Performance analysis and power allocation for a two-way amplify-and-forward relay with channel...

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Published in IET CommunicationsReceived on 15th February 2012Revised on 29th April 2012doi: 10.1049/iet-com.2012.0032

ISSN 1751-8628

Performance analysis and power allocation for atwo-way amplify-and-forward relay with channelestimation errorsC.S. Zhang1 J.H. Ge1 J. Li1,2 X.Y. Shi1

1State Key Laboratory of Integrated Service Networks (ISN), Xidian University, Xi’an 710071, People’s Republic of China2Key Laboratory of Information Coding and Transmission, Southwest Jiaotong University, Chengdu 610031,People’s Republic of ChinaE-mail: [email protected]

Abstract: This study deals with the performance analysis and power allocation of a two-way amplify-and-forward relay systemwith channel estimation errors. Exact closed-form expressions for outage probability and average symbol error rate (SER) are firstpresented. To provide more insights, their closed-form asymptotic expressions are then obtained. It is shown that the presence ofchannel estimation error causes outage probability and average SER maintain a fixed level even when a noiseless channel isadopted. These results are applied further to study the optimal power allocation problem for each node. From the perspectiveof service quality, rather than minimise the outage probability as much as possible, the goal is to use the minimum energyconsumed to satisfy the traffic requirements and thereby conserve the energy resource. Furthermore, based on the optimalsolutions to power allocation, the optimal relay location is investigated, which indicates that the system has higher energyefficiency with the relay located around the middle point of the two end nodes for any asymmetric traffic requirement. Thesimulation results verify that the derived outage probability and average SER expressions are accurate and highlight the effectof power allocation under various traffic requirements and channel estimation errors.

1 Introduction

Two-way relaying techniques have recently emerged as apromising approach to mitigate the spectral efficiency lossof conventional one-way relaying systems [1–3]. On thetwo-way relay channel, two nodes transmit their respectivesignals to the relay simultaneously and then the relaybroadcasts the received signals to both nodes. By removingits own information, called self-interference, each node canobtain its desired data. Therefore only two time intervalsare required for bi-directional transmission.

According to the function of the relay, the existingcooperative protocols can be categorised into amplify-and-forward (AF) [4] protocols and decode-and-forward (DF)[5] protocols. In the DF protocols, the relay decodes thereceived signals before determining whether to broadcastthem or not; whereas in AF protocols, the relay simplyamplifies the received signals and forwards them to thedestination terminals. Recently, the performance of the two-way AF relay systems has been well investigated fromvarious aspects. For instance, the outage probabilityperformance has been analysed in [6–9], whereas in[10–12], both the average symbol-error-rate (SER) and theoutage probability are discussed. Furthermore, for moreefficient use of the energy resource, power allocation hasbeen exploited. In [8, 9, 13], power allocation is employedto minimise the outage probability under a total power

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& The Institution of Engineering and Technology 2012

constraint. On the other hand, the provision of power levelsfor each node is investigated so as to minimise the totalenergy consumed under the restrictions of asymmetrictraffic requirement [14].

However, thus far, almost all the existing work onperformance analysis and power allocation for two-way AFsystems has been limited to the idealistic assumption ofperfect channel state information (CSI). In practice, therelay and the nodes never have perfect knowledge of theCSI. Channel estimation errors, which cannot be eliminatedtotally, can be introduced by an imperfect channelestimation algorithm or variations of the channel after it hasbeen accurately estimated. Therefore for a realistic analysis,it is important to investigate the performance and powerallocation for AF systems with imperfect CSI.

In this paper, we devote our attention to the performanceanalysis and power allocation of a two-way AF relaysystem with asymmetric traffic requirement and imperfectCSI. We first present tractable forms for the statistics of thesignal-to-noise ratios (SNRs) at the receiving terminals onthe assumption of channel estimation errors, for example,cumulative distribution function (CDF), probability densityfunction (PDF) and moment generating function (MGF),that are further applied to study performance of the two-way AF relay system. More precisely, exact closed forms ofoutage probability and average SER for the two-way AFrelay system are given. Furthermore, we extend the existing

IET Commun., 2012, Vol. 6, Iss. 12, pp. 1846–1855doi: 10.1049/iet-com.2012.0032

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work in [14] and consider power allocation for a two-way AFrelay system with asymmetric traffic requirements andchannel estimation errors. Power allocation is carried out soas to minimise the total energy consumed while satisfyingthe traffic requirements. For comparison, both asymmetricoutage probability requirements with distinct data rate andasymmetric average SER requirements are taken intoaccount. Finally, the optimum relay location in terms ofenergy efficiency is investigated.

The remainder of the paper is organised as follows.In Section 2, we describe the two-way AF relayingcommunication system. Next, in Section 3, the outageprobability and average SER performance analysis of a two-way AF relaying communication system with imperfect CSIare investigated. These results are applied to Section 4 toperform power allocation and determine the optimum relaylocation. Various simulation results and their discussionsare given in Section 5. Finally, we finish off with aconclusion in Section 6.

Notations: we use x � CN(a, b) to denote a complexGaussian random variable x with mean a and variance band z � 1(b) to represent the exponential random variable zwith parameter b. Pr[.] and |.| denote the probability andabsolute values, respectively.

2 System model

As shown in Fig. 1, we consider a two-way relay channel withtwo end nodes, A and B, and one relay node, R. In the firstphase, nodes A and B transmit their signals simultaneouslyto relay R. In the second phase, for AF relaying, the relaysimply amplifies the received signal from the first phase andforwards it to both nodes A and B.

The transmissions are subject to flat fading and additivenoise. In the sequel, hi � CN(0, s2

h,i) is the complex gain ofthe channel between nodes A, B and the relay, where i ¼ A,B and nj � CN(0, N0) is the additive white Gaussian noiseat node j, where j ¼ A, B, R. We assume that hi and nj arestatistically independent for different values of i and j, suchthat the channel coefficients and the noise samples areindependent.

In the first phase, the received signal at the relay can bewritten as

y = hA

��PA

√xA + hB

��PB

√xB + nR (1)

where xA and xB are the transmit signals with transmit powerPA and PB from the two nodes A and B, respectively. In thesecond phase, the received signals at node A via the relay are

yA = GhAhA

��PA

√xA + GhAhB

��PB

√xB + GhAnR + nA (2)

Fig. 1 System model for two-way AF relay with channel estimationerrors


The amplifying gain at the relay is set to the power-scaledgain, G, which can be considered by normalisation of thereceived signal, y. When assuming that both the nodes haveknowledge about their own symbols and can remove theirself-interference parts, for example, adaptive interferencecancellation technology, (2) can be rewritten as

yA = GhAhB

��PB

√xB + GhAnR + nA (3)

However, as stated in the previous section, all the nodes neverhave perfect knowledge of the CSI in practicalcommunication systems. Throughout this paper, we assumethat the channel is imperfectly known to the receivers, thatis, the receivers are provided with partial information of thechannel coefficients. In particular, as mentioned in [15], thechannel coefficients can be modelled as

hA = hA + eA (4)

hB = hB + eB (5)

where the estimated channel coefficients hA and hB areknown to the receivers, whereas eA and eB are not. Weassume that eA and eB are independent and identicallydistributed (i.i.d.) random variables with distributionsCN(0, s2

e,A) and CN(0, s2e,B), respectively. We further

assume that hA and hB are statistically independent of eA

and eB, respectively. As a result, hA � CN(0, s2h,A) and

hB � CN(0, s2h,B). Note that s2

h,A = s2h,A − s2

e,A and

s2h,B = s2

h,B − s2e,B. It has been shown in [16] that this

channel estimation model is valid. With these assumptions,the amplifying gain at the relay can be written as

G =��

PR

PAgA + PBgB + PAs2e,A + PBs

2e,B + N0

√(6)

where gA = |hA|2, gB = |hB|2 and PR is the transmit power ofthe relay. Then, the received signals at node A can berewritten as

yA = G(hA + eA)(hB + eB)��PB

√xB + G(hA + eA)nR + nA

= G��PB

√hAhBxB + G

��PB

√(hAeB + hBeA + eAeB)xB

+ G(hA + eA)nR + nA (7)

Then, the received SNR for signals at node A is

gA = PRPBgAgB/N20

u1,AgA + u2,AgB + bA

= PRPB

u1,Au2,AN20

u1,AgAu2,AgB

u1,AgA + u2,AgB + bA

(8)

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where

bA = PRPB

N 20

s2e,Bs

2e,A + PR

N0

s2e,A + PA

N0

s2e,A + PB

N0

s2e,B + 1,

u1,A = PRPB

N 20

s2e,B + PR

N0

+ PA

N0

( )and

u2,A = PRPB

N 20

s2e,A + PB

N0

( )

The received SNR at node B can be obtained in a similar way.From (8), we can observe that the channel estimation errors

have a significant impact on the received SNR. Furthermore,it can be deduced that the received SNR at each node isreduced by the increasing channel estimation errors. Whens2

e,A = s2e,B = 0, (8) reduces to its traditional form [9, 13].

3 Performance analysis

3.1 Outage probability

With effective SNR at the receiving terminal, the outageprobability performance of a two-way AF relay withchannel estimation errors is investigated in this part.

Lemma 1: Let X1, X2 be two independent exponent RVs withparameters b1 and b2, respectively [i.e. Xi � 1(bi), i ¼ 1, 2].Then, the CDF of X ¼ [(X1X2)/(X1 + X2 + b)], where b is aconstant, is given by

FX (x) = 1− 2e−(b2+b1)x��b1b2x(x+ b)

√K1 2

��b1b2x(x+ b)

√( )(9)

where Kv(.) is the first-order modified Bessel function definedin [17] [Eq. (9.6.22)].

Proof: See Appendix 9.1.

Theorem 1: The outage probability of node A and node B canbe written as

Pout,A = Pr[RA , RthA] = Pr[gA , gthA] = FX (hA)

= 1 − 2e−(b2,A+b1,A)hA

��b1,Ab2,AhA(hA + bA)

√× K1 2

��b1,Ab2,AhA(hA + bA)

√( )(10)

Pout,B = Pr[RB , Rth B] = Pr[gB , gth B] = FX (hB)

= 1 − 2e−(b2,B+b1,B)hB

��b1,Bb2,BhB(hB + bB)

√× K1 2


√( )(11)

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where RthA and RthB are the data rate requirements of nodes Aand B, respectively, and

gthA = 22Rth A − 1, gth B = 22Rth B − 1

b1,A = (1/(u1,As2h,A)), b2,A = (1/(u2,As

2h,B)), b1,B

= (1/(u1,Bs2h,A)), b2,B = (1/(u2,Bs

2h,B))

hA = ((u1,Au2,AN 20 )/(PRPB))gth A, hB

= ((u1,Bu2,BN 20 )/(PRPA))gthB

Proof: According to [18], gi follows exponential distributionwith parameter bi = 1/s2

h,i, [i.e. gi � 1(bi)], where i ¼ A, B.

With the help of Lemma 1, the outage probability of node Aand node B can be written as (10) and (11), respectively.

For better insights, we can represent Pout,A and Pout,B in(10) and (11) in simple closed forms by using a high SNRand low channel estimate error approximation. In thesescenarios, the term 2


√K1

(2��b1,Bb2,BhB(hB + bB)

√) converges to 1 as K1(x) ≃ 1/x for

x � 0. Then, we have (see (12) and (13))From (12) and (13), we can observe that the outage

probability will not approach zero, for any values oftransmission SNR. Therefore it is necessary to study theproperty of outage probability with an infinite SNR. When(PR/N0), (PA/N0), (PB/N0) � 1, (12) and (13) can berewritten as

Pout,A,EF = 1 − e−((s2

e,A/s2h,A

)+(s2e,B/s

2h,B

))gth,A (14)

Pout,B,EF = 1 − e−((s2

e,A/s2h,A

)+(s2e,B/s

2h,B

))gth,B (15)

Equations (14) and (15) readily reveal that the outageprobability of the two-way AF relay in a high SNR regionapproaches a non-zero constant, which is called the errorfloor (EF), if a channel estimation error exists. It indicatesthat the system will never reach a better outage probabilityperformance than EF even if a noiseless channel is adopted.From (14) and (15), we can observe that there is noconnection between the level of EF and power allocationstrategies. Furthermore, we can also see that it is therelative channel estimation error rather than the value of thechannel estimation errors that determine the EF. It is worthnoting that the EF is proximately proportional to therelative channel errors.

3.2 Average SER

In this subsection, we study the average SER performance ofa two-way AF relay with channel estimation errors. First, wederive a closed-form PDF and MGF expression for thereceived SNR. Then, based on the MGF expression, a

Pout,A ≃ 1 − e−(b2,A+b1,A)hA

= 1 − e−((s2

e,A/s2h,A

)+(s2e,B/s

2h,B

)+(N0PA/PRPBs2h,B

)+(N0/PRs2h,A

)+(N0/PBs2h,B

))gth A (12)

Pout,B ≃ 1 − e−(b2,B+b1,B)hB

= 1 − e−((s2

e,A/s2h,A

)+(s2e,B/s

2h,B

)+(N0PB/PRPAs2h,A

)+(N0/PRs2h,B

)+(N0/PAs2h,A

))gth B (13)


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closed-form SER formulation is given with M-ary quadratureamplitude modulation (M-QAM) modulation.

The average SER can be defined as the average of the SERat source nodes A and B [10] and is given by

PSER = 12(PSER,A + PSER,B) (16)

Theorem 2: The PDF of gA can be given as (see (17))

where lA = (u1,Au2,AN 20 /PRPB).

Proof: Taking the derivative of (10) given in Theorem 1 withrespect to gthA, and using the expression for the derivative ofthe Bessel function, given in [19] [Eq. (8.486.12)], we canobtain (17). A

Theorem 3: The average SER of nodes A and B for M-QAMmodulation can be written as (see (18) and (19))

where

4K

p

∫p/2

0

du− 4K2

p

∫p/4

0

du

[ ]{f (u)}

is defined as

4K

p

∫p/2

0

f (u) du− 4K2

p

∫p/4

0

f (u) du

and C is Euler’s constant.

Proof: See Appendix 9.2.The SER equations derived above can be numerically

calculated. However, the form of the SER is too complex to


deal with. Similarly, we can represent PSER,A and PSER,B in(18) and (19) with a simple closed form by using a highSNR and low channel estimate error approximation

PSER,A ≃ K(2p− K(p− 2))

pbQAM

(b1,A + b2,A)lA (20)

PSER,B ≃ K(2p− K(p− 2))

pbQAM

(b2,B + b1,B)lB (21)

Proof: See Appendix 9.3.Then, the approximate average SER can be written as

PSER = 12(PSER,A + PSER,B) (22)

When (PR/N0), (PA/N0), (PB/N0) � 1, the EF of the averageSER can be written as

PSER,EF = 1

2(PSER,A,EF + PSER,B,EF)

= K(2p− K(p− 2))

pbQAM

s2e,A

s2h,A

+s2

e,B

s2h,B

( )(23)

Equation (23) indicates that the average SER of the AF two-way relay reaches a fixed level of EF even when the SNRapproaches infinity. We can also observe that the EF can bereduced by improving the performance of channelestimation, but cannot be eliminated totally. Furthermore, itcan be deduced that nodes A and B share an identical levelof EF.

pg,A(g) = lApX (lAg)

= 2lA

��b1,Ab2,A((lAg)2 + bAlAg)

√(b2,A + b1,A)K1 2

��b1,Ab2,A((lAg)2 + bAlAg)

√( )[

+ b1,Ab2,A(2lAg+ bA)K0 2��b1,Ab2,A((lAg)2 + bAlAg)

√( )]e−(b2,A+b1,A)lAg (17)

PSER,A ≃ 4K

p

∫p/2

0

du− 4K2

p

∫p/4

0

du

[ ]lA(b1,A + b2,A)

(b1,A + b2,A)lA + (bQAM/2 sin2 u)

{

+2l2

Ab1,Ab2,A

((b1,A + b2,A)lA + (bQAM/2 sin2 u))2 (2 ln ((b1,A + b2,A)lA + (bQAM/2 sin2 u))

− ln (4l2Ab1,Ab2,A) + 2C − 2)} (18)

PSER,B ≃ 4K

p

∫p/2

0

du− 4K2

p

∫p/4

0

du

[ ]lB(b1,B + b2,B)

(b1,B + b2,B)lB + (bQAM/2 sin2 u)

{

+2l2

Bb1,Bb2,B

((b1,B + b2,B)lB + (bQAM/2 sin2 u))2 (2 ln ((b1,B + b2,B)lB + (bQAM/2 sin2 u))

− ln (4l2Bb1,Bb2,B) + 2C − 2)} (19)

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4 Power allocation and optimum relaylocation

4.1 Power allocation according to outageprobability: OPAout

In this subsection, we focus on the power allocation for eachnode so as to minimise the total energy consumed whilesatisfying the asymmetric outage probability requirements.For comparison, we also provide the total energy consumedof a direct transmission strategy as well as that of atwo-way AF relay system with uniform power allocation(PA ¼ PB ¼ PR) employed.

The energy minimisation problem can be formulated as

minPA ,PB ,PR

PA + PB + PR (24)

s.t.

Pout,A ≤ Pout,th A (a)

Pout,B ≤ Pout,th B (b)Pout,th A . Pout,A,EF (c)Pout,th B . Pout,B,EF (d)PA, PB, PR ≥ 0 (e)PA, PB, PR ≤ Pmax (f )

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

(25)

where Pout,thA and Pout,thB are the outage probabilityrequirements of nodes A and B, whereas the maximumpower, Pmax, corresponds to the maximum power that eachnode can provide [20]. As mentioned above, the outageprobability will reach a fixed level even for a considerablehigh SNR owing to the existence of channel estimationerrors. Hence, here the constraints (25c) and (25d) areincluded to insure that there exist solutions to the optimalproblem.

We assume that Pout,th A and Pout,th B both satisfy theconstraint given in (25). Neglecting the last constraint forthe moment we can obtain the following optimal solution

P∗A,out =

N0 s2h,B +

��s2

h,As2

h,B

√( )jB

P∗B,out =

N0 s2h,A +

��s2

h,As2

h,B

√( )jA

P∗R,out =

N0 s2h,A +

��s2

h,As2

h,B

√( )jB

+N0 s2

h,B +��s2

h,As2

h,B

√( )jA

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(26)

where

jA =− ln (1 − Pout,th A)

gth A

s2h,As

2h,B − (s2

h,Bs2e,A + s2

h,As2e,B)

jB =− ln (1 − Pout,th B)

gth B

s2h,As

2h,B − (s2

h,Bs2e,A + s2

h,As2e,B)

⎧⎪⎪⎨⎪⎪⎩

For details see Appendix 9.4.

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When there is no channel estimate error (s2e,A = s2

e,B = 0),(26) reduces to

P∗A,out =

gth BN0 s2h,B +

��s2

h,As2h,B

√( )Pout,th Bs

2h,As

2h,B

P∗B,out =

gth AN0 s2h,A +

��s2

h,As2h,B

√( )Pout,th As

2h,As

2h,B

P∗R,out =

N0gth B s2h,A +

��s2

h,As2h,B

√( )Pout,th Bs

2h,As

2h,B

+N0gth A s2

h,B +��s2

h,As2h,B

√( )Pout,th As

2h,As

2h,B

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(27)

which corresponds to the optimal power allocation providedin [14]. When s2

e,A = s2e,B = 0, gth A ¼ gth B and Pth A ¼

Pth B, we have

P∗A,out:P

∗B,out:P

∗R,out =

��s2

h,B

√:��s2

h,A

√:��s2

h,A

√+

��s2

h,B

√(28)

which corresponds to the optimal power allocation ratioproposed in [8].

Now, if the solution in (26) leads to a power allocation thatexceeds Pmax on any node, clipping is applied (i.e. ifP∗

R,out . Pmax, we define P∗R,out = Pmax) . Then, we can find

out the optimal solution with the same method shown inAppendix 9.4.

Substituting (26) into (24), we can obtain the minimumtotal energy consumption to satisfy the asymmetric trafficrequirements with a channel estimate error

Pproposedtotal = N0

jA + jB

jBjA

��s2

h,A

√+

��s2

h,B

√( )2

(29)

Similarly, the total power consumed of uniform powerallocation to satisfy the traffic requirement can be expressedas

Puniformtotal = 3N0 × max

(2s2h,A + s2

h,B)

jA

,(2s2

h,B + s2h,A)

jB

[ ]

(30)

Furthermore, we consider direct transmission between nodesA and B without the help of R, either of which uses half of thetime for a one-way direct transmission. Similarly, we have thefollowing assumption

hAB = hAB + eAB (31)

where hAB is the channel coefficient between nodes A and Band hAB � CN(0, s2

h,AB) is the estimated channel coefficient,

eAB � CN(0, s2e,AB) is the channel estimated error. All of

them are independent from each other. Similarly, we canobtain the minimum power consumed of the direct


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transmission strategy

Pdirecttotal = gth BN0

s2h,AB

Pout,th B − gth Bs2e,AB

+ gth AN0

s2h,BA

Pout,th A − gth As2e,BA

(32)

Equations (29), (30) and (32) all indicate that the total energyconsumed is closely related to the channel estimation errors.

4.2 Power allocation according to average SER:OPASER

In the previous part, power allocation according to outageprobability is investigated. In this subsection, forcomparison, the power allocation to minimise the totalenergy consumed under the constraints of asymmetricaverage SER requirements is performed.

The energy minimisation problem can be formulated as

minPA ,PB ,PR

PA + PB + PR (33)

st.

PSER,A ≤ PSER,th A (a)

PSER,B ≤ PSER,th B (b)PSER,th A . PSER,A,EF (c)PSER,th B . PSER,B,EF (d)PA, PB, PR ≥ 0 (e)PA, PB, PR ≤ Pmax (f )

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

(34)

where PSER,th A and PSER,th B are the average SER requirementsof nodes A and B. We assume that PSER,th A and PSER,th B bothsatisfy the constraints given in (34). Applying the sametechnique adopted by OPAout, we can obtain the followingoptimal solution

P∗A,SER =

N0V s2h,B +

��s2

h,As2

h,B

√( )PSER,th Bs

2h,As2

h,B− (s2

h,Bs2

e,A + s2h,As2

e,B)V

P∗B,SER =

N0V s2h,A +

��s2

h,As2

h,B

√( )PSER,th As

2h,As2

h,B− (s2

h,Bs2

e,A + s2h,As2

e,B)V

P∗R,SER =

N0V s2h,A +

��s2

h,As2

h,B

√( )PSER,th Bs

2h,As2

h,B− (s2

h,Bs2

e,A + s2h,As2

e,B)V

+N0V s2

h,B +��s2

h,As2

h,B

√( )PSER,th As

2h,A

s2h,B

− (s2h,Bs2

e,A + s2h,As2

e,B)V

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(35)

where V ¼ [(K(2p 2 K(p 2 2)))/pbQAM].By a comparison of (26) and (35), we can observe that

P∗R,out = P∗

R,SER, P∗A,out = P∗

A,SER and P∗B,out = P∗

B,SER if theassumptions (36) are employed. It indicates that theoptimum solution to the OPASER scheme can be obtainedby the OPAout scheme. So, for simplicity, the remainingdiscussions of the OPASER scheme are omitted. We will


mainly pay attention to the OPAout scheme, subsequently

PSER,th A = −Vln (1 − Pout,th A)

gth A

PSER,th B = −Vln (1 − Pout,th B)

gth B

⎧⎪⎪⎨⎪⎪⎩ (36)

4.3 Optimum relay location

With the OPAout scheme or the OPASER scheme beingapplied to each node, the least amount of energy consumedcan be obtained for any position. However, the total energyconsumed varies with the location of the relay. Thereforethis part is going to discuss the optimum relay location fora two-way AF relay with channel estimation errors. In thesequel, di represents the distance between node i and relayR, where i ¼ A, B. Then, we have s2

h,A = d−aA , s2

h,B = d−aB .

Let the distance between nodes A and B normalise tobecome 1. Furthermore, we have the definitions oftA = (s2

e,A/s2h,A) and tB = (s2

e,B/s2h,B), which are called

relative errors. With this knowledge, we can obtains2

h,A = (1 − tA)d−aA , s2

h,B = (1 − tB)d−aB , s2

e,A = tAd−aA and

s2e,B = tBd−a

B . Substituting them into (29) yields

Pproposedtotal (dA, dB)

= gth BN0

Pout,th B(1 − tA)(1 − tB) − (tA − 2tBtA + tB)gth B

(

+ gth AN0

Pout,th A(1 − tA)(1 − tB) − (tA − 2tBtA + tB)gth A

)

×��(1 − tA)da

B

√+

��(1 − tB)da

A

√( )2(37)

Then, the optimum relay location can be determined bysolving the following optimal problem

mindA,dB

��(1 − tA)da

B

√+

��(1 − tB)da

A

√( )2(38)

s.t. dA + dB ≥ 1, dA, dB ≥ 0 (39)

Applying Karush–Kuhn–Tucker (KKT) conditions for theabove problem, we can obtain the optimal solution aftersome involved manipulations

d∗A = 1

1 + ((1 − tB)/(1 − tA))(1/a−2)

d∗B = 1

1 + ((1 − tB)/(1 − tA))−(1/a−2)

⎧⎪⎪⎨⎪⎪⎩ (40)

For tA ¼ tB

dA = dB = 1

2(41)

which corresponds to the conclusion given by [14].Moreover, for tA = tB, in actual application, tA, tB ,, 1.Then we can obtain

dA ≃ dB ≃ 1

2(42)

For instance, when tA ¼ 0.15, tB ¼ 0.1 and a ¼ 4,

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dA ¼ 0.4929 and dB ¼ 0.5071. From (41) and (42), we cannaturally draw the conclusion that the optimum relaylocation is always near the middle point of the two endnodes although there exist channel estimation errors.Furthermore, from (40), we can observe that the optimumrelay locations d∗

A and d∗B are independent of the traffic

requirements Pout,thA and Pout,thB, which will help maintainthe system balance even with a high level of trafficasymmetry.

As mentioned above, the energy consumed changes withthe position of the relay. This implies that there exists arelay operating region in which a two-way AF relay is moreenergy efficient than direct transmission. The region ofOPAout scheme can be defined as

{(dA, dB)|Pproposedtotal (dA, dB) , Pdirect

total (dA, dB)} (43)

For simplicity, we assume that tA ¼ tB ¼ tAB ¼ t. Then, (43)can be rewritten as

(dA, dB)|��da

B

√+

��da

A

√( )2,

(gth AfB + gth BfA)wAwB

(gth AwB + gth BwA)fAfB

{ }(44)

Where fA ¼ Pout,thA(1 2 t) 2 gthAt, fB ¼ Pout,thB(1 2 t) 2gthBt, wA ¼ Pout,thA(1 2 t) 2 2tgthA and wB ¼ Pout,thB

(1 2 t) 2 2tgthB. It can be deduced that the size of theregion is inversely proportional to the relative channelestimation error t. The region of uniform power allocationcan be similarly deduced.

5 Simulation results

In this section, computer simulations are performed to verifythe accuracy of the analytical results. Meanwhile, theperformance of the proposed power allocation strategies ispresented. All channels in the cooperation model aredescribed by Rayleigh fading.

5.1 Numerical results

Here, we consider the uniform power allocation(PA ¼ PB ¼ PR) and i.i.d. channels. We let the noisevariance be N0 ¼ 1 and assume that the distance betweennodes A, B and the relay R is normalised to 1, that is,dA ¼ dB ¼ 1. Thus, the variances of the channel gain satisfys2

h,A = s2h,B = 1. Furthermore, we assume that nodes A and

B have an identical relative channel estimation error, that is,tA ¼ tB, and let tA ¼ tB ¼ t.

In Fig. 2, we present the outage probability performance ofa two-way AF relay with different channel estimation errorsby varying SNR with RthA ¼ RthB ¼ 0.5. From Fig. 2, it canbe observed that there is an excellent match between theanalytical results and the simulation results. We can alsoobserve that the presence of EF causes the outageprobability maintaining a fixed level even when the SNR issufficiently large; and the limit of the outage probability isproximately 2 × 1023 for relative channel estimation errort ¼ 0.001, whereas 2 × 1022 for t ¼ 0.01. This implies thatthe outage probability performance is very sensitive to thechannel estimation errors. For a better insight, in Fig. 3, theEF of outage probability for a different data rate is shownwith the assumption RthA ¼ RthB ¼ Rth. From Fig. 3, we cansee that the EF is 2 × 1023 for Rth ¼ 0.5, whereas6 × 1023 for Rth ¼ 1 when t ¼ 0.001. This reveals that the

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system with higher Rth requires smaller channel estimationerrors so as to achieve a common performance.Furthermore, we can also see that the provided analyticalresults are in good agreement with the simulations.

The average SER performance of a two-way AF relay withdifferent channel estimation errors by varying SNR is shownin Fig. 4. It is obvious that the analytical results match thesimulation results well. From Fig. 4, similar to Fig. 2, wecan observe that the average SER of the AF two-way relayconverges to EF for different channel estimation errors.Moreover, the EF of the average SER for differentmodulations is shown in Fig. 5. Again, the accuracy of ouranalytical results is verified, which suggests that theanalytical results can be utilised for the design of two-wayAF relay systems with imperfect CSI.

5.2 Impact of power allocation

Here, the proposed power allocation is employed for eachnode to minimise the total energy consumed whilesatisfying the asymmetric traffic requirements. We assumethat the variances of the channel estimation errors eAB and

Fig. 2 Outage probabilities of two-way AF relay with differentchannel estimation errors

Fig. 3 Error Floor of outage probability


www.ietdl.org

eBA satisfy

s2e,AB

s2h,AB

=s2

e,BA

s2h,BA

= tA = tB = t

We also presume that nodes A, B and relay R are located in astraight line with dA + dB ¼ 1. Moreover, we let the path lossexponent be a ¼ 4.

In Fig. 6, we present the total energy consumption whereasthe OPAout is adopted to each node where we also plot theenergy consumption of OPASER, uniform power allocationstrategy (PA ¼ PB ¼ PR) and direct transmission. Weconsider three cases {RthA, RthB; Pout,thA, Pout,thB; t}: case 1{0.5, 0.5; 0.1, 0.1; 0}, case 2 {1, 0.5; 0.1, 0.05; 0.001} andcase 3 {1, 0.5; 0.1, 0.05; 0.01}. The traffic requirements ofthe OPASER scheme satisfy (36). From Fig. 6, it can beobserved that there is an excellent match between theOPAout and the OPASER schemes. It is worth noting thatthey are always more energy efficient than the uniformpower allocation scheme for any cases, which is moreremarkable when the relay is located close to either node.Moreover, more than 50% of the energy on average issaved when power allocation is adopted for each node

Fig. 5 Error floor of average SER for 4QAM, 16QAM and 64QAM

Fig. 4 Average SER of two-way AF relay with different channelestimation errors for 4QAM


compared with the uniform power allocation scheme. Acomparison of case 2 and case 3 indicates that more energyis required for higher channel estimation errors so as to meetthe traffic requirements. In case 3, we can also observe that,owing to the existence of channel estimation errors, theOPAout scheme is less energy efficient than directtransmission when the relay is located close to either node.

As shown in Fig. 6, a two-way AF relay may be less energyefficient than direct transmission for certain relative channelestimation error and traffic requirements, which indicatesthat there exists a relay operating region. As defined in (43),the relay operation regions are shown in Fig. 7, withassumption (RthA, RthB) ¼ (1, 1) and (Pout,thA, Pout,thB) ¼(0.1, 0.2). Nodes A and B are located at points (0, 0) and(1, 0), respectively. From Fig. 7, we can observe that thearea of the relay operating region is inversely proportionalto the relative channel estimation errors. When the relativechannel estimation error is significantly large, there will beno relay operating region. Furthermore, it is obvious thatthe relay operating regions under the OPAout scheme arealways larger than that under the uniform power allocationstrategy for any channel estimation errors. As mentionedbefore, it is evident that the proposed power allocationstrategy will help maintain the system balance even with ahigh level of traffic asymmetry.

Fig. 6 Energy consumption under different schemes

Fig. 7 Relay operation regions under different channel estimationerrors

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6 Conclusions

In this paper, we derive the performance of the two-way AFrelay system with imperfect CSI. Applying these derivedresults, we consider the power allocation for each node soas to minimise the total energy consumed while satisfyingthe asymmetric traffic requirements. The performance andpower allocation are greatly affected by the channelestimation errors. Then, the optimum relay location isinvestigated. It has been proved that, with the proposedpower allocation adopted, the channel estimation errorshave little impact on the optimum relay location under theconsidered path-loss model. Furthermore, it has been shownthat there is an excellent match between the analyticalresults and the simulation results, and the proposed powerallocation schemes outperform the uniform power allocationscheme in terms of energy efficiency. Finally, it is worthnoting that the relay is almost best positioned at the middlepoint of the two end nodes for any asymmetric trafficrequirements despite the existence of the channel estimationerrors. Note that all the works of this paper are carried outbased on the assumption of a flat Rayleigh fading channel.Some more realistic cases, such as propagation stochasticphenomena, are left for future consideration.

7 Acknowledgments

The authors would like to thank the anonymous reviewers fortheir constructive comments and suggestions. This work wassupported by the Program for Changjiang Scholars andInnovative Research Team in University (IRT0852), theNational Basic Research Program of China(2012CB316100), the ‘111’ project (B08038), the NationalNatural Science Foundation of China (61001207, 61101144and 61101145), the Fundamental Research Funds for theCentral Universities (K50510010017) and the open researchfund of Key Laboratory of Information Coding andTransmission, Southwest Jiaotong University.

8 References

1 Rankov, B., Wittneben, A.: ‘Spectral efficiency protocols forhalf-duplex fading relay channels’, IEEE J. Sel. Areas Commun.,2006, 25, (2), pp. 379–389

2 Zhang, S.L., Liew, C.S., Lam, P.P.: ‘Hot topic: physical-layer networkcoding’. Proc. 12th Annual ACM Int. Conf. on Mobile Computingand Networking, 2006, pp. 358–365

3 Yang, Y., Ge, J.H., Ji, Y.C., Gao, Y.: ‘Performance analysis andinstantaneous power allocation for two-way opportunistic amplify-and-forward relaying’, IET Commun., 2011, 5, (10), pp. 1430–1439

4 Ping, J.J., Ting, S.: ‘Rate performance of AF two-way relaying in lowSNR region’, IEEE Commun. Lett., 2009, 13, (4), pp. 233–235

5 Hammerstrom, I., Kuhn, M., Esli, C., Zhao, J., Wittneben, A., Bauch,G.: ‘MIMO two-way relaying with transmit CSI at the relay’. Proc.IEEE Workshop Signal Processing Advanced WirelessCommunications, 2007, pp. 1–5

6 Louie, R.H.Y., Li, Y.H., Vucetic, B.: ‘Practical physical layer networkcoding for two-way relay channels: performance analysis andcomparison’, IEEE Trans. Wirel. Commun., 2010, 9, (2), pp. 764–777

7 Ju, M.C., Kim, I.-M.: ‘Relay selection with ANC and TDBC protocolsin bidirectional relay networks’, IEEE Trans. Commun., 2010, 58, (12),pp. 3500–3511

8 Yi, Z.H., Ju, M.C., Kim, I.-M.: ‘Outage probability and optimum powerallocation for analog network coding’, IEEE Trans. Wirel. Commun.,2011, 10, (2), pp. 407–412

9 Upadhyay, P.K., Prakriya, S.: ‘Performance of analog network codingwith asymmetric traffic requirements’, IEEE Commun. Lett., 2011, 15,(6), pp. 647–649

10 Hwang, K.-S., Ko, Y.-C., Alouini, M.-S.: ‘Performance bounds fortwo-way amplify-and-forward relaying based on relay path selection’.Proc. IEEE Veh. Technol. Conf., 2009, pp. 1–5

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11 Yang, J., Fan, P.Z., Duong, T.Q., Lei, X.F.: ‘Exact performance oftwo-way AF relaying in Nakagami-m fading environment’, IEEETrans. Wirel. Commun., 2011, 10, (3), pp. 980–987

12 Chen, S.P., Wang, W.B., Zhang, X., Sun, Z.: ‘Performance analysis ofOSTBC transmission in amplify-and-forward cooperative relaynetwork’, IEEE Trans. Veh. Technol., 2010, 59, (1), pp. 105–113

13 Zhang, Y.Y., Ma, Y., Tafazolli, R.: ‘Power allocation for bidirectionalAF relaying over Rayleigh fading channels’, IEEE Commun. Lett.,2010, 14, (2), pp. 145–147

14 Li, Y., Zhang, X., Peng, M.G., Wang, W.B.: ‘Power provisioning andrelay positioning for two-way relay channel with analog networkcoding’, IEEE Signal Process. Lett., 2011, 18, (9), pp. 517–520

15 Wu, Y., Patzold, M.: ‘Performance analysis of amplify-and-forwardcooperative communication systems with channel estimation errors’.Proc. IEEE Int. Conf. on Commun. Syst., 2008, pp. 1620–1624

16 Edfors, O., Sandell, M., Van de Beek, J.-J., Wilson, S.K., Borjesson,P.O.: ‘OFDM channel estimation by singular value decomposition’,IEEE Trans. Commun., 1998, 46, (7), pp. 931–939

17 Abramowitz, M., Stegun, I.A.: ‘Handbook of mathematical functionswith formulas, graphs, and mathematical tables’ (Dover, 1970, 9th edn.)

18 Simon, M.K., Alouini, M.-S.: ‘Digital communication over fadingchannels’ (Wiley, 2005, 2nd edn.)

19 Gradshteyn, I.S., Ryzhik, I.M.: ‘Table of integrals, series, and products’(Academic, 2007, 7th edn.)

20 Hasna, M.O., Alouini, M.: ‘Optimal power allocation for relayedtransmissions over Rayleigh fading channels’, IEEE Trans. Wirel.Commun., 2004, 3, (6), pp. 1999–2004

21 Su, W., Sadek, A.K., Liu, K.J.R.: ‘Cooperative communicationprotocols in wireless networks: performance analysis and optimumpower allocation’, Wirel. Pers. Commun., 2008, 44, (2), pp. 181–217

9 Appendix

9.1 Proof of Lemma 1

FX (x) = Px1x2

x1 + x2 + b≤ x

[ ]

= P[x2 ≤ x] + P x1 ≤ xx2 + b

x2 − x, x2 . x

[ ]

= 1 − e−b2x +∫1

x

b2e−b2x2 (1 − e−b1x(x2+b/x2−x)) dx2

=t=x2−x1 − b2e−(b2+b1)x

∫1

0

e−b2te−b1x(x+b/t) dt (45)

With the help of [19] [Eq. (3.471.9)] and [19] [Eq.(8.486.16)], we can obtain (9).

9.2 Proof of Theorem 3

The form of the probability of effective SNR (17) is not easyto deal with. However, in practical applications

s2e,A

s2h,A

,s2

e,B

s2h,B

≪ 1,PR

N0

,PA

N0

,PB

N0

≫ 1

We can obtain 2��b1b2((lg)2 + blg)

√� 0 and (lAg)2 ≫

bAlAg. With these assumptions and the help of [17][Eq.(9.6.9)] and [17] [Eq.(9.6.8)] the PDF of gA can berewritten as

pg,A(g) ≃ lA(b2,A + b1,A)e−(b2,A+b1,A)lAg

− 4l2Agb1,Ab2,Ae−(b2,A+b1,A)lAg ln 2lAg

��b1,Ab2,A

√( )(46)


www.ietdl.org

Thus, the MGF of gA can be given as

Mg,A(s) =∫1

0

pg,A(g)e−sg dg

≃lA(b1,A + b2,A)

lA(b1,A + b2,A) + s+

2l2Ab1,Ab2,A

((b1,A + b2,A)lA + s)2

× (2 ln ((b1,A + b2,A)lA + s) − ln (4l2Ab1,Ab2,A)

+ 2C − 2) (47)

Then, the SER of the system conditioned on the estimatedchannel coefficients can be expressed as follows [21] forM-QAM modulation

PSER ≃ 4K

p

∫p/2

0

Mg

bQAM

2 sin2 u

( )du

− 4K2

p

∫p/4

0

Mg

bQAM

2 sin2 u

( )du (48)

where

K = 1 − 1��M

√ , bQAM = 3

M − 1

Then, substituting (47) into (48) yields (18). The form ofPSER,B, likewise, can be written as (19)

9.3 Proof of (20) and (21)

Taking the derivative of (12) with respect to gthA,respectively, we can obtain

pg,A(gA) ≃lA(b2,A + b1,A) (49)

The MGF can be written as


Mg,A(s) =∫1

0

Pg,A(g)e−sg dg ≃lA(b1,A + b2,A)

s(50)

Equation 50 is corresponding to [21] (54). Then, substituting(50) into (48), we can obtain (20). Similarly, we can obtain(21).

9.4 Derivation of (26)

With the second derivative test, it is found that (24) and (25)form a convex optimisation problem. According to (12) and(13), (25a) and (25b) can be written as

PrPBs2e,A + PBN0

s2h,A

+PrPBs

2e,B + PrN0 + PAN0

s2h,B

+ln (1 − Pout,th A)PrPB

gth A

W L1 ≤ 0 (51)

PrPAs2e,A + PrN0 + PBN0

s2h,A

+PrPAs

2e,B + PAN0

s2h,B

+ln (1 − Pout,th B)PrPA

gth B

W L2 ≤ 0 (52)

Clearly, the global minimum is achieved when the inequalityconstraints in (51) and (52) are satisfied with equality. When aLagrangian multiplier method is adopted, the problem can bewritten as

L(PA, PB, PR, l1, l2) = f − l1L1 − l2L2 (53)

where f ¼ PA + PB + PR, l1 and l2 are Lagrangianmultipliers. Thus, the KKT conditions are formulated as(see (54))

It means that the optimal solution should satisfy all theequations in (54). After algebraic manipulations on (54), wecan obtain the solutions given in (26).

∂L

∂Pr

= 1 − l1

PBs2e,A

s2h,A

+PBs

2e,B + N0

s2h,B

+ln (1 − Pout,th A)PB

gth A

( )

−l2

PAs2e,A + N0

s2h,A

+PAs

2e,B

s2h,B

+ln (1 − P

out,th B)PA

gth B

( )= 0

∂L

∂PA

= 1 − l1

N0

s2h,B

− l2

Prs2e,A

s2h,A

+Prs

2e,B + N0

s2h,B

+ln (1 − P

out,th B)Pr

gth B

( )= 0

∂L

∂PB

= 1 − l1

Prs2e,A + N0

s2h,A

+Prs

2e,B

s2h,B

+ln (1 − P

out,th A)Pr

gth A

( )− l2

N0

s2h,A

= 0

∂L

∂l1

= −PrPBs

2e,A + PBN0

s2h,A

−PrPBs

2e,B + PrN0 + PAN0

s2h,B

−ln (1 − P

out,th A)PrPB

gth A

= 0

∂L

∂l2

= −PrPAs

2e,A + PrN0 + PBN0

s2h,A

−PrPAs

2e,B + PAN0

s2h,B

−ln (1 − P

out,th B)PrPA

gth B

= 0

PA, PB, PR ≥ 0, l1, l2 ≥ 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(54)

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