CN105451322A

CN105451322A - Channel allocation and power control method based on QoS in D2D network

Info

Publication number: CN105451322A
Application number: CN201510881731.6A
Authority: CN
Inventors: 朱琦; 于宝舟; 朱洪波; 杨龙祥
Original assignee: Nanjing Post and Telecommunication University
Current assignee: Beijing Yicheng Technology Co ltd
Priority date: 2015-12-03
Filing date: 2015-12-03
Publication date: 2016-03-30
Anticipated expiration: 2035-12-03
Also published as: CN105451322B

Abstract

The present invention is a channel allocation and power control method based on user QoS. The specific steps are as follows: build a system model based on the D2D user capacity maximization, and select a candidate cell that meets the requirements for the D2D user according to the minimum signal-to-interference-noise ratio requirement of the D2D user. User set; under the condition of guaranteeing the QoS requirements of cellular users and the limited user power, the Lagrangian multiplier method is used to optimize the transmission power of cellular users and D2D users; then calculate the capacity of all D2D users according to the optimal transmission power of users and Channels are allocated to the maximum capacity of D2D users, enabling D2D communication in cellular networks. This method jointly optimizes the channel allocation and power control of users, allocates the optimal transmission power for each user, and allocates the channel to the D2D user with the largest capacity, thereby reducing the energy loss of cellular users and D2D users, and improving the total capacity of D2D users .

Description

A QoS-Based Channel Allocation and Power Control Method in D2D Networks

技术领域technical field

本发明涉及一种资源分配方法，尤其涉及一种基于用户QoS的信道分配和功率控制方法，属于通信技术领域。The invention relates to a resource allocation method, in particular to a channel allocation and power control method based on user QoS, and belongs to the technical field of communication.

背景技术Background technique

随着无线通信技术的迅速发展，频谱资源短缺成为移动通信面临的挑战。在传统蜂窝网络中，不允许用户之间直接通信。通信过程由基站转接分为两个阶段：移动终端到基站，即上行链路；基站到移动终端，即下行链路。这种集中式工作方式便于对资源的干扰进行管理和控制，但资源利用效率低。在此背景下D2D通信技术应运而生。With the rapid development of wireless communication technology, the shortage of spectrum resources has become a challenge for mobile communication. In traditional cellular networks, direct communication between users is not allowed. The communication process is divided into two stages by the transfer of the base station: from the mobile terminal to the base station, that is, the uplink; from the base station to the mobile terminal, that is, the downlink. This centralized working mode facilitates the management and control of resource interference, but the resource utilization efficiency is low. In this context, D2D communication technology came into being.

D2D通信是一种在蜂窝系统控制下，允许近距离终端用户通过共享小区资源进行直接通信的新技术。在蜂窝网络中引入D2D通信，可以减轻基站负担，减小通信时延。与传统蜂窝通信相比，D2D通信仅占用一半的频谱资源。而且近距离的D2D通信可以减小传输功率，节约能耗，增加手机的续航时间。D2D通信具有明显的技术特征，相比于蓝牙和无线局域网(WLAN)等短距离通信技术，其最大的优势在于工作在授权频段，数据传输具有更高的可靠性。除此之外，D2D通信也不需要蓝牙和WLAN那样繁琐的匹配，可以为用户提供更好的体验。D2D communication is a new technology that allows short-distance terminal users to communicate directly by sharing cell resources under the control of a cellular system. Introducing D2D communication in the cellular network can reduce the burden on the base station and reduce the communication delay. Compared with traditional cellular communication, D2D communication only occupies half of the spectrum resources. Moreover, short-distance D2D communication can reduce transmission power, save energy consumption, and increase battery life of mobile phones. D2D communication has obvious technical characteristics. Compared with short-distance communication technologies such as Bluetooth and wireless local area network (WLAN), its biggest advantage is that it works in the authorized frequency band and has higher reliability of data transmission. In addition, D2D communication does not require the tedious matching of Bluetooth and WLAN, which can provide users with a better experience.

蜂窝系统在分配D2D通信资源时，可分配专用的、与小区用户正交的或复用小区内用户的频谱资源。当小区负载较低，在满足小区用户通信后有剩余的频谱资源时，可分配正交频谱资源给D2D用户，二者不会相互干扰。当小区负载较高时，D2D用户复用小区用户的资源。此时共享同一资源的蜂窝用户和D2D用户会相互干扰，基站通过控制D2D通信的分配信道和发送功率来控制对小区带来的干扰。因此为用户分配合理的资源降低用户之间的干扰，提高系统整体性能是目前的一个研究热点。本发明在保障蜂窝用户和D2D用户QoS要求、蜂窝用户发送功率受限和D2D用户总功率受限的条件下，构建了以D2D用户容量最大化为目标的优化函数。采用拉格朗日乘子法对蜂窝用户和D2D用户的发送功率进行了优化分配，并为D2D用户分配合理的信道进行通信。具有减少用户能耗开销，提高D2D用户总容量的优点。When allocating D2D communication resources, the cellular system can allocate spectrum resources that are dedicated, orthogonal to users in the cell, or multiplexed to users in the cell. When the load of the cell is low and there are remaining spectrum resources after satisfying the user communication requirements in the cell, orthogonal spectrum resources can be allocated to D2D users, and the two will not interfere with each other. When the load of the cell is high, the D2D user reuses the resources of the cell user. At this time, cellular users and D2D users sharing the same resource will interfere with each other, and the base station controls the interference to the cell by controlling the allocated channel and transmission power of D2D communication. Therefore, allocating reasonable resources to users to reduce the interference between users and improve the overall performance of the system is a current research hotspot. The present invention constructs an optimization function aiming at maximizing D2D user capacity under the conditions of guaranteeing the QoS requirements of cellular users and D2D users, limited transmitting power of cellular users and limited total power of D2D users. The transmission power of cellular users and D2D users is optimally allocated by using the Lagrangian multiplier method, and a reasonable channel is allocated for D2D users to communicate. It has the advantages of reducing user energy consumption and increasing the total capacity of D2D users.

发明内容Contents of the invention

技术问题：本发明的目的是提供一种基于QoS的信道分配和功率控制方法，通过优化蜂窝用户和D2D用户的发送功率，节约用户能量；并为D2D用户分配合理的信道，降低蜂窝用户与D2D用户间的干扰，提高蜂窝网络中D2D通信的系统性能。Technical problem: The purpose of the present invention is to provide a QoS-based channel allocation and power control method, which saves user energy by optimizing the transmission power of cellular users and D2D users; and allocates reasonable channels for D2D users to reduce cellular users and D2D users. inter-user interference to improve the system performance of D2D communication in cellular networks.

技术方案：本发明的D2D网络中基于QoS的信道分配和功率控制方法，包括以下步骤：Technical solution: The QoS-based channel allocation and power control method in the D2D network of the present invention includes the following steps:

1)分别用C＝{1,2,...,M}和D＝{1,2,...,N}表示蜂窝用户集和D2D用户集，M为蜂窝用户数，N为D2D对用户数，K表示可用信道数，定义S表示D2D用户复用蜂窝用户的信道集；1) Use C={1,2,...,M} and D={1,2,...,N} to represent the cellular user set and the D2D user set respectively, M is the number of cellular users, and N is the number of D2D pairs The number of users, K represents the number of available channels, and S is defined to represent the channel set of D2D users multiplexing cellular users;

2)计算D2D用户j(1≤j≤N)在信道k(1≤k≤K)上的信干噪比式中P_tot为D2D用户总发送功率，g_j为D2D用户对j的信道增益，N₀表示噪声功率，表示蜂窝用户的最大发送功率，h_i,j表示蜂窝用户i和D2D用户j之间的信道增益；2) Calculate the SINR of D2D user j (1≤j≤N) on channel k (1≤k≤K) where P _tot is the total transmit power of D2D users, g _j is the channel gain of D2D user pair j, N ₀ is the noise power, Indicates the maximum transmit power of a cellular user, h _{i, j} indicates the channel gain between cellular user i and D2D user j;

3)根据D2D用户j的最小信干噪比要求将满足要求的信道k归入集合Ω中，其中Ω为D2D用户的候选信道集，表示D2D用户的信干噪比阈值；3) According to the minimum SINR requirement of D2D user j Classify the channel k that meets the requirements into the set Ω, where Ω is the candidate channel set of the D2D user, Indicates the SINR threshold of the D2D user;

4)初始化t＝1和λ(t)＝2，其中t为迭代次数，λ为拉格朗日乘子；4) initialization t=1 and λ(t)=2, wherein t is the number of iterations, and λ is the Lagrange multiplier;

5)根据公式 ${\tilde{p}}_{j, k}^{d} = {[\frac{\sqrt{Δ}}{2 {g_{i, B}}^{2} \cdot B (g_{j} + B)} - \frac{A}{2 B} - \frac{A}{2 (g_{j} + B)}]}_{0}^{τ}$ 计算D2D用户j(1≤j≤N)在信道k(k∈Ω)上的发送功率，其中 ${[\cdot]}_{x}^{y} = \min (m a x (\cdot,), y), τ = \frac{1}{h_{j, B}} (\frac{g_{i, B}}{γ_{t h}^{c}} \cdot P_{m a x}^{c} - N_{0}),$ 式中h_j,B表示D2D用户j与基站间的信道增益，g_i,B表示蜂窝用户i与基站间的信道增益，为蜂窝用户的信干噪比阈值；且 $Δ = g_{j} {g_{i, B}}^{2} A (g_{j} {g_{i, B}}^{2} A + \frac{4}{λ l n 2} {g_{i, B}}^{2} B (g_{j} + B)),$ 其中 $A = N_{0} \cdot (1 + \frac{γ_{t h}^{c} \cdot h_{i, j}}{g_{i, B}}), B = \frac{γ_{t h}^{c} \cdot h_{j, B} \cdot h_{i, j}}{g_{i, B}};$ 5) According to the formula ${\tilde{p}}_{j, k}^{d} = {[\frac{\sqrt{Δ}}{2 {g_{i, B}}^{2} \cdot B (g_{j} + B)} - \frac{A}{2 B} - \frac{A}{2 (g_{j} + B)}]}_{0}^{τ}$ Calculate the transmit power of D2D user j (1≤j≤N) on channel k(k∈Ω), where ${[&Center Dot;]}_{x}^{the y} = \min (m a x (&Center Dot;,), the y), τ = \frac{1}{h_{j, B}} (\frac{g_{i, B}}{γ_{t h}^{c}} \cdot P_{m a x}^{c} - N_{0}),$ where h _j,B represents the channel gain between D2D user j and the base station, g _i,B represents the channel gain between cellular user i and the base station, is the SINR threshold for cellular users; and $Δ = g_{j} {g_{i, B}}^{2} A (g_{j} {g_{i, B}}^{2} A + \frac{4}{λ l no 2} {g_{i, B}}^{2} B (g_{j} + B)),$ in $A = N_{0} &Center Dot; (1 + \frac{γ_{t h}^{c} &Center Dot; h_{i, j}}{g_{i, B}}), B = \frac{γ_{t h}^{c} &Center Dot; h_{j, B} &Center Dot; h_{i, j}}{g_{i, B}};$

6)根据所述步骤5)中得到的D2D用户j(1≤j≤N)的发送功率将代入公式中，得到蜂窝用户i(1≤i≤M)的发送功率；6) According to the transmission power of D2D user j (1≤j≤N) obtained in step 5) Will Into the formula , get the transmit power of cellular user i (1≤i≤M);

7)根据公式计算D2D用户j(1≤j≤N)在信道k(1≤k≤K)上的容量，并将信道k分配给值最大的D2D用户j*，再将信道k归入D2D用户复用信道集合S；7) According to the formula Calculate the capacity of D2D user j (1≤j≤N) on channel k (1≤k≤K), and allocate channel k to The D2D user j* with the largest value, and then the channel k is classified into the D2D user multiplexing channel set S;

8)根据公式和t＝t+1分别迭代更新λ(t)和t，θ(t)为步长序列，且θ(t)的常用表达式为或式中η为常量；8) According to the formula and t=t+1 iteratively update λ(t) and t respectively, θ(t) is a step sequence, and the common expression of θ(t) is or In the formula, η is a constant;

9)根据所述步骤8)中得到的λ(t)，重复执行所述步骤5)至8)，直到满足条件|λ(t+1)-λ(t)|≤exp，其中exp为一数值很小的常量，且取值为exp＝10^-8。9) According to the λ(t) obtained in the step 8), repeat the steps 5) to 8), until the condition |λ(t+1)-λ(t)|≤exp is met, where exp is one It is a constant with a very small value, and its value is exp=10 ^-8 .

本发明联合优化蜂窝网络中D2D通信的功率控制和信道分配，根据蜂窝用户和D2D用户的最小信干噪比要求，通过构建以D2D用户容量最大化为目标的优化模型。采用拉格朗日乘子法求解D2D用户和蜂窝用户的最优发送功率，并根据用户最优发送功率计算D2D用户容量，将信道分配给容量值最大的D2D用户。实现了最优资源分配，能够降低蜂窝用户与D2D用户之间的干扰，减少蜂窝用户和D2D用户的能量开销，提高D2D用户容量和网络频谱利用率。The present invention jointly optimizes the power control and channel allocation of D2D communication in the cellular network, and constructs an optimization model aiming at maximizing D2D user capacity according to the minimum signal-to-interference-noise ratio requirements of cellular users and D2D users. The Lagrangian multiplier method is used to solve the optimal transmission power of D2D users and cellular users, and the D2D user capacity is calculated according to the user's optimal transmission power, and the channel is allocated to the D2D user with the largest capacity value. The optimal resource allocation is realized, the interference between cellular users and D2D users can be reduced, the energy overhead of cellular users and D2D users can be reduced, and the capacity of D2D users and the utilization rate of network spectrum can be improved.

有益效果Beneficial effect

本发明与现有技术相比，具有以下优点：Compared with the prior art, the present invention has the following advantages:

1.本发明方法以D2D用户容量最大化为优化目标，对蜂窝用户和D2D用户的发送功率进行最优分配，与等功率分配方法相比，能量分配更加优化合理，避免蜂窝用户和D2D用户的能量浪费。1. The method of the present invention takes the maximization of D2D user capacity as the optimization goal, and optimally allocates the transmission power of the cellular user and the D2D user. Compared with the equal power allocation method, the energy allocation is more optimized and reasonable, and the interference between the cellular user and the D2D user is avoided. Energy wasted.

2.不同于传统的资源分配方法，本发明提出的联合优化信道分配和功率控制方法可以为用户分配最优资源，从而大大提高频谱利用率，也提高了网络性能。2. Unlike traditional resource allocation methods, the joint optimization channel allocation and power control method proposed by the present invention can allocate optimal resources for users, thereby greatly improving spectrum utilization and network performance.

3.本发明的信道分配和功率控制方法在保障蜂窝用户和D2D用户的QoS条件下，可以提高不同蜂窝用户数下的D2D用户容量，降低用户的功耗和费用，从而为用户提供更好地服务。3. The channel allocation and power control method of the present invention can improve the capacity of D2D users with different numbers of cellular users under the condition of ensuring the QoS of cellular users and D2D users, and reduce the power consumption and cost of users, thereby providing users with better Serve.

附图说明Description of drawings

图1为本发明方法的流程示意图。Fig. 1 is a schematic flow chart of the method of the present invention.

图2为本发明方法的网络模型示意图。Fig. 2 is a schematic diagram of a network model of the method of the present invention.

图3为不同资源分配方案下的D2D用户容量比较图。FIG. 3 is a comparison diagram of D2D user capacity under different resource allocation schemes.

图4为随着D2D对间距离增大的D2D用户容量变化图。Fig. 4 is a graph of D2D user capacity change with the increase of the distance between D2D pairs.

具体实施方法Specific implementation method

下面结合实施例和说明书附图对发明的技术方案进行详细说明：Below in conjunction with embodiment and accompanying drawing, the technical solution of the invention is described in detail:

本发明的共享蜂窝系统上行链路资源的D2D通信模型如附图2所示，它由一个位于小区中心的基站，随机分布于小区的M个蜂窝用户和N对D2D用户组成，并分别用C＝{1,2,..,M}和D＝{1,2,...,N}表示蜂窝用户集和D2D用户集。定义包括M×K个变量和N×K个变量的数组，分别用于存放蜂窝用户与D2D用户的发送功率，并初始化数组P_C＝Ο_M×K和P_D＝Ο_N×K。蜂窝用户i(1≤i≤M)占用信道k(1≤k≤K)进行通信，D2D用户复用已经分配给蜂窝用户的K个信道，其中K表示可用信道数，定义S为D2D用户复用蜂窝用户信道集。The D2D communication model of the present invention for sharing the uplink resources of the cellular system is shown in Figure 2. It consists of a base station located in the center of the cell, M cellular users randomly distributed in the cell and N pairs of D2D users. ={1,2,..,M} and D={1,2,...,N} represent a cellular user set and a D2D user set. Define an array including M×K variables and N×K variables, which are respectively used to store the transmit power of cellular users and D2D users, and initialize the arrays P _C =Ο _M×K and _PD =Ο _N×K . Cellular user i (1≤i≤M) occupies channel k (1≤k≤K) for communication, and D2D users reuse K channels that have been allocated to cellular users, where K represents the number of available channels, and S is defined as D2D user multiplex Use the cellular user channel set.

在信道k上，蜂窝用户i向基站发送信号x_c,i，D2D用户j的发送信号为x_d,j，基站接收到蜂窝用户i的信号与D2D用户j的接收信号分别为：On channel k, cellular user i sends a signal x _c, i to the base station, and the signal sent by D2D user j is x _d,j , the signal received by the base station from cellular user i and the received signal from D2D user j are respectively:

${y the y}_{c c,, i i} = = \sqrt{{p p}_{i i,, k k}^{c c}} {g g}_{i i,, B B} {x x}_{c c . . i i} + + \sqrt{{p p}_{j j,, k k}^{c c}} {g g}_{j j,, B B} {x x}_{c c,, j j} + + {n no}_{11} - - - - - - ((11))$

${y the y}_{d d,, j j} = = \sqrt{{p p}_{j j,, k k}^{d d}} {g g}_{j j} {x x}_{d d,, j j} + + \sqrt{{p p}_{i i,, k k}^{c c}} {h h}_{i i,, j j} {x x}_{c c . . i i} + + {n no}_{22} - - - - - - ((22))$

其中和分别是共享信道k的蜂窝用户i的发送功率和D2D对j发送端的发送功率。g_i,B表示蜂窝用户i与基站间的信道增益，g_i,B可以表示为g_i,B＝κ·d_i,B ^-α·|g₀|²，式中，d_i,B是蜂窝用户i发送端到基站之间的距离，κ和α分别是信道衰落常数和信道衰落因子，g₀服从均值为0方差为1的复指数分布。同样的，可以得到D2D用户j发送端与接收端的信道增益g_j，D2D用户j与基站间的信道增益h_j,B，占用同一信道的蜂窝用户i与D2D用户j之间的信道增益h_i,j。n₁和n₂分别是蜂窝用户到基站链路和D2D接收端链路上的加性高斯白噪声。不失一般性，假设所有通信链路都具有相同的噪声功率N₀。in and are the transmit power of cellular user i sharing channel k and the transmit power of D2D pair j sender, respectively. g _i,B represents the channel gain between cellular user i and the base station, g _i,B can be expressed as g _i,B =κ·d _i,B ^-α ·|g ₀ | ² , where d _i,B is The distance between the sending end of cellular user i and the base station, κ and α are the channel fading constant and channel fading factor respectively, g ₀ obeys the complex exponential distribution with mean 0 and variance 1. Similarly, the channel gain g _j between the sending end and the receiving end of D2D user j, the channel gain h _j,B between D2D user j and the base station, and the channel gain h _i between cellular user i and D2D user j occupying the same channel can be obtained _,j . n ₁ and n ₂ are the additive white Gaussian noise on the cellular user-to-base station link and the D2D receiver link, respectively. Without loss of generality, it is assumed that all communication links have the same noise power N ₀ .

根据计算D2D用户j(1≤j≤N)在信道k(1≤k≤K)上的信干噪比，将满足条件的蜂窝用户i占用的信道k归入集合Ω，否则不归入集合Ω，其中Ω为D2D用户的候选信道集，即 $Ω_{j} = {k | Γ_{j, k}^{d} > γ_{t h}^{d}, k = i, i = 1, 2, ..., M}, &ForAll; j &Element; D$ 且Ω＝{Ω_j|j＝1,2,...,N}。P_tot表示D2D用户的总发送功率，是蜂窝用户的最大发送功率，表示D2D用户的信干噪比阈值。本发明以最大化D2D用户容量，构建以下优化目标问题：according to Calculating the SINR of D2D user j (1≤j≤N) on channel k (1≤k≤K) will satisfy the condition The channel k occupied by the cellular user i belongs to the set Ω, otherwise it does not belong to the set Ω, where Ω is the candidate channel set of the D2D user, that is $Ω_{j} = {k | Γ_{j, k}^{d} > γ_{t h}^{d}, k = i, i = 1, 2, ..., m}, &ForAll; j &Element; D.$ And Ω={Ω _j |j=1, 2, . . . , N}. P _tot represents the total transmission power of D2D users, is the maximum transmit power of the cellular user, Indicates the SINR threshold of D2D users. In order to maximize the D2D user capacity, the present invention constructs the following optimization target problem:

$\underset{{{{p p}_{i i,, k k}^{c c},, {p p}_{j j,, k k}^{d d},, {π π}_{j j,, k k}}}}{max max} \underset{k k &Element; &Element; Ω Ω}{Σ Σ} \underset{j j &Element; &Element; D D.}{Σ Σ} {log log}_{22} ((11 + + \frac{{p p}_{j j,, k k}^{d d} {g g}_{j j}}{{N N}_{00} + + {p p}_{i i,, k k}^{c c} {h h}_{i i,, j j}})) \cdot &Center Dot; {π π}_{j j,, k k} - - - - - - ((33))$

Subjectto:Subject to:

$00 \leq \leq {p p}_{i i,, k k}^{c c} \leq \leq {P P}_{m m a a x x}^{c c},, &ForAll; &ForAll; i i &Element; &Element; C C - - - - - - ((44))$

${p p}_{j j,, k k}^{d d} \cdot &Center Dot; {h h}_{j j,, B B} - - \frac{{p p}_{i i,, k k}^{c c} \cdot &Center Dot; {g g}_{i i,, B B}}{{γ γ}_{t t h h}^{c c}} + + {N N}_{00} \leq \leq 00 - - - - - - ((55))$

$\underset{k k &Element; &Element; Ω Ω}{Σ Σ} \underset{j j &Element; &Element; D D.}{Σ Σ} {π π}_{j j,, k k} \cdot &Center Dot; {p p}_{j j,, k k}^{d d} \leq \leq {P P}_{t t o o t t} - - - - - - ((66))$

$\underset{j j &Element; &Element; D D.}{Σ Σ} {π π}_{j j,, k k} \leq \leq 11,, {π π}_{j j,, k k} &Element; &Element; {{00,, 11}},, k k = = 11,, 22,, ... ...,, K K - - - - - - ((77))$

其中和分别为在信道k上蜂窝用户i和D2D用户j的发送功率，π_j,k是信道选择因子，π_j,k＝1表示信道k被D2D用户j复用，π_j,k＝0则表示信道k未被D2D用户j复用，是蜂窝用户i在信道k上的信干噪比，表示蜂窝用户的信干噪比阈值。式(5)是由蜂窝用户最小信干噪比要求转化而得。in and are the transmit powers of cellular user i and D2D user j on channel k respectively, π _j,k is the channel selection factor, π _j,k = 1 means that channel k is multiplexed by D2D user j, π _j,k = 0 means Channel k is not multiplexed by D2D user j, is the SINR of cellular user i on channel k, Indicates the SINR threshold for cellular users. Equation (5) is the minimum signal-to-interference-noise ratio requirement of cellular users Transformed.

问题(3)相对于约束条件(6)的拉格朗日函数为：The Lagrangian function of problem (3) relative to constraint (6) is:

$L L (({{{p p}_{i i,, k k}^{c c},, {p p}_{j j,, k k}^{d d},, {π π}_{j j,, k k}}},, λ λ)) = = \underset{k k &Element; &Element; Ω Ω}{Σ Σ} \underset{j j &Element; &Element; D D.}{Σ Σ} {log log}_{22} ((11 + + \frac{{p p}_{j j,, k k}^{d d} {g g}_{j j}}{{N N}_{00} + + {p p}_{i i,, k k}^{c c} {h h}_{i i,, j j}})) {π π}_{j j,, k k} - - λ λ ((\underset{k k &Element; &Element; Ω Ω}{Σ Σ} \underset{j j &Element; &Element; D D.}{Σ Σ} {π π}_{j j,, k k} \cdot &Center Dot; {p p}_{j j,, k k}^{d d} - - {P P}_{t t o o t t})) - - - - - - ((88))$

其中λ是拉格朗日乘子，拉格朗日对偶函数可以表示为：where λ is the Lagrangian multiplier, and the Lagrangian dual function can be expressed as:

其中的定义为G_k(λ)为：one of them defined as G _k (λ) is:

对于给定λ，式(15)可以分解为K个独立问题，每个问题对应一个给定的信道k，即式(16)中的问题。表示将信道k分配给D2D用户j时G_k(λ)的值，此时π_j,k＝1。那么对于给定λ，原优化函数可以转化为如下表达式：For a given λ, Equation (15) can be decomposed into K independent problems, and each problem corresponds to a given channel k, which is the problem in Equation (16). Indicates the value of G _k (λ) when channel k is allocated to D2D user j, and at this time π _j,k =1. Then for a given λ, the original optimization function can be transformed into the following expression:

${G G}_{k k}^{j j} ((λ λ)) = = \underset{{{{p p}_{i i,, k k}^{c c},, {p p}_{j j,, k k}^{d d}}}}{max max} {log log}_{22} ((11 + + \frac{{p p}_{j j,, k k}^{d d} \cdot \cdot {g g}_{j j}}{{N N}_{00} + + {p p}_{i i . . k k}^{c c} \cdot \cdot {h h}_{i i,, j j}})) - - λ λ \cdot \cdot {p p}_{j j,, k k}^{d d} - - - - - - ((1111))$

Subjectto:Subject to:

$00 \leq \leq {p p}_{i i,, k k}^{c c} \leq \leq {P P}_{m m a a x x}^{c c} - - - - - - ((1212))$

${p p}_{j j,, k k}^{d d} \cdot \cdot {h h}_{j j,, B B} - - \frac{{p p}_{i i,, k k}^{c c} \cdot \cdot {g g}_{i i,, B B}}{{γ γ}_{t t h h}^{c c}} + + {N N}_{00} \leq \leq 00 - - - - - - ((1313))$

上述问题中的和是耦合的，可以通过分解方法对该问题求解。对于给定的可以发现问题(11)的目标函数是的单调减函数。这说明问题(11)的目标函数在取得最小值时可以达到最大值。约束(12)和(13)重写为更加紧凑的形式：of the above questions and are coupled, the problem can be solved by decomposition method. for a given It can be found that the objective function of problem (11) is The monotonically decreasing function of . This shows that the objective function of problem (11) is The maximum value can be reached when the minimum value is taken. Constraints (12) and (13) are rewritten in a more compact form:

$\frac{{γ γ}_{t t h h}^{c c}}{{g g}_{i i,, B B}} (({p p}_{j j,, k k}^{d d} \cdot &Center Dot; {h h}_{j j,, B B} + + {N N}_{00})) \leq \leq {p p}_{i i,, k k}^{c c} \leq \leq {P P}_{max max}^{c c} - - - - - - ((1414))$

因此，最优的为：Therefore, the optimal for:

${\overset{~ ~}{p p}}_{i i,, k k}^{c c} = = \frac{{γ γ}_{t t h h}^{c c}}{{g g}_{i i,, B B}} (({p p}_{j j,, k k}^{d d} \cdot &Center Dot; {h h}_{j j,, B B} + + {N N}_{00})) - - - - - - ((1515))$

则将式(15)代入问题(11)的目标函数中可以简化问题(11)如下：Then, substituting equation (15) into the objective function of problem (11) can simplify problem (11) as follows:

${G G}_{k k}^{j j} ((λ λ)) = = \underset{{p p}_{j j,, k k}^{d d}}{max max} {log log}_{22} ((11 + + \frac{{p p}_{j j,, k k}^{d d} \cdot &Center Dot; {g g}_{j j}}{{N N}_{00} ((11 + + \frac{{γ γ}_{t t h h}^{c c} \cdot &Center Dot; {h h}_{i i,, j j}}{{g g}_{i i,, B B}})) + + \frac{{γ γ}_{t t h h}^{c c} \cdot &Center Dot; {h h}_{j j,, B B} \cdot \cdot {h h}_{i i,, j j}}{{g g}_{i i,, B B}} \cdot &Center Dot; {p p}_{j j,, k k}^{d d}})) - - λ λ \cdot &Center Dot; {p p}_{j j,, k k}^{d d} - - - - - - ((1616))$

Subjectto:Subject to:

${p p}_{j j,, k k}^{d d} \leq \leq \frac{11}{{h h}_{j j,, B B}} ((\frac{{g g}_{i i,, B B}}{{γ γ}_{t t h h}^{c c}} \cdot &Center Dot; {P P}_{max max}^{c c} - - {N N}_{00})) - - - - - - ((1717))$

约束条件(17)由约束条件(14)转化而得，为了证明上述问题为凸问题，求公式(16)中对的二阶导数，可得：Constraint condition (17) is transformed from constraint condition (14). In order to prove that the above problem is a convex problem, find the formula (16) right The second derivative of , we can get:

$\frac{{d d}^{22} {G G}_{k k}^{j j} ((λ λ))}{d d {(({p p}_{j j,, k k}^{d d}))}^{22}} = = \frac{- - A A \cdot &Center Dot; {g g}_{j j} \cdot &Center Dot; ((A A \cdot &Center Dot; ((22 \cdot &Center Dot; B B + + {g g}_{j j})) + + 22 \cdot &Center Dot; B B ((B B + + {g g}_{j j})) \cdot &Center Dot; {p p}_{j j,, k k}^{d d}))}{{((A A + + B B \cdot &Center Dot; {p p}_{j j,, k k}^{d d} + + {g g}_{j j} \cdot &Center Dot; {p p}_{j j,, k k}^{d d}))}^{22} \cdot \cdot {((A A + + B B \cdot \cdot {p p}_{j j,, k k}^{d d}))}^{22}} - - - - - - ((1818))$

其中：in:

$A A = = {N N}_{00} \cdot &Center Dot; ((11 + + \frac{{γ γ}_{t t h h}^{c c} \cdot &Center Dot; {h h}_{i i,, j j}}{{g g}_{i i,, B B}})) - - - - - - ((1919))$

$B B = = \frac{{γ γ}_{t t h h}^{c c} \cdot &Center Dot; {h h}_{j j,, B B} \cdot &Center Dot; {h h}_{i i,, j j}}{{g g}_{i i,, B B}} - - - - - - ((2020))$

式(18)恒小于0，所以是的凸函数，可以通过凸优化方法得到最优解。Equation (18) is always less than 0, so yes The convex function of , the optimal solution can be obtained by convex optimization method.

根据KKT条件，求出对的一阶导数并令其等于0可得：According to the KKT conditions, find right and setting it equal to 0 gives:

$\frac{11}{ln ln 22} \cdot &Center Dot; \frac{{g g}_{j j} {g g}_{i i,, B B}^{22} \cdot &Center Dot; A A}{{g g}_{i i,, B B} \cdot &Center Dot; A A + + {g g}_{i i,, B B} (({g g}_{j j} + + B B)) \cdot &Center Dot; {p p}_{j j,, k k}^{d d}} \cdot &Center Dot; \frac{11}{{g g}_{i i,, B B} \cdot &Center Dot; A A + + {g g}_{i i,, B B} \cdot \cdot B B \cdot \cdot {p p}_{j j,, k k}^{d d}} - - λ λ = = 00 - - - - - - ((21 twenty one))$

由式(21)得到D2D用户的最优发送功率：The optimal transmit power of D2D users can be obtained from formula (21):

${\overset{~ ~}{p p}}_{j j,, k k}^{d d} = = {[[\frac{\sqrt{Δ Δ}}{22 {g g}_{i i,, B B}^{22} \cdot \cdot B B (({g g}_{j j} + + B B))} - - \frac{A A}{22 B B} - - \frac{A A}{22 (({g g}_{j j} + + B B))}]]}_{00}^{τ τ} - - - - - - ((22 twenty two))$

其中 ${[\cdot]}_{x}^{y} = m i n (m a x (\cdot, x), y), τ = \frac{1}{h_{j, B}} (\frac{g_{i, B}}{γ_{t h}^{c}} \cdot P_{m a x}^{c} - N_{0})$ 且Δ有如下表达式：in ${[&Center Dot;]}_{x}^{the y} = m i no (m a x (&Center Dot;, x), the y), τ = \frac{1}{h_{j, B}} (\frac{g_{i, B}}{γ_{t h}^{c}} &Center Dot; P_{m a x}^{c} - N_{0})$ And Δ has the following expression:

$Δ Δ = = {g g}_{j j} {g g}_{i i,, B B}^{22} \cdot \cdot A A \cdot \cdot (({g g}_{j j} {g g}_{i i,, B B}^{22} \cdot &Center Dot; A A + + \frac{44}{λ λ \cdot \cdot l l n no 22} \cdot \cdot {g g}_{i i,, B B}^{22} \cdot \cdot B B (({g g}_{j j} + + B B)))) - - - - - - ((23 twenty three))$

将式(22)代入式(15)中可以得到蜂窝用户i在信道k上的最优发送功率在求得最优的和之后，可以求得为：Substituting Equation (22) into Equation (15), the optimal transmit power of cellular user i on channel k can be obtained in pursuit of the best and after, can be obtained as:

${G G}_{k k}^{j j} ((λ λ)) = = {log log}_{22} ((11 + + \frac{{\overset{~ ~}{p p}}_{j j,, k k}^{d d} \cdot \cdot {g g}_{j j}}{{N N}_{00} + + {\overset{~ ~}{p p}}_{i i . . k k}^{c c} \cdot \cdot {h h}_{i i,, j j}})) - - λ λ \cdot &Center Dot; {\overset{~ ~}{p p}}_{j j,, k k}^{d d} - - - - - - ((24 twenty four))$

通过求得信道k的所有候选D2D用户对应的值，将信道k分配给值最大的D2D用户j*，即且π_j*,k＝1，并将信道k归入集合S。而且在信道k上，对于D2D用户j≠j*，其π_j,k＝0,则可得最优的蜂窝用户和D2D用户的发送功率集和在求得所有K个信道的之后，对偶函数G(λ)可以通过式(9)求得。最后，需要求得最大化G(λ)的最优λ≥0。初始化t＝1和λ(t)＝2，并且λ根据下式进行迭代更新：By finding all candidate D2D users corresponding to channel k value, assign channel k to The D2D user j* with the largest value, namely And π _j*,k =1, and the channel k is classified into the set S. And on channel k, for D2D user j≠j*, its π _j,k =0, Then the optimal transmit power set of cellular users and D2D users can be obtained and After obtaining all K channels Afterwards, the dual function G(λ) can be obtained by formula (9). Finally, it is necessary to obtain the optimal λ≥0 that maximizes G(λ). Initialize t=1 and λ(t)=2, and λ is iteratively updated according to the following formula:

$λ λ ((t t + + 11)) = = λ λ ((t t)) - - θ θ ((t t)) (({P P}_{t t o o t t} - - \underset{k k}{Σ Σ} \underset{j j}{Σ Σ} {\overset{~ ~}{p p}}_{j j,, k k}^{d d})) - - - - - - ((2525))$

式中的θ(t)＞0是一系列的步长序列，一些流行的θ(t)选取规则为和其中η为常数。In the formula, θ(t)>0 is a series of step sequences, and some popular θ(t) selection rules are and where n is a constant.

本发明基于QoS的信道分配和功率控制方法的具体流程如附图1所示。The specific flow of the QoS-based channel allocation and power control method of the present invention is shown in FIG. 1 .

综上所述，本发明在考虑蜂窝用户和D2D用户的QoS要求，蜂窝用户功率受限和D2D用户总功率受限的条件下，以最大化D2D用户容量为优化目标，对蜂窝用户和D2D用户的发送功率进行了最优分配，并为D2D用户分配合理的信道。如附图3所示是本发明提出的信道分配和功率控制方法与等功率资源分配方法的D2D用户容量对比图。从图中可以看出本发明方法可以获得更好的系统性能；如附图4所示是本发明提出的随着D2D用户距离增大时的D2D用户容量，并与其他资源分配方法相比较的效果图，从图中可以看出本方法可以进一步提高D2D用户容量。In summary, the present invention considers the QoS requirements of cellular users and D2D users, the limited power of cellular users and the limited total power of D2D users, and takes maximizing the capacity of D2D users as the optimization goal. The transmit power is optimally allocated, and reasonable channels are allocated for D2D users. As shown in FIG. 3 , it is a comparison diagram of D2D user capacity between the channel allocation and power control method proposed by the present invention and the equal power resource allocation method. It can be seen from the figure that the method of the present invention can obtain better system performance; as shown in Figure 4, it is the D2D user capacity proposed by the present invention as the D2D user distance increases, and compared with other resource allocation methods Effect diagram, it can be seen from the diagram that this method can further improve the D2D user capacity.

Claims

A kind of channel allocation based on QoS and Poewr control method in 1.D2D network, it is characterized in that, the method comprises the following steps:

1) use C={1 respectively, 2 ..., M} and D={1,2 ..., N} represents that phone user collects and D2D user's collection, and M be phone user's number, N be D2D to number of users, K represents number of available channels, defines S and represents the channel set of the multiplexing phone user of D2D user;

2) D2D user j (1≤j≤N) Signal to Interference plus Noise Ratio on channel k (1≤k≤K) is calculated p in formula _totfor the total transmitted power of D2D user, g _jfor D2D user is to the channel gain of j, N ₀represent noise power, represent the maximum transmit power of phone user, h _i,jrepresent the channel gain between phone user i and D2D user j;

3) according to the minimum Signal to Interference plus Noise Ratio requirement of D2D user j be included in set omega by the channel k met the demands, wherein Ω is the candidate channel set of D2D user, represent the Signal to Interference plus Noise Ratio threshold value of D2D user;

4) initialization t=1 and λ (t)=2, wherein t is iterations, and λ is Lagrange multiplier;

5) according to formula ${\tilde{p}}_{j, k}^{d} = {[\frac{\sqrt{Δ}}{2 {g_{i, B}}^{2} \cdot B (g_{j} + B)} - \frac{A}{2 B} - \frac{A}{2 (g_{j} + B)}]}_{0}^{τ}$ Calculate D2D user j (1≤j≤N) transmitted power on channel k (k ∈ Ω), wherein ${[\cdot]}_{x}^{y} = m i n (m a x (\cdot, x), y), τ = \frac{1}{h_{j, B}} (\frac{g_{i, B}}{γ_{t h}^{c}} \cdot P_{m a x}^{c} - N_{0}),$ H in formula _j,Brepresent the channel gain between D2D user j and base station, g _i,Brepresent the channel gain between phone user i and base station, for the Signal to Interference plus Noise Ratio threshold value of phone user; And $Δ = g_{j} {g_{i, B}}^{2} A (g_{j} {g_{i, B}}^{2} A + \frac{4}{λ l n 2} {g_{i, B}}^{2} B (g_{j} + B)),$ Wherein $A = N_{0} \cdot (1 + \frac{γ_{t h}^{c} \cdot h_{i, j}}{g_{i, B}}), B = \frac{γ_{t h}^{c} \cdot h_{j, B} \cdot h_{i, j}}{g_{i, B}};$

6) according to described step 5) in the transmitted power of D2D user j (1≤j≤N) that obtains will substitute into formula in, obtain the transmitted power of phone user i (1≤i≤M);

7) according to formula calculate D2D user j (1≤j≤N) capacity on channel k (1≤k≤K), and channel k is distributed to be worth maximum D2D user j*, then channel k is included into D2D user's multipling channel S set;

8) according to formula with t=t+1 respectively iteration upgrade λ (t) and t, θ (t) they are step series, and the conventional expression formula of θ (t) is or in formula, η is constant;

9) according to described step 8) in the λ (t) that obtains, repeat described step 5) to 8), until satisfy condition | λ (t+1)-λ (t) |≤exp, wherein exp is the constant that a numerical value is very little, and value is exp=10 ^-8.