Simultaneous Information and Energy Transfer: A Two-User MISO Interference Channel Case Chao Shen⋆, Wei-Chiang Li† and Tsung-Hui Chang‡ ⋆Institute of Information Science, Beijing Jiaotong University, Beijing, China 100044 †Institute of Commun. Eng., National Tsing Hua University, Hsinchu, Taiwan 30013 ‡Department of Elect. and Computer Eng., University of California, Davis, CA, USA 95616 Abstract—This paper considers the sum rate maximization energy harvesting (EH) and information detection (ID) either 3 problem of a two-user multiple-input single-output interference overthe time domain(i.e.,TDMA) oroverthe powerdomain 1 channel with receivers that can scavenge energy from the radio (i.e.,powersplitting).In[6],therateperformanceachievedby 0 signals transmitted by the transmitters. We first study the 2 optimaltransmissionstrategyforanidealscenariowherethetwo theidealreceiverwhichimplementsEHandIDsimultaneously n receivers can simultaneously decode the information signal and serves as an upper bound of the two practical schemes. a harvest energy. Then, considering the limitations of the current In this paper, we consider the transmission design prob- J circuit technology, we propose two practical schemes based on lem for a two-user multiple-input single-output interference TDMA, where, at each time slot, the receiver either operates in 7 channel (MISO-IFC), assuming energy harvesting receivers. theenergyharvestingmodeorintheinformationdetectionmode. 2 It is interestingto note that, despite that the cross-linksignals Optimaltransmissionstrategiesforthetwopracticalschemesare respectively investigated. Simulation results show that the three are interference which limits the achievable sum rate, they ] T schemesexhibitinterestingtradeoffbetweenachievablesumrate are helpfulin boosting the energyharvesting of the receivers. I and energy harvesting requirement, and do not dominate each We firstconsidertheidealreceiverswhichcansimultaneously . other in terms of maximum achievable sum rate. s perform ID and EH. We formulate the design problem by c IndexTerms—Energyharvesting,Energytransfer,Interference maximizing the sum rate of the two transmitter-receiverpairs [ channel, Transmitter Optimization subject to minimum energy harvesting constraints, i.e., con- 1 I. INTRODUCTION straints on the minimum amount of energy to be harvested. v Recently, energy harvesting has been considered as a The considered problem is, however, intrinsically difficult to 2 promising technique with great potential for prolonging the handle. We present an analysis to characterize the optimal 0 life time of the battery-powered mobile devices or for im- solution structure, showing that transmit beamforming is an 3 6 plementing self-sustained communication systems. Transmis- optimal strategy. . sion designs with energy harvesting constraints imposed on Wefurtherproposetwopracticalschemes.Thefirstscheme, 1 0 the transmitter have been studied in [1]–[3] (and references whichwecallTDMAschemeA,dividesthetransmissiontime 3 therein). Specifically, assuming that the transmitter is able into two time slots. Both receivers perform EH in the first 1 to harvest energy from some external energy sources, the timeslotandsubsequentlyperformIDinthesecondtimeslot. : v work in [1] investigated the optimal power allocation scheme The second scheme, which we call TDMA scheme B, again i for minimizing the transmission completion time in a point- divides the transmission time into two time slots, but in each X to-point single-input and single-output (SISO) channel. In time slot, one receiver performsEH while the other performs r a [2], the throughput maximization problem was studied for a ID.We respectivelypresentefficientoptimizationmethodsfor relay network with energy harvesting transmitters and relays. solving the transmission design problems associated with the Transmission designs for an interference channel (IFC) with two practical schemes. transmitter energy harvesting constraints were also studied in Simulation results are presented to compare the achievable [3]. sum rates of the three proposed schemes. Intriguingly, we Insomeotherworks[4]–[6],ontheotherhand,thereceivers observe that the scheme with ideal receivers may not be were assumed to be able to scavenge energy from the radio idealin termsofsum ratemaximization.Instead,thepractical signals transmitted by the transmitters. The assumption made TDMA schemeA mayoutperformthe idealschemewhenthe thereisthatthereceivercansimultaneouslydetectinformation system is interference limited. Besides, the TDMA scheme B bits and harvest energy from the received signal. Under mayalsoyieldahighersumratethanTDMAschemeAwhen this assumption, the works in [4] and [5] investigated the oneofthereceiversrequiresmuchmoreenergythantheother. optimaltradeoffbetween informationand energytransferin a We should mention that these results are very different from SISO flat-fading channel and in a frequency-selective fading those in [6] where the idealreceiveralways performsthe best channel, respectively. Considering the fact that simultaneous owing to the absence of interference. informationdetectionandenergyharvestingcannotbefulfilled Notations: Tr(X) represents the trace of matrix X. X 0 by current circuit technologies, the work in [6] proposed two means that matrix X is positive semidefinite (PSD). (cid:23)x practical schemes where the receiver separates the modes for denotes the Euclidean norm of vector x. The orthogoknakl Transmitter 1 Transmitter 2 optimal transmission strategies of S and S so that the sum 1 2 rate of the two transmitter-receiver pairs can be maximized while their energy harvesting requirements are satisfied at h h the same time. In the next section, we first study an ‘ideal’ 11 22 situation that the receiver can simultaneously operate in the h h 21 12 IDmodeandEHmode.Inthesubsequentsections,wefurther investigatetwo practicalTDMA schemes where each receiver Receiver 1 Receiver 2 either operates in the ID mode or in the EH mode. Fig. 1: A two-user MISO-IFC system for simultaneous infor- mation and energy transfer. III. OPTIMALTRANSMISSIONSTRATEGYFORIDEAL RECEIVERS projectionontothecolumnspaceofatallmatrixXisdenoted Letus first considerthe ideal situation thatthe receivercan by ΠX , X(XHX)−1XH, and the projection onto the simultaneously decode the information bits and harvest the orthogonalcomplement of the column space of X is denoted energy. Under such assumption, we consider the following by Π⊥X ,I ΠX where I is an identity matrix. problem formulation: − II. SIGNAL MODELANDPROBLEM STATEMENT (P) max r (S ,S )+r (S ,S ) (5a) 1 1 2 2 1 2 S1(cid:23)0,S2(cid:23)0 We consider a two-user MISO-IFC, as shown in Fig. 1, s.t. hHS h +hHS h E , (5b) whereeachtransmitterisequippedwith Nt antennasandeach 11 1 11 21 2 21 ≥ 1 tvrhee1cec,et2coivhre.artnTrnhahenaelssmrvaeeicstcetiteniovdgreldfebryosaimngttrneatanrnlansnamas.tmiLtrteieetctrteexiriv,iei∈artnoiCdirNsehctgiekidivveee∈nnrotbkCey,Ntfhtoerdsiei,gnknoat∈el hTTH2rr2((SSS212h))2≤≤2+PP12h,,H12S1h12 ≥E2, (((555dce))) { } where (5b) and (5c) are the energyharvesting constraints(we y =hHx +hHx +n , k =i, (1) i ii i ki k i 6 have set γ = ∆ = 1 for notational simplicity), and (5d) where n (0,σ2) is the additive Gaussian noise. and (5e) are the individual power constraints. Since, when Differien∼t fCroNm theiconventionalMISO IFC [7], we assume E1 = E2 = 0, problem (P) reduces to the well-known sum in the paper that the receivers can either extract information rate maximization problem in interference channels [7], (P) or harvest energy from the received signal y [4]–[6], which is difficult to solve in general . However, an explicit solution i we refer to as the information detection (ID) mode and the structurefor S1 andS2 in (P) can be obtained,as we show in energy harvesting (EH) mode, respectively. Assume that x the following proposition: i contains the information intended for receiver i which is Gdi.eeac.u,osxdseiiasn∼xeiCnNcboy(d0es,dinSgwil)ie.thMuzsoeerrreoodvmeeteer,acntaisoasnnudminecotthvheaartiIaDenaccemhmordeaect.reiixTvehSreini, PdaenrndoopTtoers(ittShieo⋆2n)o=p1tiAmPs2as.luMmsoeolurtethiooavnter(p,Pat)hireisroeffee(axPsis)i.btslTeah,eapnnadiTrrlo(eStf⋆1(()SS⋆1⋆1=,,SSP⋆2⋆21)) satisfying the achievable information rates of the two receivers are respectively given by S⋆ =(a h +b h )(a h +b h )H, (6a) 1 1 11 1 12 1 11 1 12 r (S ,S )=log 1+ hH11S1h11 , (2) S⋆2 =(a2h21+b2h22)(a2h21+b2h22)H, (6b) 1 1 2 hHS h +σ2 (cid:18) 21hH2S21h 1(cid:19) for some ai,bi ∈C, i=1,2. r (S ,S )=log 1+ 22 2 22 . (3) 2 1 2 hHS h +σ2 Proposition 1 implies that beamforming is an optimal (cid:18) 12 1 12 2(cid:19) transmission strategy of (P), and that the beamforming di- Alternatively, the receiver i may choose to harvest energy rection of transmitter i should lie in the range space of from yi, i.e., operating in the EH mode. In particular, it can [h ,h ], for i = 1,2. By (6), the search of S and S i1 i2 1 2 be assumed that the total harvested RF-band energy during a in (P) reduces to the search of a and b , over the ellipsoid i i symboltransmissioninterval∆isproportionaltothepowerof a 2 h 2 + b 2 h 2 = P , for all i = 1,2. We should i i1 i i2 i thereceivedbasebandsignal,e.g.,forreceiver i,theharvested | | k k | | k k mention here that the optimal beamformingsolution structure energy, denoted by , can be expressed as Ei in (6) is reminiscent of that in the traditional MISO IFC =γ∆(hHS h +hHS h ), i=1,2, (4) without energy harvesting constraints [7], Ei 1i 1 1i 2i 2 2i whereγisaconstantaccountingfortheenergyconversionloss A. Proof of Proposition 1 in the transducer [6]. It should be noted that current practical Without loss of generality, we assume that h ∦ h and 11 12 circuitsdonotallowthereceivertosimultaneouslydecodethe h ∦ h . We prove by contradiction that Tr(S⋆) = P for 21 22 i i information bits and harvest the energy [6]. i = 1,2. Suppose that Tr(S⋆) < P , then there exists some 1 1 Suppose that receiver i desires to harvest a total amountof ǫ>0 and energy E for i = 1,2. Our interest lies in investigating the S′ =S⋆+ǫΠ⊥ h hHΠ⊥ i 1 1 h12 11 11 h12 such that Tr(S′) = P . Note that (S′,S⋆) is feasible to (P). modes,(P)maynotbetheoptimaldesignformulationinterms 1 1 1 2 Moreover, since h ∦h , we have r (S′,S⋆)>r (S⋆,S⋆) of sum rate maximization. The reason is that the cross-link 11 12 1 1 2 1 1 2 andr (S′,S⋆)=r (S⋆,S⋆),whichcontradictstheoptimality signalpowerhHS h ,thoughboostingtheenergyharvesting 2 1 2 2 1 2 ik i ik of (S⋆,S⋆). Hence, it must be that Tr(S⋆) = P ; similarly, of receiver k, also degrades the achievable information rate. 1 2 1 1 one can show that Tr(S⋆)=P . Therefore, when the cross-link interference is strong (e.g., 2 2 Next, we show that S⋆ and S⋆ lie in the range space of interference dominated scenario), it might be better to split 1 2 H , [h h ] and H , [h h ], respectively, i.e., the ID and EH modes, which is also preferred by the current 1 11 12 2 21 22 Π⊥ S⋆Π⊥ =0fori=1,2.Onecansee that,foranyS 0, practical circuits. Hi i Hi (cid:23) hHik(ΠHiSΠHHi)hik =hHikShik, (7) IV. PRACTICAL SCHEMES ANDTHEIROPTIMAL Tr(ΠHiSΠHHi)≤Tr(S), (8) Inthe section,TbRaAseNdSMonIStShIeOiNdeSaToRfATTDEGMIEAS,we presenttwo for i,k 1,2 , where the equality in (7) holds be- practical schemes where each of the receivers operates either cause ΠX∈X{= X}for all X ∈ Cm×n. Therefore, (S⋆1,S⋆2) in the ID mode or in the EH mode. The associated optimal is an optimal solution to problem (P) if and only if transmission strategies of S and S are also investigated. 1 2 (sΠupHp1oSse⋆1ΠthHa1t,SΠ⋆Hd2oSe⋆2sΠnHo2t)liiesionptthimearlantogepsrpoabcleemof(PH).,Nio.ew., A. TDMA scheme A 1 1 Tr(Π⊥ S⋆Π⊥ )>0. Then, In the first practicalscheme, which we call TDMA scheme H1 1 H1 A,thetransmissionintervalisdividedintotwotimeslots–one Tr(ΠH1S⋆1ΠHH1)=Tr(S⋆1)−Tr(Π⊥H1S⋆1Π⊥H1)<Tr(S⋆1)≤P1, dedicated for the EH mode and the other for the ID mode. liwisehniiocnhttoihmpetpirmlaineaslgettohsa(ptPaΠ)c.eHAo1nSfa⋆1HloΠgHo.u1silsy,noonteopcatinmsahl,oawndthtahteSre⋆2bmyuSs⋆1t (AS1u−pispαodse)esfcrtrahicbatetidoαnasoffrfaotchltleioowtnims:oefisthfeortitmimeeissloftor2.tiTmDeMsAlotsc1heamnde 2 Whatremainstoprove(6)istoshowthatthereexistsapair • Timeslot1(EHmode):Boththetworeceiversoperatein of (S⋆,S⋆) that are of rank one. It is not difficult to see that the EH mode. The goal is to guarantee the two receivers 1 2 (P) is equivalent to the following problems toachievetheirrespectiveenergyharvestingrequirements E and E in α fraction of the time, i.e., hHS h 1 2 Smsi.(cid:23)atx0. hloHikgS(cid:18)i1hi+k+Γ⋆kiΓii⋆k+kiσ≥ii2iE(cid:19)k, ((99ba)) αα··((hhH1H212SS12hh1212++hhH2H112SS21hh2112))≥≥EE12,. ((1111ba)) Γ⋆ki+hHiiSihii ≥Ei, (9c) • Timeslot2 (IDmode):Both thetwo receiversoperatein hHS h Γ⋆ , (9d) the ID mode. It is aimed to maximize the sum rate: ik i ik ≤ ik Tr(Si)≤Pi, (9e) S1(cid:23)m0,aSx2(cid:23)0 (1−α)(r1(S1,S2)+r2(S1,S2)) (12a) where Γ⋆ki=hHkiS⋆khki, i,k∈{1,2}and i6=k. Let usfocuson s.t. Tr(S1)≤P1, Tr(S2)≤P2. (12b) the case of i=1, k =2, and rewrite (9) as Problem(12)istheclassicalsumratemaximizationproblem max hHS h (10a) in MISO IFC, and there exist several efficient algorithms for S1(cid:23)0 11 1 11 handling (12); see, e.g., [7], [9]. s.t. hH12S1h12 ≥E2−Γ⋆22, (10b) Sincetimeslot1isonlyforenergyharvestinganddoesnot hHS h Γ⋆ , (10c) contribute to the information rate, it is desirable to spend as 12 1 12 ≤ 12 least as possible time for the EH mode, i.e., to minimize the hHS h E Γ⋆ , (10d) 11 1 11 ≥ 1− 21 time fraction α. Mathematically, this can be expressed as the Tr(S1) P1. (10e) following design problem ≤ toSounpepoeqseuathliatytΓc⋆1o2n=straEin2t−. ΓIn⋆22t.hTathecnas(e1,0(b1)0a)ndha(s10ocn)lymethrgreees β∈R,Sm1(cid:23)a0x,S2(cid:23)0 β (13a) inequalityconstraints.Accordingto[8,Theorem3.2],problem s.t. hH11S1h11+hH21S2h21 ≥βE1, (13b) c(O1on0n)stthtrheaeinnosthh(ea1rs0ahbna)naodnp,dtiim(f1aΓ0lc⋆12s)om>luutsiEot2nbe−Si⋆1nΓas⋆2uc2tc,ihvtehtehfnoatrorSna⋆en.koT(fhSte⋆1hr)eef≤otwr1eo., ThrH1(2SS11)h≤12P+1h, H2T2Sr(2Sh22)2≤≥Pβ2E,2, ((1133dc)) 1 theeffectivenumberofinequalitiesin(10)isagainthree.Thus where β , 1/α. Note that if the optimal β of (13) is less rank(S⋆) 1 by [8, Theorem 3.2]. Analogously,for the case than one (i.e., optimal α > 1), then it implies that the 1 ≤ ofi=2,k =1,onecanshowthatproblem(9)hasanoptimal energy harvesting requirements (11) cannot be fulfilled even S⋆ with rank(S⋆) 1. The proof is thus complete. (cid:4) the receivers operate in the EH mode for the whole symbol 2 2 ≤ Itisimportanttoremarkthat,while(P)isidealinthesense transmission period. In that case, we declare that TDMA thatthe receivercansimultaneouslyoperatein the ID andEH scheme A is not feasible. Whileproblem(13)isalinearprogram,whichcanbesolved by the bisection or golden search methods. Due to space by the off-the-shelf solvers, we show next that the optimal limitations, we omit the proof here; it will be presented in solution of (13) can actually be obtained via solving a simple our future publication. Note that Proposition 2 also implies one-dimensionalproblem. that beamforming is optimal to the TDMA scheme A. Proposition 2 Denote (β⋆,S⋆,S⋆) as the optimalsolution of B. TDMA scheme B 1 2 (13). Let Different from TDMA scheme A, in each time slot of w⋆ =arg max min β (w),β (w) , (14) TDMA scheme B, one receiver operates in the ID mode and 1 2 0≤w≤1 { } the other receiver operates in the EH mode. Specifically, where • Time slot 1: Receiver 1 operates in the ID mode and β (w)=(P hHv (w)2+P hHv (w)2)/E , (15a) receiver 2 operates in the EH mode. The objective is to 1 1| 11 1 | 2| 21 2 | 1 maximize the information rate of receiver 1 while guar- β (w)=(P hHv (w)2+P hHv (w)2)/E , (15b) 2 1| 12 1 | 2| 22 2 | 2 anteeingthe energyharvestingrequirementofreceiver2. and v (w) CNt and v (w) CNt are the principal eigen- The design problem is given by 1 2 avencdtowrshofhth∈He/twEom+a(t1ricesww)hh∈11hhHH11//EE1,+re(s1p−ecwtiv)hel1y2.hTH1h2e/nE2 max αlog 1+ hH11S1h11 (21a) 21 21 1 − 22 22 2 S1(cid:23)0,S2(cid:23)0 (cid:18) hH21S2h21+σ12(cid:19) S⋆1 =P1v1(w⋆)v1H(w⋆), (16a) s.t. hH12S1h12+hH22S2h22 ≥E2/α, (21b) S⋆2 =P2v2(w⋆)v2H(w⋆), (16b) Tr(S1) P1, Tr(S2) P2, (21c) ≤ ≤ and β⋆ =min β (w⋆),β (w⋆) . { 1 2 } • Time slot 2: The operation modes of the two receivers are exchanged: Proof: Let us denote β1 =(hH11S1h11+hH21S2h21)/E1, (17a) max (1 α)log 1+ hH22S2h22 (22a) β2 =(hH12S1h12+hH22S2h22)/E2. (17b) S1(cid:23)0,S2(cid:23)0 − (cid:18) hH12S1h12+σ22(cid:19) s.t. hHS h +hHS h E /(1 α), (22b) Then (13) can be written as 11 1 11 21 2 21≥ 1 − Tr(S ) P , Tr(S ) P . (22c) 1 1 2 2 ≤ ≤ max min β ,β (18a) 1 2 S1(cid:23)0,S2(cid:23)0 { } First of all, it is easy to see that s.t. constraints in (17a), (17b), (18b) Tr(S1) P1, Tr(S2) P2. (18c) Lemma 1 The TDMA scheme B is feasible if and only if ≤ ≤ Define the following set E E 1 + 2 1. (23) P h 2+P h 2 P h 2+P h 2 ≤ , (β ,β ) (17a), (17b), . (19) 1k 11k 2k 21k 1k 12k 2k 22k P ( 1 2 (cid:12) Si (cid:23)0,Tr(Si)≤Pi,i=1,2, ) Proof:TDMAschemeBisfeasibleif andonlyifboth(21) (cid:12) and (22) are feasible. Problem (21) is feasible if and only if Itcanbeverified(cid:12)that isconvex,and,moreover,theoptimal (cid:12) P there exists some α [0,1] such that tuple (β ,β ) of (18) must lie on the Pareto boundary of . ∈ 1 2 P Therefore, there must exist a value 0 w⋆ 1 such that max hHS h +hHS h (S⋆1,S⋆2) of (18) is also optimal to the f≤ollowin≤g problem E2 ≤α· Tr(S1S)≤1(cid:23)P01,,TSr2((cid:23)S02)≤P2 12 1 12 22 2 22! max w⋆β1+(1 w⋆)β2 (20a) =α (P h 2+P h 2), (24) S1(cid:23)0,S2(cid:23)0 − · 1k 12k 2k 22k s.t. constraints in (17a), (17b), (20b) where the equality is obtained by [6, Proposition 2.1]. Simi- Tr(S1) P1, Tr(S2) P2. (20c) larly, one can show that (22) is feasible if and only if ≤ ≤ By (20) and by [6, Proposition 2.1], (S⋆1,S⋆2) is thus given E1 (1 α) (P1 h11 2+P2 h21 2). (25) by the forms in (16). By substituting (16) into (17), defining ≤ − · k k k k β (w⋆) and β (w⋆) as in (15), and by (18), we obtain the Combining(24)and (25) givesrise to (23). Conversely,given 1 2 optimal objective value of (13) as (23), one can show that there exists α [0,1] such that (24) and (25) hold with equalities. ∈ (cid:4) β⋆ =min β (w⋆),β (w⋆) . { 1 2 } Problem (21) and problem (22) are quasi-convexproblems. Since for any 0 w 1, the corresponding solution of While quasi-convex problems can be solved by the bisection (20) is feasible to (1≤8), th≤e optimal w⋆ is given by (14). (cid:4) technique,weshowthat(21)andproblem(22)canactuallybe Onecanfurthershowthatthefunction min β (w),β (w) recast as convex problems, by applying the Charnes-Cooper 1 2 { } in (14) is unimodal, and thus (14) can be efficiently solved transformation [10]. To illustrate this, let us take (21) as an example.Considerthefollowingconvexsemidefiniteprogram sets of channel realizations: (SDP) Channel realization 1: X1(cid:23)0,mXa2(cid:23)x0,y≥0 αlog 1+hH11X1h11 (26a) h11 = 00.0.4690482−00.1.1829162jj ,h12 = 00.7.0330669−00.6.1469762jj , s.t. hHX(cid:0)h +yσ2 =1(cid:1), (26b) (cid:20)− − (cid:21) (cid:20)− − (cid:21) 21 2 21 1 0.4320 0.3112j 0.5634+0.2935j hH12X1h12+hH22X2h22 ≥yE2/α, (26c) h21 =(cid:20)−−0.4142−−0.0515j(cid:21),h22 =(cid:20)−0.0672−0.2515j(cid:21), Tr(X ) yP , Tr(X ) yP , (26d) 1 ≤ 1 2 ≤ 2 wherej=√ 1.Thenormsofthechannelvectorsare h = 11 Notethattheoptimaly⋆of(26)mustbepositive;otherwisewe 0.5464, h − = 0.9925, h = 0.6765, h =k0.68k65, 12 21 22 have X⋆ =X⋆ =0 which violates (26b). Moreover,consider respectivkely. Tkhe noise varkiancekσ2 is set tok0.1.k 1 2 the following correspondence: Channel realization2: The direct-link channelsh and h 11 22 arethesameasthoseforChannelrealization1,andthecross- y =1/(hHS h +σ2)>0, (27a) 21 2 21 1 link channels are given by X =yS , X =yS . (27b) 1 1 2 2 0.8948 0.7956j 0.5291 0.3811j Then,onecanshowthat(S1,S2)isfeasibleto(21)ifandonly h12 = 0.0452− 0.2047j , h21 = −0.5073−0.0630j , if (X ,X ,y) is feasible to (26). Furthermore, the objective (cid:20)− − (cid:21) (cid:20)− − (cid:21) 1 2 valueachievedby(S1,S2)in(21)isthesameastheobjective whose norms are kh12k = 1.2156 and kh21k = 0.8286. The value achieved by (X ,X ,y) in (26). Therefore, the two noisevarianceσ2issetto0.001.Onecanseethat,forChannel 1 2 problems (21) and (26) are equivalent, and one actually can realization 2, interference will be the major factor that limits obtain(S⋆,S⋆)of(21)bysolvingtheconvexproblem(26).In the sum rate. 1 2 addition,byapplying[8,Theorem3.2],onecanshowthat(26) In the simulations, both problem (P) and problem (12) for hasrank-oneoptimal(X⋆,X⋆),implyingthatbeamformingis TDMA scheme A are solved by an exhaustive search method 1 2 also optimal to the TDMA scheme B. similar to that in [9]. The optimal time fraction α of TDMA It is also possible to obtain a closed-form solution to scheme A is solved by Proposition 2. For TDMA scheme problem (21), if the channel condition favors receiver 2 to B, the associated optimal time fraction α is obtained via harvest the energy: exhaustive search over [0,1]. The SDP problem (26) for both time slot 1 and slot 2 are solved by CVX [11]. Lemma 2 Consider problem (21) and assume that In Figure 2, we present the simulation results of sum rate versus (E ,E ) of the three transmission schemes under 1 2 P1|h12hˆ11|2+P2|hH22hˆ21⊥|2 ≥E2/α, (28) Channel realization 1. Figure 2(a) displays the comparison results between problem (P) and TDMA scheme A; while where hˆ , h / h and hˆ , Π⊥ h / Π⊥ h . 11 11 k 11k 21⊥ h21 22 k h21 22k Figure 2(b) shows the comparison results between TDMA Then(S⋆1,S⋆2)=(P1hˆ11hˆH11,P2hˆ21⊥hˆH21⊥)isoptimalto(21). schemeAandTDMAschemeB.WecanobservefromFigure 2(a)thatproblem(P),whichideallyassumesthatthereceivers Proof: Consider the following optimization problem can simultaneously decode the information bits and harvests hHS h the energy, exhibits a higher sum rate then TDMA scheme A max log 1+ 11 1 11 (29a) S1(cid:23)0,S2(cid:23)0 (cid:18) hH21S2h21+σ12(cid:19) forall valuesof (E1,E2). Note that, when (E1,E2)=(0,0), s.t. Tr(S ) P , Tr(S ) P , (29b) the two schemes coincides, thus they have the same sum rate 1 1 2 2 ≤ ≤ at that point. From Figure 2(b), we see that the two practical which is obtained by removing (21b) from (21). Since the schemes, TDMA scheme A and TDMA scheme B, do not objective function is strictly increasing w. r. t. hHS h dominate each other in terms of sum rate, though TDMA 11 1 11 and strictly decreasing w. r. t. hHS h , it can be easily scheme A has a higher sum rate for most of the values 21 2 21 seen that (S⋆1,S⋆2) = (P1hˆ11hˆH11,P2hˆ21⊥hˆH21⊥) is optimal of (E1,E2). It is interesting to see that TDMA scheme B to (29). Substituting this (S⋆,S⋆) into (21b), we see that performs better when either of the two receivers requests a 1 2 (S⋆,S⋆) is also feasible to (21) owing to the premise of (28). higher energy than the other. 1 2 Consequently, (S⋆,S⋆) is also optimal to (21). (cid:4) Figure 3 presents the simulation results under Channel 1 2 realization2.ComparingFigure3(a)withFigure2(a),onecan V. SIMULATIONRESULTS observe that, in the interference-dominated scenario, TDMA In this section, we present some simulation results to scheme A may even yield a higher sum rate than problem comparethethreetransmissionschemes,namely,problem(P), (P). This implies that the ‘ideal’ formulation (P) may not be TDMAschemeAandTDMAschemeB.Weassumethateach ‘ideal’ in terms of sum rate maximization when interference transmitterhastwoantennas(N =2),andsetP =P =1and dominates the system performance. Similarly, we see from t 1 2 σ2 , σ2 = σ2. The channel vectors are randomly generated Figure 3(b) that TDMA scheme B can outperform TDMA 1 2 followingcomplexGaussiandistribution.Specifically,we will scheme A when E >>E or E >>E ; otherwise TDMA 1 2 2 1 present simulation comparison results of the following two scheme A exhibits a higher sum rate. use)3.5 TDMA scheme A (P) scheme el 3 e)3.5 nn TDMA scheme B ate (bits/per channel us12..12355 TDMA scheme A Sum rate (bits/per cha012...0555012 m r0.5 1.5 0.2 Su 00 1 E0.4 1.5 0.2 E10.4 0.6 0 0.5 E2 1 0.6 0 0.5 E2 1 (a) Problem(P)versusTDMAschemeA (b) TDMAschemeAversusTDMAschemeB Fig. 2: Sum rate versus (E ,E ) under Channel realization 1. 1 2 hannel use)15 TDMA scheme A (P) scheme nnel use)15 TDMA scheme A TDMA scheme B ate (bits/per c1005 e (bits/per cha105 um r 0 m rat 0 S 0.2 u 0 S 0.4 0.2 0.6 2 0.4 E1 0.8 1 0 0.5 1E2 1.5 E01.6 0.8 1 0 0.5 1E2 1.5 2 (a) Problem(P)versusTDMAschemeA (b) TDMAschemeAversusTDMAschemeB Fig. 3: Sum rate versus (E ,E ) under Channel realization 2. 1 2 In summary, we have investigated the transmission design [3] K. 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