2 +  y  −  x  →  0  as  c  →  0 , and (31)returns to the simple water-filling solution.We note that this case happens when γ  2 P  r ≥  ( α 2 − β  2 ) P  s .  (35)2) If   − ν  j /c + Q r ( t )  <  0 , (30) cannot be equal. Thus,  P  r =0 . Since  P  s ≥  0 , relation (35) can be satisfied only when P  s = 0 . This subcase is summarized in case 3.3) If   − ν  j /c + Q r ( t ) = 0 , and suppose  P  r >  0 , (30) shouldbe equal. But it cannot happen for (30). Hence,  P  r hasto be zero. This subcase is also summarized in case 3. Case 2 : if   τ  ⋆ = 1 , we have  R 1  ≤  R 2 , the KKT conditions of (27) are given by β  2 ln2(1 + 2 β  2 P  s /θ  + 2 γ  2 P  r /θ )  ≤  µ l  +  c [ P  s − Q s ( t )] , with equality if   P  s >  0 ,  (36) γ  2 ln2(1 + 2 β  2 P  s /θ  + 2 γ  2 P  r /θ )  ≤  ν  j  +  c [ P  r − Q r ( t )] , with equality if   P  r >  0 .  (37)1) If   { µ l + c [ P  s − Q s ( t )] } /β  2 <  { ν  j  + c [ P  r − Q r ( t )] } /γ  2 ,(37) cannot be equal, thus  P  r = 0 , and  P  s is given by P  s =  f  ( θ,c,µ l ,Q s ( t ) ,β  2 , 1) .  (38)Equation (38) also reduces to the water-filling solutionas  c  →  0 . Note that such a solution always satisfies R 1  ≤  R 2 .2) If   { µ l + c [ P  s − Q s ( t )] } /β  2 >  { ν  j  + c [ P  r − Q r ( t )] } /γ  2 ,(36) cannot be equal, thus  P  s = 0 . Since  P  r ≥  0 , R 1  ≤  R 2  can be satisfied only when  P  r = 0 . Thissubcase is summarized in case 3.3) If   { µ l + c [ P  s − Q s ( t )] } /β  2 =  { ν  j  + c [ P  r − Q r ( t )] } /γ  2 ,there are two possibilities:For the first case, both (36) and (37) achieve equality.After some manipulations, we obtain that β  2 P  s +  γ  2 P  r =  f  ( θ ( β  4 +  γ  4 ) ,c,β  2 µ l  +  γ  2 ν  j ,β  2 Q s ( t ) +  γ  2 Q r ( t ) ,β  4 +  γ  4 , 1) .  (39)We note that as  c  approaches 0, one can obtain β  2 P  s +  γ  2 P  r →  θ 2   β  4 +  γ  4 ln2( β  2 µ l  +  γ  2 ν  j )  − 1  † . Since the condition of this subcase limits to  µ l /β  2 = ν  j /γ  2 , we have P  s +  γ  2 P  r /β  2 →  θ 2   1ln2 µ l −  1 β  2  †  just like [4, Eq. 49]. Therefore, when  c  = 0 , the valuesof   P  s and  P  r are not unique and are hard to recover.Hence, some difficulties arise in the dual iterations (24).Let  a  =  β  2 P  s + γ  2 P  r and substituting it into the condi-tion { µ l + c [ P  s − Q s ( t )] } /β  2 =  { ν  j + c [ P  r − Q r ( t )] } /γ  2 ,we obtain P  s =  − γ  4 µ l  +  β  2 γ  2 ν  j ( β  4 +  γ  4 ) c  +  γ  4 Q s ( t ) − β  2 γ  2 Q r ( t ) +  β  2 aβ  4 +  γ  4 P  r =  − β  4 ν  j  +  β  2 γ  2 µ l ( β  4 +  γ  4 ) c  +  β  4 Q r ( t ) − β  2 γ  2 Q s ( t ) +  γ  2 aβ  4 +  γ  4 This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings. 978-1-4244-4148-8/09/$25.00 ©2009  This case happens when 0  ≤  γ  2 P  r ≤  ( α 2 − β  2 ) P  s .  (40)If none of (36) and (37) achieves equality, we have  P  s = P  r = 0 , which is summarized in case 3. Case 3 : if   0  < τ  ⋆ <  1 , we have  R 1  =  R 2 , the KKT conditionsof (27) are given by τβ  2 ln2(1 + 2 β  2 P  s /θ  + 2 γ  2 P  r /θ ) + (1 − τ  ) α 2 ln2(1 + 2 α 2 P  s /θ ) ≤  µ l  +  c [ P  s − Q s ( t )] ,  with equality if   P  s >  0 ,  (41) τγ  2 ln2(1 + 2 β  2 P  s /θ  + 2 γ  2 P  r /θ )  ≤  ν  j  +  c [ P  r − Q r ( t )] , with equality if   P  r >  0 .  (42)Since  R 1  =  R 2 , we have γ  2 P  r = ( α 2 − β  2 ) P  s .  (43)Two subcases need to be considered:1) If   P  r >  0  and  P  s >  0 , (41) and (42) achieve equality.From (41)-(43), we derive P  s = f   θ,c,µ l + ν  j ( α 2 − β  2 ) /γ  2 ,Q s ( t )+ Q r ( t )( α 2 − β  2 ) /γ  2 ,α 2 , 1 + ( α 2 − β  2 ) 2 /γ  4  P  r = P  s ( α 2 − β  2 )] /γ  2 ,τ   =  α 2 [ ν  j  +  c ( P  r − Q r ( t ))] γ  2 [ µ l + c ( P  s − Q s ( t ))]+( α 2 − β  2 )[ ν  j + c ( P  r − Q r ( t ))] This case happens when  0  < τ <  1 . The powerallocation variable  P  s also reduces to the water-fillingsolution in this case as  c  →  0 .2)  P  r =  P  s = 0 . It happens when all of previous casesare not satisfied. C. Channel resource allocation We now fix the power allocation variables and performchannel resource allocation. By using the dual decompositionmethod as in (17), and auxiliary variables as in (27), the KKTconditions of channel resource allocation are given by C  ( β  2 m P  sm /θ m ) −  β  2 m P  sm /θ m ln2(1 +  β  2 m P  sm /θ m )  ≤  ε with equality if   θ m  >  0 ,  (44) (1 − τ  ) C  (2 α 2 mj P  smj /θ mj ) − (1 − τ  )2 α 2 mj P  smj /θ mj ln2(1 + 2 α 2 mj P  smj /θ mj )+  τC  (2( β  2 mj P  smj  +  γ  2 mj P  rmj ) /θ mj ) − τ  2( β  2 mj P  smj  +  γ  2 mj P  rmj ) /θ mj ln2[1 + 2( β  2 mj P  smj  +  γ  2 mj P  rmj ) /θ mj ]  ≤  2 ε with equality if   θ mj  >  0 .  (45)where  ε  ≥  0  is the dual variable for the channel resourceconstraint. Considering the different cases of   τ  ⋆ given in (28),one can show that (45) is equivalent to C  (2min { α 2 mj P  smj ,β  2 mj P  smj  +  γ  2 mj P  rmj } /θ mj ) − 2min { α 2 mj P  smj ,β  2 mj P  smj  +  γ  2 mj P  rmj } /θ mj ln2[1+2min { α 2 mj P  smj ,β  2 mj P  smj + γ  2 mj P  rmj } /θ mj ]  ≤  2 ε, (46) -1 0 1 2 3 4 5 6-10123456 BSRelay 1Relay 2User 1User 2User 3User 4(0,5)(0,0)(1,2.5)(2.5,1)(3,4)(4,3)(5,0) Fig. 1. The topology of an uplink F/TDMA cellular relay network. the proof of which is omitted for space limitation. The leftside of (44) is zero only when  P  sm  = 0  or  θ m  =  ∞ , the leftside of (46) is zero only when  P  smj  =  P  rmj  = 0  or  θ mj  =  ∞ .Therefore,  ε  should be positive since there is at least one link with positive power allocation.When  P  sm  >  0 , the left side of (44) grows to infinite when θ m  →  0 . Thus,  θ m  = 0  only if   P  sm  = 0 . For (46), we havethat  θ mj  = 0  happens only if   P  smj  =  P  rmj  = 0 . One can set aminimal value for  θ mj  and  θ m  to avoid the problem that theobjective function has no definition for  θ mj  = 0  or  θ m  = 0 .Let  θ 0 ( a,b )  be the root of equation C  ( a/θ ) −  a/θ ln2(1 +  a/θ ) =  b  (47)For the case  θ m ,θ mj  >  0 , the optimal value of   θ m  isgiven by  θ 0 ( β  2 m P  sm ,ε )  and the optimal value of   θ mj  isgiven by  θ 0 (2min { α 2 mj P  smj ,β  2 mj P  smj  +  γ  2 mj P  rmj } , 2 ε ) . Thefamous rapidly convergent Newton’s method [14, Section5.5.3], which requires only several iterations, is used to solve(47) and the dual variable  ε  is obtained by bisection method.By performing power allocation and channel resource allo-cation iteratively, the joint optimal power and channel resourceallocation solution is derived. The proof for this is quite similarwith the one given in [4, pp. 3440]. It is omitted here for spacelimitation.V. N UMERICAL  R ESULTS In this section, we present numerical results to demonstratethe performance of the proposed joint optimal power andchannel resource allocation for F/TDMA DF relay networks.We compare our method with two other schemes:Scheme 1 is F/TDMA cellular network without the assis-tance of relay nodes. The optimal power and channel resourceallocation is utilized. Scheme 2 is F/TDMA relay network withoptimal power allocation and equal channel resource allocationamong the relay links. Our joint power and channel resourceallocation scheme for F/TDMA DF relay networks is denotedas Scheme 3.We consider an uplink F/TDMA cellular relay network with4 users, 2 relay nodes, and 1 base station, whose topology is This full text paper was peer reviewed at the direction of IEEE Communications Society subject matter experts for publication in the IEEE "GLOBECOM" 2009 proceedings. 978-1-4244-4148-8/09/$25.00 ©2009
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