IEEE COMMUNICATIONS LETTERS, VOL. 15, NO. 9, SEPTEMBER 2011
959
Optimal Power Allocation for the Two-Way Relay Channel with Data Rate Fairness Mylene Pischella and Didier Le Ruyet, Member, IEEE
Abstract—This letter studies data rate fairness on the twoway relay channel. It analytically determines the optimal power values at both nodes and at the relay, that lead to a maximization of the sum rate under the fairness constraint. Amplify-andForward (AF) and Decode-and-Forward (DF) relaying protocols are considered. For AF, the optimization problem is turned into a single-variable convex optimization problem. For DF, rate balancing between Multiple Access and broadcast phases must be performed prior to setting nodes powers. Both optimized protocols are compared with reference AF and DF in terms of data rates through numerical simulations. Index Terms—Two-way relay channel, power optimization, fairness, amplify-and-forward, decode-and-forward.
I. I NTRODUCTION
T
WO-WAY relaying techniques achieve higher spectral efficiency than conventional one-way relaying techniques [1] [2]. On the two-way relay channel with two nodes and one relay, bi-directional transmission requires two time intervals. In the first time interval, both nodes jointly transmit their data to the relay. The channel is then equivalent to a Multiple Access Channel (MAC). In the second time interval, the relay broadcasts the received data to both nodes. Each node removes its own information, called self-interference, and decodes the other node’s data. The relaying protocol may be Amplifyand-Forward (AF) [3] [4], Decode-and-Forward (DF) [2], or new protocols making use of recent advances on physical layer network coding [5]. Bi-directional relaying may be an interesting transmission scheme for providing symmetrical data rates, since both nodes may play an identical role in the transmission, provided that adequate power control is performed. It could thus be used for real-time applications like voice or video-conferencing. In this letter, we investigate fairness in achievable data rates for the SISO two-way relay channel. The objective is to determine the optimal power allocation at both nodes and at the relay that maximizes the sum rate, under a sum power constraint, and under the following fairness constraint: the rate from Node 𝐴 to Node 𝐵, 𝑅𝐴𝐵 , must be equal to the rate from Node 𝐵 to Node 𝐴, 𝑅𝐵𝐴 . To the best of our knowledge, this problem has not been solved yet. AF and DF relaying protocols are studied, in Sections II and III respectively. We provide analytical solutions, and compare the performances with equal power allocation in Section IV.
Manuscript received April 12, 2011. The associate editor coordinating the review of this letter and approving it for publication was N.-D. Dao. The authors are with the CEDRIC/LAETITIA Laboratory, Conservatoire National des Arts et M´etiers, 75003 Paris, France (e-mail:
[email protected]). Digital Object Identifier 10.1109/LCOMM.2011.070711.110789
The following notations are used: 𝑃𝐾 is the transmit power and n𝐾 is an additive white Gaussian noise with variance 𝜎 2 , with 𝐾 ∈ {𝐴, 𝐵, 𝑅} corresponding to one of the Nodes or the relay. We consider a narrowband Time Division Duplex system, where the channel coherence time is larger than two time intervals, and where channel reciprocity holds, thanks to accurate calibration of the radio-frequency electronic circuitry [6]. The channel between Node 𝐾 ∈ {𝐴, 𝐵} and the relay is denoted as ℎ𝐾 . The sum power used in the two time intervals is set to a fixed value, 𝑃sum = 𝑃𝐴 + 𝑃𝐵 + 𝑃𝑅 . II. A MPLIFY- AND -F ORWARD In the first time interval, Nodes 𝐴 and 𝐵 jointly transmit to the relay. Data are transmitted using a Gaussian codebook. The relay receives y𝑅 = ℎ𝐴 x𝐴 + ℎ𝐵 x𝐵 + n𝑅 . In the second time interval, the relay broadcasts 𝛾y𝑅 , with an amplification factor 𝛾: √ 𝑃𝑅 (1) 𝛾= 2 ∣ℎ𝐴 ∣ 𝑃𝐴 + ∣ℎ𝐵 ∣2 𝑃𝐵 + 𝜎 2 Node 𝐴 receives y𝐴 = 𝛾ℎ𝐴 2 x𝐴 + 𝛾ℎ𝐴 ℎ𝐵 x𝐵 + 𝛾ℎ𝐴 n𝑅 + n𝐴 . The self-interference term, 𝛾ℎ𝐴 2 x𝐴 , is removed. Then the Signal to Noise Ratio (SNR) at Node 𝐴 for the remaining signal y ˆ𝐴 = 𝛾ℎ𝐴 ℎ𝐵 x𝐵 + 𝛾ℎ𝐴 n𝑅 + n𝐴 is: SNR𝐴 =
2
2
𝛾 2 ∣ℎ𝐴 ∣ ∣ℎ𝐵 ∣ 𝑃𝐵 2
𝛾 2 ∣ℎ𝐴 ∣ 𝜎 2 + 𝜎 2
(2)
Similarly, the SNR at Node 𝐵 is: SNR𝐵 =
𝛾 2 ∣ℎ𝐵 ∣2 ∣ℎ𝐴 ∣2 𝑃𝐴 2
𝛾 2 ∣ℎ𝐵 ∣ 𝜎 2 + 𝜎 2
(3)
We will now determine the optimal power values, 𝑃𝐴 , 𝑃𝐵 and 𝑃𝑅 that lead to fairness: SNR𝐴 = SNR𝐵 = SNR, and then maximize the SNR. The corresponding sum rate is: 𝑅sum = 1 1 2 log2 (1 + SNR𝐵 ) + 2 log2 (1 + SNR𝐴 ) = log2 (1 + SNR). 2
A. Case ∣ℎ𝐴 ∣ = ∣ℎ𝐵 ∣
2
From eq. (2) and (3), SNR𝐴 = SNR𝐵 if 𝑃𝐴 = 𝑃𝐵 . Let us denote ∣ℎ∣2 = ∣ℎ𝐴 ∣2 = ∣ℎ𝐵 ∣2 and 𝑃 = 𝑃𝐴 = 𝑃𝐵 . By replacing 𝑃𝑅 by 𝑃sum − 2𝑃 , and using eq. (1), we can express the fair SNR as the following function of 𝑃 : SNR =
∣ℎ∣4 𝑃 (𝑃sum − 2𝑃 ) 𝜎 4 + 𝜎 2 ∣ℎ∣2 𝑃sum
(4)
The SNR is maximized when 𝑔 ′ (𝑃 ∗ ) = 0, where 𝑔(𝑃 ) = 4 ∣ℎ∣ 𝑃 (𝑃sum − 2𝑃 ). The optimal value is 𝑃 ∗ = 𝑃4sum . Consequently, the optimal power allocation achieving fairness when 𝑃sum ∗ ∣ℎ𝐴 ∣2 = ∣ℎ𝐵 ∣2 is: 𝑃𝐴∗ = 𝑃𝐵∗ = 𝑃sum 4 and 𝑃𝑅 = 2 .
c 2011 IEEE 1089-7798/11$25.00 ⃝
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IEEE COMMUNICATIONS LETTERS, VOL. 15, NO. 9, SEPTEMBER 2011 𝑎𝑃 +𝑏𝑃 2
B. Case ∣ℎ𝐴 ∣2 ∕= ∣ℎ𝐵 ∣2
𝐴 𝐴 Let us study the convexity of function of 𝑓 (𝑃𝐴 ) = 𝑐+𝑑𝑃 𝐴 2 2 with 𝑎 = 𝑃sum , 𝑏 = −2, 𝑐 = 𝜎 + ∣ℎ𝐵 ∣ 𝑃sum and 2 2 𝑑 = 2(∣ℎ𝐴 ∣ − ∣ℎ𝐵 ∣ ). Its second derivative is:
In that case, SNR𝐴 = SNR𝐵 is equivalent to : 𝛾2 =
2
𝑃𝐴 − 𝑃𝐵
(5)
2
∣ℎ𝐵 ∣ 𝑃𝐵 − ∣ℎ𝐴 ∣ 𝑃𝐴
𝑃𝑅 can be expressed as a function of 𝑃𝐴 and 𝑃𝐵 , using (1) and (5): ) ( (𝑃𝐴 − 𝑃𝐵 ) ∣ℎ𝐴 ∣2 𝑃𝐴 + ∣ℎ𝐵 ∣2 𝑃𝐵 + 𝜎 2 𝑃𝑅 = (6) 2 2 ∣ℎ𝐵 ∣ 𝑃𝐵 − ∣ℎ𝐴 ∣ 𝑃𝐴 2
We can notice that there is no solution if ∣ℎ𝐵 ∣ 𝑃𝐵 = ∣ℎ𝐴 ∣2 𝑃𝐴 . In all other cases, the fair SNR is equal to: SNR =
2
(𝑃𝐴 − 𝑃𝐵 ) ∣ℎ𝐴 ∣ ∣ℎ𝐵 ∣ ( ) 2 2 ∣ℎ𝐵 ∣ − ∣ℎ𝐴 ∣ 𝜎 2
2
(7)
Two different cases must be distinguished. 1) Case ∣ℎ𝐵 ∣2 > ∣ℎ𝐴 ∣2 : The optimization problem corresponding to sum rate maximization is then: max{𝑃𝐴 ,𝑃𝐵 } log2 (1 + SNR)
s.t. ∣ℎ𝐵 ∣2 𝑃𝐵 − ∣ℎ𝐴 ∣2 𝑃𝐴 > 0
(𝐶1 )
s.t. 𝑃𝐴 > 𝑃𝐵 s.t. 𝑃𝐴 + 𝑃𝐵 + 𝑃𝑅 = 𝑃sum
(𝐶2 ) (𝐶3 )
(8)
where constraints (𝐶1 ) and (𝐶2 ) are necessary for 𝛾 2 , 𝑃𝑅 and the SNR to be positive. Using (𝐶3 ) and eq. (6), 𝑃𝐵 can be expressed as functions of 𝑃𝐴 . We obtain: 𝑃𝐵 =
2
(𝜎 2 + ∣ℎ𝐴 ∣ 𝑃sum )𝑃𝐴 2
2
2
𝜎 2 + ∣ℎ𝐵 ∣ 𝑃sum + 2(∣ℎ𝐴 ∣ − ∣ℎ𝐵 ∣ )𝑃𝐴
(9)
Eq. (7) shows that maximizing the SNR is equivalent to maximizing the difference (𝑃𝐴 − 𝑃𝐵 ): ) ( ) 2 2 ( 𝑃sum 𝑃𝐴 − 2𝑃𝐴2 ∣ℎ𝐵 ∣ − ∣ℎ𝐴 ∣ 𝑃𝐴 − 𝑃𝐵 = 2 2 2 𝜎 2 + ∣ℎ𝐵 ∣ 𝑃sum + 2(∣ℎ𝐴 ∣ − ∣ℎ𝐵 ∣ )𝑃𝐴 ) ( = ∣ℎ𝐵 ∣2 − ∣ℎ𝐴 ∣2 𝑓 (𝑃𝐴 ) (10) 2
2
Since ∣ℎ𝐵 ∣ > ∣ℎ𝐴 ∣ , this corresponds to maximizing 𝑓 (𝑃𝐴 ). Constraint (𝐶1 ) can also be turned into a function of 𝑃𝐴 , by replacing 𝑃𝐵 with its expression in (9): 𝑃𝐴
0. The denominator of 𝑓 (𝑃𝐴 ) is necessarily positive from eq. (11). Thus, (𝐶2 ) is equivalent to 𝑃𝐴 < 𝑃2sum . This constraint is stronger than (11) 2 2 −(𝜎2 +∣ℎ𝐴 ∣2 𝑃sum ) 𝐵 ∣ 𝑃sum = < 0. since 𝑃2sum − 2𝜎∣ℎ+∣ℎ 2 2 2(∣ℎ𝐵 ∣2 −∣ℎ𝐴 ∣2 ) ( 𝐵 ∣ −∣ℎ𝐴 ∣ ) The initial optimization problem (8) can consequently be written as the following optimization problem in 𝑃𝐴 : max𝑃𝐴
𝑃sum 𝑃𝐴 − 2𝑃𝐴2
𝜎 2 + ∣ℎ𝐵 ∣2 𝑃sum + 2(∣ℎ𝐴 ∣2 − ∣ℎ𝐵 ∣2 )𝑃𝐴 𝑃sum s.t. 𝑃𝐴 < 2
(𝐶1′ ) (12)
′′
𝑓 (𝑃𝐴 ) = −
2𝑐(𝑎𝑑 − 𝑏𝑐) (𝑑𝑃𝐴 + 𝑐)3 2
2
Since 2𝑐(𝑎𝑑− 𝑏𝑐) = 4(𝜎 2 + ∣ℎ𝐵 ∣ 𝑃sum )(𝜎 2 + ∣ℎ𝐴 ∣ 𝑃sum ) > 0, ′′ 𝑓 is of the opposite sign of 𝑔(𝑃𝐴 ) = (𝑑𝑃𝐴 + 𝑐)3 . 𝑔(𝑃𝐴 ) > 0 2 2 𝐵 ∣ 𝑃sum if 𝑃𝐴 < 2𝜎∣ℎ+∣ℎ . This is always true since constraint ( 𝐵 ∣2 −∣ℎ𝐴 ∣2 ) (11) holds. Consequently, in the allowed variation area for ′′ 𝑃𝐴 , 𝑓 (𝑃𝐴 ) < 0, so function 𝑓 is concave. The optimization problem (12) is thus convex, and has a unique global optimum. ′ It is equal to the solution of 𝑓 (𝑃𝐴∗ ) = 0 that verifies constraint 2 ′ 𝑏𝑑𝑃𝐴 +2𝑏𝑐𝑃𝐴 +𝑎𝑐 , the potential solutions (𝐶1′ ). Since 𝑓 (𝑃𝐴 ) = (𝑑𝑃𝐴 +𝑐)2 are the two roots of the numerator. Both roots are real, but only one of them fulfills constraint (𝐶1′ ). The global optimum of problem (12) is: 𝑃𝐴∗ = ( ) √ 2 2 2 𝜎 2 + ∣ℎ𝐵 ∣ 𝑃sum − (𝜎 2 + ∣ℎ𝐵 ∣ 𝑃sum )(𝜎 2 + ∣ℎ𝐴 ∣ 𝑃sum ) ) ( 2 ∣ℎ𝐵 ∣2 − ∣ℎ𝐴 ∣2 (13) The optimal values of 𝑃𝐵 , deduced from eq. (9), is: 𝑃𝐵∗ = ) √ ( − 𝜎 2 + ∣ℎ𝐴 ∣2 𝑃sum + (𝜎 2 + ∣ℎ𝐵 ∣2 𝑃sum )(𝜎 2 + ∣ℎ𝐴 ∣2 𝑃sum ) ) ( 2 2 2 ∣ℎ𝐵 ∣ − ∣ℎ𝐴 ∣ (14) Equations (13) and (14) show that 𝑃𝐴∗ + 𝑃𝐵∗ = 𝑃2sum . This implies that the optimal relay power is always equal to 𝑃𝑅∗ = 𝑃sum 2 . Thus, the relay gets twice more power than Nodes 𝐴 and 𝐵, in order to achieve data rate fairness. 2 2 2) Case ∣ℎ𝐴 ∣ > ∣ℎ𝐵 ∣ : That case can be treated similarly to the previous one, by writting the initial optimization problem (8) as a convex optimization problem in 𝑃𝐵 . The expression of the global optimum, 𝑃𝐵∗ , is still (14). Consequently, 𝑃𝐴∗ and 𝑃𝑅∗ are also unchanged. III. D ECODE - AND -F ORWARD The rate region of the two-way relay channel, using Decodeand-Forward, has been determined in [2]: { ( ) ( )} 1 𝑃𝐴 𝑃𝑅 min log 2 1 + ∣ℎ𝐴 ∣2 2 , log2 1 + ∣ℎ𝐵 ∣2 2 2 𝜎 𝜎 { ( ) ( )} 1 2 𝑃𝐵 2 𝑃𝑅 ≤ min log2 1 + ∣ℎ𝐵 ∣ ∣ , log 1 + ∣ℎ 𝐴 2 2 𝜎2 𝜎2 ( ) 1 2 𝑃𝐴 2 𝑃𝐵 𝑅𝐴𝐵 + 𝑅𝐵𝐴 ≤ log2 1 + ∣ℎ𝐴 ∣ + ∣ℎ𝐵 ∣ 2 𝜎2 𝜎2 (15)
𝑅𝐴𝐵 ≤ 𝑅𝐵𝐴
where the last inequality is due to the joint data rate requirement on the MAC. In order to maximize the sum rate under the sum power and fairness constraints, 𝑃𝑅 is set to the minimum value such that the broadcast link does not decrease each point-to-point
PISCHELLA and LE RUYET: OPTIMAL POWER ALLOCATION FOR THE TWO-WAY RELAY CHANNEL WITH DATA RATE FAIRNESS
60
We can notice that in the particular case when ∣ℎ𝐴 ∣2 = ∣ℎ𝐵 ∣2 , eq. (16) simplifies to the Equal Power Allocation (EPA) solution: 𝑃𝐴∗ = 𝑃𝐵∗ = 𝑃𝑅∗ = 𝑃3sum .
P (σ2/|h |2=10) A
B
P (σ2/|h |2=10) B
50
B
2
2
P (σ /|h | =10) R
B
Power value (Psum=100)
2
2
IV. S IMULATION R ESULTS
P (σ /|h | =1) A
40
B
2
2
PB (σ /|hB| =1) 2
2
2
2
A. Power variations for Amplify-and-Forward
PR (σ /|hB| =1) 30
P (σ /|h | =0.1) A
B
2
B
B
2
2
PR (σ /|hB| =0.1)
10
0 0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2
κ
Fig. 1.
𝑃𝐴 , 𝑃𝐵 and 𝑃𝑅 variations as functions of 𝜅2 .
B. Point-to-point data rate
7
6
AF, Power optimization Average AF, PA=PB=Psum/4
Point−to−point rate (bits/s/Hz)
Minimum AF,PA=PB=Psum/4 5
4
Maximum AF, P =P =P /4 A
B
sum
DF, Power optimization Average DF, PA=PB=Psum/3 Minimum DF, PA=PB=Psum/3 Maximum DF, P =P =P /3 A
B
sum
3
2
1
0 5
Fig. 2.
The variations of the power values depending on 𝜅2 = 2 for different levels of ∣ℎ𝜎 ∣2 , are given on Fig. 1. The 𝐵 optimal relay power is always 𝑃𝑅∗ = 𝑃sum 2 . Consequently, in order to achieve fairness, power must be equally shared between both transmission phases. Besides, in the MAC phase, power is split between nodes so that the node with best channel conditions gets the least power. Power allocation not only depends on the channel gains and on 𝑃sum , but also on the noise variance, due to noise amplification at the relay. ∣ℎ𝐴 ∣2 , ∣ℎ𝐵 ∣2
2
P (σ /|h | =0.1) 20
961
10
15
20
25 30 2 Psum/σ (dB)
35
40
45
50
Point-to-point data rate comparison.
rate. { From the first two equations of (15), it is equal to: 𝑃𝑅 = } ∣ℎ𝐴 ∣2 ∣ℎ𝐵 ∣2 . max ∣ℎ 2 𝑃𝐴 , 2 𝑃𝐵 ∣ℎ𝐴 ∣ 𝐵∣ Then, data rate fairness 𝑅𝐴𝐵 = 𝑅𝐵𝐴 can only be achieved 2 2 if ∣ℎ𝐴 ∣ 𝑃𝐴 = ∣ℎ𝐵 ∣ 𝑃𝐵 . By introducing this constraint and 𝑃𝑅 expression in the sum power, we obtain the optimal power allocation: 𝑃sum 𝑃sum ∗ ∗ ∗ ∗ 𝑃𝐴∗ = 2 , 𝑃𝐵 = 2 , 𝑃𝑅 = max {𝑃𝐴 , 𝑃𝐵 } 𝐷 ∣ℎ𝐴 ∣ 𝐷 ∣ℎ𝐵 ∣ { } 1 1 1 1 with 𝐷 = (16) 2 + 2 + max 2, 2 ∣ℎ𝐴 ∣ ∣ℎ𝐵 ∣ ∣ℎ𝐴 ∣ ∣ℎ𝐵 ∣ ( ) 𝑃sum = 𝑅𝐵𝐴 = 12(log2 1 + )𝐷𝜎 The individual rates are: 2 . ( 𝑅𝑃𝐴𝐵 ) 𝑃sum sum However, since log2 1 + 𝐷𝜎 ≥ 12 log2 1 + 2 𝐷𝜎 2 2 , the MAC joint data rate constraint always sets the limit for the achievable sum data rate. It is finally equal to : ( ) 1 𝑃sum 𝑅sum = 2𝑅𝐴𝐵 = 2𝑅𝐵𝐴 = log2 1 + 2 (17) 2 𝐷𝜎 2
The point-to-point data rates for AF and DF with and without power optimization as a function of 𝑃𝜎sum 2 is represented on Fig. 2. The reference cases correspond to the optimum 2 2 values when ∣ℎ𝐴 ∣ = ∣ℎ𝐵 ∣ : 𝑃𝐴 = 𝑃𝐵 = 𝑃4sum and 𝑃𝑅 = 𝑃sum 2 with AF, and EPA with DF. Fig. 2 also shows the minimum and maximum data rates with these power allocations. With power optimization, higher data rates are reached with AF than with DF when 𝑃𝜎sum 2 exceeds 11 dB. The average data rate is higher with power optimization than in the reference case with AF, but the opposite holds for DF. This is due to the unfairness of DF with EPA. On the contrary, even without power optimization, AF is quite fair, with low differences between the minimum and maximum data rates. V. C ONCLUSIONS This letter has determined the analytical solutions for maximizing the sum rate while achieving fairness on the two-way relay channel, with AF and DF. Numerical results have shown that fair AF is more efficient than fair DF, and that fairness also improves the average data rate for AF, compared to reference cases. R EFERENCES [1] B. Rankov and A. Wittneben, “Achievable rate regions for the two-way relay channel,” in Proc. IEEE ISIT, July 2006. [2] I. Hammerstrom, M. Kuhn, C. Esli, J. Zhao, A. Wittneben, and G. Bauch, “MIMO two-way relaying with transmit CSI at the relay,” in Proc. IEEE SPAWC, June 2007. [3] J. Ping and S. H. Ting, “Rate performance of AF two-way relaying in low SNR region,” IEEE Commun. Lett., vol. 13, pp. 233-235, Apr. 2009. [4] Y. Zhang, Y. Ma, and R. Tafazolli, “Power allocation for bidirectional AF relaying over Rayleigh fading channels,” IEEE Commun. Lett., vol. 14, pp. 145-147, Feb. 2010. [5] M. P. Wilson and K. Narayanan, “Power allocation strategies and lattice based coding schemes for bi-directional relaying,” in Proc. IEEE ISIT, July 2009. [6] M. Guillaud, D. T. M. Slock, and R. Knopp, “A practical method for wireless channel reciprocity exploitation through relative calibration,” in Proc. IEEE ISSPA, Aug.-Sep. 2005.
