CN104539334A

CN104539334A - Method for forming wave beam supporting multiple users to high-speed communication based on SWIPT

Info

Publication number: CN104539334A
Application number: CN201410804039.9A
Authority: CN
Inventors: 李国兵; 严飞; 吕刚明; 张国梅
Original assignee: Xian Jiaotong University
Current assignee: Xian Jiaotong University
Priority date: 2014-12-19
Filing date: 2014-12-19
Publication date: 2015-04-22

Abstract

The present invention discloses a beamforming method supporting multi-user pair high-speed communication based on SWIPT, comprising the following steps: 1) developing a new multi-antenna relay, constructing a new beamforming scheme according to the new multi-antenna relay, Then obtain the mathematical model of the maximum rate solution according to the new beamforming scheme constructed, the mathematical model of the maximum rate is non-convex; 2) by introducing the Taylor series, the mathematical model of the maximum rate solution obtained in step 1) is transformed into Convex problem, then set the initial value, and based on the initial value, maximize the transmission rate through an iterative method, and then form a beam that supports multi-user pair high-speed communication based on SWIPT according to the maximum transmission rate. The present invention can form beams supporting multi-user pair high-speed communication based on SWIPT by obtaining the maximum transmission rate.

Description

A beamforming method supporting multi-user pair high-speed communication based on SWIPT

技术领域technical field

本发明属于无线中继网络领域，涉及一种基于SWIPT下支持多用户对高速通信的波束形成方法。The invention belongs to the field of wireless relay networks, and relates to a beam forming method supporting multi-user pair high-speed communication based on SWIPT.

背景技术Background technique

中继协作通信技术在提高边缘小区的频谱效率的同时，还可以增大小区的覆盖范围和实现盲区覆盖，已成为目前研究热点之一。相关研究主要集中在单向的中继系统。然而，单向中继系统会导致频谱效率的降低。针对这一问题，研究的焦点开始转向双向中继。随之而来，在双向中继系统中，在追求传输速率的最大化的同时，往往受到中继系统中功率的限制。Relay cooperative communication technology can not only improve the spectral efficiency of edge cells, but also increase the coverage of cells and realize coverage in blind areas, which has become one of the current research hotspots. Related research mainly focuses on the one-way relay system. However, a one-way relay system will result in a reduction in spectral efficiency. In response to this problem, the focus of research began to turn to two-way relay. Subsequently, in the two-way relay system, while pursuing the maximum transmission rate, it is often limited by the power of the relay system.

以往是以单向的中继系统作为研究目标，或者双向中继系统中，当前的研究仅限于一对信号源到目的节点发送的最大速率求解问题的解决，结果表明在没有能量收集且信噪比较大的情况下，是可以逼近最优值的，但中继系统中多天线并没有被充分利用，从而不能准确的形成多用户对高速通信的波束。In the past, the one-way relay system was used as the research target, or the two-way relay system. The current research is limited to the solution of the maximum rate of a pair of signal sources sent to the destination node. The results show that there is no energy harvesting and signal noise In the case of a relatively large value, it can approach the optimal value, but the multi-antenna in the relay system is not fully utilized, so that the multi-user beam for high-speed communication cannot be accurately formed.

发明内容Contents of the invention

本发明的目的在于克服上述现有技术的缺点，提供了一种基于SWIPT下支持多用户对高速通信的波束形成方法，该方法可以通过获取最大化传输速率来形成基于SWIPT下支持多用户对高速通信的波束。The purpose of the present invention is to overcome the above-mentioned shortcoming of prior art, provide a kind of beam forming method that supports multi-user to high-speed communication under SWIPT, this method can form the beamforming method that supports multi-user to high-speed communication under SWIPT by obtaining the maximum transmission rate Communication beams.

为达到上述目的，本发明所述的基于SWIPT下支持多用户对高速通信的波束形成方法包括以下步骤：In order to achieve the above object, the beamforming method for supporting multi-user high-speed communication based on SWIPT of the present invention comprises the following steps:

1)开拓新的多天线中继，根据所述新的多天线中继构建新的波束成形方案，然后根据构建的新的波束成型方案得最大速率求解的数学模型，所述最大速率的数学模型为非凸问题；1) Develop a new multi-antenna relay, construct a new beamforming scheme according to the new multi-antenna relay, and then obtain a mathematical model for solving the maximum rate according to the new beamforming scheme constructed, and the mathematical model of the maximum rate is a non-convex problem;

2)通过引入泰勒级数将步骤1)得到的最大速率求解的数学模型转化为凸问题，再设置初始值，并在初始值的基础上通过迭代的方法得最大化传输速率，然后根据所述最大化传输速率形成基于SWIPT下支持多用户对高速通信的波束。2) Convert the mathematical model of the maximum rate solution obtained in step 1) into a convex problem by introducing Taylor series, then set the initial value, and obtain the maximum transmission rate by an iterative method on the basis of the initial value, and then according to the Maximize the transmission rate to form beams that support multi-user pair high-speed communication based on SWIPT.

步骤1)中的最大速率求解的数学模型为：The mathematical model of the maximum rate solution in step 1) is:

$\begin{matrix} ((P P 11)) & \underset{w w}{max max imize imize} {R R}_{Σ Σ} \\ s the s . . t t . . & {P P}_{EH EH} &GreaterEqual; &Greater Equal; {P P}_{th the th} \\ {P P}_{s the s} {| | | | {WH WH}_{r r} | | | |}^{22} + + {σ σ}_{r r}^{22} {| | | | W W | | | |}^{22} \leq \leq {P P}_{r r} \end{matrix} - - - - - - ((11))$

其中，P_th最小的收获功率，P_r最大传输功率，W_∈ ^N×N为中继处线性波束形成系数，R_∑为总传输速率，γ_k为目的节点处的信干比，E[|x_k|²]＝P_s，K为中继网络中单天线源节点的数量，x_k为第k个源节点发出的信号，E为中继节点接收到EH节点处的能量，S_k为第k个源节点，P_EH为EH节点处收获的能量，H_r：＝[h_r1，...，h_rk，...，h_rK]，h_rk为从第k个源节点S_k到中继节点的衰落系数，为第k对用户中发送端到中继的噪声功率。Among them, the minimum harvesting power of P _th , the maximum transmission power of P _r , W _∈ ^N×N is the linear beamforming coefficient at the relay, R _∑ is the total transmission rate, γ _k is the signal-to-interference ratio at the destination node, E[|x _k | ² ]=P _s , K is the number of single-antenna source nodes in the relay network, x _k is the signal sent by the kth source node, and E is The relay node receives the energy at the EH node, S _k is the kth source node, P _EH is the energy harvested at the EH node, H _r :=[h _r1 ,...,h _rk ,...,h _rK ], h _rk is the fading coefficient from the kth source node S _k to the relay node, is the noise power from the transmitter to the relay in the kth pair of users.

步骤2)的具体过程为：The concrete process of step 2) is:

式(1)转化为：Formula (1) transforms into:

$\begin{matrix} ((P P 22)) & \underset{w w}{max max imize imize} {Σ Σ}_{k k = = 11}^{K K} \frac{11}{22} {log log}_{22} ((11 + + \frac{{P P}_{s the s} {w w}^{H h} {G G}_{kk kk} w w}{{P P}_{s the s} \underset{l l &NotEqual; &NotEqual; k k}{Σ Σ} {w w}^{H h} {G G}_{kl kl} w w + + {σ σ}_{r r}^{22} {w w}^{H h} {G G}_{rk rk} w w + + {σ σ}_{k k}^{22}})) \\ s the s . . t t . . & {P P}_{s the s} {w w}^{H h} {G G}_{eh eh} w w + + {σ σ}_{r r}^{22} {w w}^{H h} {G G}_{ee ee} w w &GreaterEqual; &Greater Equal; {P P}_{th the th} \\ {P P}_{s the s} {w w}^{H h} {G G}_{rk rk} w w + + {σ σ}_{r r}^{k k} {w w}^{H h} w w \leq \leq {P P}_{r r} \end{matrix} - - - - - - ((22))$

其中 $w = vec (W), G_{kk} = ({h_{rk}}^{*} &CircleTimes; {h_{k}}^{*}) ({h_{rk}}^{T} &CircleTimes; {h_{k}}^{T}), G_{kl} = ({h_{rl}}^{*} &CircleTimes; {h_{k}}^{*}) ({h_{rl}}^{T} &CircleTimes; {h_{k}}^{T}), G_{rk} = (I &CircleTimes; {h_{k}}^{*}) (I &CircleTimes; {h_{k}}^{T}),$ $G_{eh} = ({H_{r}}^{*} &CircleTimes; {h_{e}}^{*}) ({H_{r}}^{T} &CircleTimes; {h_{e}}^{T}), G_{ee} = (I &CircleTimes; {h_{e}}^{*}) (I &CircleTimes; {h_{e}}^{T}),$ 为第k对用户中中继到接收端的噪声功率，h_rk为从S_k到中继的衰落系数，h_k为从中继到D_k信道的衰落系数，h_rl为H_r中第l个列向量，I为单位矩阵；in $w = vec (W), G_{kk} = ({h_{rk}}^{*} &CircleTimes; {h_{k}}^{*}) ({h_{rk}}^{T} &CircleTimes; {h_{k}}^{T}), G_{kl} = ({h_{rl}}^{*} &CircleTimes; {h_{k}}^{*}) ({h_{rl}}^{T} &CircleTimes; {h_{k}}^{T}), G_{rk} = (I &CircleTimes; {h_{k}}^{*}) (I &CircleTimes; {h_{k}}^{T}),$ $G_{eh} = ({h_{r}}^{*} &CircleTimes; {h_{e}}^{*}) ({h_{r}}^{T} &CircleTimes; {h_{e}}^{T}), G_{ee} = (I &CircleTimes; {h_{e}}^{*}) (I &CircleTimes; {h_{e}}^{T}),$ is the noise power from the relay to the receiver in the k-th pair of users, h _rk is the fading coefficient from S _k to the relay, h _k is the fading coefficient from the relay to D _k channel, h _rl is the lth column in H _r Vector, I is the identity matrix;

引入则式(2)转换为introduce Then formula (2) is transformed into

$\begin{matrix} ((P P 33)) & \underset{x x}{max max imize imize} {Σ Σ}_{k k - - 11}^{K K} \frac{11}{22} {log log}_{22} ((11 + + \frac{{P P}_{s the s} Tr Tr (({G G}_{kk kk} X x))}{{P P}_{s the s} \underset{l l &NotEqual; &NotEqual; k k}{Σ Σ} Tr Tr (({G G}_{kl kl} X x)) + + {σ σ}_{r r}^{22} Tr Tr (({G G}_{rk rk} X x)) + + {σ σ}_{k k}^{22}})) \\ s the s . . t t . . & {P P}_{s the s} Tr Tr (({G G}_{eh eh} X x)) + + {σ σ}_{r r}^{22} Tr Tr (({G G}_{ee ee} X x)) &GreaterEqual; &Greater Equal; {P P}_{th the th} \\ {P P}_{s the s} Tr Tr (({G G}_{rk rk} X x)) + + {σ σ}_{r r}^{22} Tr Tr ((X x)) \leq \leq {P P}_{r r} \\ X x &PlusMinus; &PlusMinus; 00 \\ rank rank ((X x)) = = 11 \end{matrix} - - - - - - ((33))$

其中，w＝vec(W) $G_{kk} = ({h_{rk}}^{*} &CircleTimes; {h_{k}}^{*}) ({h_{rk}}^{T} &CircleTimes; {h_{k}}^{T}),$ Tr为矩阵求迹运算；in, w=vec(W) $G_{kk} = ({h_{rk}}^{*} &CircleTimes; {h_{k}}^{*}) ({h_{rk}}^{T} &CircleTimes; {h_{k}}^{T}),$ Tr is a matrix trace operation;

然后去掉式(3)中的限制条件rank(X)＝1，并引入泰勒展开级数，得Then remove the constraint condition rank(X)=1 in formula (3), and introduce the Taylor expansion series, get

$\begin{matrix} ((P P 44)) & \underset{X x,, {α α}_{k k},, {β β}_{k k},, &ForAll; &ForAll; k k}{min min imize imize} - - {Σ Σ}_{k k = = 11}^{K K} \frac{11}{22} log log (({α α}_{k k})) + + {Σ Σ}_{k k = = 11}^{K K} \frac{11}{22} [[log log (({β β}_{k k}^{((i i))})) + + \frac{11}{{β β}_{k k}^{((i i))}} (({β β}_{k k} - - {β β}_{k k}^{((i i))}))]] \\ s the s . . t t . . & ((11)),, ((22)),, ((33)) \\ {P P}_{s the s} Tr Tr (({G G}_{k k} X x)) + + {σ σ}_{r r}^{k k} Tr Tr (({G G}_{rk rk} X x)) + + {σ σ}_{k k}^{22} = = {α α}_{k k},, &ForAll; &ForAll; k k \\ {P P}_{s the s} \underset{l l &NotEqual; &NotEqual; k k}{Σ Σ} Tr Tr (({G G}_{kl kl} X x)) + + {σ σ}_{r r}^{22} Tr Tr (({G G}_{rk rk} X x)) + + {σ σ}_{k k}^{22} = = {β β}_{k k},, &ForAll; &ForAll; k k \end{matrix} - - - - - - ((44))$

其中，是β_k迭代的初值，β_k为迭代的变量，为β_k第(i-1)次迭代的优化解，i＝(1，2，...)，求解式(4)，得最大传输速率，然后根据最大传输速率形成基于SWIPT下支持多用户对高速通信的波束。in, is the initial value of β _k iteration, β _k is the variable of iteration, is the optimized solution of the (i-1) iteration of β _k , i=(1, 2, ...), solve the formula (4), get the maximum transmission rate, and then form the multi-user support based on SWIPT according to the maximum transmission rate Beams for high-speed communications.

本发明具有以下有益效果：The present invention has the following beneficial effects:

本发明所述的基于SWIPT下支持多用户对高速通信的波束形成方法在操作时，先开拓新的多天线中继，并根据新的多天线中继构建新的波束成形方案，再根据所述新的波束成形方案得到非凸的最大速率求解的数学模型，然后该非凸的最大速率求解的数学模型转换为凸问题，然后求解得到最大化传输速率，再根据所述最大化传输速率完成基于SWIPT下支持多用户对高速通信的波束形成，操作方便，简单，可行性强。During the operation of the beamforming method for supporting multi-user pair high-speed communication based on SWIPT described in the present invention, a new multi-antenna relay is developed first, and a new beamforming scheme is constructed according to the new multi-antenna relay, and then according to the described The new beamforming scheme obtains the mathematical model of the non-convex maximum rate solution, and then converts the non-convex maximum rate solution mathematical model into a convex problem, and then solves it to obtain the maximum transmission rate, and then completes based on the maximum transmission rate based on the SWIPT supports multi-user beamforming for high-speed communication, which is easy to operate, simple and highly feasible.

附图说明Description of drawings

图1为多路双向中继原理图；Figure 1 is a schematic diagram of a multi-way bidirectional relay;

图2为不同源到目的节点对数K下的平均总速率分布图；Fig. 2 is the distribution figure of the average total rate under different source-to-destination node logarithms K;

图3当K＝4时迭代次数与传输速率的变化关系图。Fig. 3 is a diagram of the relationship between the number of iterations and the transmission rate when K=4.

具体实施方式Detailed ways

下面结合附图对本发明做进一步详细描述：The present invention is described in further detail below in conjunction with accompanying drawing:

考虑一个两跳的放大转发中继网络：K个单天线源节点，K个单天线目的节点，N个天线的中继节点，一个单天线的EH。第k个源节点发送信号S_K经过中继到目的节点D_K，其中，k＝1，2，...，K，信号传输和能量传输被分为两个阶段，在第一个阶段，所有的源节点向中继发送信号，在第二阶段，所有的中继节点将接收到的信号S_k转发到目的节点D_k，同时收获EH节点处的能量E。Consider a two-hop amplifying and forwarding relay network: K single-antenna source nodes, K single-antenna destination nodes, N-antenna relay nodes, and a single-antenna EH. The kth source node sends a signal S _K to the destination node D _K through relay, where k=1, 2, ..., K, signal transmission and energy transmission are divided into two stages, in the first stage, All source nodes send signals to the relay, and in the second stage, all relay nodes forward the received signal S _k to the destination node D _k and harvest the energy E at the EH node.

用户D_k处收到的信号为：The signal received at user _Dk is:

y_k＝h_k ^TW H_rx+h_k ^TWn_r+n_k y _k ＝h _k ^T W H _r x+h _k ^T Wn _r +n _k

这里(·)^T代表着矩阵的转置，h_k∈^N×1为从中继到D_k信道的衰落系数，H_r∈^N×K定义为：H_r：＝[h_r1，k...，h_rk，...，h_rK]，h_rk为从S_k到中继的衰落系数，W_∈ ^N×N为中继处线性波束形成系数，x＝[x₁，...，x_k，...，x_K]^T中x_K为第k个源节点S_k发出的信号，满足E[|x_k|²]＝P_s，是独立加性白噪声，CN(μ，σ²)代表均值为μ方差为σ²的复高斯分布；Here (·) ^T represents the transpose of the matrix, h _k ∈ ^N×1 is the fading coefficient from the relay to D _k channel, H _r ∈ ^N×K is defined as: H _r :=[h _r1 , k... , h _rk ,..., h _rK ], h _rk is the fading coefficient from S _k to the relay, W _∈ ^N×N is the linear beamforming coefficient at the relay, x=[x ₁ ,...,x _k ,...,x _K ] x _K in ^T is the signal sent by the kth source node S _k , satisfying E[|x _k | ² ]=P _s , is an independent additive white noise, CN(μ, σ ² ) represents a complex Gaussian distribution with a mean of μ and a variance of σ ² ;

总传输速率R_∑为：The total transmission rate R _∑ is:

${R R}_{Σ Σ} = = {Σ Σ}_{k k = = 11}^{K K} \frac{11}{22} log log ((11 + + {γ γ}_{k k}))$

其中目的节点处的信干比γ_k为The signal-to-interference ratio γ _k at the destination node is

${γ γ}_{k k} = = \frac{{P P}_{s the s} {| | {h h}_{k k}^{T T} W W {h h}_{rk rk} | |}^{22}}{{P P}_{s the s} {Σ Σ}_{l l &NotEqual; &NotEqual; k k} {| | {h h}_{k k}^{T T} W W {h h}_{rl rl} | |}^{22} + + {| | | | {h h}_{k k}^{T T} W W | | | |}^{22} {σ σ}_{r r}^{22} + + {σ σ}_{k k}^{22}};;$

同理，EH节点处接收到的信号为：Similarly, the signal received at the EH node is:

y_e＝h_e ^TW H_rx+h_e ^TWn_r+n_e，y _e = h _e ^T W H _r x + h _e ^T Wn _r + n _e ,

因此在EH节点处收获的能量表达如下：The energy harvested at the EH node is thus expressed as follows:

${P P}_{Eh Eh} = = {P P}_{s the s} {| | | | {h h}_{e e}^{T T} {WH WH}_{r r} | | | |}^{22} + + {σ σ}_{r r}^{22} {| | | | {h h}_{e e}^{T T} W W | | | |}^{22} + + {σ σ}_{e e}^{22} . .$

参考图1，本发明所述的基于SWIPT下支持多用户对高速通信的波束形成方法包括以下步骤：With reference to Fig. 1, the beamforming method that supports multi-user to high-speed communication based on SWIPT of the present invention comprises the following steps:

其中，P_th最小的收获功率，P_r最大传输功率，W∈^N×N为中继处线性波束形成系数，R_∑为总传输速率，γ_k为目的节点处的信干比，E[|x_k|²]＝P_s，K为中继网络中单天线源节点的数量，x_K为第k个源节点发出的信号，E为中继节点接收到EH节点处的能量，S_k为第k个源节点，P_EH为EH节点处收获的能量，H_r：＝[h_r1，...，h_rk，...，h_rK]，h_rk为从第k个源节点S_k到中继节点的衰落系数，为第k对用户中发送端到中继的噪声功率。Among them, P _th is the minimum harvesting power, P _r is the maximum transmission power, W ∈ ^N×N is the linear beamforming coefficient at the relay, R _∑ is the total transmission rate, γ _k is the signal-to-interference ratio at the destination node, E[|x _k | ² ]=P _s , K is the number of single-antenna source nodes in the relay network, x _K is the signal sent by the kth source node, and E is The relay node receives the energy at the EH node, S _k is the kth source node, P _EH is the energy harvested at the EH node, H _r :=[h _r1 ,...,h _rk ,...,h _rK ], h _rk is the fading coefficient from the kth source node S _k to the relay node, is the noise power from the transmitter to the relay in the kth pair of users.

步骤2)的具体过程为：The concrete process of step 2) is:

式(1)转化为：Formula (1) transforms into:

引入则式(2)转换为introduce Then formula (2) is transformed into

以下将通过仿真来具体进行说明：The following will be explained in detail through simulation:

仿真条件如下：源节点处的传输功率P_s＝10dB，中继处天线数N＝4；用户对数K分别为1，2和4；EH节点处所需的最小充电功率为：P_th＝0dB；为了不失一般性，信道系数仍然服从方差为1的瑞利衰落，并且和均为1。The simulation conditions are as follows: the transmission power at the source node P _s =10dB, the number of antennas at the relay station N=4; the user logarithms K are 1, 2 and 4 respectively; the minimum charging power required at the EH node is: P _th = 0dB; without loss of generality, the channel coefficients still obey Rayleigh fading with variance 1, and and Both are 1.

如图2所示，当K＝1时，本发明和max-min波束成形的与最优值的速率曲线几乎一致，随着K的增加，即使用户对数一样，本发明远也优于其它两个方案。同时，当K从1到2时，本发明和max-min下的总速率都会增加，但当K从2增加到4时，总速率减小，这是因为用户对之间的干扰造成的，在波束形成过程中不可避免，并引起总速率的损失。从图2中可看出，本发明要远优于max-min。As shown in Figure 2, when K=1, the speed curve of the present invention and max-min beamforming is almost consistent with the optimal value, and as K increases, even if the logarithm of users is the same, the present invention is far better than The other two options. At the same time, when K is from 1 to 2, the total rate under the present invention and max-min will both increase, but when K increases from 2 to 4, the total rate decreases, which is caused by the interference between user pairs, Unavoidable during beamforming and causes a loss in overall rate. It can be seen from Fig. 2 that the present invention is far superior to max-min.

参考图3，K＝4，假设为同一信道，初值X和都随机产生，基于迭代指数i下的每个用户对应的传输速率，以求得该方案下的收敛速率。从图中知，在前四次迭代中，用户对的传输速率收敛的很快，而在此之后的迭代中变化几乎可以忽略，收敛越快，复杂度越低，说明本发明的可行性越强。Referring to Figure 3, K=4, assuming the same channel, the initial value X and are randomly generated, based on the transmission rate corresponding to each user under the iteration index i, to obtain the convergence rate under the scheme. It is known from the figure that in the first four iterations, the transmission rate of the user pair converges very quickly, and the change in the subsequent iterations is almost negligible. The faster the convergence and the lower the complexity, the more feasible the present invention is. powerful.

Claims

1. A wave beam forming method for supporting multi-user to high-speed communication based on SWIPT is characterized by comprising the following steps:

1) developing a new multi-antenna relay, constructing a new beam forming scheme according to the new multi-antenna relay, and then obtaining a mathematical model for solving the maximum rate according to the constructed new beam forming scheme, wherein the mathematical model for the maximum rate is a non-convex problem;

2) converting the mathematical model solved by the maximum rate obtained in the step 1) into a convex problem by introducing a Taylor series, then setting an initial value, obtaining a maximum transmission rate by an iterative method on the basis of the initial value, and then forming a wave beam supporting multi-user to high-speed communication based on SWIPT according to the maximum transmission rate.

2. The beamforming method for supporting multi-user-high-speed communication based on SWIPT according to claim 1, wherein the mathematical model of the maximum rate solution in step 1) is as follows:

<math> <mrow> <mrow> <mo>(</mo> <mi>P</mi> <mn>1</mn> <mo>)</mo> </mrow> <munder> <mrow> <mi>max</mi> <mi>imize</mi> </mrow> <mi>w</mi> </munder> <msub> <mi>R</mi> <mi>Σ</mi> </msub> </mrow> </math>

<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>s</mi> <mo>.</mo> <mi>t</mi> <mo>.</mo> <msub> <mi>P</mi> <mi>EH</mi> </msub> <mo>&GreaterEqual;</mo> <msub> <mi>P</mi> <mi>th</mi> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>P</mi> <mi>s</mi> </msub> <msup> <mrow> <mo>|</mo> <mo>|</mo> <msub> <mi>WH</mi> <mi>r</mi> </msub> <mo>|</mo> <mo>|</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <msup> <mrow> <mo>|</mo> <mo>|</mo> <mi>W</mi> <mo>|</mo> <mo>|</mo> </mrow> <mn>2</mn> </msup> <mo>≤</mo> <msub> <mi>P</mi> <mi>r</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>

wherein, P_thMinimum harvest power, P_rMaximum transmission power, W is the linear beam forming coefficient at the relay, R_∑As a result of the total transmission rate,γ_kfor the signal-to-interference ratio at the destination node, E [ | x [ ]_k|²]＝P_sK is the number of single antenna source nodes in the relay network, x_KFor the signal sent by the kth source node, E is the energy received by the relay node at the EH node, S_kIs the kth source node, P_EHFor the energy harvested at the EH node, H_r：＝[h_r1，...，h_rk，...，h_rK]，h_rkTo be from the kth source node S_kThe fading coefficient to the relay node is,the transmit-to-relay noise power in the kth pair of users.

3. The beamforming method supporting multi-user-to-high-speed communication based on SWIPT according to claim 2, wherein the specific process of step 2) is as follows:

the formula (1) is converted into:

<math> <mrow> <mrow> <mo>(</mo> <mi>P</mi> <mn>2</mn> <mo>)</mo> </mrow> <munder> <mrow> <mi>max</mi> <mi>imize</mi> </mrow> <mi>w</mi> </munder> <munderover> <mi>Σ</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </munderover> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mfrac> <mrow> <msub> <mi>P</mi> <mi>s</mi> </msub> <msup> <mi>w</mi> <mi>H</mi> </msup> <msub> <mi>G</mi> <mi>kk</mi> </msub> <mi>w</mi> </mrow> <mrow> <msub> <mi>P</mi> <mi>s</mi> </msub> <munder> <mi>Σ</mi> <mrow> <mi>l</mi> <mo>&NotEqual;</mo> <mi>k</mi> </mrow> </munder> <msup> <mi>w</mi> <mi>H</mi> </msup> <msub> <mi>G</mi> <mi>kl</mi> </msub> <mi>w</mi> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <msup> <mi>w</mi> <mi>H</mi> </msup> <msub> <mi>G</mi> <mi>rk</mi> </msub> <mi>w</mi> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>k</mi> <mn>2</mn> </msubsup> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> </math>

<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>s</mi> <mo>.</mo> <mi>t</mi> <mo>.</mo> <msub> <mi>P</mi> <mi>s</mi> </msub> <msup> <mi>w</mi> <mi>H</mi> </msup> <msub> <mi>G</mi> <mi>eh</mi> </msub> <mi>w</mi> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <msup> <mi>w</mi> <mi>H</mi> </msup> </mtd> <mtd> <msub> <mi>G</mi> <mi>ee</mi> </msub> <mi>w</mi> <mo>&GreaterEqual;</mo> <msub> <mi>P</mi> <mi>th</mi> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>P</mi> <mi>s</mi> </msub> <msup> <mi>w</mi> <mi>H</mi> </msup> <msub> <mi>G</mi> <mi>rk</mi> </msub> <mi>w</mi> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <msup> <mi>w</mi> <mi>H</mi> </msup> </mtd> <mtd> <mi>w</mi> <mo>≤</mo> <msub> <mi>P</mi> <mi>r</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> </math>

wherein

<math> <mrow> <mi>w</mi> <mo>=</mo> <mi>vec</mi> <mrow> <mo>(</mo> <mi>W</mi> <mo>)</mo> </mrow> <mo>,</mo> <msub> <mi>G</mi> <mi>kk</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mi>rk</mi> </msub> <mo>*</mo> </msup> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>k</mi> </msub> <mo>*</mo> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mi>rk</mi> </msub> <mi>T</mi> </msup> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>k</mi> </msub> <mi>T</mi> </msup> <mo>)</mo> </mrow> <mo>,</mo> <msub> <mi>G</mi> <mi>kl</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mi>rl</mi> </msub> <mo>*</mo> </msup> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>k</mi> </msub> <mo>*</mo> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mi>rl</mi> </msub> <mi>T</mi> </msup> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>k</mi> </msub> <mi>T</mi> </msup> <mo>)</mo> </mrow> <mo>,</mo> <msub> <mi>G</mi> <mi>rk</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <mi>I</mi> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>k</mi> </msub> <mo>*</mo> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>I</mi> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>k</mi> </msub> <mi>T</mi> </msup> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </math>

<math> <mrow> <msub> <mi>G</mi> <mi>eh</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msup> <msub> <mi>H</mi> <mi>r</mi> </msub> <mo>*</mo> </msup> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>e</mi> </msub> <mo>*</mo> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msup> <msub> <mi>H</mi> <mi>r</mi> </msub> <mi>T</mi> </msup> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>e</mi> </msub> <mi>T</mi> </msup> <mo>)</mo> </mrow> <mo>,</mo> <msub> <mi>G</mi> <mi>ee</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <mi>I</mi> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>e</mi> </msub> <mo>*</mo> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>I</mi> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>e</mi> </msub> <mi>T</mi> </msup> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </math>

For the k-th pair of users, the noise power, h, relayed to the receiving end_rkIs a slave S_kFading coefficient to relay, h_kTo go from relay to D_kFading coefficient of the channel, h_rlIs H_rThe ith column vector, I is an identity matrix;

introduction ofThen the formula (2) is converted into

<math> <mrow> <mrow> <mo>(</mo> <mi>P</mi> <mn>3</mn> <mo>)</mo> </mrow> <munder> <mrow> <mi>max</mi> <mi>imize</mi> </mrow> <mi>x</mi> </munder> <munderover> <mi>Σ</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </munderover> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mfrac> <mrow> <msub> <mi>P</mi> <mi>s</mi> </msub> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>kk</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <mi>s</mi> </msub> <munder> <mi>Σ</mi> <mrow> <mi>l</mi> <mo>&NotEqual;</mo> <mi>k</mi> </mrow> </munder> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>kl</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>rk</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>k</mi> <mn>2</mn> </msubsup> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> </math>

<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>s</mi> <mo>.</mo> <mi>t</mi> <mo>.</mo> <msub> <mi>P</mi> <mi>s</mi> </msub> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>ek</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>ee</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>&GreaterEqual;</mo> <msub> <mi>P</mi> <mi>th</mi> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>P</mi> <mi>s</mi> </msub> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>rk</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <mi>Tr</mi> <mrow> <mo>(</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>≤</mo> <msub> <mi>P</mi> <mi>r</mi> </msub> </mtd> </mtr> <mtr> <mtd> <mi>X</mi> <mo>&PlusMinus;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mi>rank</mi> <mrow> <mo>(</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>=</mo> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> </math>

Wherein,w＝vec(W)

<math> <mrow> <msub> <mi>G</mi> <mi>kk</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mi>rk</mi> </msub> <mo>*</mo> </msup> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>k</mi> </msub> <mo>*</mo> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mi>rk</mi> </msub> <mi>T</mi> </msup> <mo>&CircleTimes;</mo> <msup> <msub> <mi>h</mi> <mi>k</mi> </msub> <mi>T</mi> </msup> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </math>

tr is matrix tracing calculation;

then removing the limiting condition rank (X) 1 in the formula (3) and introducing a Taylor expansion series to obtain

<math> <mrow> <mrow> <mo>(</mo> <mi>P</mi> <mn>4</mn> <mo>)</mo> </mrow> <munder> <mrow> <mi>min</mi> <mi>imize</mi> </mrow> <mrow> <mi>X</mi> <mo>,</mo> <msub> <mi>α</mi> <mi>k</mi> </msub> <mo>,</mo> <msub> <mi>β</mi> <mi>k</mi> </msub> <mo>,</mo> <mo>&ForAll;</mo> <mi>k</mi> </mrow> </munder> <mo>-</mo> <munderover> <mi>Σ</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </munderover> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mi>log</mi> <mrow> <mo>(</mo> <msub> <mi>α</mi> <mi>k</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <munderover> <mi>Σ</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </munderover> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mo>[</mo> <mi>log</mi> <mrow> <mo>(</mo> <msubsup> <mi>β</mi> <mi>k</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> </msubsup> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mn>1</mn> <msubsup> <mi>β</mi> <mi>k</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> </msubsup> </mfrac> <mrow> <mo>(</mo> <msub> <mi>β</mi> <mi>k</mi> </msub> <mo>-</mo> <msubsup> <mi>β</mi> <mi>k</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> </msubsup> <mo>)</mo> </mrow> <mo>]</mo> </mrow> </math>

<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>s</mi> <mo>.</mo> <mi>t</mi> <mo>.</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>,</mo> <mrow> <mo>(</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>,</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <msub> <mi>P</mi> <mi>s</mi> </msub> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>k</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>rk</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>k</mi> <mn>2</mn> </msubsup> <mo>=</mo> <msub> <mi>α</mi> <mi>k</mi> </msub> <mo>,</mo> <mo>&ForAll;</mo> <mi>k</mi> </mtd> </mtr> <mtr> <mtd> <msub> <mi>P</mi> <mi>s</mi> </msub> <munder> <mi>Σ</mi> <mrow> <mi>l</mi> <mo>&NotEqual;</mo> <mi>k</mi> </mrow> </munder> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>kl</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>r</mi> <mn>2</mn> </msubsup> <mi>Tr</mi> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mi>rk</mi> </msub> <mi>X</mi> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>σ</mi> <mi>k</mi> <mn>2</mn> </msubsup> <mo>=</mo> <msub> <mi>β</mi> <mi>k</mi> </msub> <mo>,</mo> <mo>&ForAll;</mo> <mi>k</mi> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> </math>

Wherein, is beta_kInitial value of iteration, beta_kFor the variables of the iteration to be considered,is beta_kAnd (3) solving an optimized solution of the (i-1) th iteration, namely i is (1, 2.), solving the formula (4) to obtain the maximum transmission rate, and then forming a beam supporting multi-user to high-speed communication based on SWIPT according to the maximum transmission rate.