Artificial Noise Aided Secrecy Information and Power Transfer in OFDMA Systems

0.68 MB·

by  Meng Zhang

1
Artificial Noise Aided Secrecy Information and
Power Transfer in OFDMA Systems
Meng Zhang, Student Member, IEEE, Yuan Liu, Member, IEEE, and Rui Zhang, Senior Member, IEEE
Abstract—In this paper, we study simultaneous wireless in-
fdoirvmisiaotnionmuanltdippleowacecrestrsa(nOsfFeDrM(SAW)IsPyTst)eimnsowrtihthogtohnealcofreexqisuteennccye Data Energy
6 of information receivers (IRs) and energy receivers (ERs). The
1 IRs are served with best-effort secrecy data and the ERs
0 harvest energy with minimum required harvested power. To
2 enhancethephysicallayersecurityforIRsandyetsatisfyenergy
harvesting requirements for ERs, we propose a new frequency-
n domain artificial noise (AN) aided transmission strategy. With
a the new strategy, we study the optimal resource allocation for
J the weighted sum secrecy rate maximization for IRs by power Base Station Information Receiver Energy Receiver

7 andsubcarrierallocationatthetransmitter.Thestudiedproblem
is shown to be a mixed integer programming problem and thus
Fig. 1. System model of OFDMA-based SWIPT,where each receiver is a
] non-convex, while we propose an efficient algorithm for solving
T potential eavesdropper tootherreceivers.
it based on the Lagrange duality method. To further reduce
I the computational complexity, we also propose a suboptimal
.
s algorithm of lower complexity. The simulation results illustrate
c the effectiveness of proposed algorithms as compared against On the other hand, due to the increasing importance of in-
[ formationsecurity,substantial researcheffortshave beended-
other heuristic schemes.
icated to information-theoretic physical layer security [10]–
1 Index Terms—Physical layer security, simultaneous wireless
v information and power transfer (SWIPT), artificial noise (AN), [17],asacomplementarysolutiontothetraditionalcryptogra-
5 orthogonal frequency-division multiple access (OFDMA), re- phybasedencryptionappliedin the upperlayers. The authors
3 source allocation. in [11] considered physical layer security in an OFDMA
4
system, with the goal of maximizing the sum rate of best-
1
I. INTRODUCTION effort information users subject to the individual secrecy rate
0
. ORTHOGONAL frequency division multiple access requirementsofsecureusers.In[13],theorthogonalfrequency
1
0 (OFDMA) has many advantages such as flexibility in division multiplexing (OFDM) based wiretap channel was
6 resource allocation and robustness against multipath channel considered and the achievable secrecy rate with Gaussian
1 fading, and thereforehas become a well established multiple- inputs was studied. Artificial noise (AN) is a well-known
v: access technique for multiuser wireless communications sys- approach for improving physical layer security by degrading
eavesdroppers’ channel condition [14], [15]. In [14], in order
i tems.
X Recently, simultaneous wireless information and power to assist secrecy information transmission, AN is transmitted
r transfer (SWIPT) provides an appealing solution to prolong into the null space of the channels of legitimate users to
a
the operation time of energy-limited wireless nodes [2]–[9]. interferewiththeeavesdroppers.In[15],theauthorsproposed
SWIPTsystemsenabletheuserstoharvestenergyanddecode a time-domain AN design by exploiting temporal degrees of
information from the same received signal, thus making most freedom from the cyclic prefix in OFDM modulated signals,
efficient use of the wireless spectrum for both information even with a single antenna at the transmitter. In [17], the
and energy transfer. SWIPT has drawn a great amount of authors studied robust transmission schemes for the multiple-
research interests. For instance, two practical schemes for input single-output (MISO) wiretap channels.
SWIPT,namelypowersplitting(PS) andtimeswitching(TS), Ahandfulofworkshavebeeninvestigatedthesecureissues
wereproposedin[2]and[3].WithTSappliedateachreceiver, inSWIPTsystems[18]–[21].Sincetheenergyreceivers(ERs)
the received signal is either processed for energy harvesting need to be deployed much closer to the access points than
or for informationdecoding.When PS is used at the receiver, the information receivers (IRs) due to their much higher
the signal is split into two streams, for information decoding received power requirement [9], they are inevitably capable
and energy harvesting, respectively. The authors in [2] and of eavesdropping the messages to the IRs. Moreover, AN
[3] also investigatedthe achievablerate-energytradeoffsfor a also plays a role of energy signal, i.e., besides interfering
multiple-input multiple-output (MIMO) SWIPT system and a with the eavesdroppers to facilitate secure communication,
single-inputsingle-output(SISO)SWIPTsystem,respectively. AN is a new source for wireless power transfer as well. In
SWIPT systems in fading channels were studied by dynamic [18], [19], the authors studied the secrecy communication
time switching (DTS) and dynamic power splitting (DPS) in in SWIPT by properly designing the beamforming vectors
[4] and [5], respectively. at the multi-antenna transmitter. Secrecy communication in
2
Secrecy Key
OFDMA Transmitter Information Receiver k for IR k
Information P\S & S\P & Artificial Noise
IFFT FFT
signal source CP Incertion CP Removal Removal
(cid:894)(cid:18)(cid:14)(cid:127)k,n(cid:895)(cid:393)k,n (cid:448)k
Artificial Recevier
Noise signal
Generator processor
(cid:127) p
k,n k,n
Fig.2. Blockdiagram ofanOFDMAtransmitter andreceiver withANgeneration andremoval procedure.
SWIPToverfadingchannelswasalsostudiedin[20].In[21], algorithms. Finally, we conclude the paper in Section VI.
the authors studied the secure OFDMA-based systems with
a power splitter applied at each user terminal to coordinate II. SYSTEMMODELAND PROBLEMFORMULATION
the secure transmission and energy harvesting. However, AN
We consider a downlink OFDMA-based SWIPT system
aidedOFDMA-basedSWIPTsystemswithsecrecyconstraints
with secrecy constraints as shown in Fig. 1. The system
have not yet been investigated in the literature. In a secure
consists of one base station (BS) with a single antenna, K
OFDMA system without AN, only the user with the largest
single-antenna receivers and N SCs. The set of receivers is
channel gain over each subcarrier (SC) can receive secure
denoted by = 1,...,K , among which K receivers are
1
information [11]. Thus, the new method of using AN not K { }
IRs given by the set and the rest K receivers are ERs
1 2
only achieves the secrecy information and wireless power K
givenbytheset ,i.e., = .Notethatthereceivers
2 1 2
transfer at the same time, but also leads to new resource K K ∪K K
(both IRs and ERs) are considered to be separated and can
allocation solutions different from the conventional secure
only decode information or harvest energy at a time, unlike
OFDMA system without AN.
the co-located receiversconsideredin [6], [7]. The set of SCs
Motivated by the aforementioned reasons, in this paper,
is denoted as = 1,...,N . Furthermore, we assume that
we study the optimal resource allocation in the AN aided N { }
for each IR, all other receivers (IRs and ERs) are potential
secure OFDMA systems with SWIPT as shown in Fig. 1,
eavesdroppers,similar to the case considered in [11]. The BS
where two types of receivers are assumed, i.e., IRs and ERs.
is assumed to knowthe channelstate information(CSI) of all
Our goal is to maximize the weighted sum secrecy rate of
receivers.This ispracticallyvalid since the IRs andERs need
the IRs subject to minimum harvested power requirements
to help the BS in obtaining their individual CSI for receiving
of individual ERs. We propose a new frequency-domain AN
requiredinformationandenergy,respectively.We assumethat
method in OFDMA-based SWIPT to facilitate both secrecy
theOFDM symbolsaretime slottedso thatthelengthofeach
information transmission and energy transfer to IRs and ERs,
time slot is comparable to the channel coherence time, i.e.,
respectively. Specifically, as shown in Fig. 2, independent
the channel impulse response can be treated as time invariant
AN is added over each SC at the transmitter and only the
during each time slot. As a result, the BS can accurately
desired IR is able to cancel it using the corresponding key
estimate CSI of all receivers on all SCs.
before decoding the information1. The formulated problem is
Weproposeafrequency-domainANgenerationandremoval
a mixed integer programming problem and thus non-convex.
method for OFDMA-based SWIPT, similar to that in [20]
We propose an efficient algorithm based on the Lagrange
over the time domain. The scheme is illustrated in Fig. 2 and
duality method, which solves the problem asymptotically
described as follows. A large ensemble of sequences used to
optimally when the number of SCs becomeslarge. Moreover,
generate Gaussian distributed AN are pre-stored at the BS2,
asuboptimalalgorithmisalsoproposedtotradeoffcomplexity
and the indices of the sequences are regarded as the keys.
and performance.
After SC allocation to IRs, the BS first randomly picks N
The remainder of this paper is organized as follows. In
sequences (each corresponds to one SC) from the ensemble
Section II, we introduce the system model of the OFDMA-
and transmits each of their indices (keys) to the IR to whom
based SWIPT with secrecy constraints, and present the prob-
thecorrespondingSCisassigned.Astherandomsequence(or
lem formulation. The problem is solved by the Lagrange
AN) is only known to the intended IR but unknownto all the
duality method in Section III. In Section IV, we propose the
otherreceivers,anypotentialeavesdroppercannothaveaccess
suboptimal algorithm of lower complexity. In Section V, we
totherandomsequenceusedateachSC.Moreover,inorderto
providethe numericalresults on the performanceof proposed
preventthe eavesdropperfromdecodingthe randomsequence
1Note that the key-assisted approach is normally exclusively used for bylong-termobservationofthesignal,theBSrandomlypicks
cryptography,whilephysical-layermethodsaretraditionallyadoptedwhenthe new random sequences and transmits the corresponding keys
shared keys are not available. However, some recent works (e.g. [22], [23]) in a secretmannerto thedesired IRsfromtime to time, using
have considered applying physical-layer security to enhance cryptographic
e.g. the method proposed in [24] by exploiting the channel
secrecy, showing the potentials to benefit from both types of secrecy ap-
proaches.Hence,undersuchmulti-layersecurityframework,itisalsopossible
to jointly consider the key-assisted physical-layer security and cryptography 2Notethatintheliterature, theANisusuallyassumedtofollowGaussian
design,whichisleftforourfuturework. distribution (e.g.[14]–[16]).
3
independenceandreciprocity.Specifically,theIRsendsapilot The achievable secrecy rate for IR k on SC n is thus given
signal to the BS, and then the BS sends a random key and by [25]
modulates it over the phase of the transmitted signal with
Rs =[r re ]+
the received channel phase pre-compensated. In this way, the k,n k,n− k,n
intendedIRisabletodecodethekeywhilethechannelphases (1 αk,n)hk,n 2pk,n
= log 1+ − | |
between the BS and other receivers are different from that 2 σ2
(cid:20) (cid:18) (cid:19)
between the BS and the intended IR. Thus, the key can be (1 α )β 2p +
confidentiallytransmittedwithoutbeingeavesdroppedbyother −log2 1+ α − βk,n 2|pk,n|+kσ,2n , (8)
receivers. (cid:18) k,n| k,n| k,n (cid:19)(cid:21)
The transmit signal comprises the transmitted data symbol for all k 1 and n , where []+ ,max(0, ).
∈K ∈N · ·
sk,n from the BS to IR k on SC n and the AN bearing signal Lemma 1. Rs in (8) can be further expressed as
z forIR k, k and n . Itis assumed thats and k,n
k,n 1 k,n
∈K ∈N
zk,n are independent circularly symmetric complex Gaussian Rs = 0, if 0≤pk,n ≤[Xk,n(αk,n)]+,
(dCenSoCtGed)braynsdom variab(l0es,1w)iathndzezro mean an(0d,u1n),itwvhairciahnacree, k,n ( rk,n−rke,n ≥0, if pk,n >[Xk,n(αk,n)]+,
k,n k,n (9)
∼CN ∼CN
also independent over n.
where
The transmitted signal to IR k at SC n is given by
Xk,n = (1−αk,n)pk,nsk,n+√αk,npk,nzk,n, (1) k,n(αk,n), ασk2,n |hk1,n|2 − |βk1,n|2 if αk,n 6=0 ,
q X (sgn (cid:16)βk,n 2 hk,n 2 (cid:17) if αk,n =0
where pk,n 0 is the total power at SC n and 0 αk,n 1 | | −| | ×∞ (10)
is the transm≥it power splitting ratio at the BS-side≤to gene≤rate and sgn(x)= x/x(cid:0)if x=0 and sgn((cid:1)x)=1 if x=0.
AN to be added at SC n. | | 6
Proof: Please refer to Appendix A.
Let h denote the complex channel coefficient from the
k,n
BS to receiver k at SC n, and βk,n denote the eavesdrop- Remark 1. Note that the traditional AN scheme (without AN
per’s complex channel coefficient. Here, we let βk,n 2 = cancelation, e.g. [14], [17]) is ineffective for the considered
| |
maxk′∈K,k′6=k hk′,n 2, indicating that the considered eaves- SISO systems, i.e., without cancelling AN in the intended IRs,
| |
dropperofreceiverkisthereceiverofthelargestchannelgain AN cannot achieve a higher secrecy rate compared to the
amongalltheotherreceiversonSCn.Thedownlinkreceived transmissionwithoutAN.ThedetailscanbefoundinAppendix
signal at IR k on SC n and that at a potential eavesdropper B.
who is wiretapping IR k over SC n are respectively given by
The weighted sum (secrecy) rate of all K IRs is given by
1
Yk,n =hk,nXk,n+vk, (2) Rs = w x Rs , (11)
sum k k,n k,n
E =β X +e , (3)
k,n k,n k,n k kX∈K1 nX∈N
wherew isthepositiveweightofIRk andx isthebinary
wherethenoise v ande areassumedto beindependentand k k,n
k k
identically distributed (i.i.d.) as (0,σ2). SC allocation variable with xk,n = 1 representing SC n is
CN allocated to IR k and x = 0 otherwise. Note that in the
With the aforementioned scheme, the AN can be canceled k,n
considered system, the ERs can harvest energy from all SCs
at the desired IR at each SC but not possibly at any of the
while the IRs need orthogonal SC assignment for avoiding
potential eavesdroppers. From (1)-(3), the received signals at
mutualinterference.In addition, if the power allocated on SC
IR k after AN cancelationand the “best” eavesdropperon SC
n is given by p , then ER l can harvest ζ p h 2 on SC n
n are further expressed as n l n| l,n|
regardless of which receiver it is allocated to. Notice that if
p > 0 and α = 1 for any SC n, then this SC is used
Y =h (1 α )p s +v , (4) k,n k,n
k,n k,n k,n k,n k,n k
− onlyforenergytransfer,i.e.,there is noinformationsentover
q
Ek,n =βk,n (1 αk,n)pk,nsk,n+βk,n√αk,npk,nzk,n+ek. the SC. As a result, we only need to focus on the cases that
− SCs are allocated to IRs without loss of generality.
q (5)
Thus, the harvested power at each ER l is given by
2
∈K
Here we can write the achievable informationrate of IR k on
SC n, which is given by Q =ζ x p h 2, (12)
l l k,n k,n l,n
!| |
(1 α )h 2p nX∈N kX∈K1
rk,n =log2 1+ − k,nσ|2k,n| k,n . (6) where 0<ζl <1 denotes the energy harvesting efficiency.
(cid:18) (cid:19) An example of the energy utilization at receivers in an
The decodableinformationrate of the “best” eavesdropperon OFDMA-based SWIPT system with secrecy constraints is
SC n is given by shown in Fig. 3, with K = 2 and K = 1. As it is shown,
1 2
the AN does not interfere with the intended receiver but all
(1 α )β 2p
re =log 1+ − k,n | k,n| k,n . (7) other receivers. In addition, the ER is able to harvest energy
k,n 2(cid:18) σ2+αk,n|βk,n|2pk,n(cid:19) from both information signal and AN signal.
4
Power Power Power First, the Lagrangian of problem (13a) is given by
(P,α,X,λ,γ)
L
= w x Rs γ x p P
1 (cid:258) n (cid:258) N SC Index 1 (cid:258) n (cid:258) N SC Index 1 (cid:258) n (cid:258) N SC Index k k,n k,n− k,n k,n− max!
kX∈K1 nX∈N kX∈K1nX∈N
InDfoercmodaitniogn AN Efrnoemrg yIn Hfoarrmveasttiionng Energfryo Hma rAvNesting + λl(Ql−Q¯l)
Unintended lX∈K2
Information AN Canelation = w x Rs γ x p
k k,n k,n− k,n k,n
kX∈K1 nX∈N kX∈K1nX∈N
Fig.3. AnexampleofpowerutilizationforanOFDMA-basedSWIPTsystem
oftwoIRsandoneER. + xk,npk,n λlζl hl,n 2
! | |
nX∈N kX∈K1 lX∈K2
λQ¯ +γP , (14)
l l max
Our goal is to maximize the weighted sum rate of the −
lX∈K2
IRs by optimizing transmit power and SC allocation as well
where λ=[λ ,λ ,...,λ ] and γ are the Lagrange multipli-
as transmit power splitting ratio at each SC, subject to the 1 2 K2
ers (dual variables) corresponding to the minimum required
harvested power constraints of all ERs. The problem can be
harvested power constraints and the total transmit power
mathematically formulated as
constraint, respectively.
We then define for given X as the set of all possible
power allocations oPf P that satisfy 0 p P for
k,n peak
Pm,Xax,αRssum (13a) pxok,snsib=le1Xantdhaptks,antis=fy0cownhsetrnaixnkts,n(1=3f)0≤,anSda(1s3t≤gh)e, asentdof aalsl
s.t. Ql ≥Q¯l,∀l∈K2, (13b) thesetofallfeasibleαthatsatisfy(13e).Then,wecanobAtain
p x P (13c) the dual function for problem (13a) as
k,n k,n max
≤
kX∈K1nX∈N g(λ,γ)= max (P,α,X,λ,γ). (15)
0 pk,n Ppeak, n ,k 1 (13d) P∈P(X),α∈A,X∈SL
≤ ≤ ∀ ∈N ∈K
0 α 1, n ,k (13e) The dual problem is then given by
k,n 1
≤ ≤ ∀ ∈N ∈K
xk,n 0,1 , n ,k 1 (13f) min g(λ,γ). (16)
∈{ } ∀ ∈N ∈K
λ(cid:23)0,γ≥0
x 1, n , (13g)
k,n
≤ ∀ ∈N From (14), we can observe that the maximization in (16)
kX∈K1 canbedecomposedintoN independentsubproblems.Accord-
ingly, we can rewrite the Lagrangian as
Xwhe,re {Pxk,,n}{pdke,nno}tedsenthoetesSCtheallpoocwateironalfloocraItiRosn, oavnedrαSC,s, L(P,α,X,λ,γ)=nX∈NLn(Pn,αn,Xn)
Q{¯αkd,ne}nodteensottheesthhaervtreasntesdmiptopwoewrecrosnpslittrtainingtofvoerrESRCsl.In(13.bI)n, − λlQ¯l+γPmax, (17)
l ∈ K2 lX∈K2
(13c) and (13d), P and P represent the total power
max peak where
constraint over all SCs and the peak power constraint over
each SC, respectively. Finally, (13f) and (13g) constrain that n(Pn,αn,Xn)
L
any SC can only be assigned to at most one IR.
, x w Rs γp +p λ ζ h 2 .
k,n( k k,n− k,n k,n l l| l,n| !)
kX∈K1 lX∈K2
(18)
Since x 0,1 and x =1, there exists a k∗
k,n ∈{ } k∈K1 k,n ∈
such that
III. OPTIMALSOLUTION K1 P
1, if k =k∗,
x∗ = , n , (19)
k,n ( 0, otherwise ∀ ∈N
Problem (13a) is a mixed integer programming and thus is the optimal solution to maximize .
is NP-hard and non-convex. As shown in [26], [27], the Hence, with given λ and γ, the mLaximization of can be
L
dualitygapbecomeszero in OFDM-basedresourceallocation attained by selecting
problemsincluding problem(13a) as the numberof SCs goes
to infinity due to the so-called time-sharing condition. This k∗ =argmax w Rs +p λ ζ h 2 γ
implies that problem (13a) can be solved by the Lagrange k∈K1( k k,n k,n l l| l,n| − !)
duality method asymptotically optimally. lX∈K2 (20)
5
for each SC n, and the optimal (p∗ ,α∗ ) can be solved by p 1 1 σ2. Substituting it into
assuming k = k∗ and then solvingk,tnhe fko,nllowing subproblem (2k3,n) to≥elim(cid:16)in|βakte,n|α2k−,n,|hwk,en|h2a(cid:17)ve
for each SC n,
a p2 +b p +c =0, (30)
2 k,n 2 k,n 2
max ′ (P ,α )
Pn∈P(X),αn∈ALn n n where
,wkRks,n+pk,n λlζl|hl,n|2−γ!. a2 =ln2|βk,n|4|hk,n|2Ωn, (31)
lX∈K2 (21) b2 =wk|βk,n|4|hk,n|2+ln2Ωn|βk,n|2σ2, (32)
c =σ2 β 2 h 2w (1 β 2)
2 k,n k,n k k,n
| | | | −| |
+ln2Ω2(β 2+ h 2) . (33)
A. Joint Optimization of Power Allocation and Transmit (cid:8) n | k,n| | k,n|
Power Splitting Ratio
Similarly,wedefineΦ asthesetofall(cid:9)non-negativereal
2
We cannot directly express the partial derivative of Rs roots to (30) that satisfy [ (α )]+ <p P .
k,n Xk,n k,n k,n ≤ peak
in (8) with respect to pk,n or αk,n. However, as we have We further define Ψ2 as the set of all feasible candidates
discussed in Lemma 1, Rks,n = 0 when 0 ≤ pk,n ≤ for (p∗k,n,α∗k,n) in subregion i as follows:
[ (α )]+ and Rs > 0 when p > [ (α )]+.
IXnke,anchkr,engion, Rks,n iks,ndifferentiable wki,tnh respXeckt,nto pkk,n,n or Ψ2 ,{(pk,n,αk,n)|pk,n ∈Φ2,αk,n =α∗k,n(pk,n)}
(αpk∗,n.,Hαe∗nc)e,inwaellfirrestgifionnds.thTehesnet,wofeaslellfeecats(ibpl∗e c,aαn∗did)ataesstfhoer ∪ Ppeak,α∗k,n(Ppeak) , (34)
k,n k,n k,n k,n
on1e)acRheigeivoinngIt(hpek,lnar>ge[stkv,anl(uαeko,nf)L]+′n):in (21). • wFohreresuαb∗kr,eng((cid:0)ipokn,n)ii,is oαb∗kt,ani(npekd,ni)n(cid:1)(2=2). 0 and pk,n <
X
1 1 σ2 (which can be true only when
Lemma 2. Theoptimalαk,n withgivenpk,n forproblem(21) |βk,n|2 − |hk,n|2
is given by (cid:16)hk,n 2 > βk,n 2(cid:17)). The set of all feasible candidates
| | | |
for (p∗ ,α∗ ) in this case is given by Ψ (α = 0)
1 σ2 1 1 + obtainke,dnviak,(n23). 1 k,n
α∗ (p )= + <1,
k,n k,n (cid:20)2 2pk,n (cid:18)|hk,n|2 − |βk,n|2(cid:19)(cid:21) 2) Region II (0 pk,n [ k,n(αk,n)]+): As we have
(22) discussed, Rs =≤0 in thi≤s caXse, which can be true only
k,n
for all k ∈K1 and n∈N. washen|hk,n|2 <|βk,n|2. The Lagrangiancan thusbe rewritten
Proof: Please refer to Appendix C.
On the other hand, by deriving the partial derivative of ′
with respect to pk,n and equating it to zero, we have Ln L′n(Pn,αn)=pk,n λlζl|hl,n|2−γ!, (35)
lX∈K2
a1p3k,n+b1p2k,n+c1pk,n+d1 =0, (23) which is a linear function of pk,n and is regardless of αk,n.
Here,wesetα∗ =0forconvenience.Thefeasiblecandidate
where k,n
(pˆ ,αˆ ) in this region can thus be obtained as3
k,n k,n
a =ln2h 2(α2 α )β 4Ω , (24)
1 | k,n| k,n− k,n | k,n| n (pˆ ,αˆ )= min [ (α =0)]+,P ,0
b =(α2 α )β 4 h 2w k,n k,n {Xk,n k,n peak}
1 k,n− k,n | k,n| | k,n| k =(P ,0). (36)
+ln2β 2σ2 (α2 1)h 2 β 2α Ω , (cid:0) peak (cid:1)
| k,n| k,n− | k,n| −| k,n| k,n n
(25) It is observed that the feasibility of the above two regions
(cid:2) (cid:3)
c =ln2(α 1)(h 2 β 2)σ4Ω is determined by the channel conditions and the peak power
1 k,n k,n k,n n
− | | −| | constraint.FivescenariosareillustratedinFig.4andexplained
+2(α2 α )β 2 h 2w σ2, (26)
k,n− k,n | k,n| | k,n| k as follows:
d =(α 1)(h 2 β 2)w σ4 ln2σ6Ω , (27)
1 k,n− | k,n| −| k,n| k − n 1) In scenario (a), hk,n 2 < βk,n 2 and Ppeak >
Ωn =−γ+ λlζl|hk,n|2. (28) Xk,n(αk,n =1). Bo|th R|egion II|and|subregion i in Re-
lX∈K2 gionI arefeasible.Notethatcurveα∗k,n(pk,n)andcurve
p = (α ) intersect at ( (α =1),1).
Wrooetsfirtsot(d2e3fi)ntehaΦt1s(aαtiskf,ny)[ask,nth(eαks,ent)]o+f<allpkn,onn-nePgpaetiavkewreitahl 2) Ink,nscenXakri,no (bk,)n, |hk,n|2 < |Xβkk,,nn|2k,annd Ppeak ≤
X ≤ (α =1). Only Region II is feasible.
given α . Then, we define another set Ψ (α ) as follows: k,n k,n
k,n 1 k,n 3) XIn scenario (c), h 2 = β 2. α∗ = 1 in this
Ψ (α ), (p ,α )p Φ (α ) . (29) scenario so only s|ukb,rne|gion i|ink,Rne|gionkI,nis fea2sible.
1 k,n k,n k,n k,n 1 k,n
{ | ∈ }
folTloowfiinngd tfweaosisbulebrceagniodnidsa.tes for (p∗k,n,α∗k,n), we consider the acc3oNrdoitnegthtoatthheerSeCwaelloascsautimonepPolilc∈yKa2sλwlζel|whikll,nd|i2scu>ssγin. Tlahteisr,iSsCbencawusilel
• For subregion i, we remove the [·]+ operator bLenacllaoncnaotetdbetopoIsRitikveornelgyaridfleLssnoifsppko,nsi.tivAes.aIfrePsullt∈,Kw2eλiglζnlo|rhekt,hne|2ca≤seγof,
of α∗k,n(pk,n) in (22) and assume that Pl∈K2λlζl|hk,n|2≤γ withoutlossofgenerality.
6
a* (p )
k,n k,n
1 1 1
Region I: a* (p )
k,n k,n
(cid:160) subregion i
2
a Region II a Region II a
k,n Region I: k,n k,n (cid:160)
subregion i p = 2
kk,,,,nn kk,,,,nn
(pˆk,n,aˆk,n) (pˆk,n,aˆk,n)
kkkk,,,,nnnn(((((aakkkk,,,,nnnn==1P)k,n Ppeak Pk,Pnpeak kkkk,,,,nnnn(((((aakkkk,,,,nnnn==1) Pk,n Ppeak
(a) (b) (c)
Region I:
subregion ii
1
1
Region I:
a* (p )
subregion i k,n k,n
a Region I:
k,n (cid:160) ak,n subregion ii
2
(cid:160)1(ak,n=0) (cid:182)k,nPk,n Ppeak (cid:160)1(ak,n =0) PkP,npeak (cid:182)k,n
(d)
(e)
Fig.4. Fivescenarios offeasibleregions, whereφk,n=(cid:16)|βk1,n|2 − |hk1,n|2(cid:17).
4) In scenario (d), hk,n 2 > βk,n 2 and Ppeak > Algorithm 1 Joint optimization of p∗k,n and α∗k,n
| | | |
|βk1,n|2 − |hk1,n|2 . Two subregions in Region I are 1: if |hk,n|2 >|βk,n|2 then
5) (cid:16)Ifneassibcelen.ario (e),(cid:17)|hk,n|2 > |βk,n|2 and Ppeak ≤ 32:: if PCpoemakpu>te(cid:16)|βk=1,nΨ|21−(α|kh,kn1,n=|2(cid:17)0)theΨn2 via (23) and (30).
|βk1,n|2 − |hk1,n|2 . Only subregion i in Region I is 4: else F ∪
f(cid:16)easible. (cid:17) 5: Compute =Ψ1(αk,n =0) via (23).
F
6: end if
aboNveextd,iswcuessdieonnosteasFfolalsowthse: feasible set by combining the 87:: elsCeoimf |phukt,en|2 ==|Ψβk2,nv|i2at(h3e0n).
F
9: else if Ppeak > k,n(αk,n =1) then
X
Ψ2∪{(Ppeak,0)}, if Ppeak >Xk,n(αk,n =1) 1110:: elsCeompute F =Ψ2∪{(Ppeak,0)} via (30).
 and |hk,n|2 <|βk,n|2, 12: Set = (Ppeak,0) .
F { }
F =ΨΨ{(22P,∪peΨak1,(0α)}k,,n =0), iiiafffn|PPdhppk|eeh,aankkk|,2n>≤|=2X<|k|ββ,kn|kβ1,,(nnkα||,22nk,,−|n2,=|hk11,n)|2 1134:: eCnodmipfute (p∗k,n,α∗k,n) according to (37).
and h 2(cid:16)> β 2, (cid:17) B. Subcarrier Allocation
Ψ1(αk,n =0), iafnPdp||ehakkk,,nn≤||2(cid:16)>|β||kβ1,nkk|,,2nn−||2.|hk1,n|2(cid:17) SCSuasbssitgitnumt1ine,ngtthpifeolkoicp=ytimiks∗agl=ipv∗kea,nnrgbamyndaxα∗k,n into′ (Lp∗′n, t,hαe∗op)timal
k∈K1Ln k,n k,n
Given SC n being allocated to IR k, the jointly optimized x∗ = and max ′ (p∗ ,α∗ )>0, . (38)
(p∗k,n,α∗k,n) is obtained as k,n  0, otherwisek∈K1Ln k,n k,n

(p∗ ,α∗ )=arg max (p ,α ). (37)
k,n k,n (pk,n,αk,n)∈FLn k,n k,n C. Dual Update
According to [28], the dual problem is always convex;
The above algorithm is summarized as Algorithm 1. hence, the subgradientmethod can be used to update the dual
7
Algorithm 2 Optimal Algorithm for Problem (13a) Algorithm 3 Suboptimal Algorithm for Problem (13a)
1: repeat 1: Set 1 = , and p=min Ppeak,Pmax/N .
2: Jointly optimize p∗k,n and α∗k,n for all k ∈K1 and n∈ 2: for ENach E∅R l do { }
3: NSacoclavocerdcoSinrCgdintaolglo(t3co8atA)i.olgnoxri∗kth,nmfo1r. all k ∈ K1 and n ∈ N 543::: rCeopFmeinapdtuteunQasls=ignζelpdPSCn∈nN1th|halt,nh|a2s. the largest channel
4: Updateλandγ accordingto(39)and(40),respectively. gain for ER l.
5: until λ and γ converges. 6: Set 1 1 n and assign SC n to IR k having
N ← N ∪
the largest channel gain.
7: Determine the optimal transmit power splitting ratio
variables to the optimal ones by an iterative procedure: α∗k,n by using (22).
8: Compute Ql Ql+ζlphl,n 2.
λtl+1 = λtl −ξl Ql−Q¯l +,∀l ∈K2, (39) 9: until Ql ≥Q¯l ← | |
+ 10: end for
(cid:2) (cid:0) (cid:1)(cid:3)
γt+1 = γt ν Pmax xk,npk,n , (40) 11: for The rest N2 of unassigned SCs do
" − − !# 12: Determine the optimal transmit power splitting ratios
nX∈NkX∈K1 α∗ via (22), for all k and n .
pwrhoepreerlyt d≥esi0gniesdthpeosiitteivreatisotenp-isnidzeexs., [ξ1,...,ξK2] and ν are 13: nSokl,nve SCbyalulsoicnagtiognreevdayr∈iambKlee1thxo∗kd,.n fo∈rNal2l k ∈ K1 and
2
Note that the subgradient method is an iterative method 14: end ∈forN
for solving convex optimization problems in general, and the
proposed algorithm is a direct application of the subgradient
method to our problem. Thus the convergenceand optimality Inthesecondstage,weconsiderthefollowingproblemthat
of the proposed algorithm can be guaranteed. is simplified from problem (13a).
maxRs (41)
D. Complexity X,α sum
s.t. (13e) (13g).
The complexity of this iterative algorithm is analyzed as −
follows. For each SC, (K ) computations are needed for Note that all ERs’ constraints on required harvested power
2
solving Ω in (28) andO (K ) computations are needed for are removed as they are already achieved after the first
n 1
searchingthebestIR.SinOcetheoptimizationisindependentat stage. The simplified problem (41) for power allocation, SC
each SC, the complexity is (KN) for each iteration. Last, assignmentanddeterminingtransmitpowersplittingratioscan
the complexity of subgradieOnt based updates is polynomial be regarded as a special case of problem (13a). Accordingly,
in the number of dual variables K + 1 [28]. As a result, we can obtain the optimal transmit power splitting ratios by
2
the overall complexity of the proposed algorithm for solving (22). After that, the problem is reduced to a SC assignment
problem (13a) is ((K +1)qKN), where q is a positive problem for weighted sum secrecy rate maximization, which
2
constant. Note thatOthe complexity is polynomial. can be optimally solved by a greedy algorithm, i.e., each SC
Finally, we summarize the overall algorithm for solving is assigned to the IR havingthe largestweightedsecrecy rate.
problem (13a) in Algorithm 2. Note that the ERs can harvestadditionalenergyfrom the SCs
assigned to the IRs in the second stage.
The above suboptimal algorithm is summarized in Algo-
IV. SUBOPTIMAL SOLUTION rithm 3. The complexity order of the first stage is (K N )
2 1
O
The complexity of the optimal algorithm becomes high as and the complexity order of the second stage is (K1N2).
O
K1,K2 and/orN increases,mainlyduetotheupdatingofthe Thusthe total complexityis thus given as O(K2N1+K1N2)
Lagrangemultipliersλandγ.Byeliminatingthedualupdates, which is upper-bounded by (KN) and is much lower than
O
in this section, we present an efficient suboptimal algorithm that of Algorithm 2.
which significantly reduces the complexity.
We design a two-stage algorithm by assuming equal power V. NUMERICAL RESULTS
allocation, i.e., p = min P , P /N , k ,n Inthissection,weevaluatetheperformanceoftheproposed
k,n peak max 1
{ } ∀ ∈ K ∈
.Herewedropindexk andnofp forbrevity.Inthefirst algorithms through extensive simulations. In the simulation
k,n
N
stage, for each unsatisfied ER k, we select the SC at which setup,asinglecellwithradiusof200meters(m)isconsidered.
ER k has the largest channel gain among all unsatisfied ERs The BS is located at the centre of the cell. The carrier fre-
andthenassignthisSCtotheIRkthathasthelargestchannel quencyis 900 MHz and the bandwidthis 1 MHz. We assume
gain among all IRs. The above process is repeated until the thenoisepowerσ2 = 83dBm,andantennagainstobe0dB.
−
minimum harvested power of all the ERs are satisfied. We The peak transmit power constraint is set to be P = .
peak
∞
denote N as the number of SCs assigned in this stage given We consider K = 4 IRs that are randomly located in the
1 1
inthe set ,andN asthenumberofunassignedSCsin the cell with distance to the BS uniformly distributed. For each
1 2
N
set . IR, we set w = 1, k , i.e., we consider the sum
2 k 1
N ∀ ∈ K
8
x 10−5
5.5 0.08
5 Watt)0.06 PPmax==3373ddBBmm
4.5 wer ( Pmmaaxx=28dBm
o0.04
gap (bps/ Hz))3.534 Transmit P0.020
uality 2.5 Watt) 10 20 Subc3a0rrier Inde4x0 50 60
D 2 on (0.04
1.5 ormati0.03 PPmax==3373ddBBmm
0.518 16 24 32 40 48 56 64 Split for Inf0.02 Pmmaaxx=28dBm
Number of SCs er 0.01
w
o
Fig.5. DualitygapversusnumberofSCs. mit P 0 10 20 30 40 50 60
s
n Subcarrier Index
a
Tr
secrecy rate of all IRs. We also consider K = 4 ERs that
2
are uniformly distributed within the circle of radius of 2 m Fig.6. TransmitpowerandpowersplitforinformationsourceoneachSC.
around the BS.4 For each ER, we set ζ = 60%, l .
l 2
∀ ∈ K
Thechannelcoefficientsconsistofbothlarge-scalefadingand 3.5
small-scale fading. The path loss exponentis set to be 3. The Optimal Algorithm
Suboptimal Algorithm
small-scale fading is modeled as Rayleigh fading and each 3 α=0.5
FSA
channel realization is composed of 8 i.i.d. Rayleigh fading z) NoAN
H2.5
paths. We also assume that all ERs have the same harvested s/
p
power requirement, i.e., Q¯l =Q¯, l 2. e (b 2
For performance comparison, w∀e∈alsKo consider the follow- Rat
m-
ing benchmarking schemes. First, the fixed transmit power u1.5
S
splitting ratio with α =0.5, k ,n is considered y
k,n 1 c
∀ ∈K ∈N e
for complexity reduction, while the power and SC allocation cr 1
e
S
is still optimized as in Algorithm 2. In this case we drop the
0.5
indexk and n of α forbrevity.Second,the SC assignment
k,n
is fixed (FSA) while the power allocation and transmit power
0
splittingarejointlyoptimizedasinAlgorithm2.Last,wealso
0 0.2 0.4 0.6 0.8 1 1.2 1.4
consider the scheme without using AN (NoAN). It is worth Energy Harvesting Requirement (mW)
noting that NoAN performs the same as the traditional AN
scheme(ANschemewithoutcancelling)aswehavediscussed Fig.7. Achievable secrecy rateRssum versusrequiredharvested powerQ¯.
in Remark 1.
First, the duality gaps with different number of SCs N are
sahnodwbnecinomFeigs.s5m.aIltleisr oabssNerveindcrtheaastedsu.aFliotrytghaepciassveeoryf sNma=ll nαo∗ki,sne≈po21wienrtihserheilgahtivSeNlyRsrmegailolna.nIdnroeusrulstismiunlahtiiognhspeσktu,2np,atnhde
c6o4n,stihdeergeadptobebceomneegslisgmibalell.er than 1×10−5 bps/Hz, thus is thuIsntFhieg.op7t,imthaelssuomlutisoencrαec∗ky,nra≈te21R. s versus the harvested
sum
We also show the allocated transmit power and power split power requirement Q¯ is shown with Pmax = 37 dBm and
for informationsourceoverSCs in Fig. 6, with Q¯ =100 µW. N = 64. First, for all schemes (except NoAN), the sum
First, we observe that for the case where P =37dBm, the secrecy rate is observed to decrease with increasing Q¯. It is
max
allocated power on each SC is almost uniform, which shows also observed that the suboptimal algorithm and the optimal
that the suboptimal algorithm that allocates power uniformly algorithm outperform FSA and NoAN and the suboptimal
overSCsmayperformclosertotheoptimalalgorithmasP algorithm incurs at most 30% loss in secrecy rate compared
max
increases. In addition, we observe that the power used for to the optimal algorithm. An interesting observation is that
information source is approximately one half of the power the scheme with α = 0.5 performs closely to the proposed
allocated on each corresponding SC, i.e., the optimal α optimal algorithm, which is in accordance to our previous
k,n
0.5. This is because, according to (22), we have the optim≈al discussion that α∗k,n ≈ 21 in the high SNR region. The poor
performance of FSA compared to the proposed algorithms
4We consider ERs in general closer to the BS than IRs to receive larger indicatesthat dynamicSC allocation providessignificantgain
power(versusthatofIRsusedfordecoding information againstbackground intermsofsumsecrecyrate.Moreover,allconsideredschemes
noise only). However, under this circumstance, ERs in general have better
with AN achieve significant rate-energy gains compared to
channel conditions than IRs, and as a result they are more capable of
eavesdropping theinformation sentbytheBS[18]. NoAN, which has almost zero sum secrecy rate even if there
9
3.5 3.5
Optimal Algorithm
Suboptimal Algorithm
3 α=0.5 3
FSA
z) NoAN z)
H2.5 H2.5
s/ s/
p p
b b
e ( 2 e ( 2 Optimal Algorithm
at at Suboptimal Algorithm
um-R1.5 um-R1.5 αF=S0A.5
S S NoAN
y y
c c
e e
cr 1 cr 1
e e
S S
0.5 0.5
0 0
26 27 28 29 30 31 32 33 34 35 36 37 5 10 15 20 25 30 35 40 45 50 55 60
Total Transmit Power (mW) Number of ERs
Fig.8. AchievablesecrecyrateRssum versustotaltransmitpowerconstraint Fig.9. Achievable secrecy rateRssum versusthenumberofERs.
Pmax.
we maximize the weighted sum secrecy rate for IRs subject
isnoharvestedpowerrequirement.Thisisbecausewithoutthe to individual harvested power constraints of ERs by jointly
effectiveaidoftheAN,thesecrecyrateoneachSCispositive optimizing transmit power and SC allocation as well as
only when it is assigned to the receiver of largest channel transmit power splitting ratios over SCs for AN signals. We
gain [11]. However, in our simulation setup, the ERs possess proposedan algorithmbased onthe Lagrangedualityto solve
muchbetterchannelgainscomparedtotheIRs,duetoshorter the formulatedproblemwith polynomialtime complexity.We
distances to the BS. As a result, hk,n 2 < βk,n 2 is almost also proposed a suboptimal algorithm with lower complexity.
| | | |
true for all n ,k 1, and hence no secrecy information Through extensive simulations, we showed that the proposed
∈N ∈K
can be transmitted at all. This demonstrates the effectiveness algorithms outperform other heuristically designed schemes
of the proposed frequency-domainAN aided approach. with or without using the AN.
Fig. 8 demonstrates the sum secrecy rate Rs versus the
sum
totaltransmitpowerPmax,withtheharvestedpowerconstraint APPENDIX A
set as Q¯ = 100 µW and N = 64. Compared with FSA PROOF OF LEMMA1
and NoAN, bothproposedoptimalandsuboptimalalgorithms
We consider the following two cases:
performbetter. Inaddition,itcan beobservedthatsuboptimal
1) α =0: Equating r re to zero, we obtain
algorithm performs more closely to the optimal algorithm as k,n 6 k,n− k,n
the total transmit power increases, which collapses to the
h 2p β 2p
observation from Fig. 7 that the allocated power on SCs | k,n| k,n = | k,n| k,n . (42)
is more uniformly distributed as transmit power increases. σ2 αk,n βk,n 2pk,n+σ2
| |
Moreover, the scheme with α = 0.5 is also observed to We thushavep =0 or p = (α ). However,
k,n k,n k,n k,n
X
perform very closely to the optimal algorithm. p is always non-negative, so p = (α )>0
k,n k,n k,n k,n
X
Fig. 9 illustrates the sum secrecy rate versus the numberof can be true only when h 2 < β 2. Thus, it is
k,n k,n
ERs, with the harvested power requirement set as Q¯ = 100 shown that r re =| 0 h|as one| roo|t at p = 0,
k,n − k,n k,n
µW, Pmax = 37 dBm and N = 64. First, we observe that when hk,n 2 βk,n 2, and two roots at pk,n = 0 and
with the increasing number of ERs, the sum secrecy rate of p =| |(α≥ |), w|hen h 2 < β 2.
k,n k,n k,n k,n k,n
IRs for all schemes decreases. This is because when a new For breXvity, we define x| , |α ,| y ,| p , h ,
k,n k,n
aEtRanisyaSdCdend.iAnsthaerseysustlet,ms,ec|βrekc,ny|2inmfoarymainticorneaissemfoorrealelasIiRlys |βhk,n|22/σ2ha<ndg,git f,ollo|wβsk,nth|a2t/σ2. When |hk,n|2 <
k,n
eavesdropped. In addition, with more ERs, more power will | | ⇔
be allocated to the SCs for satisfying the requirements of the ∂(rk,n−rke,n) , ∂f
ERs but not necessarily achieving the maximum sum secrecy ∂pk,n (cid:12) ∂y y=Xk,n(x)
rate for IRs. It is also observed that FSA becomes infeasible g (cid:12)(cid:12)pk,n=Xk,n(αk,n) gx (cid:12)(cid:12)
whenthenumberofERsislargerthan55,whiletheproposed =ln2[(1 g/h)/(cid:12)(cid:12)x 1] − ln2[(1 g/(cid:12)h) 1]
algorithms perform with noticeably higher sum secrecy rate. − − − −
h(1 x)
+ −
ln2((h/g 1)(x 1)/x+1)
VI. CONCLUSION − −
hx(g h)(1 x)
This paper studies the optimal resource allocation for = − −
ln2(g h+hx)
OFDMA-based SWIPT with secrecy constraints. With a pro- −
0. (43)
posed frequency-domainAN generation and removalmethod,
≥
10
Hence, r re 0 when 0 p (α ) energy harvesting constraints similar to problem (13a), we
k,n − k,n ≤ ≤ k,n ≤ Xk,n k,n
and h 2 < β 2,whichisequivalentto0 p consider the following problem
k,n k,n k,n
| | | | ≤ ≤
[ (α )]+.Ontheotherhand,r re >0when
Xk,n k,n k,n− k,n max Rs,NC (48)
i)pk,n > k,n(αk,n)and hk,n 2 < βk,n 2 orii)pk,n > α k,n
0 and hkX,n 2 βk,n 2, w| hich| is e|quiva|lent to pk,n > kX∈K1nX∈N
[ (α| )|]+.≥ | | s.t. (13b) (13g).
k,n k,n −
X
2) αk,n =0: In this case, we have We can show the decomposed Lagrangian on each SC
n
L
of problem (48) is obtained in (18) by replacing Rs with
0, if β 2 h 2 k,n
Rks,n =(rk,n−rke,n >0, if ||βkk,,nn||2 ≥<||hkk,,nn||2 . Rks,,NnC and
(44) ∂ ∂Rs,NC
Forcing [ ]+ + when β 2 h 2 and Ln =w k,n 0. (49)
k,n k,n k,n k
X → ∞ | | ≥ | | ∂α ∂α ≤
[ ]+ =0when β 2 < h 2,(44)isequivalently k,n k,n
k,n k,n k,n
wXritten as | | | | Thus,the solutionα∗ =0, k,nalso holdsoptimalityfor
k,n ∀
problem (48).
0, if 0 p [ ]+
Rs = ≤ k,n ≤ Xk,n .
k,n (rk,n−rke,n >0, if pk,n >[Xk,n]+ APPENDIX C
(45)
PROOF OF LEMMA2
Combining the above two cases, we can finally conclude
By applying the KKT (Karush-Kuhn-Tucker) conditions
thatRs =0 when0 p [ (α )]+, while Rs =
k,n ≤ k,n ≤ Xk,n k,n k,n [28], we obtain
r re >0 when p >[ (α )]+.
k,Tnh−e pkro,nof is thus comkp,nletedX. k,n k,n 1 σ2 1 1 1
α∗ (p )= + , (50)
k,n k,n 2 2p h 2 − β 2
APPENDIX B (cid:20) k,n (cid:18)| k,n| | k,n| (cid:19)(cid:21)0
OPTIMALTRANSMIT POWERSPLITTING RATIO FOR for all k ∈K1,n∈N, where [·]ba ,min{max{·,a},b}.
TRADITIONAL AN SCHEME When hk,n 2 < βk,n 2, pk,n > [ k,n(αk,n)]+ =
| | | | X
(α ), we thus have
When the AN cannot be cancelled at the intended IR, the Xk,n k,n
secrecy rate in (8) should be rewritten as 1 (β 2 h 2)σ2 1
α∗ = + | k,n| −| k,n|
Rks,,NnC =[rk,n−rke,n]+ k,n (cid:20)2 2|βk,n|2|hk,n|2pk,n (cid:21)0
= log 1+ (1−αk,n)|hk,n|2pk,n < 1 + (|βk,n|2−|hk,n|2)σ2
(cid:20)−lo2g2(cid:18) 1+α(αk1,n−|hαβk,kn,n|2)p2|βkp,kn,n+|2+pσkσ2,2n(cid:19) +. (46) = 221 + α2|∗k2β,nk,n|2|hk,n|2Xk,n(αk,n)
(cid:18) k,n| k,n| k,n (cid:19)(cid:21) <1. (51)
We first consider the problem max Rs,NC by focusing
αk,n k,n When h 2 β 2, p > [ (α )]+ = 0, we
on the following two cases: | k,n| ≥ | k,n| k,n Xk,n k,n
thus have
1) For the case that h 2 > β 2, we have Rs,NC >0
and | k,n| | k,n| k,n 1 σ2 1 1 1 1
α∗ = + < . (52)
∂Rks,,NnC k,n (cid:20)2 2pk,n (cid:18)|hk,n|2 − |βk,n|2(cid:19)(cid:21)0 2
∂α To conclude the above two cases, we have α∗ < 1 is
k,n k,n
1 (hk,n 2 βk,n 2)σ2pk,n always true for pk,n ≥[Xk,n(αk,n)]+. Thus the optimal α∗k,n
= | | −| | with given p is rewritten as
− ln2(α h 2p +σ2)(α β 2p +σ2) k,n
k,n k,n k,n k,n k,n k,n
| | | |
0. (47) 1 σ2 1 1 +
≤ α∗ (p )= + , (53)
Thus, we have that Rs,NC is monotonically non- k,n k,n (cid:20)2 2pk,n (cid:18)|hk,n|2 − |βk,n|2(cid:19)(cid:21)
k,n
increasingwith respectto αk,n andthe optimalsolution for all k ∈K1,n∈N.
is given by α∗ =0, k,n. The proof is thus completed.
k,n ∀
2) For the case that h 2 β 2, we have Rs,NC =0
| k,n| ≤| k,n| k,n
regardless of α . REFERENCES
k,n
Combining the above two cases, we conclude that α∗ = [1] M. Zhang, Y. Liu, and R. Zhang, “Secrecy information and power
k,n
0, k,n,is alwaysoptimalto maximizethe secrecyrate using transfer in OFDMA systems,” in Proc. IEEE Global Commun.Conf.
∀ (Globecom), 2015.
traditional AN scheme without cancelation at the receiver,
[2] R. Zhang and C. K. Ho, “MIMO broadcasting for simultaneous wire-
i.e., the traditional AN scheme performs no better than the less information and power transfer,” IEEE Trans. Wireless Commun.,
transmission without AN. vol.12,no.5,pp.1989–2001, May2013.
[3] X. Zhou, R. Zhang, and C. K. Ho, “Wireless information and power
In addition, to show α∗ = 0, k,n is also the optimal
k,n ∀ transfer: architecture design and rate-energy tradeoff,” IEEE Trans.
solution to the sum secrecy rate maximization problem under Commun.,vol.61,no.11,pp.4757–4767,November2013.