CN101047417A

CN101047417A - Selection preprocess method for downlink link antenna of multi-user MIMO system

Info

Publication number: CN101047417A
Application number: CNA2007100720750A
Authority: CN
Inventors: 张林波; 刘彤; 田园; 张曙; 田野; 赵旦峰
Original assignee: Harbin Engineering University
Current assignee: Harbin Engineering University
Priority date: 2007-04-20
Filing date: 2007-04-20
Publication date: 2007-10-03
Anticipated expiration: 2027-04-20
Also published as: CN101047417B

Abstract

本发明提供了一种多用户MIMO系统下行链路的天线选择预处理方法，包括如下步骤，步骤1：建立多用户MIMO系统下行链路模型，确定基站与每个用户间的信道传输矩阵及发送符号的空时正交分组码与空间复用组合的编码方式；步骤2：通过基站与每个用户之间的信道传输矩阵，确定每个用户的预处理矩阵的基；步骤3：利用基站与每个用户之间的信道传输矩阵与步骤2中确定的每个用户的预处理矩阵的基确定每个用户的预处理矩阵的最优酉阵；步骤4：将多用户MIMO系统下行链路模型转换为等效的矢量形式，使用线形迫零的方法实现信号检测。本发明能够在不显著提高方法复杂度的前提下，大大提高多用户MIMO系统下行链路的性能。The present invention provides a multi-user MIMO system downlink antenna selection preprocessing method, including the following steps, step 1: establish a multi-user MIMO system downlink model, determine the channel transmission matrix and transmission matrix between the base station and each user Combination of space-time orthogonal block code and spatial multiplexing of symbols; Step 2: Determine the basis of each user's preprocessing matrix through the channel transmission matrix between the base station and each user; Step 3: Use the base station and each user The basis of the channel transmission matrix between each user and the preprocessing matrix of each user determined in step 2 determines the optimal unitary matrix of the preprocessing matrix of each user; step 4: the multi-user MIMO system downlink model Convert to the equivalent vector form, and use the linear zero-forcing method to realize signal detection. The invention can greatly improve the downlink performance of the multi-user MIMO system without significantly increasing the complexity of the method.

Description

A preprocessing method for downlink antenna selection in multi-user MIMO system

(一)技术领域(1) Technical field

本发明属于无线通信技术领域，特别涉及一种下行链路的天线选择预处理方法。The invention belongs to the technical field of wireless communication, and in particular relates to a downlink antenna selection preprocessing method.

(二)背景技术(2) Background technology

MIMO(Multiple Input Multiple Output)系统在概念上非常简单，任何一个无线通信系统，只要是在无线链路的两端使用多根天线，或者天线阵列，就构成了一个MIMO系统。在多径环境里，MIMO系统可以在不增加带宽的情况下成倍地提高通信系统的容量，并且通过空时编码的方法能克服信道衰落问题。The MIMO (Multiple Input Multiple Output) system is very simple in concept. Any wireless communication system that uses multiple antennas or antenna arrays at both ends of the wireless link constitutes a MIMO system. In a multipath environment, the MIMO system can double the capacity of the communication system without increasing the bandwidth, and the channel fading problem can be overcome by the method of space-time coding.

但对于多用户MIMO系统而言，除了要克服信道衰落的问题以外，更需要解决的问题是用户间的共道干扰(CCI，Co-Channel Interference)问题，若不去除用户间的共道干扰，将会对目标信号的检测与估计产生很大的影响，导致系统性能急剧下降。就多用户MIMO系统的上行链路而言，采用预处理技术会增加移动台的设计难度，在工程上不容易实现，因此在基站可以使用多用户检测(MUD，Multi User Detection)算法去除这种干扰。对于多用户MIMO系统的下行链路，多用户检测技术会增加移动终端的复杂度和成本，实现上有很大的难度。目前常用的去除下行链路干扰的办法是采用Lai-U Choi，Ross D.Murch.A TransmitPreprocessing technique for multiuser MIMO Systems Using a Decomposition Approach.IEEETransactions on Wireless Communications.2004，3(1)：20-24P中提出的预处理算法，具体的工作原理如图1所示：首先执行101模块，将获得的基站与每个用户之间的信道信息组成信道传输矩阵的形式；接着利用101模块输出的信道传输矩阵，在模块102中分别计算每个用户的预处理矩阵的基；在模块103中分别为每个用户产生任意酉阵；104模块为数据源，将基站发送给每个用户的数据进行调制，输出发送符号；105模块将104模块输出的发送符号通过空时编码器进行编码，输出发送符号矩阵；模块102的输出与模块103的输出以及模块105输出的发送符号矩阵相乘以后由基站的天线单元106发送出去；经过信道传播，由每个用户的天线单元107接收；天线单元107接收到的符号分别由每个用户的检测单元108进行检测处理。这种传统的预处理方法可以将多用户之间的共道干扰完全消除，从而将多用户通信系统转换为并行的多个独立单用户通信系统，实现下行链路的多用户传输。但该方法存在的缺点是：仅仅考虑了如何去除用户间的共道干扰问题，没有考虑接收信噪比对解调的影响，导致系统性能不佳。However, for multi-user MIMO systems, in addition to overcoming the problem of channel fading, the problem that needs to be solved is the problem of co-channel interference (CCI, Co-Channel Interference) between users. If the co-channel interference between users is not removed, It will have a great impact on the detection and estimation of the target signal, resulting in a sharp decline in system performance. As far as the uplink of the multi-user MIMO system is concerned, the use of preprocessing technology will increase the design difficulty of the mobile station, which is not easy to implement in engineering, so the multi-user detection (MUD, Multi User Detection) algorithm can be used in the base station to remove this interference. For the downlink of a multi-user MIMO system, the multi-user detection technology will increase the complexity and cost of the mobile terminal, and it is very difficult to implement. The commonly used method to remove downlink interference is to use Lai-U Choi, Ross D.Murch.A TransmitPreprocessing technique for multiuser MIMO Systems Using a Decomposition Approach.IEEETransactions on Wireless Communications.2004, 3(1):20-24P The specific working principle of the proposed preprocessing algorithm is shown in Figure 1: firstly, the 101 module is executed, and the obtained channel information between the base station and each user is formed into the form of a channel transmission matrix; then, the channel transmission matrix output by the 101 module is used , calculate the basis of the preprocessing matrix of each user in module 102 respectively; Produce arbitrary unitary matrix for each user in module 103 respectively; 104 modules are data source, the data that base station sends to each user is modulated, output Transmit symbol; module 105 encodes the transmit symbol output by module 104 through a space-time encoder, and outputs a transmit symbol matrix; the output of module 102 is multiplied by the output of module 103 and the transmit symbol matrix output by module 105, and then the antenna unit of the base station 106 and sent out; propagated through the channel, and received by the antenna unit 107 of each user; the symbols received by the antenna unit 107 are respectively detected and processed by the detection unit 108 of each user. This traditional preprocessing method can completely eliminate the co-channel interference among multiple users, thereby converting the multi-user communication system into parallel multiple independent single-user communication systems, and realizing downlink multi-user transmission. However, the disadvantage of this method is that it only considers how to remove the co-channel interference between users, and does not consider the influence of the received signal-to-noise ratio on demodulation, resulting in poor system performance.

(三)发明内容(3) Contents of the invention

本发明的目的在于提供一种能够在不显著提高方法复杂度的前提下，大大提高系统性能的多用户MIMO系统下行链路的天线选择预处理方法。The purpose of the present invention is to provide a multi-user MIMO system downlink antenna selection preprocessing method that can greatly improve system performance without significantly increasing the complexity of the method.

本发明的目的是这样实现的：The purpose of the present invention is achieved like this:

步骤1：建立多用户MIMO系统下行链路模型，确定基站与每个用户间的信道传输矩阵及发送符号的空时正交分组码与空间复用组合的编码方式；Step 1: Establish the downlink model of the multi-user MIMO system, determine the channel transmission matrix between the base station and each user and the coding method of the space-time orthogonal block code and spatial multiplexing combination of the transmitted symbols;

步骤2：通过基站与每个用户之间的信道传输矩阵，确定每个用户的预处理矩阵的基；Step 2: Determine the basis of each user's preprocessing matrix through the channel transmission matrix between the base station and each user;

步骤3：利用基站与每个用户之间的信道传输矩阵与步骤2中确定的每个用户的预处理矩阵的基相乘，并将乘积进行奇异值分解，确定每个用户的预处理矩阵的最优酉阵；Step 3: Use the channel transmission matrix between the base station and each user to multiply the basis of each user's preprocessing matrix determined in step 2, and perform singular value decomposition on the product to determine the basis of each user's preprocessing matrix optimal unitary array;

步骤4：利用空时正交分组码与空间复用组合的编码方式的特点将多用户MIMO系统下行链路模型转换为等效的矢量形式，并使用线形迫零的方法实现信号检测。Step 4: Convert the downlink model of the multi-user MIMO system into an equivalent vector form by using the characteristics of the space-time orthogonal block code and space multiplexing combination coding method, and use the linear zero-forcing method to realize signal detection.

本发明还有这样一些技术特征：The present invention also has some technical characteristics:

1、所述的步骤1中建立多用户MIMO系统下行链路模型的步骤为：设定基站发射符号所需的天线数m为偶数，并且满足 $m < M - \max {Σ_{i = 1, i &NotEqual; k}^{K} N_{i}, k = 1,2, \cdot \cdot \cdot, K}$ ，这里M表示基站具有的天线数，K为移动终端用户数，并且第k个用户具有N_k根接收天线，k＝1，…，K；1. The step of establishing the downlink model of the multi-user MIMO system in the step 1 is: setting the number of antennas m required by the base station to transmit symbols as an even number, and satisfying $m < m - \max {Σ_{i = 1, i &NotEqual; k}^{K} N_{i}, k = 1,2, &Center Dot; &Center Dot; &Center Dot;, K}$ , where M represents the number of antennas that the base station has, K is the number of mobile terminal users, and the kth user has N _k receiving antennas, k=1,...,K;

2、所述的步骤1中建立多用户MIMO系统下行链路模型后在系统频分双工的情况下，基站通过各个用户的反馈信道获得每个用户的信道信息，并将信道信息组成信道传输矩阵；2. After the downlink model of the multi-user MIMO system is established in step 1, under the condition of frequency division duplexing of the system, the base station obtains the channel information of each user through the feedback channel of each user, and forms the channel information into a channel for transmission matrix;

3、所述的步骤1中建立多用户MIMO系统下行链路模型后在系统时分双工的情况下，基站利用信道的对称性，直接计算出每个用户的信道信息，并将信道信息组成信道传输矩阵。3. After the downlink model of the multi-user MIMO system is established in step 1, under the condition of system time division duplex, the base station uses the symmetry of the channel to directly calculate the channel information of each user, and form the channel information into a channel transfer matrix.

4、所述的步骤1中基站为每个用户发送的数据经过调制之后，得到发送符号，并通过空时正交分组码与空间复用组合的编码方式，得到基站为每个用户发送的符号矩阵，由基站的发射天线在两个相邻时隙发送出去；4. After the data sent by the base station for each user in the step 1 is modulated, the transmitted symbols are obtained, and the symbols transmitted by the base station for each user are obtained through the encoding method of space-time orthogonal block code and space multiplexing combination The matrix is sent by the transmitting antenna of the base station in two adjacent time slots;

5、所述的步骤2的具体步骤为：基于已知的基站与每个用户之间的信道传输矩阵，利用递推的方法，得到每个用户的预处理矩阵的基。5. The specific steps of step 2 are: based on the known channel transmission matrix between the base station and each user, using a recursive method to obtain the basis of the preprocessing matrix of each user.

6、所述的步骤3的具体步骤为：将基站与每个用户之间的信道传输矩阵与每个用户各自的预处理矩阵的基相乘并进行奇异值分解后，得到U∑D三个矩阵乘积的形式；最后选择矩阵D的前m列为每个用户的预处理矩阵的最优酉阵，m为基站发送符号矩阵所需的发射天线数。6. The specific steps of step 3 are: after multiplying the channel transmission matrix between the base station and each user with the basis of each user's respective preprocessing matrix and performing singular value decomposition, three U∑D The form of the matrix product; finally select the first m columns of the matrix D to be the optimal unitary matrix of each user's preprocessing matrix, and m is the number of transmitting antennas required for the base station to send the symbol matrix.

7、所述的步骤4的具体步骤为：将基站与各用户之间的信道传输矩阵与预处理矩阵的乘积作为等效信道传输矩阵，利用空时正交分组码与空间复用组合的编码方式的特点，得到每个用户的输入输出符号间的等效矢量关系，在每个用户的接收端，用线性迫零方法检测信号。。7. The specific steps of step 4 are: the product of the channel transmission matrix and the preprocessing matrix between the base station and each user is used as the equivalent channel transmission matrix, and the encoding of the combination of space-time orthogonal block code and space multiplexing is used. According to the characteristics of the method, the equivalent vector relationship between the input and output symbols of each user is obtained, and the signal is detected by the linear zero-forcing method at the receiving end of each user. .

本发明不仅可以彻底去除用户间的干扰，实现下行链路中的多用户传输，而且与传统预处理方法相比，本发明的方法能够在不显著提高方法复杂度的前提下，大大提高多用户MIMO系统下行链路的性能。The present invention can not only completely remove the interference between users and realize the multi-user transmission in the downlink, but also compared with the traditional preprocessing method, the method of the present invention can greatly improve the multi-user transmission without significantly increasing the complexity of the method. Performance of MIMO system downlink.

本发明的创新之处在于：能够在不显著提高方法复杂度的前提下，大大提高多用户MIMO系统下行链路的性能。多用户MIMO系统使用本发明的下行链路天线选择预处理方法，不仅可以去除用户间的干扰，实现下行链路的多用户传输；而且通过选择预处理矩阵的最优酉阵，使得信号检测的后验信噪比下限最大，从而提高多用户MIMO系统的性能。以基站具有9根发射天线，共有3个移动终端用户，并且每个用户具有2根接收天线的多用户MIMO系统为例，在误比特率为2×10^-4时，与传统预处理方法相比，使用本发明的下行链路天线选择预处理方法使得多用户MIMO系统的性能提高了约7.0dB。The innovation of the present invention lies in that it can greatly improve the downlink performance of the multi-user MIMO system without significantly increasing the complexity of the method. The multi-user MIMO system uses the downlink antenna selection preprocessing method of the present invention, which can not only remove the interference between users, and realize the multi-user transmission of the downlink; and by selecting the optimal unitary matrix of the preprocessing matrix, the signal detection The lower bound of the a posteriori SNR is maximized, thereby improving the performance of the multi-user MIMO system. Taking a multi-user MIMO system with 9 transmitting antennas in the base station and 3 mobile terminal users in total, and each user has 2 receiving antennas as an example, when the bit error rate is 2×10 ^-4 , it is comparable to the traditional preprocessing method Compared with that, using the downlink antenna selection preprocessing method of the present invention improves the performance of the multi-user MIMO system by about 7.0dB.

(四)附图说明(4) Description of drawings

图1是传统的多用户MIMO系统下行链路的工作原理图；FIG. 1 is a working principle diagram of a traditional multi-user MIMO system downlink;

图2是本发明中确定每个用户预处理矩阵的基的工作流程图；Fig. 2 is the work flowchart of determining the basis of each user's preprocessing matrix among the present invention;

图3是本发明中确定每个用户预处理矩阵的最优酉阵的工作流程图；Fig. 3 is the work flowchart of determining the optimal unitary array of each user's preprocessing matrix among the present invention;

图4是本发明的多用户MIMO系统下行链路的工作原理图；Fig. 4 is a working principle diagram of the multi-user MIMO system downlink of the present invention;

图5是多用户MIMO系统下行链路使用传统预处理方法和本发明的天线选择预处理方法后获得的系统性能比较图。Fig. 5 is a comparison diagram of system performance obtained after using the traditional preprocessing method and the antenna selection preprocessing method of the present invention in the downlink of the multi-user MIMO system.

(五)具体实施方式(5) Specific implementation methods

下面结合附图和具体实施例对本发明作进一步的说明：The present invention will be further described below in conjunction with accompanying drawing and specific embodiment:

结合图1，传统的多用户MIMO系统下行链路的工作原理是：101是信道信息源单元，将获得的基站与每个用户之间的信道信息组成信道传输矩阵的形式；102是预处理矩阵基的计算单元，利用模块101的输出求得每个用户的预处理矩阵的基；103是任意酉阵的生成单元；104是数据源单元；105是空时编码单元，将104模块的输出以空时正交分组码与空间复用组合的编码方式进行编码；106是发射天线阵列单元；107是接收天线阵列单元；108是单用户检测单元。H⁽¹⁾，H⁽²⁾，…，H^(K)是单元101的输出信号，V⁽¹⁾，V⁽²⁾，…，V^(K)是单元102的输出信号。In conjunction with Figure 1, the working principle of the traditional multi-user MIMO system downlink is: 101 is the channel information source unit, which forms the channel transmission matrix from the obtained channel information between the base station and each user; 102 is the preprocessing matrix The calculation unit of basis utilizes the output of module 101 to obtain the basis of the preprocessing matrix of each user; 103 is the generating unit of any unitary matrix; 104 is a data source unit; 105 is a space-time coding unit, and the output of module 104 is obtained by Space-time orthogonal block code and space multiplexing are combined for coding; 106 is a transmitting antenna array unit; 107 is a receiving antenna array unit; 108 is a single user detection unit. H ⁽¹⁾ ^, H ⁽² ⁾ ^, ^.

结合图4，本发明的多用户MIMO系统下行链路的工作原理是：401是信道信息源，将获得的基站与每个用户之间的信道信息组成信道传输矩阵的形式；402是本发明多用户下行链路天线选择预处理单元，将模块101的输出按照图2和图3中的具体工作流程，求得每个用户的天线选择预处理矩阵；403是数据源单元；404是空时编码单元，将403模块的输出以空时正交分组码与空间复用组合的编码方式进行编码；405是发射天线阵列单元；406是接收天线阵列单元；407是单用户检测单元。H⁽¹⁾，H⁽²⁾，…，H^(K)是单元401的输出信号，T⁽¹⁾，T⁽²⁾，…，T^(K)是单元402的输出信号。4, the working principle of the downlink of the multi-user MIMO system of the present invention is: 401 is a channel information source, and the channel information between the obtained base station and each user is formed into a channel transmission matrix; 402 is the multi-user MIMO system of the present invention. The user downlink antenna selection preprocessing unit obtains the antenna selection preprocessing matrix of each user according to the output of module 101 according to the specific workflow in Figure 2 and Figure 3; 403 is the data source unit; 404 is the space-time coding unit, which encodes the output of module 403 with the combination of space-time orthogonal block code and space multiplexing; 405 is a transmitting antenna array unit; 406 is a receiving antenna array unit; 407 is a single user detection unit. H ⁽¹⁾ , H ⁽²⁾ , . . . , H ^(K) are output signals of unit 401 , T ⁽¹⁾ , T ⁽²⁾ ^, .

本发明的一个具体实施例如下所述，假设基站的发射天线数为9，用户数为3，且每个用户的接收天线数为2。A specific embodiment of the present invention is as follows, assuming that the number of transmitting antennas of the base station is 9, the number of users is 3, and the number of receiving antennas of each user is 2.

步骤1：基站获得每个用户的信道信息，并将它组成信道传输矩阵{H^(k)}，k＝1，2，3。说明如下：Step 1: The base station obtains the channel information of each user, and forms it into a channel transmission matrix {H ^(k) }, k=1, 2, 3. described as follows:

以第1个用户为例：h_1，1 ⁽¹⁾为基站第1根发射天线和用户1的第1根接收天线之间的信道；h_2，1 ⁽¹⁾为基站第1根发射天线和用户1的第2根接收天线之间的信道；依此类推，h_1，9 ⁽¹⁾为基站第9根发射天线和用户1的第1根接收天线之间的信道；h_2，9 ⁽¹⁾为基站第9根发射天线和用户1的第2根接收天线之间的信道。依次类推，可得基站与3个用户之间的信道传输矩阵分别为：Take the first user as an example: h _1,1 ⁽¹⁾ is the channel between the first transmitting antenna of the base station and the first receiving antenna of user 1; h _2,1 ⁽¹⁾ is the first transmitting antenna of the base station and the channel between the second receiving antenna of user 1; and so on, h _{1, 9} ⁽¹⁾ is the channel between the ninth transmitting antenna of the base station and the first receiving antenna of user 1; h _{2, 9} ⁽¹⁾ is the channel between the ninth transmitting antenna of the base station and the second receiving antenna of user 1. By analogy, the channel transmission matrices between the base station and the three users can be obtained as follows:

${H h}^{((k k))} = = [\begin{matrix} {h h}_{1,1 1,1}^{((k k))} & {h h}_{1,2 1,2}^{((k k))} & {h h}_{1,3 1,3}^{((k k))} & {h h}_{1,4 1,4}^{((k k))} & {h h}_{1,5 1,5}^{((k k))} & {h h}_{1,6 1,6}^{((k k))} & {h h}_{1,7 1,7}^{((k k))} & {h h}_{1,8 1,8}^{((k k))} & {h h}_{1,9 1,9}^{((k k))} \\ {h h}_{2,1 2,1}^{((k k))} & {h h}_{2,2 2,2}^{((k k))} & {h h}_{2,3 2,3}^{((k k))} & {h h}_{2,4 2,4}^{((k k))} & {h h}_{2,5 2,5}^{((k k))} & {h h}_{2,6 2,6}^{((k k))} & {h h}_{2,7 2,7}^{((k k))} & {h h}_{2,8 2,8}^{((k k))} & {h h}_{2,9 2,9}^{((k k))} \end{matrix}],, k k = = 1,2,3 1,2,3 - - - - - - ((11))$

基站为第k个用户发送的数据经过16-QAM调制之后，进行串并转换分成4路，得到4个发送符号s_k，1、s_k，2、s_k，3和s_k，4，并进一步通过空时正交分组码与空间复用组合的编码方式，得到基站为每个用户发送的符号矩阵，如公式(2)所示。发送符号由基站的4根发射天线在两个相邻时隙发送出去。After the data sent by the base station for the k-th user is modulated by 16-QAM, it is serial-to-parallel converted and divided into 4 channels to obtain 4 transmission symbols s _k,1 , s _k,2 , s _k,3 and s _k,4 , and Further, through the combination of space-time orthogonal block code and space multiplexing, the symbol matrix sent by the base station for each user is obtained, as shown in formula (2). The transmitted symbols are transmitted by the 4 transmitting antennas of the base station in two adjacent time slots.

公式(2)中点划线区分的是同一信道条件下相邻两个时隙发射的符号矢量。假设基站发射的符号总能量为E_s，第k个用户接收端的噪声矩阵n^(k)中的每个元素都是独立同分布的，均值为0，每一维的功率谱密度为N₀/2的复高斯随机变量。第k个用户的预处理矩阵为T^(k)，接收到的符号矩阵为x^(k)，则对于第k个用户(k＝1，2，3)，输入输出关系的复基带表达式为： $x^{(k)} = \sqrt{\frac{E_{s}}{Km}} H^{(k)} T^{(k)} s^{(k)} + Σ_{i = 1, i &NotEqual; k}^{3} (\sqrt{\frac{E_{s}}{Km}} H^{(k)} T^{(i)} s^{(i)}) + n^{(k)}, k = 1,2,3 - - - (3)$ The dotted line in formula (2) distinguishes the symbol vectors transmitted in two adjacent time slots under the same channel condition. Assuming that the total energy of the symbol transmitted by the base station is E _s , each element in the noise matrix n ^(k) of the kth user receiving end is independent and identically distributed, with a mean value of 0, and the power spectral density of each dimension is N ₀ / A complex Gaussian random variable of 2. The preprocessing matrix of the kth user is T ^(k) , and the received symbol matrix is x ^(k) , then for the kth user (k=1, 2, 3), the complex baseband expression of the input-output relationship is : $x^{(k)} = \sqrt{\frac{{E.}_{the s}}{Km}} h^{(k)} T^{(k)} {the s}^{(k)} + Σ_{i = 1, i &NotEqual; k}^{3} (\sqrt{\frac{{E.}_{the s}}{Km}} h^{(k)} T^{(i)} {the s}^{(i)}) + {no}^{(k)}, k = 1,2,3 - - - (3)$

对于每个用户而言，为了去除其它2个用户的干扰，需要预处理矩阵满足 $H^{(i)} T^{(k)} \overset{i &NotEqual; k}{=} 0 (1 \leq i, k \leq 3)$ 由于任何一个矩阵都可以分解为基矩阵与任意酉阵的乘积形式，所以首先在步骤2中确定3个用户的预处理矩阵的基。For each user, in order to remove the interference of the other 2 users, the preprocessing matrix needs to satisfy $h^{(i)} T^{(k)} \overset{i &NotEqual; k}{=} 0 (1 \leq i, k \leq 3)$ Since any matrix can be decomposed into the form of the product of the base matrix and any unitary matrix, first determine the bases of the preprocessing matrices of the three users in step 2.

步骤2：采用下面几个步骤分别求得3个用户预处理矩阵的基：Step 2: Use the following steps to obtain the basis of the three user preprocessing matrices respectively:

结合图2，以第1个用户为例，首先执行201模块，初始化k＝1；然后在202模块，初始化i＝1(如果k＝1，则初始化i＝2)；接着在模块203将基站与第2个用户之间的信道传输矩阵H⁽²⁾进行奇异值分解得到H⁽²⁾＝U∑D，选择矩阵D中与零奇异值对应的列向量组成矩阵Z，矩阵Z就是基站与第2个用户之间的信道传输矩阵H⁽²⁾的基；下一步执行模块204得到i＝3并且满足i≠1；接着在模块205中判断i是否大于3；由于i≤3，则执行模块206将基站与第3个用户之间的信道传输矩阵H⁽³⁾与203模块中求得的矩阵Z相乘并进行奇异值分解得到H⁽³⁾Z＝U∑D，选择矩阵D中与零奇异值对应的列向量组成矩阵W，矩阵W就是H⁽³⁾Z的基；下一步执行模块207将矩阵Z更新为Z＝ZW；接着执行模块204得到i＝4并且满足i≠1；然后执行模块205再判断i是否大于3；由于i＞3，则执行模块208得到第1个用户的预处理矩阵的基V⁽¹⁾＝Z。In conjunction with Fig. 2, taking the first user as an example, first execute module 201, initialize k=1; then in module 202, initialize i=1 (if k=1, then initialize i=2); The channel transmission matrix H ⁽²⁾ between the second user and the second user is subjected to singular value decomposition to obtain H ⁽²⁾ = U∑D, and the column vector corresponding to zero singular value in the matrix D is selected to form a matrix Z, and the matrix Z is the base station and The basis of the channel transmission matrix H ⁽²⁾ between the 2nd user; Next step execution module 204 obtains i=3 and satisfies i≠1; Then judges whether i is greater than 3 in module 205; Since i≤3, then execute Module 206 multiplies the channel transmission matrix H ⁽³⁾ between the base station and the third user with the matrix Z obtained in module 203 and performs singular value decomposition to obtain H ⁽³⁾ Z=U∑D, select matrix D The column vector corresponding to zero singular value forms matrix W, and matrix W is exactly the basis of H ⁽³⁾ Z; Next execution module 207 updates matrix Z to Z=ZW; Then execution module 204 obtains i=4 and satisfies i≠1 ; Then the execution module 205 judges whether i is greater than 3; since i>3, the execution module 208 obtains the basis V ⁽¹⁾ =Z of the preprocessing matrix of the first user.

第2个用户和第3个用户同样采用上述的过程分别求得它们的预处理矩阵的基V⁽²⁾和V⁽³⁾。The second user and the third user also use the above process to obtain the bases V ⁽²⁾ and V ⁽³⁾ of their preprocessing matrices respectively.

确定了预处理矩阵的基以后，需要在步骤3中选定预处理矩阵的最优酉阵，使得信号检测的后验信噪比下限最大。After the basis of the preprocessing matrix is determined, it is necessary to select the optimal unitary matrix of the preprocessing matrix in step 3, so that the lower limit of the a posteriori signal-to-noise ratio of the signal detection is the largest.

步骤3：利用基站与3个用户之间的信道传输矩阵以及步骤2中确定的3个用户的预处理矩阵的基，确定3个用户的预处理矩阵的最优酉阵。具体求得预处理矩阵的最优酉阵的步骤如下：Step 3: Using the channel transmission matrix between the base station and the 3 users and the bases of the 3 users' preprocessing matrices determined in step 2, determine the optimal unitary matrix of the 3 users' preprocessing matrices. The specific steps to obtain the optimal unitary matrix of the preprocessing matrix are as follows:

结合图3，以第1个用户为例，首先执行模块301，初始化k＝1；接着在302模块中将基站与第1个用户之间的信道传输矩阵H⁽¹⁾与第1个用户的预处理矩阵的基V⁽¹⁾相乘，并进行奇异值分解得到H⁽¹⁾V⁽¹⁾＝U∑D；最后在模块303中选择A⁽¹⁾为矩阵D的前4列，确定第1个用户的预处理矩阵的最优酉阵。In conjunction with FIG. 3 , taking the first user as an example, module 301 is first executed, and k=1 is initialized; then in module 302, the channel transmission matrix H (1) between the base station and the first user is combined with the channel transmission matrix H ⁽¹⁾ of the first user The basis V ⁽¹⁾ of preprocessing matrix is multiplied, and carries out singular value decomposition and obtains H ⁽¹⁾ V ⁽¹⁾ =U∑D; Finally, in module 303, select A ⁽¹⁾ to be the first 4 columns of matrix D, determine The optimal unitary matrix of the preprocessing matrix of the first user.

第2个用户和第3个用户同样采用上述的过程分别求得它们的预处理矩阵的最优酉阵A⁽²⁾和A⁽³⁾。The second user and the third user also use the above process to obtain the optimal unitary matrices A ⁽²⁾ and A ⁽³⁾ of their preprocessing matrices respectively.

经过步骤2和步骤3后，就可以确定多用户MIMO系统中每个用户下行链路的天线选择预处理矩阵T^(k)＝V^(k)A^(k)(k＝1，2，3)。After step 2 and step 3, the antenna selection preprocessing matrix T ^(k) = V ^(k) A ^(k) (k = 1, 2, 3) for each user downlink in the multi-user MIMO system can be determined .

步骤4：将3个用户的预处理矩阵V^(k)A^(k)带入到公式(3)中，得到第k个用户接收到的符号矩阵为：Step 4: Bring the preprocessing matrix V ^(k) A ^(k) of the three users into the formula (3), and the symbol matrix received by the kth user is obtained as:

$[\begin{matrix} {x x}_{1,1 1,1}^{((k k))} & {x x}_{1,2 1,2}^{((k k))} \\ {x x}_{2,1 2,1}^{((k k))} & {x x}_{2,2 2,2}^{((k k))} \end{matrix}] = = {H h}^{((k k))} {V V}^{((k k))} {A A}^{((k k))} [\begin{matrix} {s the s}_{k k,, 11} & - - {s the s}_{k k,, 44}^{* *} \\ {s the s}_{k k,, 22} & {s the s}_{k k,, 33}^{* *} \\ {s the s}_{k k,, 33} & {- - s the s}_{k k,, 22}^{* *} \\ {s the s}_{k k,, 22} & {s the s}_{k k,, 11}^{* *} \end{matrix}] + + [\begin{matrix} {n no}_{1,1 1,1}^{((k k))} & {n no}_{1,2 1,2}^{((k k))} \\ {n no}_{2,1 2,1}^{((k k))} & {n no}_{2,2 2,2}^{((k k))} \end{matrix}],, k k = = 1,2,3 1,2,3 - - - - - - ((44))$

定义等效信道传输矩阵为 ${\overset{&OverBar;}{H}}^{(k)} = H^{(k)} V^{(k)} A^{(k)} = [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(k)} & {\overset{&OverBar;}{h}}_{1,2}^{(k)} & {\overset{&OverBar;}{h}}_{1,3}^{(k)} & {\overset{&OverBar;}{h}}_{1,4}^{(k)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(k)} & {\overset{&OverBar;}{h}}_{2,2}^{(k)} & {\overset{&OverBar;}{h}}_{2,3}^{(k)} & {\overset{&OverBar;}{h}}_{2,4}^{(k)} \end{matrix}]$ ，利用空时正交分组码与空间复用组合的编码方式的特点，通过下面的变换Define the equivalent channel transfer matrix as ${\overset{&OverBar;}{h}}^{(k)} = h^{(k)} V^{(k)} A^{(k)} = [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(k)} & {\overset{&OverBar;}{h}}_{1,2}^{(k)} & {\overset{&OverBar;}{h}}_{1,3}^{(k)} & {\overset{&OverBar;}{h}}_{1,4}^{(k)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(k)} & {\overset{&OverBar;}{h}}_{2,2}^{(k)} & {\overset{&OverBar;}{h}}_{2,3}^{(k)} & {\overset{&OverBar;}{h}}_{2,4}^{(k)} \end{matrix}]$ , using the characteristics of the coding method of space-time orthogonal block code and space multiplexing combination, through the following transformation

${\overset{&OverBar; &OverBar;}{H h}}^{((k k))} = = [\begin{matrix} {\overset{&OverBar; &OverBar;}{h h}}_{1,1 1,1}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{1,2 1,2}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{1,3 1,3}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{1,4 1,4}^{((k k))} \\ {(({\overset{&OverBar; &OverBar;}{h h}}_{1,4 1,4}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{1,3 1,3}^{((k k))}))}^{* *} & {(({\overset{&OverBar; &OverBar;}{h h}}_{1,2 1,2}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{1,1 1,1}^{((k k))}))}^{* *} \\ {\overset{&OverBar; &OverBar;}{h h}}_{2,1 2,1}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{2,2 2,2}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{2,3 2,3}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{2,4 2,4}^{((k k))} \\ {(({\overset{&OverBar; &OverBar;}{h h}}_{2,4 2,4}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{2,3 2,3}^{((k k))}))}^{* *} & {(({\overset{&OverBar; &OverBar;}{h h}}_{2,2 2,2}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{2,1 2,1}^{((k k))}))}^{* *} \end{matrix}],, k k = = 1,2,3 1,2,3 - - - - - - ((55))$

得到第k个用户的输入输出符号间的等效矢量关系为：The equivalent vector relationship between the input and output symbols of the kth user is obtained as follows:

公式(6)中，定义 ${\overset{&OverBar;}{x}}^{(k)} = [\begin{matrix} x_{1,1}^{(k)} \\ {(x_{1,2}^{(k)})}^{*} \\ x_{2,1}^{(k)} \\ {(x_{2,2}^{(k)})}^{*} \end{matrix}]$ 为第k个用户的接收符号矢量。在接收端，用4×4维的线性均衡矩阵G^(k)＝( H^(k))^_乘以接收符号矢量

，然后将所有接收到的信号轮流作为期望信号，选择迫零矢量

{\overset{&RightArrow;}{w}}_{i}^{(k)} = {[G^{(k)}]}_{i}

。这里i＝1，2，3，4，[G^(k)]_i为线性均衡矩阵G^(k)的第i行。因此，基站发送给第k个用户的第i个符号的判决值__k，i为

In formula (6), define

{\overset{&OverBar;}{x}}^{(k)} = [\begin{matrix} x_{1,1}^{(k)} \\ {(x_{1,2}^{(k)})}^{*} \\ x_{2,1}^{(k)} \\ {(x_{2,2}^{(k)})}^{*} \end{matrix}]

is the received symbol vector of the kth user. At the receiving end, multiply the received symbol vector by the 4×4-dimensional linear equalization matrix G ^(k) = ( H ^(k) ) ^_

, and then take all the received signals as the desired signal in turn, and choose the zero-forcing vector

{\overset{&Right Arrow;}{w}}_{i}^{(k)} = {[G^{(k)}]}_{i}

. Here i=1, 2, 3, 4, [G ^(k) ] _i is the i-th row of the linear equalization matrix G ^(k) . Therefore, the decision value_k _,i of the i-th symbol sent by the base station to the k-th user is

以下举例来说明整个过程(假设条件见具体实施方式)。在步骤1中，首先基站获得3个用户的信道信息，并将它组成信道传输矩阵{H^(k)}，k＝1，2，3。假设基站与第1个用户之间的信道传输矩阵为：The following examples illustrate the whole process (assuming conditions refer to the detailed description). In step 1, firstly, the base station obtains channel information of 3 users, and forms it into a channel transmission matrix {H ^(k) }, k=1, 2, 3. Suppose the channel transmission matrix between the base station and the first user is:

${H h}^{((11))} = = {[\begin{matrix} 0.0000 0.0000 + + 1.0207 1.0207 i i & - - 0.2248 0.2248 - - 0.2482 0.2482 i i \\ 0.7743 0.7743 + + 0.4407 0.4407 i i & - - 1.3251 1.3251 + + 0.5650 0.5650 i i \\ 0.3028 0.3028 + + 0.6653 0.6653 i i & 0.6333 0.6333 - - 0.7015 0.7015 i i \\ 0.5169 0.5169 + + 0.1499 0.1499 i i & 0.4086 0.4086 + + 0.1682 0.1682 i i \\ 0.0285 0.0285 - - 0.7126 0.7126 i i & 0.4788 0.4788 - - 0.5247 0.5247 i i \\ 0.4023 0.4023 + + 0.7653 0.7653 i i & - - 0.1808 0.1808 - - 0.0930 0.0930 i i \\ - - 0.2669 0.2669 + + 0.2757 0.2757 i i & - - 0.2092 0.2092 + + 0.0622 0.0622 i i \\ - - 1.0431 1.0431 - - 0.4493 0.4493 i i & - - 0.1655 0.1655 - - 0.3957 0.3957 i i \\ 0.0838 0.0838 + + 0.3137 0.3137 i i & 0.2226 0.2226 - - 0.6717 0.6717 i i \end{matrix}]}^{T T}$

基站与第2个用户之间的信道传输矩阵为：The channel transmission matrix between the base station and the second user is:

${H h}^{((22))} = = {[\begin{matrix} 0.5524 0.5524 - - 0.8608 0.8608 i i & 0.4023 0.4023 - - 0.0292 0.0292 i i \\ - - 0.5810 0.5810 - - 0.7979 0.7979 i i & - - 0.1878 0.1878 - - 0.9541 0.9541 i i \\ - - 0.8399 0.8399 - - 0.1864 0.1864 i i & - - 1.5573 1.5573 + + 0.6742 0.6742 i i \\ 0.6974 0.6974 + + 0.0910 0.0910 i i & - - 0.3667 0.3667 + + 0.4642 0.4642 i i \\ 0.2315 0.2315 - - 0.8258 0.8258 i i & 0.1655 0.1655 - - 0.3257 0.3257 i i \\ 0.0152 0.0152 - - 0.1856 0.1856 i i & - - 0.7099 0.7099 - - 0.8578 0.8578 i i \\ - - 0.6697 0.6697 - - 0.9330 0.9330 i i & - - 0.2648 0.2648 + + 0.6585 0.6585 i i \\ - - 0.8385 0.8385 + + 0.0080 0.0080 i i & - - 0.7466 0.7466 - - 0.4562 0.4562 i i \\ 1.0412 1.0412 + + 0.5697 0.5697 i i & 0.0394 0.0394 + + 0.1638 0.1638 i i \end{matrix}]}^{T T}$

基站与第3个用户之间的信道传输矩阵为：The channel transmission matrix between the base station and the third user is:

${H h}^{((33))} = = {[\begin{matrix} 0.6999 0.6999 - - 0.0848 0.0848 i i & 0.9472 0.9472 - - 0.0462 0.0462 i i \\ 0.2074 0.2074 + + 0.3432 0.3432 i i & 1.0458 1.0458 - - 0.4211 0.4211 i i \\ 0.8074 0.8074 - - 0.1058 0.1058 i i & - - 0.4838 0.4838 - - 0.3074 0.3074 i i \\ - - 0.9135 0.9135 - - 0.0561 0.0561 i i & - - 0.0516 0.0516 - - 1.0855 1.0855 i i \\ - - 0.2338 0.2338 - - 0.4288 0.4288 i i & - - 0.5965 0.5965 - - 0.9527 0.9527 i i \\ 0.3520 0.3520 + + 0.3319 0.3319 i i & 1.0525 1.0525 - - 0.6389 0.6389 i i \\ - - 0.3864 0.3864 + + 0.0254 0.0254 i i & - - 0.5987 0.5987 + + 0.4437 0.4437 i i \\ - - 0.1742 0.1742 + + 0.3786 0.3786 i i & 0.4688 0.4688 + + 0.3909 0.3909 i i \\ - - 0.6040 0.6040 - - 0.1440 0.1440 i i & - - 0.8495 0.8495 - - 1.4526 1.4526 i i \end{matrix}]}^{T T}$

基站发送给第1个用户的数据经过16-QAM调制之后，进行串并转换分成4路，得到4个发送符号s_1，1＝0.9487-0.9487i、s_1，2＝-0.9487-0.3162i、s_1，3＝0.3162-0.9487i和s_1，4＝-0.9487-0.3162i，按照空时正交分组码与空间复用组合的编码方式，得到基站为第1个用户发送的符号矩阵s⁽¹⁾为：After the data sent by the base station to the first user is modulated by 16-QAM, it is serial-to-parallel converted and divided into 4 channels to obtain 4 transmission symbols s _1,1 =0.9487-0.9487i, s _1,2 =-0.9487-0.3162i, s _1,3 ＝0.3162-0.9487i and s _{1, 4} ＝-0.9487-0.3162i, according to the encoding method of space-time orthogonal block code and space multiplexing combination, the symbol matrix s ^{( 1)} as:

${s the s}^{((11))} = = [\begin{matrix} 0.9487 0.9487 - - 0.9487 0.9487 i i & 0.9487 0.9487 - - 0.3162 0.3162 i i \\ - - 0.9487 0.9487 - - 0.3162 0.3162 i i & 0.3162 0.3162 + + 0.9487 0.9487 i i \\ 0.3162 0.3162 - - 0.9487 0.9487 i i & 0.9487 0.9487 - - 0.3162 0.3162 i i \\ - - 0.9487 0.9487 - - 0.3162 0.3162 i i & 0.9487 0.9487 + + 0.9487 0.9487 i i \end{matrix}]$

基站发送给第2个用户的数据经过16-QAM调制之后，进行串并转换分成4路，得到4个发送符号s_2，1＝-0.3162-0.3162i、s_2，2＝0.9487+0.9487i、s_2，3＝-0.3162+0.9487i和s_2，4＝-0.3162+0.9487i，按照空时正交分组码与空间复用组合的编码方式，得到基站为第2个用户发送的符号矩阵s⁽²⁾为：After the data sent by the base station to the second user is modulated by 16-QAM, it is serial-to-parallel converted and divided into 4 channels to obtain 4 transmission symbols s _2,1 =-0.3162-0.3162i, s _2,2 =0.9487+0.9487i, s _2,3 ＝-0.3162+0.9487i and s _2,4 ＝-0.3162+0.9487i, according to the encoding method of space-time orthogonal block code and space multiplexing combination, obtain the symbol matrix s sent by the base station for the second user ⁽²⁾ is:

${s the s}^{((22))} = = [\begin{matrix} - - 0.3162 0.3162 - - 0.3162 0.3162 i i & 0.3162 0.3162 + + 0.9487 0.9487 i i \\ 0.9487 0.9487 + + 0.9487 0.9487 i i & - - 0.3162 0.3162 - - 0.9487 0.9487 i i \\ - - 0.3162 0.3162 + + 0.9487 0.9487 i i & - - 0.9487 0.9487 + + 0.9487 0.9487 i i \\ - - 0.3162 0.3162 + + 0.9487 0.9487 i i & - - 0.3162 0.3162 + + 0.3162 0.3162 i i \end{matrix}]$

基站发送给第3个用户的数据经过16-QAM调制之后，进行串并转换分成4路，得到4个发送符号s_3，1＝-0.9487+0.3162i、s_3，2＝-0.3162-0.3162i、s_3，3＝0.9487-0.9487i和s_3，4＝0.9487-0.3162i，按照空时正交分组码与空间复用组合的编码方式，得到基站为第3个用户发送的符号矩阵s⁽³⁾为：The data sent by the base station to the third user is modulated by 16-QAM, then serial-to-parallel converted and divided into 4 channels to obtain 4 transmission symbols s _3,1 =-0.9487+0.3162i, s _3,2 =-0.3162-0.3162i , s _3, 3 = 0.9487-0.9487i and s _{3, 4} = 0.9487-0.3162i, according to the encoding method of space-time orthogonal block code and space multiplexing combination, the symbol matrix s ^{( 3)} as:

${s the s}^{((33))} = = [\begin{matrix} - - 0.9487 0.9487 + + 0.3162 0.3162 i i & - - 0.9487 0.9487 - - 0.3162 0.3162 i i \\ - - 0.3162 0.3162 - - 0.3162 0.3162 i i & 0.9487 0.9487 + + 0.9487 0.9487 i i \\ 0.9487 0.9487 - - 0.9487 0.9487 i i & 0.3162 0.3162 - - 0.3162 0.3162 i i \\ 0.9487 0.9487 - - 0.3162 0.3162 i i & - - 0.9487 0.9487 - - 0.3162 0.3162 i i \end{matrix}]$

步骤2：利用基站与第1个用户之间的信道传输矩阵，按照图2中的具体工作流程，确定第1个用户预处理矩阵的基V⁽¹⁾为：Step 2: Using the channel transmission matrix between the base station and the first user, according to the specific workflow in Figure 2, determine the basis V ⁽¹⁾ of the first user's preprocessing matrix as:

${V V}^{((11))} = = [\begin{matrix} - - 0.2400 0.2400 + + 0.1176 0.1176 i i & 0.1248 0.1248 - - 0.0741 0.0741 i i & - - 0.2161 0.2161 + + 0.4329 0.4329 i i & 0.2462 0.2462 + + 0.1422 0.1422 i i & - - 0.1640 0.1640 - - 0.4017 0.4017 i i \\ - - 0.2196 0.2196 + + 0.0054 0.0054 i i & - - 0.4242 0.4242 - - 0.1289 0.1289 i i & - - 0.0026 0.0026 + + 0.3335 0.3335 i i & - - 0.2831 0.2831 + + 0.0816 0.0816 i i & - - 0.0301 0.0301 + + 0.5299 0.5299 i i \\ 0.0559 0.0559 + + 0.1834 0.1834 i i & - - 0.1128 0.1128 - - 0.2498 0.2498 i i & - - 0.0620 0.0620 + + 0.2370 0.2370 i i & 0.0921 0.0921 - - 0.1055 0.1055 i i & 0.4826 0.4826 - - 0.1811 0.1811 i i \\ 0.2487 0.2487 - - 0.2427 0.2427 i i & - - 0.320 0.320 + + 0.0691 0.0691 i i & - - 0.1842 0.1842 + + 0.1061 0.1061 i i & - - 0.4231 0.4231 - - 0.0558 0.0558 i i & 0.1273 0.1273 + + 0.1119 0.1119 i i \\ 0.7332 0.7332 + + 0.0700 0.0700 i i & 0.0251 0.0251 - - . . 04580458 i i & - - 0.1375 0.1375 + + 0.0530 0.0530 i i & 0.1127 0.1127 + + 0.1497 0.1497 i i & 0.0064 0.0064 + + 0.1641 0.1641 i i \\ 0.0134 0.0134 + + 0.1201 0.1201 i i & 0.7770 0.7770 + + 0.0613 0.0613 i i & 0.1286 0.1286 + + 0.1072 0.1072 i i & - - 0.0857 0.0857 - - 0.0046 0.0046 i i & 0.1104 0.1104 + + 0.2369 0.2369 i i \\ - - 0.0621 0.0621 - - 0.2406 0.2406 i i & 0.1288 0.1288 - - 0.1116 0.1116 i i & 0.6554 0.6554 - - 0.1137 0.1137 i i & - - 0.1245 0.1245 + + 0.0156 0.0156 i i & - - 0.0647 0.0647 - - 0.0226 0.0226 i i \\ 0.0748 0.0748 - - 0.1855 0.1855 i i & - - 0.1021 0.1021 + + 0.1006 0.1006 i i & 0.0415 0.0415 - - 0.1003 0.1003 i i & 0.7207 0.7207 + + 0.0329 0.0329 i i & 0.0025 0.0025 + + 0.2250 0.2250 i i \\ - - 0.1804 0.1804 + + 0.1775 0.1775 i i & - - 0.1357 0.1357 - - 0.1522 0.1522 i i & 0.1480 0.1480 + + 0.1537 0.1537 i i & 0.1581 0.1581 - - 0.1639 0.1639 i i & 0.2397 0.2397 + + 0.1659 0.1659 i i \end{matrix}]$

利用基站与2个用户之间的信道传输矩阵，按照图2中的具体工作流程，确定第2个用户预处理矩阵的基V⁽²⁾为：Using the channel transmission matrix between the base station and the two users, according to the specific workflow in Figure 2, determine the basis V ⁽²⁾ of the second user's preprocessing matrix as:

${V V}^{((22))} = = [\begin{matrix} 0.0483 0.0483 + + 0.3714 0.3714 i i & - - 0.3485 0.3485 + + 0.0713 0.0713 i i & 0.0271 0.0271 + + 0.0639 0.0639 i i & 0.2443 0.2443 - - 0.0671 0.0671 i i & - - 0.1834 0.1834 + + 0.4403 0.4403 i i \\ 0.3638 0.3638 - - 0.1072 0.1072 i i & - - 0.2079 0.2079 - - 0.0848 0.0848 i i & - - 0.0739 0.0739 - - 0.0699 0.0699 i i & 0.0528 0.0528 - - 0.3173 0.3173 i i & 0.1989 0.1989 - - 0.3039 0.3039 i i \\ 0.1093 0.1093 + + 0.1346 0.1346 i i & - - 0.2182 0.2182 - - 0.1453 0.1453 i i & 0.1512 0.1512 - - 0.2142 0.2142 i i & 0.3206 0.3206 - - 0.1500 0.1500 i i & 0.0443 0.0443 - - 0.2082 0.2082 i i \\ 0.1797 0.1797 - - 0.1191 0.1191 i i & 0.2054 0.2054 + + 0.1355 0.1355 i i & - - 0.0627 0.0627 - - 0.2949 0.2949 i i & 0.2140 0.2140 + + 0.1624 0.1624 i i & 0.0577 0.0577 - - 0.3418 0.3418 i i \\ 0.7003 0.7003 - - 0.0650 0.0650 i i & 0.0348 0.0348 - - 0.1462 0.1462 i i & - - 0.0836 0.0836 + + 0.1226 0.1226 i i & - - 0.0464 0.0464 + + 0.0956 0.0956 i i & 0.3382 0.3382 + + 0.0070 0.0070 i i \\ 0.0580 0.0580 + + 0.0986 0.0986 i i & 0.8085 0.8085 + + 0.0551 0.0551 i i & - - 0.0114 0.0114 + + 0.0264 0.0264 i i & 0.0316 0.0316 - - 0.1758 0.1758 i i & - - 0.0733 0.0733 + + 0.1698 0.1698 i i \\ 0.0259 0.0259 - - 0.1562 0.1562 i i & 0.0897 0.0897 + + 0.0180 0.0180 i i & 0.8551 0.8551 - - 0.0242 0.0242 i i & - - 0.0049 0.0049 - - 0.0902 0.0902 i i & - - 0.0923 0.0923 - - 0.1036 0.1036 i i \\ - - 0.0798 0.0798 - - 0.1900 0.1900 i i & 0.0748 0.0748 + + 0.0007 0.0007 i i & - - 0.0577 0.0577 + + 0.0755 0.0755 i i & 0.7565 0.7565 - - 0.0297 0.0297 i i & - - 0.0305 0.0305 - - 0.0930 0.0930 i i \\ - - 0.1810 0.1810 - - 0.1769 0.1769 i i & 0.0168 0.0168 - - 0.0166 0.0166 i i & - - 0.2387 0.2387 + + 0.0679 0.0679 i i & - - 0.0178 0.0178 - - 0.1002 0.1002 i i & 0.5454 0.5454 - - 0.0118 0.0118 i i \end{matrix}]$

利用基站与3个用户之间的信道传输矩阵，按照图2中的具体工作流程，确定第3个用户预处理矩阵的基V⁽³⁾为：Using the channel transmission matrix between the base station and the three users, according to the specific workflow in Figure 2, determine the basis V ⁽³⁾ of the third user’s preprocessing matrix as:

${V V}^{((33))} = = [\begin{matrix} 0.1211 0.1211 + + 0.3462 0.3462 i i & - - 0.1446 0.1446 + + 0.1071 0.1071 i i & - - 0.1346 0.1346 - - 0.2235 0.2235 i i & 0.6125 0.6125 - - 0.1003 0.1003 i i & - - 0.0234 0.0234 - - 0.0553 0.0553 i i \\ 0.2583 0.2583 - - 0.2235 0.2235 i i & 0.2307 0.2307 - - 0.1191 0.1191 i i & - - 0.1848 0.1848 + + 0.0844 0.0844 i i & - - 0.0970 0.0970 - - 0.2173 0.2173 i i & 0.2036 0.2036 - - 0.2807 0.2807 i i \\ 0.0098 0.0098 - - 0.1720 0.1720 i i & - - 0.2258 0.2258 - - 0.0982 0.0982 i i & - - 0.2355 0.2355 - - 0.1284 0.1284 i i & - - 0.1400 0.1400 - - 0.1549 0.1549 i i & - - 0.0204 0.0204 + + 0.1627 0.1627 i i \\ - - 0.0761 0.0761 + + 0.1776 0.1776 i i & - - 0.1365 0.1365 - - 0.4680 0.4680 i i & 0.1156 0.1156 + + 0.5673 0.5673 i i & 0.0643 0.0643 - - 0.0384 0.0384 i i & - - 0.2830 0.2830 - - 0.2732 0.2732 i i \\ 0.7563 0.7563 + + 0.0141 0.0141 i i & - - 0.0475 0.0475 + + 0.0223 0.0223 i i & 0.1123 0.1123 + + 0.0731 0.0731 i i & - - 0.0371 0.0371 + + 0.2980 0.2980 i i & - - 0.0515 0.0515 - - 0.0467 0.0467 i i \\ 0.0926 0.0926 + + 0.0723 0.0723 i i & 0.7319 0.7319 + + 0.0272 0.0272 i i & 00 . . 16941694 + + 0.0220 0.0220 i i & 0.0698 0.0698 - - 0.1070 0.1070 i i & - - 0.1728 0.1728 + + 0.0510 0.0510 i i \\ 0.0106 0.0106 - - 0.1448 0.1448 i i & 0.1961 0.1961 - - 0.0122 0.0122 i i & 00 . . 57785778 + + 00 . . 05760576 i i & - - 0.0981 0.0981 - - 0.1202 0.1202 i i & 0.2370 0.2370 - - 0.0943 0.0943 \\ - - 0.0861 0.0861 - - 0.2467 0.2467 i i & - - 0.0525 0.0525 - - 0.1204 0.1204 i i & 0.0556 0.0556 + + 0.2067 0.2067 i i & 0.6094 0.6094 + + 0.0568 0.0568 i i & - - 0.0099 0.0099 - - 0.0397 0.0397 i i \\ - - 0.0151 0.0151 + + 0.0791 0.0791 i i & - - 0.0893 0.0893 - - 0.0029 0.0029 i i & 0.2157 0.2157 + + 0.0829 0.0829 i i & 0.0615 0.0615 - - 0.0052 0.0052 i i & 0.7682 0.7682 - - 0.0178 0.0178 i i \end{matrix}]$

步骤3：利用基站与第1个用户之间的信道传输矩阵以及步骤2中确定的第1个用户的预处理矩阵的基，通过图3所示的流程，得到第1个用户预处理矩阵的最优酉阵A⁽¹⁾为：Step 3: Using the channel transmission matrix between the base station and the first user and the basis of the preprocessing matrix of the first user determined in step 2, through the process shown in Figure 3, get the preprocessing matrix of the first user The optimal unitary matrix A ⁽¹⁾ is:

$A^{(1)} = [\begin{matrix} 0.3038 - 0.4416 i & - 0.1796 + 0.2077 i & - 0.1800 + 0.3768 i & - 0.4384 + 0.3409 i \\ - 0.0666 + 0.0878 i & - 0.2094 + 0.1106 i & 0.2032 - 0.3913 i & 0.2691 + 0.3385 i \\ 0.4210 + 0.2725 i & 0.3637 + 0.1857 i & 0.6757 + 0.0605 i & - 0.1988 - 0.1627 i \\ 0.6081 - 0.0487 i & - 0.0683 - 0.2963 i & - 0.0011 + 0.3112 i & 0.6312 + 0.1510 i \\ 0.1094 + 0.2548 i & - 0.7714 + 0.1195 i & 0.2213 - 0.1586 i & - 0.0592 + 0.1176 i \end{matrix}]$ 利用基站 $A^{(1)} = [\begin{matrix} 0.3038 - 0.4416 i & - 0.1796 + 0.2077 i & - 0.1800 + 0.3768 i & - 0.4384 + 0.3409 i \\ - 0.0666 + 0.0878 i & - 0.2094 + 0.1106 i & 0.2032 - 0.3913 i & 0.2691 + 0.3385 i \\ 0.4210 + 0.2725 i & 0.3637 + 0.1857 i & 0.6757 + 0.0605 i & - 0.1988 - 0.1627 i \\ 0.6081 - 0.0487 i & - 0.0683 - 0.2963 i & - 0.0011 + 0.3112 i & 0.6312 + 0.1510 i \\ 0.1094 + 0.2548 i & - 0.7714 + 0.1195 i & 0.2213 - 0.1586 i & - 0.0592 + 0.1176 i \end{matrix}]$ use base station

与第2个用户之间的信道传输矩阵以及步骤2中确定的第2个用户的预处理矩阵的基，通过图3所示的流程，得到第2个用户预处理矩阵的最优酉阵A⁽²⁾为：The channel transmission matrix between the second user and the base of the second user's preprocessing matrix determined in step 2, through the process shown in Figure 3, the optimal unitary matrix A of the second user's preprocessing matrix is obtained ⁽²⁾ is:

$A^{(2)} = [\begin{matrix} - 0.0173 - 0.3651 i & 0.0267 + 0.1805 i & 0.1746 + 0.5405 i & 0.3169 + 0.3633 i \\ 0.0674 + 0.0029 i & 0.1502 - 0.4472 i & 0.4974 + 0.4423 i & - 0.4640 + 0.1580 i \\ 0.4044 + 0.0368 i & - 0.0777 + 0.7293 i & 0.4345 + 0.0166 i & - 0.2094 - 0.2210 i \\ 0.6482 - 0.0638 i & 0.2196 - 0.2213 i & - 0.0115 + 0.0535 i & 0.6013 - 0.0648 i \\ - 0.0540 + 0.5194 i & 0.2938 - 0.1511 i & 0.2046 + 0.0214 i & 0.0950 - 0.2445 i \end{matrix}]$ 利用基站与第3个用户之间的信道传输矩阵以及步骤2中确定的第3个用户的预处理矩阵的基，通过图3所示的流程，得到第3个用户预处理矩阵的最优酉阵A⁽³⁾为： $A^{(2)} = [\begin{matrix} - 0.0173 - 0.3651 i & 0.0267 + 0.1805 i & 0.1746 + 0.5405 i & 0.3169 + 0.3633 i \\ 0.0674 + 0.0029 i & 0.1502 - 0.4472 i & 0.4974 + 0.4423 i & - 0.4640 + 0.1580 i \\ 0.4044 + 0.0368 i & - 0.0777 + 0.7293 i & 0.4345 + 0.0166 i & - 0.2094 - 0.2210 i \\ 0.6482 - 0.0638 i & 0.2196 - 0.2213 i & - 0.0115 + 0.0535 i & 0.6013 - 0.0648 i \\ - 0.0540 + 0.5194 i & 0.2938 - 0.1511 i & 0.2046 + 0.0214 i & 0.0950 - 0.2445 i \end{matrix}]$ Using the channel transmission matrix between the base station and the third user and the basis of the preprocessing matrix of the third user determined in step 2, through the process shown in Figure 3, the optimal unitary of the preprocessing matrix of the third user is obtained Array A ⁽³⁾ is:

$A^{(3)} = [\begin{matrix} - 0.1120 - 0.3297 i & 0.0436 + 0.4380 i & 0.0141 + 0.5974 i & 02037 + 0.3165 i \\ - 0.3840 + 0.0791 i & 0.1041 - 0.3392 i & 0.3127 + 0.2883 i & - 0.3466 - 0.0393 i \\ 0.2741 - 0.4021 i & - 0.3728 + 0.2940 i & 0.6527 - 0.1142 i & - 0.1326 - 0.0630 i \\ - 0.2933 + 0.1254 i & - 0.4462 + 0.0008 i & - 0.0395 - 0.0520 i & 0.7423 - 0.0903 i \\ - 0.0977 - 0.6139 i & - 0.1944 - 0.4668 i & - 0.1020 - 0.0911 i & 0.0399 - 0.3928 i \end{matrix}]$ 由步骤2和步骤3，得到3个用户的预处理矩阵分别为T^(k)＝V^(k)A^(k)(k＝1，2，3)。 $A^{(3)} = [\begin{matrix} - 0.1120 - 0.3297 i & 0.0436 + 0.4380 i & 0.0141 + 0.5974 i & 02037 + 0.3165 i \\ - 0.3840 + 0.0791 i & 0.1041 - 0.3392 i & 0.3127 + 0.2883 i & - 0.3466 - 0.0393 i \\ 0.2741 - 0.4021 i & - 0.3728 + 0.2940 i & 0.6527 - 0.1142 i & - 0.1326 - 0.0630 i \\ - 0.2933 + 0.1254 i & - 0.4462 + 0.0008 i & - 0.0395 - 0.0520 i & 0.7423 - 0.0903 i \\ - 0.0977 - 0.6139 i & - 0.1944 - 0.4668 i & - 0.1020 - 0.0911 i & 0.0399 - 0.3928 i \end{matrix}]$ From step 2 and step 3, the preprocessing matrices of the three users are respectively T ^(k) =V ^(k) A ^(k) (k=1, 2, 3).

步骤4：定义 ${\overset{&OverBar;}{H}}^{(1)} = H^{(1)} V^{(1)} A^{(1)} = [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(1)} & {\overset{&OverBar;}{h}}_{1,2}^{(1)} & {\overset{&OverBar;}{h}}_{1,3}^{(1)} & {\overset{&OverBar;}{h}}_{1,4}^{(1)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(1)} & {\overset{&OverBar;}{h}}_{2,2}^{(1)} & {\overset{&OverBar;}{h}}_{2,3}^{(1)} & {\overset{&OverBar;}{h}}_{2,4}^{(1)} \end{matrix}]$ ，得到 H⁽¹⁾为：Step 4: Define ${\overset{&OverBar;}{h}}^{(1)} = h^{(1)} V^{(1)} A^{(1)} = [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(1)} & {\overset{&OverBar;}{h}}_{1,2}^{(1)} & {\overset{&OverBar;}{h}}_{1,3}^{(1)} & {\overset{&OverBar;}{h}}_{1,4}^{(1)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(1)} & {\overset{&OverBar;}{h}}_{2,2}^{(1)} & {\overset{&OverBar;}{h}}_{2,3}^{(1)} & {\overset{&OverBar;}{h}}_{2,4}^{(1)} \end{matrix}]$ , get H ⁽¹⁾ as:

${\overset{&OverBar;}{H}}^{(1)} = [\begin{matrix} - 1.8798 + 0.0000 i & - 0.9655 - 0.0000 i & - 0.0000 & - 0.0000 + 0.0000 i \\ 0.5905 - 1.3342 i & - 0.5034 + 1.1376 i & - 0.0000 + 0.0000 i & - 0.0000 - 0.0000 i \end{matrix}]$ 定义 ${\overset{&OverBar;}{h}}^{(1)} = [\begin{matrix} - 1.8798 + 0.0000 i & - 0.9655 - 0.0000 i & - 0.0000 & - 0.0000 + 0.0000 i \\ 0.5905 - 1.3342 i & - 0.5034 + 1.1376 i & - 0.0000 + 0.0000 i & - 0.0000 - 0.0000 i \end{matrix}]$ definition

${\overset{&OverBar;}{H}}^{(2)} = H^{(2)} V^{(2)} A^{(2)} = [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(2)} & {\overset{&OverBar;}{h}}_{1,2}^{(2)} & {\overset{&OverBar;}{h}}_{1,3}^{(2)} & {\overset{&OverBar;}{h}}_{1,4}^{(2)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(2)} & {\overset{&OverBar;}{h}}_{2,2}^{(2)} & {\overset{&OverBar;}{h}}_{2,3}^{(2)} & {\overset{&OverBar;}{h}}_{2,4}^{(2)} \end{matrix}]$ ，得到 H⁽²⁾为： ${\overset{&OverBar;}{h}}^{(2)} = h^{(2)} V^{(2)} A^{(2)} = [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(2)} & {\overset{&OverBar;}{h}}_{1,2}^{(2)} & {\overset{&OverBar;}{h}}_{1,3}^{(2)} & {\overset{&OverBar;}{h}}_{1,4}^{(2)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(2)} & {\overset{&OverBar;}{h}}_{2,2}^{(2)} & {\overset{&OverBar;}{h}}_{2,3}^{(2)} & {\overset{&OverBar;}{h}}_{2,4}^{(2)} \end{matrix}]$ , get H ⁽²⁾ as:

${\overset{&OverBar;}{H}}^{(2)} = [\begin{matrix} - 1.9430 - 0.0000 i & 1.1383 - 0.0000 i & 0.0000 + 0.0000 i & 0.0000 + 0.0000 i \\ - 1.5232 + 0.7810 i & - 1.1498 + 0.5896 i & 0.0000 + 0.0000 i & - 0.0000 + 0.0000 i \end{matrix}]$ 定义 ${\overset{&OverBar;}{h}}^{(2)} = [\begin{matrix} - 1.9430 - 0.0000 i & 1.1383 - 0.0000 i & 0.0000 + 0.0000 i & 0.0000 + 0.0000 i \\ - 1.5232 + 0.7810 i & - 1.1498 + 0.5896 i & 0.0000 + 0.0000 i & - 0.0000 + 0.0000 i \end{matrix}]$ definition

${\overset{&OverBar;}{H}}^{(3)} = H^{(3)} V^{(3)} A^{(3)} = [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(3)} & {\overset{&OverBar;}{h}}_{1,2}^{(3)} & {\overset{&OverBar;}{h}}_{1,3}^{(3)} & {\overset{&OverBar;}{h}}_{1,4}^{(3)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(3)} & {\overset{&OverBar;}{h}}_{2,2}^{(3)} & {\overset{&OverBar;}{h}}_{2,3}^{(3)} & {\overset{&OverBar;}{h}}_{2,4}^{(3)} \end{matrix}]$ ，得到 H⁽³⁾为： ${\overset{&OverBar;}{h}}^{(3)} = h^{(3)} V^{(3)} A^{(3)} = [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(3)} & {\overset{&OverBar;}{h}}_{1,2}^{(3)} & {\overset{&OverBar;}{h}}_{1,3}^{(3)} & {\overset{&OverBar;}{h}}_{1,4}^{(3)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(3)} & {\overset{&OverBar;}{h}}_{2,2}^{(3)} & {\overset{&OverBar;}{h}}_{2,3}^{(3)} & {\overset{&OverBar;}{h}}_{2,4}^{(3)} \end{matrix}]$ , get H ⁽³⁾ as:

${\overset{&OverBar;}{H}}^{(3)} = [\begin{matrix} - 0.4313 + 0.0000 i & 1.4048 + 0.0000 i & - 0.0000 - 0.0000 i & - 0.0000 - 0.0000 i \\ - 2.4258 + 1.3022 i & - 0.1939 + 0.1041 i & 0 + 0.0000 i & - 0.0000 - 0.0000 i \end{matrix}]$ 在假设接收端无噪声的情况下，利用公式(4)得到第1个用户接收到的符号矩阵x⁽¹⁾为： ${\overset{&OverBar;}{h}}^{(3)} = [\begin{matrix} - 0.4313 + 0.0000 i & 1.4048 + 0.0000 i & - 0.0000 - 0.0000 i & - 0.0000 - 0.0000 i \\ - 2.4258 + 1.3022 i & - 0.1939 + 0.1041 i & 0 + 0.0000 i & - 0.0000 - 0.0000 i \end{matrix}]$ Assuming that the receiving end is noise-free, the symbol matrix x ⁽¹⁾ received by the first user can be obtained by formula (4):

${x x}^{((11))} = = [\begin{matrix} - - 0.8674 0.8674 + + 2.0887 2.0887 i i & - - 2.0887 2.0887 - - 0.3215 0.3215 i i \\ 0.1317 0.1317 - - 2.7459 2.7459 i i & - - 1.1002 1.1002 - - 1.5704 1.5704 i i \end{matrix}]$

在假设接收端无噪声的情况下，利用公式(4)得到第2个用户接收到的符号矩阵x⁽²⁾为：Assuming that the receiving end is noise-free, the symbol matrix x ⁽²⁾ received by the second user can be obtained by formula (4):

${x x}^{((22))} = = [\begin{matrix} 1.6943 1.6943 + + 1.6943 1.6943 i i & - - 0.9744 0.9744 - - 2.9232 2.9232 i i \\ - - 0.9215 0.9215 - - 0.2968 0.2968 i i & - - 0.2996 0.2996 - - 0.2936 0.2936 i i \end{matrix}]$

在假设接收端无噪声的情况下，利用公式(4)得到第3个用户接收到的符号矩阵x⁽³⁾为：Assuming that the receiving end is noise-free, the symbol matrix x ⁽³⁾ received by the third user can be obtained by formula (4):

${x x}^{((33))} = = [\begin{matrix} - - 0.0351 0.0351 - - 0.5806 0.5806 i i & 1.7419 1.7419 + + 1.4691 1.4691 i i \\ 1.9838 1.9838 - - 1.9740 1.9740 i i & 2.4304 2.4304 - - 0.5534 0.5534 i i \end{matrix}]$

利用空时正交分组码与空间复用组合的编码方式的特点，将 H⁽¹⁾通过公式(5)中的变换，得到H⁽¹⁾为：Utilizing the characteristics of the coding method of space-time orthogonal block code and space multiplexing combination, H ⁽¹⁾ is transformed in formula (5), and H ⁽¹⁾ is obtained as:

${\overset{&OverBar;}{H}}^{(1)} = [\begin{matrix} - 1.8798 + 0.0000 i & - 0.9655 - 0.0000 i & - 0.0000 & - 0.0000 + 0.0000 i \\ - 0.0000 - 0.0000 i & 0.0000 & - 09655 + 0.0000 i & 1.8798 + 0.0000 i \\ 0.5905 - 1.3342 i & - 0.5034 + 1.1376 i & - 0.0000 + 0.0000 i & - 0.0000 - 0.0000 i \\ - 0.0000 + 0.0000 i & 0.0000 + 0.0000 i & - 0.5034 - 1.1376 i & - 0.5905 - 1.3342 i \end{matrix}]$ 将 H⁽²⁾通过公式(5)中的变换，得到 H⁽²⁾为： ${\overset{&OverBar;}{h}}^{(1)} = [\begin{matrix} - 1.8798 + 0.0000 i & - 0.9655 - 0.0000 i & - 0.0000 & - 0.0000 + 0.0000 i \\ - 0.0000 - 0.0000 i & 0.0000 & - 09655 + 0.0000 i & 1.8798 + 0.0000 i \\ 0.5905 - 1.3342 i & - 0.5034 + 1.1376 i & - 0.0000 + 0.0000 i & - 0.0000 - 0.0000 i \\ - 0.0000 + 0.0000 i & 0.0000 + 0.0000 i & - 0.5034 - 1.1376 i & - 0.5905 - 1.3342 i \end{matrix}]$ Put H ⁽²⁾ through the transformation in formula (5), and get H ⁽²⁾ as:

${\overset{&OverBar;}{H}}^{(2)} = [\begin{matrix} - 1.9430 - 0.0000 i & 1.1383 - 0.0000 i & 0.0000 + 0.0000 i & 0.0000 + 0.0000 i \\ 0.0000 - 0.0000 i & - 0.0000 + 0.0000 i & 1.1383 + 0.0000 i & 1.9430 - 0.0000 i \\ - 1.5231 + 0.7810 i & - 1.1498 + 0.5896 i & 0.0000 + 0.0000 i & - 0.0000 + 0.0000 i \\ - 0.0000 - 0.0000 i & - 0.0000 + 0.0000 i & - 1.1498 - 0.5896 i & 1.5231 + 0.7810 i \end{matrix}]$ 将 H⁽³⁾通过公式(5)中的变换，得到 H⁽³⁾为： ${\overset{&OverBar;}{h}}^{(2)} = [\begin{matrix} - 1.9430 - 0.0000 i & 1.1383 - 0.0000 i & 0.0000 + 0.0000 i & 0.0000 + 0.0000 i \\ 0.0000 - 0.0000 i & - 0.0000 + 0.0000 i & 1.1383 + 0.0000 i & 1.9430 - 0.0000 i \\ - 1.5231 + 0.7810 i & - 1.1498 + 0.5896 i & 0.0000 + 0.0000 i & - 0.0000 + 0.0000 i \\ - 0.0000 - 0.0000 i & - 0.0000 + 0.0000 i & - 1.1498 - 0.5896 i & 1.5231 + 0.7810 i \end{matrix}]$ Put H ⁽³⁾ through the transformation in formula (5), and get H ⁽³⁾ as:

${\overset{&OverBar;}{H}}^{(3)} = [\begin{matrix} - 0.4313 + 0.0000 i & 1.4048 + 0.0000 i & - 0.0000 - 0.0000 i & - 0.0000 - 0.0000 i \\ - 0.0000 + 0.0000 i & 0.0000 - 0.0000 i & 1.4048 - 0.0000 i & 0.4313 + 0.0000 i \\ - 2.4258 + 1.3022 i & - 0.1939 + 0.1041 i & 0 + 0.0000 i & - 0.0000 - 0.0000 i \\ - 0.0000 + 0.0000 i & 0 + 0.0000 i & - 0.1939 - 0.1041 i & 2.4258 + 1.3022 i \end{matrix}]$ 将第1个用户接收到的符号矩阵x⁽¹⁾变换成公式(6)中所示的矢量形式，得到第1个用户的接收符号矢量

为：

{\overset{&OverBar;}{h}}^{(3)} = [\begin{matrix} - 0.4313 + 0.0000 i & 1.4048 + 0.0000 i & - 0.0000 - 0.0000 i & - 0.0000 - 0.0000 i \\ - 0.0000 + 0.0000 i & 0.0000 - 0.0000 i & 1.4048 - 0.0000 i & 0.4313 + 0.0000 i \\ - 2.4258 + 1.3022 i & - 0.1939 + 0.1041 i & 0 + 0.0000 i & - 0.0000 - 0.0000 i \\ - 0.0000 + 0.0000 i & 0 + 0.0000 i & - 0.1939 - 0.1041 i & 2.4258 + 1.3022 i \end{matrix}]

Transform the symbol matrix x ⁽¹⁾ received by the first user into the vector form shown in formula (6), and obtain the received symbol vector of the first user

for:

${\overset{&RightArrow; &Right Arrow;}{x x}}^{((11))} = = [\begin{matrix} - - 0.8674 0.8674 + + 2.0887 2.0887 i i \\ - - 2.0887 2.0887 + + 0.3215 0.3215 i i \\ 0.1317 0.1317 - - 2.7459 2.7459 i i \\ - - 1.1002 1.1002 + + 1.5704 1.5704 i i \end{matrix}]$

将第2个用户接收到的符号矩阵x⁽²⁾变换成公式(6)中所示的矢量形式，得到第2个用户的接收符号矢量

为：Transform the symbol matrix x ⁽²⁾ received by the second user into the vector form shown in formula (6), and obtain the received symbol vector of the second user

for:

${\overset{&RightArrow; &Right Arrow;}{x x}}^{((22))} = = [\begin{matrix} 1.6943 1.6943 + + 1.6943 1.6943 i i \\ - - 0.9744 0.9744 + + 2.9232 2.9232 i i \\ - - 0.9215 0.9215 - - 0.2968 0.2968 i i \\ - - 0.2996 0.2996 + + 0.2936 0.2936 i i \end{matrix}]$

将第3个用户接收到的符号矩阵x⁽³⁾变换成公式(6)中所示的矢量形式，得到第3个用户的接收符号矢量

为：Transform the symbol matrix x ⁽³⁾ received by the third user into the vector form shown in formula (6), and obtain the received symbol vector of the third user

for:

${\overset{&RightArrow; &Right Arrow;}{x x}}^{((33))} = = [\begin{matrix} - - 0.0351 0.0351 - - 0.5806 0.5806 i i \\ 1.7419 1.7419 - - 1.4691 1.4691 i i \\ 1.9837 1.9837 - - 1.9740 1.9740 i i \\ 2.4304 2.4304 + + 0.5534 0.5534 i i \end{matrix}]$

在用户1的接收端，得到线性均衡矩阵G⁽¹⁾＝( H⁽¹⁾)^_为：At the receiving end of user 1, the linear equalization matrix G ⁽¹⁾ = ( H ⁽¹⁾ ) ^_ is obtained as:

在用户2的接收端，得到线性均衡矩阵G⁽²⁾＝( H⁽²⁾)^_为：

At the receiving end of user 2, the linear equalization matrix G ⁽²⁾ = ( H ⁽²⁾ ) ^_ is obtained as:

在用户3的接收端，得到线性均衡矩阵G⁽³⁾＝( H⁽³⁾)^_为：

At the receiving end of user 3, the linear equalization matrix G ⁽³⁾ = ( H ⁽³⁾ ) ^_ is obtained as:

用4×4维的线性均衡矩阵G⁽¹⁾＝( H⁽¹⁾)^_乘以第1个用户的接收符号矢量

{\overset{&RightArrow;}{w}}_{i}^{(1)} = {[G^{(1)}]}_{i}

。这里i＝1，2，3，4，[G⁽¹⁾]_i为线性均衡矩阵G⁽¹⁾的第i行。因此得到基站发送给第1个用户的4个符号的判决值__1，1、__1，2、__1，3和__1，4为：

Multiply the received symbol vector of the first user by the 4×4-dimensional linear equalization matrix G ⁽¹⁾ = ( H ⁽¹⁾ ) ^_

{\overset{&Right Arrow;}{w}}_{i}^{(1)} = {[G^{(1)}]}_{i}

. Here i=1, 2, 3, 4, [G ⁽¹⁾ ] _i is the i-th row of the linear equalization matrix G ⁽¹⁾ . Therefore, the judgment values_1,1, _{_1,2} , _{_1,3} _{and_1,4} of the 4 symbols sent by the base station to the first _user are:

$[\begin{matrix} {\overset{^^}{s the s}}_{1,1 1,1} \\ {\overset{^^}{s the s}}_{1,2 1,2} \\ {\overset{^^}{s the s}}_{1,3 1,3} \\ {\overset{^^}{s the s}}_{1,4 1,4} \end{matrix}] = = {G G}^{((11))} {\overset{&RightArrow; &Right Arrow;}{x x}}^{((11))} = = [\begin{matrix} 0.9486 0.9486 - - 0.9488 0.9488 i i \\ - - 0.9485 0.9485 - - 0.3163 0.3163 i i \\ 0.3163 0.3163 - - 0.9486 0.9486 i i \\ 0.9487 0.9487 - - 0.3163 0.3163 i i \end{matrix}]$

用4×4维的线性均衡矩阵G⁽²⁾＝( H⁽²⁾)^_乘以第2个用户的接收符号矢量

{\overset{&RightArrow;}{w}}_{i}^{(2)} = {[G^{(2)}]}_{i}

。这里i＝1，2，3，4，[G⁽²⁾]_i为线性均衡矩阵G⁽²⁾的第i行。因此得到基站为第2个用户发送的4个符号的判决值__2，1、__2，2、__2，3和__2，4为：Multiply the received symbol vector of the second user by the 4×4-dimensional linear equalization matrix G ⁽²⁾ = ( H ⁽²⁾ ) ^_

{\overset{&Right Arrow;}{w}}_{i}^{(2)} = {[G^{(2)}]}_{i}

. Here i=1, 2, 3, 4, [G ⁽²⁾ ] _i is the i-th row of the linear equalization matrix G ⁽²⁾ . Therefore, the judgment values_2,1, _{_2,2} , _{_2,3} _{and_2,4} of the 4 symbols sent by the base station _for the second user are:

$[\begin{matrix} {\overset{^^}{s the s}}_{2,1 2,1} \\ {\overset{^^}{s the s}}_{2,2 2,2} \\ {\overset{^^}{s the s}}_{2,3 2,3} \\ {\overset{^^}{s the s}}_{2,4 2,4} \end{matrix}] = = {G G}^{((22))} {\overset{&RightArrow; &Right Arrow;}{x x}}^{((22))} = = [\begin{matrix} - - 0.3162 0.3162 - - 0.3162 0.3162 i i \\ 0.9487 0.9487 + + 0.9487 0.9487 i i \\ - - 0.3163 0.3163 + + 0.9488 0.9488 i i \\ - - 0.3162 0.3162 + + 0.9488 0.9488 i i \end{matrix}]$

用4×4维的线性均衡矩阵G⁽³⁾=( H⁽³⁾)^_乘以第3个用户的接收符号矢量

{\overset{&RightArrow;}{w}}_{i}^{(3)} = {[G^{(3)}]}_{i}

。这里i＝1，2，3，4，[G⁽³⁾]_i为线性均衡矩阵G⁽³⁾的第i行。因此得到基站发送给第3个用户的4个符号的判决值__3，1、__3，2、__3，3和__3，4为：Multiply the received symbol vector of the third user by the 4×4-dimensional linear equalization matrix G ⁽³⁾ =( H ⁽³⁾ ) ^_

{\overset{&Right Arrow;}{w}}_{i}^{(3)} = {[G^{(3)}]}_{i}

. Here i=1, 2, 3, 4, [G ⁽³⁾ ] _i is the i-th row of the linear equalization matrix G ⁽³⁾ . Therefore, the judgment values_3,1, _{_3,2} , _{_3,3} _{and_3,4} of the 4 symbols sent by the base station _to the third user are:

$[\begin{matrix} {\overset{^^}{s the s}}_{3,1 3,1} \\ {\overset{^^}{s the s}}_{3,2 3,2} \\ {\overset{^^}{s the s}}_{3,3 3,3} \\ {\overset{^^}{s the s}}_{3,4 3,4} \end{matrix}] = = {G G}^{((33))} {\overset{&RightArrow; &Right Arrow;}{x x}}^{((33))} = = [\begin{matrix} - - 0.9488 0.9488 + + 0.3162 0.3162 i i \\ - - 0.3162 0.3162 - - 0.3163 0.3163 i i \\ 0.9487 0.9487 - - 0.9486 0.9486 i i \\ 0.9487 0.9487 - - 0.3162 0.3162 i i \end{matrix}]$

上述过程不失一般性，可以将用户数和天线数进行扩展。The above process does not lose generality, and the number of users and the number of antennas can be extended.

假设多用户MIMO系统的基站具有M根天线，并共有K个移动终端用户，并设第k个用户具有N_k根天线，k＝1，…，K。It is assumed that the base station of the multi-user MIMO system has M antennas, and there are K mobile terminal users in total, and it is assumed that the kth user has N _k antennas, k=1,...,K.

在步骤1中，假定下行信道为窄带平坦瑞利衰落信道，并且h_i，j ^(k)代表基站的第j根发射天线到第k个用户第i根接收天线之间的信道衰落系数，则基站与第k个用户之间的信道传输矩阵表示为：In step 1, it is assumed that the downlink channel is a narrow-band flat Rayleigh fading channel, and h _{i, j} ^(k) represents the channel fading coefficient between the j-th transmit antenna of the base station and the i-th receive antenna of the k-th user, then The channel transmission matrix between the base station and the kth user is expressed as:

基站发射的数据经过调制之后，进行串并转换分成m路信号流，由基站的m根发射天线发射出去。要求m为偶数，并且满足 $m < M - \max {Σ_{i = 1, i &NotEqual; k}^{K} N_{i,} k = 1,2, \cdot \cdot \cdot, k}$ 。基站发射给第k个用户的符号按照空时正交分组码与空间复用组合的编码方式，得到具体的发射符号矩阵为：After the data transmitted by the base station is modulated, it is serial-to-parallel converted and divided into m signal streams, which are transmitted by the m transmitting antennas of the base station. Requires m to be an even number, and satisfies $m < m - \max {Σ_{i = 1, i &NotEqual; k}^{K} N_{i,} k = 1,2, &Center Dot; &Center Dot; &Center Dot;, k}$ . The symbol transmitted by the base station to the kth user is encoded according to the combination of space-time orthogonal block code and space multiplexing, and the specific transmitted symbol matrix is obtained as:

公式(8)中点划线区分的是基站再同一信道条件下相邻两个时隙发射的符号矢量。假设基站发射的信号总能量为E_s，第k个用户接收端的噪声矩阵n^(k)中的每个元素都是独立同分布的，均值为0，每一维的功率谱密度为N₀/2的复高斯随机变量。设第k个用户的预处理为T^(k)，接收到的符号矩阵为x^(k)，则对于第k个用户，k∈[1，K]，输入输出关系的复基带表达式为：The dotted line in formula (8) distinguishes the symbol vectors transmitted by the base station in two adjacent time slots under the same channel condition. Assuming that the total energy of the signal transmitted by the base station is E _s , each element in the noise matrix n ^(k) of the kth user receiving end is independent and identically distributed, with a mean value of 0, and the power spectral density of each dimension is N ₀ / A complex Gaussian random variable of 2. Suppose the preprocessing of the kth user is T ^(k) and the received symbol matrix is x ^(k) , then for the kth user, k∈[1, K], the complex baseband expression of the input-output relationship is:

${x x}^{((k k))} = = \sqrt{\frac{{E E.}_{s the s}}{Km Km}} {H h}^{((k k))} {T T}^{((k k))} {s the s}^{((k k))} + + {Σ Σ}_{i i = = 11,, i i &NotEqual; &NotEqual; k k}^{K K} ((\sqrt{\frac{{E E.}_{s the s}}{Km Km}} {H h}^{((k k))} {T T}^{((i i))} {s the s}^{((i i))})) + + {n no}^{((k k))},, k k &Element; &Element; [[11,, K K]] - - - - - - ((99))$

对于每个用户而言，为了去除其它K-1个用户的干扰，需要预处理矩阵满足 $H^{(i)} T^{(k)} \overset{i &NotEqual; k}{=} 0 (1 \leq i, k \leq K)$ 。由于任何一个矩阵都可以分解为基矩阵与任意酉阵的乘积形式，所以首先在步骤2中确定预处理矩阵的基。For each user, in order to remove the interference of other K-1 users, the preprocessing matrix needs to satisfy $h^{(i)} T^{(k)} \overset{i &NotEqual; k}{=} 0 (1 \leq i, k \leq K)$ . Since any matrix can be decomposed into the product form of the basis matrix and any unitary matrix, the basis of the preprocessing matrix is firstly determined in step 2.

在步骤2中，采用下面几个步骤求得预处理矩阵的基：In step 2, the following steps are used to obtain the basis of the preprocessing matrix:

图2是本发明中确定每个用户预处理矩阵的基的工作流程图。不失一般性，首先执行201模块，初始化k；然后在202模块，初始化i＝1(如果k＝1，则初始化i＝2)；接着在模块203将基站与第i个用户之间的信道传输矩阵H⁽ⁱ⁾进行奇异值分解得到H⁽ⁱ⁾＝U∑D，选择矩阵D中与零奇异值对应的列向量组成矩阵Z，矩阵Z就是基站与第i个用户之间的信道传输矩阵H⁽ⁱ⁾的基；下一步执行模块204令i＝i+1并且满足i≠k；接着在模块205中判断i是否大于K；如果i≤K，则执行模块206将基站与第i个用户之间的信道传输矩阵H⁽ⁱ⁾与203模块中求得的矩阵Z相乘并进行奇异值分解得到H⁽ⁱ⁾Z＝U∑D，选择矩阵D中与零奇异值对应的列向量组成矩阵W，矩阵W就是H⁽ⁱ⁾Z的基；下一步执行模块207将矩阵Z更新为Z＝ZW；接着再执行模块204令i＝i+1并且满足i≠k；然后执行模块205再判断i是否大于K；如果i＞K，则执行模块208得到第k个用户的预处理矩阵的基V^(k)＝Z。Fig. 2 is a flow chart of determining the basis of each user's preprocessing matrix in the present invention. Without loss of generality, first execute module 201 to initialize k; then in module 202, initialize i=1 (if k=1, then initialize i=2); then in module 203, the channel between the base station and the i-th user The transmission matrix H ⁽ⁱ⁾ is subjected to singular value decomposition to obtain H ⁽ⁱ⁾ = U∑D, and the column vector corresponding to the zero singular value in the matrix D is selected to form a matrix Z, and the matrix Z is the channel transmission between the base station and the i-th user The base of matrix H ⁽ⁱ⁾ ; next step execution module 204 makes i=i+1 and satisfies i≠k; then judges whether i is greater than K in module 205; If i≤K, then execution module 206 will base station and i-th The channel transmission matrix H ⁽ⁱ⁾ between the users is multiplied with the matrix Z obtained in the 203 module and carried out singular value decomposition to obtain H ⁽ⁱ⁾ Z=U∑D, select the column corresponding to the zero singular value in the matrix D Vector forms matrix W, and matrix W is exactly the basis of H ⁽ⁱ⁾ Z; Next step execution module 207 is updated matrix Z to Z=ZW; Then execution module 204 makes i=i+1 and satisfies i≠k; Execution module then 205 judges whether i is greater than K; if i>K, then the execution module 208 obtains the basis V ^(k) =Z of the kth user's preprocessing matrix.

对于所有K个用户，均采用上述步骤2中的方法得到自己的预处理矩阵的基。For all K users, use the method in step 2 above to obtain the basis of their own preprocessing matrix.

步骤3：利用基站与每个用户之间的信道传输矩阵以及步骤2中确定的每个用户的预处理矩阵的基，确定每个用户的预处理矩阵的最优酉阵。具体求得预处理矩阵的最优酉阵的步骤如下：Step 3: Using the channel transmission matrix between the base station and each user and the basis of each user's preprocessing matrix determined in step 2, determine the optimal unitary matrix of each user's preprocessing matrix. The specific steps to obtain the optimal unitary matrix of the preprocessing matrix are as follows:

图3是本发明中确定每个用户预处理矩阵的最优酉阵的工作流程图。不失一般性，首先执行301模块，初始化k；接着在302模块中将矩阵H^(k)V^(k)进行奇异值分解得到H^(k)V^(k)＝U∑D；最后在模块303中选择A^(k)为矩阵D的前m列，确定第k个用户的预处理矩阵的最优酉阵。Fig. 3 is a working flow chart of determining the optimal unitary matrix of each user's preprocessing matrix in the present invention. Without loss of generality, first execute module 301 to initialize k; then in module 302, matrix H ^(k) V ^(k) is subjected to singular value decomposition to obtain H ^(k) V ^(k) = U∑D; finally in module 303 Select A ^(k) as the first m columns of the matrix D, and determine the optimal unitary matrix of the kth user's preprocessing matrix.

对于所有K个用户，均采用上述步骤3中的方法得到自己的预处理矩阵的最优酉阵。For all K users, use the method in step 3 above to obtain the optimal unitary matrix of their own preprocessing matrix.

经过步骤2和步骤3后，就可以确定多用户MIMO系统中每个用户下行链路的天线选择预处理矩阵T^(k)＝V^(k)A^(k)(k∈[1，K])。After step 2 and step 3, the antenna selection preprocessing matrix T ^(k) = V ^(k) A ^(k) (k∈[1, K]) for each user downlink in the multi-user MIMO system can be determined .

步骤4：将第k个用户的预处理矩阵V^(k)A^(k)带入到公式(9)中，得到第k个用户接收到的符号矩阵为：Step 4: Bring the preprocessing matrix V ^(k) A ^(k) of the kth user into formula (9), and obtain the symbol matrix received by the kth user as:

$[\begin{matrix} {x x}_{1.1 1.1}^{((k k))} & {x x}_{1,2 1,2}^{((k k))} \\ {x x}_{2,1 2,1}^{((k k))} & {x x}_{2,2 2,2}^{((k k))} \\ . . & . . \\ . . & . . \\ . . & . . \\ {x x}_{{N N}_{k k},, 11}^{((k k))} & {x x}_{{N N}_{k k},, 22}^{((k k))} \end{matrix}] = = {H h}^{((k k))} {V V}^{((k k))} {A A}^{((k k))} [\begin{matrix} {s the s}_{k k,, 11} & - - {s the s}_{k k,, m m}^{* *} \\ {s the s}_{k k,, 22} & {s the s}_{k k,, m m - - 11}^{* *} \\ {s the s}_{k k,, 33} & - - {s the s}_{k k,, m m - - 22}^{* *} \\ {s the s}_{k k,, 44} & {s the s}_{k k,, m m - - 33}^{* *} \\ . . & . . \\ . . & . . \\ . . & . . \\ {S S}_{k k,, m m - - 11} & {- - S S}_{k k,, 22}^{* *} \\ {S S}_{k k,, m m} & {S S}_{k k,, 11} \end{matrix}] + + [\begin{matrix} {n no}_{1,1 1,1}^{((k k))} & {n no}_{1,2 1,2}^{((k k))} \\ {n no}_{2,1 2,1}^{((k k))} & {n no}_{2,2 2,2}^{((k k))} \\ . . & . . \\ . . & . . \\ . . & . . \\ {n no}_{{N N}_{k k},, 11}^{((k k))} & {n no}_{{N N}_{k k},, 22}^{((k k))} \end{matrix}],, k k &Element; &Element; [[11,, K K]] - - - - - - ((1010))$

定义等效信道传输矩阵为 ${\overset{&OverBar;}{H}}^{(k)} = H^{(k)} V^{(k)} A^{(k)} = \begin{matrix} \end{matrix} [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(k)} & {\overset{&OverBar;}{h}}_{1,2}^{(k)} & . . . & {\overset{&OverBar;}{h}}_{1, m}^{(k)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(k)} & {\overset{&OverBar;}{h}}_{2,2}^{(k)} & . . . & {\overset{&OverBar;}{h}}_{2, m}^{(k)} \\ . & . & . & . \\ . & . & . & . \\ . & . & . & . \\ {\overset{&OverBar;}{h}}_{N_{k}, 1}^{(k)} & {\overset{&OverBar;}{h}}_{N_{k}, 2}^{(k)} & . . . & {\overset{&OverBar;}{h}}_{N_{k}, m}^{(k)} \end{matrix}]$ ，并利用空时正交分组码与空间复用组合的编码方式，作n如下变换得Define the equivalent channel transfer matrix as ${\overset{&OverBar;}{h}}^{(k)} = h^{(k)} V^{(k)} A^{(k)} = \begin{matrix} \end{matrix} [\begin{matrix} {\overset{&OverBar;}{h}}_{1,1}^{(k)} & {\overset{&OverBar;}{h}}_{1,2}^{(k)} & . . . & {\overset{&OverBar;}{h}}_{1, m}^{(k)} \\ {\overset{&OverBar;}{h}}_{2,1}^{(k)} & {\overset{&OverBar;}{h}}_{2,2}^{(k)} & . . . & {\overset{&OverBar;}{h}}_{2, m}^{(k)} \\ . & . & . & . \\ . & . & . & . \\ . & . & . & . \\ {\overset{&OverBar;}{h}}_{N_{k}, 1}^{(k)} & {\overset{&OverBar;}{h}}_{N_{k}, 2}^{(k)} & . . . & {\overset{&OverBar;}{h}}_{N_{k}, m}^{(k)} \end{matrix}]$ , and use the combination of space-time orthogonal block code and space multiplexing to transform n as follows:

${\overset{&OverBar; &OverBar;}{H h}}^{((k k))} = = [\begin{matrix} {\overset{&OverBar; &OverBar;}{h h}}_{1,1 1,1}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{1,2 1,2}^{((k k))} & . . . . . . & {\overset{&OverBar; &OverBar;}{h h}}_{11,, m m - - 11}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{11,, m m}^{((k k))} \\ {(({\overset{&OverBar; &OverBar;}{h h}}_{11,, m m}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{11,, m m - - 11}^{((k k))}))}^{* *} & . . . . . . & {(({\overset{&OverBar; &OverBar;}{h h}}_{1,2 1,2}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{1,1 1,1}^{((k k))}))}^{* *} \\ {\overset{&OverBar; &OverBar;}{h h}}_{2,1 2,1}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{2,2 2,2}^{((k k))} & . . . . . . & {\overset{&OverBar; &OverBar;}{h h}}_{22,, m m - - 11}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{22,, m m}^{((k k))} \\ {(({\overset{&OverBar; &OverBar;}{h h}}_{22,, m m}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{22,, m m - - 11}^{((k k))}))}^{* *} & . . . . . . & {(({\overset{&OverBar; &OverBar;}{h h}}_{2,2 2,2}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{2,1 2,1}^{((k k))}))}^{* *} \\ . . & . . & . . & . . & . . \\ . . & . . & . . & . . & . . \\ . . & . . & . . & . . & . . \\ {\overset{&OverBar; &OverBar;}{h h}}_{{N N}_{k k},, 11}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{{N N}_{k k},, 22}^{((k k))} & . . . . . . & {\overset{&OverBar; &OverBar;}{h h}}_{{N N}_{k k},, m m - - 11}^{((k k))} & {\overset{&OverBar; &OverBar;}{h h}}_{{N N}_{k k},, m m}^{((k k))} \\ {(({\overset{&OverBar; &OverBar;}{h h}}_{{N N}_{k k},, m m}^{((k k))}))}^{* *} & - - {(({\overset{&OverBar; &OverBar;}{h h}}_{{N N}_{k k},, m m - - 11}^{((k k))}))}^{* *} & . . . . . . & {(({\overset{&OverBar; &OverBar;}{h h}}_{{N N}_{k k},, 22}^{((k k))}))}^{* *} & {- - (({\overset{&OverBar; &OverBar;}{h h}}_{{N N}_{k k},, 11}^{((k k))}))}^{* *} \end{matrix}],, k k &Element; &Element; [[11,, k k]] - - - - - - ((1111))$

则第k个用户的输入输出符号间的等效矢量关系为：Then the equivalent vector relationship between the input and output symbols of the kth user is:

公式(12)中，定义 ${\overset{&RightArrow;}{x}}^{(k)} = [\begin{matrix} x_{1,1}^{(k)} \\ {(x_{1,2}^{(k)})}^{*} \\ x_{2,1}^{(k)} \\ {(x_{2,2}^{(k)})}^{*} \\ . \\ . \\ . \\ x_{N_{k, 1}}^{(k)} \\ {(x_{N_{k}, 2}^{(k)})}^{*} \end{matrix}]$ 为第k个用户的接收符号矢量。在接收端，用m×2N_k维的线性均衡矩阵G^(k)＝( H^(k))^_乘以第k个用户的接收符号矢量，然后将所有接收到的信号轮流作为期望信号，选择迫零矢量 ${\overset{&RightArrow;}{w}}_{i}^{(k)} = {[G^{(k)}]}_{i}$ 。这里的i＝1，2，…，m，[G^(k)]_i为线性均衡矩阵G^(k)的第i行。因此，基站发送给第k个用户的第i个符号的判决值__k，i为

In formula (12), define

{\overset{&Right Arrow;}{x}}^{(k)} = [\begin{matrix} x_{1,1}^{(k)} \\ {(x_{1,2}^{(k)})}^{*} \\ x_{2,1}^{(k)} \\ {(x_{2,2}^{(k)})}^{*} \\ . \\ . \\ . \\ x_{N_{k, 1}}^{(k)} \\ {(x_{N_{k}, 2}^{(k)})}^{*} \end{matrix}]

is the received symbol vector of the kth user. At the receiving end, use the m×2N _k -dimensional linear equalization matrix G ^(k) = ( H ^(k) ) ^_ to multiply the received symbol vector of the kth user , and then take all the received signals as the desired signal in turn, and choose the zero-forcing vector

{\overset{&Right Arrow;}{w}}_{i}^{(k)} = {[G^{(k)}]}_{i}

. Here i=1, 2, ..., m, [G ^(k) ] _i is the i-th row of the linear equalization matrix G ^(k) . Therefore, the decision value_k _,i of the i-th symbol sent by the base station to the k-th user is

结合图5，其中曲线1是基站具有9根发射天线，共有3个移动终端用户，并且每个用户具有2根接收天线的多用户MIMO系统下行链路使用传统预处理方法的性能曲线，曲线2是该多用户MIMO系统使用本发明预处理方法的性能曲线。可以看到，与传统预处理方法相比，本发明的预处理方法是通过选择预处理矩阵的最优酉阵，使得信号检测的后验信噪比下限最大，从而显著提高多用户MIMO系统的性能。例如在误比特率为2×10^-4时，与传统预处理方法相比，使用天线选择预处理方法使得整个多用户MIMO系统的性能提高了约7.0dB。Combining with Figure 5, curve 1 is the downlink performance curve of a multi-user MIMO system with 9 transmit antennas in the base station, a total of 3 mobile terminal users, and each user has 2 receive antennas using traditional preprocessing methods, curve 2 is the performance curve of the multi-user MIMO system using the preprocessing method of the present invention. It can be seen that compared with the traditional preprocessing method, the preprocessing method of the present invention makes the lower limit of the posterior signal-to-noise ratio of signal detection the largest by selecting the optimal unitary matrix of the preprocessing matrix, thereby significantly improving the performance of the multi-user MIMO system. performance. For example, when the bit error rate is 2×10 ^-4 , compared with the traditional preprocessing method, the antenna selection preprocessing method improves the performance of the entire multi-user MIMO system by about 7.0 dB.

以上所述，仅为本发明中的一个具体实施例，但本发明的保护范围并不局限于此，任何熟悉该技术的人在本发明所揭露的技术范围内，可轻易想到的变换或替换，都应涵盖在本发明的包含范围之内。因此，本发明的保护范围应该以权利要求书的保护范围为准。The above is only a specific embodiment of the present invention, but the scope of protection of the present invention is not limited thereto, and any person familiar with the technology can easily conceive of changes or replacements within the technical scope disclosed in the present invention , should be covered within the scope of the present invention. Therefore, the protection scope of the present invention should be determined by the protection scope of the claims.

Claims

1. A multi-user MIMO system downlink antenna selection preprocessing method is characterized in that: it comprises the following steps:

Step 1: Establish the downlink model of the multi-user MIMO system, obtain the channel transmission matrix between the base station and each user and the coding method of the space-time orthogonal block code and spatial multiplexing combination of the transmitted symbols;

Step 2: Determine the basis of each user's preprocessing matrix through the channel transmission matrix between the base station and each user;

Step 3: Use the channel transmission matrix between the base station and each user to multiply the basis of each user's preprocessing matrix determined in step 2, and perform singular value decomposition on the product to determine each user's preprocessing matrix The optimal unitary matrix of ;

Step 4: Using the characteristics of the space-time orthogonal block code and space multiplexing combination coding method, the downlink model of the multi-user MIMO system is converted into an equivalent vector form, and the signal detection is realized by using the linear zero-forcing equalization method.

2. A multi-user MIMO system downlink antenna selection preprocessing method according to claim 1, characterized in that: the step of establishing a multi-user MIMO system downlink model in said step 1 is: setting the base station The number of antennas m required to transmit a symbol is even and satisfies

m < m - \max {Σ_{i = 1, i &NotEqual; k}^{K} N_{i}, k = 1,2, \cdot \cdot \cdot, K},

Here M represents the number of antennas of the base station, K is the number of mobile terminal users, and the kth user has N _k receiving antennas, k=1,...,K.

3. A multi-user MIMO system downlink antenna selection preprocessing method according to claim 1, characterized in that: after the multi-user MIMO system downlink model is established in the step 1, the system frequency division duplex In the case of , the base station obtains the channel information of each user through the feedback channel of each user, and forms the channel information into a channel transmission matrix.

4. A multi-user MIMO system downlink antenna selection preprocessing method according to claim 1, characterized in that: after the multi-user MIMO system downlink model is established in the step 1, the system time division duplex In this case, the base station directly calculates the channel information of each user by using the symmetry of the channel, and forms the channel information into a channel transmission matrix.

5. A multi-user MIMO system downlink antenna selection preprocessing method according to claim 1, characterized in that: after the data sent by the base station for each user in the step 1 is modulated, the transmitted symbols are obtained, And through the combination of space-time orthogonal block code and space multiplexing, the symbol matrix sent by the base station for each user is obtained, which is sent out by the transmitting antenna of the base station in two adjacent time slots.

6. A multi-user MIMO system downlink antenna selection preprocessing method according to claim 1, characterized in that: the specific step of step 2 is: based on the known relationship between the base station and each user The channel transmission matrix, using the recursive method, obtains the basis of the preprocessing matrix of each user.

7. A multi-user MIMO system downlink antenna selection preprocessing method according to claim 1, characterized in that: the specific step of step 3 is: the channel transmission matrix between the base station and each user Multiply with the basis of each user's respective preprocessing matrix and perform singular value decomposition to obtain the form of the product of three matrices U∑D; finally select the first m columns of the matrix D as the optimal preprocessing matrix for each user unitary matrix, m is the number of transmitting antennas required by the base station to transmit the symbol matrix.

8. A multi-user MIMO system downlink antenna selection preprocessing method according to claim 1, characterized in that: the specific step of step 4 is: the channel transmission matrix between the base station and each user and The product of the preprocessing matrix is used as the equivalent channel transmission matrix, and the equivalent vector relationship between the input and output symbols of each user is obtained by using the characteristics of the space-time orthogonal block code and the space multiplexing combination coding method. At the receiving end, a linear zero-forcing method is used to detect the signal.