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

Abstract

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.

Appendices

Appendix I

Considering the minimum rate constraints, we can obtain

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

(13)

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

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

(14)

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

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

(15)

where

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

(16)

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

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

• Case 1: f(α) has no solution in [α₁, α₂].

• Case 2: f(α) has one solution in [α₁, α₂], then the unique value is extreme point of f(α).

• Case 3: f(α) has two solutions in [α₁, α₂].

  Due to α₃, α₄ ∈ (0, 1), hence

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

(17)

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

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

(18)

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

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

(19)

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

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

(20)

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

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

(21)

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

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

(22)

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.