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Energy-Efficient Power Allocation Scheme for Uplink Distributed Antenna System with D2D Communication


In this paper, the performance of uplink distributed antenna system (DAS) with Device-to-Device (D2D) communication is investigated over composite Rayleigh fading channels, and an energy-efficient power allocation (PA) scheme is developed for D2D communication underlaying DAS. Firstly, we establish the uplink DAS model with D2D communication. Then, the optimization problem for energy efficiency (EE) maximization subject to maximum total power constraint and the minimal rate constraints of cellular user and D2D user is formulated. Based on the pseudo-concave of objective function in optimization problem, we propose an optimal PA scheme with the bisection method to obtain the optimal solution of the optimization problem. The simulation results demonstrate the effectiveness of our proposed scheme. The proposed optimal PA scheme can achieve better EE performance than the conventional equal PA scheme, and the same EE as the PA scheme based on two-dimensional search method but with lower complexity.

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The authors would like to thank the anonymous reviewers for their valuable comments which improve the quality of this paper greatly. This work is supported by National Natural Science Foundation of China (61571225), Natural Science Foundation of Jiangsu Province in China (BK20181289), and Open Research Fund of National Mobile Communications Research Laboratory of Southeast University (2017D03).

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Correspondence to Xiangbin Yu.

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Appendix I

Considering the minimum rate constraints, we can obtain

$$ {\displaystyle \begin{array}{l}{R}_c={\log}_2\left(1+\frac{m_1\alpha P}{m_2\left(1-\alpha \right)P+1}\right)\ge {R}_{c,\min}\\ {}{R}_d={\log}_2\left(1+\frac{m_3\left(1-\alpha \right)P}{m_4\alpha P+1}\right)\ge {R}_{d,\min}\end{array}} $$

From (13), it is easily obtained the lower bound and the upper bound of α

$$ {\displaystyle \begin{array}{l}\alpha \ge \frac{r_1+{m}_2{r}_1P}{m_1P+{m}_2{r}_1P}\\ {}\alpha \le \frac{m_3P-{r}_2}{m_3P+{m}_4{r}_2P}\end{array}} $$

where \( {r}_1={2}^{R_{c,\min }}-1,{r}_2={2}^{R_{d,\min }}-1 \).

Let \( {\alpha}_1=\frac{r_1+{m}_2{r}_1P}{m_1P+{m}_2{r}_1P},{\alpha}_2=\frac{m_3P-{r}_2}{m_3P+{m}_4{r}_2P} \), we can get α ∈ [α1, α2].

Appendix II

For the given P, let \( {\left.\frac{\partial {\eta}_{EE}\left(\alpha \right)}{\partial \alpha}\right|}_{P={P}_{\mathrm{min}}}=0 \), we can get the quadratic equation

$$ {n}_1{\alpha}^2+{n}_2\alpha +{n}_3=0 $$


$$ {\displaystyle \begin{array}{l}{n}_1=\left({k}_1+{k}_1{k}_2\right)\left({k}_4^2-{k}_3{k}_4\right)+\left({k}_1{k}_2-{k}_2^2\right)\left({k}_3+{k}_3{k}_4\right)\\ {}{n}_2=\left({k}_3{k}_4-{k}_3+2{k}_4\right)\left({k}_1+{k}_1{k}_2\right)\\ {}\kern1em -\left({k}_3+{k}_3{k}_4\right)\left({k}_1+{k}_1{k}_2-2{k}_2-2{k}_2^2\right)\\ {}{n}_3=\left({k}_1+{k}_1{k}_2\right)\left(1+{k}_3\right)-\left({k}_3+{k}_3{k}_4\right){\left(1+{k}_2\right)}^2\\ {}{k}_1={m}_1P,{k}_2={m}_2P,{k}_3={m}_3P,{k}_4={m}_4P\end{array}} $$

Let f(α) = n1α2 + n2α + n3, we discuss the candidate solutions for f(α) = 0 below.

Set α3, α4 as solutions for f(α) = 0, next we judge the symbol with upper and lower bounds of [α1, α2].

  1. (1)

    If f(α1)f(α2) ≤ 0, there must be the only solution in [α1, α2]. We set a unique value α3 ∈ [α1, α2].

  2. (2)

    If f(α1)f(α2) > 0, there are three cases of solutions.

  3. Case 1:

    f(α) has no solution in [α1, α2].

  4. Case 2:

    f(α) has one solution in [α1, α2], then the unique value is extreme point of f(α).

  5. Case 3:

    f(α) has two solutions in [α1, α2].

    Due to α3, α4 ∈ (0, 1), hence

$$ {\displaystyle \begin{array}{c}0<\frac{-{n}_2+\sqrt{n_2^2-4{n}_1{n}_3}}{2{n}_1}<1\\ {}0<\frac{-{n}_2-\sqrt{n_2^2-4{n}_1{n}_3}}{2{n}_1}<1\end{array}} $$

Then we can get the following formula by \( \varDelta ={n}_2^2-4{n}_1{n}_3>0 \)

$$ {\displaystyle \begin{array}{c}{k}_2+{k}_4\left(1+{k}_2\right)<{k}_3\\ {}{k}_4+{k}_2\left(1+{k}_4\right)<{k}_1\end{array}} $$

However, the symbol of n1 is uncertain so we have to discuss it under the following two cases.

$$ {n}_1>0 $$

From (17), we obtain

$$ {\displaystyle \begin{array}{c}{n}_2<\sqrt{n_2^2-4{n}_1{n}_3}<2{n}_1+{n}_2\\ {}{n}_2<-\sqrt{n_2^2-4{n}_1{n}_3}<2{n}_1+{n}_2\end{array}} $$

And then we further get n1 > 0, n2 < 0, 2n1 + n2 > 0 and n3 > 0. Moreover from (16), 2n1 + n2 = 2(1 + k4).

(k2k3 + k1(−k3 + k4 + k2k4)) ≈ k2k3 + k1(−k3 + k4 + k2k4) < k2(k3 − k1). Due to 2n1 + n2 > 0, there must be k3 > k1.

Furthermore, according to \( \sqrt{n_2^2-4{n}_1{n}_3}<2{n}_1+{n}_2 \), we derive

$$ {n}_1+{n}_2+{n}_3>0 $$

And from (16),

n1 + n2 + n3 = (1 + k4)(−k3 + k1(1 + k2 − k3 + k4 + k2k4)) ≈  − k3 + k1(1 + k2 − k3 + k4 + k2k4) < k1 − k3 < 0, which is in contradiction to (20). Thus, the case (i) does not exist.

$$ {n}_1<0 $$

Similarly from (17), we get

$$ {\displaystyle \begin{array}{c}2{n}_1+{n}_2<\sqrt{n_2^2-4{n}_1{n}_3}<{n}_2\\ {}2{n}_1+{n}_2<-\sqrt{n_2^2-4{n}_1{n}_3}<{n}_2\end{array}} $$

And then we further get n1 < 0, n2 > 0, 2n1 + n2 < 0, n3 < 0. By (18), we have

$$ {\displaystyle \begin{array}{c}{k}_2{k}_3+{k}_3{k}_4+{k}_2{k}_3{k}_4<{k}_3^2\\ {}{k}_2{k}_3+{k}_3{k}_4+{k}_2{k}_3{k}_4<{k}_1{k}_3\end{array}} $$

Combined with (16) and (22), we can derive n3 ≈ k1(1 + k3) − (1 + k2)k3(1 + k4) > k1 − k3. Because of n3 < 0, then k1 < k3.

Meanwhile, n2 ≈ k2k3(1 + k4) + k1(k4 − k3) < k4(k1 − k3) < 0, but we have derived n2 > 0. This result is conflictive. Thus, the case (ii) does not exist. From the above, the case 3 does not exist.

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Wang, G., Yu, X. & Teng, T. Energy-Efficient Power Allocation Scheme for Uplink Distributed Antenna System with D2D Communication. Mobile Netw Appl 26, 1225–1232 (2021).

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  • Device-to-Device Communication
  • Distributed Antenna System
  • Energy Efficiency
  • Power Allocation