Relaying Energy Allocation Scheme Based on Multi-User SWIPT Relaying System

Abstract

This paper investigates the relay energy allocation scheme based on the time switching (TS) operation strategy in multi-user simultaneous wireless information and power transfer (SWIPT) relaying system, where the relay is energy-constrained and utilizes the energy harvested in the energy harvesting mode to amplify and forward the information of the users. In the multi-user relaying system, the distances of the receivers of the users to the relay may be different. Thus, the total information rate maximization model is proposed and the corresponding energy allocation scheme is derived. After comparing the information rates of users, it is found that the information rates of users who are far from the relay are significantly lower than those of users who are close to the relay, i.e., “far-near” phenomenon. For the “far-near” problem, we propose the common information rate maximization model, and derive the corresponding energy allocation scheme. In this model, the information rates of all users are equal, which ensures the fairness of the information rate in multi-user relaying communication system. The simulation results show that the energy allocation scheme based on the common information rate maximization model can effectively solve the “far-near” problem in multi-user relaying communication system.

Appendices

Appendix 1

Notation: f¹(x)|_x represents the derivative of f(x) with respect tox; f²(x)|_x denotes the derivative of f¹(x)|_x with respect tox

In (11), taking the first and second derivatives of R_i(τ) with respect toτ_i, we obtain

$$ \nabla {R}_i\left(\uptau \right)=\frac{1}{K}\cdotp \frac{1}{\ln 2}\cdotp \frac{\gamma_i}{1+{\gamma}_i\cdotp {\tau}_i} $$

(16)

$$ {\nabla}^2{R}_i\left(\uptau \right)=\frac{1}{K}\cdotp \frac{1}{\ln 2}\cdotp \frac{-{\gamma_i}^2}{{\left(1+{\gamma}_i\cdotp {\tau}_i\right)}^2} $$

(17)

The Hessian matrix of R_i(τ) is defined as

$$ {\nabla}^2{R}_i\left(\uptau \right)=\left[\ {d}_{j,j}^{(i)}\ \right],\kern0.5em 1\le j\le K $$

(18)

where ${d}_{j,j}^{(i)}$ represents the entry for row j, column j of∇²R_i(τ), and

$$ {d}_{j,j}^{(i)}=\left\{\begin{array}{c}\frac{1}{K}\cdotp \frac{1}{\ln 2}\cdotp \frac{-{\gamma_i}^2}{{\left(1+{\gamma}_i\cdotp {\tau}_i\right)}^2}\kern0.75em ,\kern0.75em j=i\\ {}\kern3.5em 0\kern3.5em ,\kern1em \mathrm{others}\end{array}\right. $$

(19)

For any real vectorv = [v₁, ⋯v_K]^T, we have

$$ {\mathrm{v}}^T{\nabla}^2{R}_i\left(\uptau \right)\mathrm{v}=\frac{1}{K\cdotp \ln 2}\cdotp \frac{-{\gamma_i}^2}{{\left(1+{\gamma}_i\cdotp {\tau}_i\right)}^2}\cdotp {v}_i^2\le 0 $$

(20)

i.e., ∇²R_i(τ)is a negative semidefinite matrix for any i; therefore, R_i(τ)is a convex function ofτ = [τ₁, ⋯τ_K].

Appendix 2

The Lagrangian of the problem P₁ can be expressed as

$$ {L}_{sum}\left(\uptau, \upsilon \right)={R}_{sum}\left(\uptau \right)-\upsilon \left({\sum}_{i=1}^K{\tau}_i-1\right) $$

(21)

where υ represents the lagrangian multiplier. Taking the first derivatives of L_sum(τ, υ) with respect to υ andτ_i, respectively, we have

$$ {\left.{L}_{sum}{\left(\tau, \upsilon \right)}^1\right|}_v={\sum}_{i=}^K{\tau}_i-1 $$

(22)

$$ {\left.{L}_{sum}{\left(\tau, \upsilon \right)}^1\right|}_{ti}=\frac{\partial }{\partial_{\tau i}}{R}_{sum}\left(\tau \right)-\upsilon, \circ \circ \circ i=1,\cdots, K $$

(23)

Making L_sum(τ, υ)¹⃓_v = 0 andL_sum(τ, υ)¹⃓_ti = 0, we have

$$ {\sum}_{i=1}^K{\tau}_i-1=0 $$

(24)

$$ \frac{1}{K}\cdotp \frac{1}{\ln 2}\cdotp \frac{\gamma_i}{1+{\gamma}_i\cdotp {\tau}_i}-\upsilon =0,\kern0.5em i=1,\cdots, K $$

(25)

According to (24), we obtain

$$ {\sum}_{i=1}^K{\tau}_i=1 $$

(26)

According to (25), we obtain

$$ \frac{\gamma_i}{1+{\gamma}_i\cdotp {\tau}_i}= K\upsilon \cdotp \ln 2=\mathrm{constant} $$

(27)

So, we have

$$ \frac{\gamma_1}{1+{\gamma}_1\cdotp {\tau}_1}=\frac{\gamma_2}{1+{\gamma}_2\cdotp {\tau}_2}=\cdots =\frac{\gamma_K}{1+{\gamma}_K\cdotp {\tau}_K} $$

(28)

According to$\frac{\gamma_1}{1+{\gamma}_1\cdotp {\tau}_1}=\frac{\gamma_2}{1+{\gamma}_2\cdotp {\tau}_2}$, we have

$$ {\tau}_2=\frac{1}{\gamma_1}-\frac{1}{\gamma_2}+{\tau}_1 $$

(29)

Similarly, we can obtain

$$ {\tau}_i=\frac{1}{\gamma_1}-\frac{1}{\gamma_i}+{\tau}_1,\kern0.5em i=2,\cdots, K $$

(30)

Combined with${\sum}_{i=1}^K{\tau}_i=1$, we obtain

$$ {\tau}_1+\left(\frac{1}{\gamma_1}-\frac{1}{\gamma_2}+{\tau}_1\right)+\cdots +\left(\frac{1}{\gamma_1}-\frac{1}{\gamma_K}+{\tau}_1\right)=1 $$

(31)

Further, we have

$$ K{\tau}_1+\frac{K-1}{\gamma_1}-\sum \limits_{j=2}^K\frac{1}{\gamma_j}=1 $$

(32)

So, we have

$$ {\tau}_1=\frac{1}{K}-\frac{K-1}{K{\gamma}_1}+\frac{1}{K}\sum \limits_{j=2}^K\frac{1}{\gamma_j} $$

(33)

By substituting (33) into (29), we obtain

$$ {\tau}_2=\frac{1}{K}-\frac{K-1}{K{\gamma}_2}+\frac{1}{K}\sum \limits_{\begin{array}{c}j=1\\ {}j\ne 2\end{array}}^K\frac{1}{\gamma_j} $$

(34)

Similarly, we have

$$ {\tau}_i=\frac{1}{K}-\frac{K-1}{K{\gamma}_i}+\frac{1}{K}{\sum}_{\begin{array}{c}j=1\\ {}j\ne i\end{array}}^K\frac{1}{\gamma_j},\kern0.5em i=1,\cdots, K $$

(35)