CN102711047B

CN102711047B - Optimal opportunistic multicast method under mechanism of equal transmission rate in multimedia multicast technology

Info

Publication number: CN102711047B
Application number: CN201210151565.0A
Authority: CN
Inventors: 邓建国; 王小鹏; 陈志刚; 孟卫博; 钱鹏; 杜威
Original assignee: Xian Jiaotong University
Current assignee: Xian Jiaotong University
Priority date: 2012-05-16
Filing date: 2012-05-16
Publication date: 2015-04-29
Anticipated expiration: 2032-05-16
Also published as: CN102711047A

Abstract

The invention discloses an optimal opportunistic multicast method under the medium transmission rate mechanism of multimedia multicast technology, which specifically includes: 1) In the case of each fixed transmission rate r _k , the calculation requires n times that all users can receive correctly The probability of data P _n ^k ; 2) After obtaining P _n ^k , calculate the equivalent transmission rate corresponding to each transmission rate r _k ; 3) Select the transmission rate corresponding to the maximum equivalent transmission rate as the transmission rate of the base station send data. Under the condition of known channel distribution and almost no feedback data, the present invention provides an equal transmission rate opportunity multicast mechanism, and proposes how to obtain the optimal transmission rate according to the knowledge of probability theory and random process R's scheme. This solution greatly reduces the feedback and effectively improves the system capacity, making it more suitable for practical systems, and the present invention considers various scenarios and optimal solutions in various situations, making the solution more versatile .

Description

An Optimal Opportunistic Multicast Method under Medium Sending Rate Mechanism in Multimedia Multicast Technology

技术领域technical field

本发明属于无线通信技术领域，尤其是一种多媒体多播技术(MBMS，Multimedia Broadcast Multicast Service)中等发送速率的机会多播方案。该方案可以降低反馈，提高系统的性能。The invention belongs to the technical field of wireless communication, in particular to an opportunistic multicast scheme of medium transmission rate in multimedia multicast technology (MBMS, Multimedia Broadcast Multicast Service). This scheme can reduce the feedback and improve the performance of the system.

背景技术Background technique

随着大屏幕、高性能、低耗电便携终端的逐步普及，人们对数据业务的需求也越来越大，移动用户已经不再满足通过手机接收新闻、天气、交通和股价等这些简单的数据业务，他们希望能利用手中的移动终端收看电视节目和享受高速的流媒体业务，如视频点播、电视广播、视频会议、网上教育、互动游戏等。所以实现高速率的多媒体业务组播和广播业务，是未来移动数据发展的趋势。With the gradual popularization of large-screen, high-performance, and low-power portable terminals, people's demand for data services is also increasing. Mobile users are no longer satisfied with receiving simple data such as news, weather, traffic, and stock prices through mobile phones. They hope to use their mobile terminals to watch TV programs and enjoy high-speed streaming media services, such as video-on-demand, TV broadcasting, video conferencing, online education, and interactive games. Therefore, realizing high-speed multimedia multicast and broadcast services is the trend of future mobile data development.

无线通信系统中不同的用户由于地理位置，环境等因素的影响，使得用户的信道状态存在较大差异，这也使得系统拥有多用户分集增益。在多媒体多播系统中，同样的数据要发送给很多个用户，所以系统又存在多播增益，为了使得系统在多用户分集增益与多播增益之间有更好的折衷，在2004年提出了机会多播的概念，基站每次只选择一部分信道条件好的用户接收多播业务数据，由于无线信道时变的特点，所以每个用户都有一定的机会能够正确接收数据。经过若干次的重传或者使用纠删码就可以使所有用户都能正确收到数据。这就是机会多播。In the wireless communication system, due to the influence of factors such as geographical location and environment, the channel states of different users are quite different, which also makes the system have multi-user diversity gain. In a multimedia multicast system, the same data has to be sent to many users, so there is a multicast gain in the system. In order to make the system have a better compromise between multi-user diversity gain and multicast gain, in 2004 proposed With the concept of opportunistic multicast, the base station only selects some users with good channel conditions to receive multicast service data each time. Due to the time-varying characteristics of wireless channels, each user has a certain chance to receive data correctly. After several retransmissions or the use of erasure codes, all users can receive the data correctly. This is opportunistic multicasting.

目前大多关于机会多播的方案共同点是所有用户每次都要将用户的信噪比反馈回基站，然后基站再对反馈来的信噪比进行排序，最后基站根据相关理论选择合适的用户进行调度并发送数据。这种方法的缺点是基站每次发送数据前需要大量的反馈，且需要对数据进行一些处理，并且每次发送的速率都不一样，这会对数据处理方面造成一些不便。也有文献在已知信道分布的条件下，通过概率论相关知识研究得到了使用RS码情况下的信噪比门限，在一定长度上降低了反馈，但是文献并没有考虑用户的公平性，这样会导致某些用户得不到服务，并且文献只考虑了所有用户的信道独立同分布，缺乏对更一般信道模型的讨论。基于以上考虑，一个好的机会多播方案应该是简单实用，且能够降低反馈且在一定程度上提高系统性能的方案。At present, most of the schemes about opportunistic multicasting have in common that all users have to feed back the signal-to-noise ratio of users to the base station every time, and then the base station sorts the fed-back signal-to-noise ratio, and finally the base station selects the appropriate user according to relevant theories. Schedule and send data. The disadvantage of this method is that the base station needs a lot of feedback before sending data each time, and some data processing is required, and the sending rate is different each time, which will cause some inconvenience in data processing. There are also literatures under the condition of known channel distribution, through the research of relevant knowledge of probability theory, the SNR threshold of using RS code is obtained, which reduces the feedback in a certain length, but the literature does not consider the fairness of users, which will cause As a result, some users cannot be served, and the literature only considers the channel independent and identical distribution of all users, lacking the discussion of the more general channel model. Based on the above considerations, a good opportunistic multicast solution should be simple and practical, and can reduce feedback and improve system performance to a certain extent.

发明内容Contents of the invention

本发明的目的是为了克服现有的多媒体多播技术中机会多播方案的不合理性，在几乎没有反馈的情况下，提出一种多媒体多播技术中等发送速率机制下的最优机会多播方案，该方案能有效的降低反馈，提高系统性能，更加适用于实际系统。The purpose of the present invention is to overcome the irrationality of the opportunistic multicast scheme in the existing multimedia multicast technology, and propose an optimal opportunistic multicast under the medium transmission rate mechanism of the multimedia multicast technology under the condition of almost no feedback This scheme can effectively reduce feedback, improve system performance, and is more suitable for practical systems.

本发明的目的是通过以下技术方案来解决的：The purpose of the present invention is solved by the following technical solutions:

该种多媒体多播技术中等发送速率机制下的最优机会多播方法，具体包括以下步骤：The optimal opportunistic multicast method under the medium transmission rate mechanism of the multimedia multicast technology specifically includes the following steps:

1)在每个固定的发送速率r_k情况下，计算需要n次所有用户都能正确接收数据的概率P_n ^k；1) In the case of each fixed transmission rate r _k , calculate the probability P _n ^k that all users can receive data correctly for n times;

2)求得P_n ^k后，计算每个发送速率r_k所对应的等效传输速率；2) After obtaining P _n ^k , calculate the equivalent transmission rate corresponding to each sending rate r _k ;

3)选择等效传输速率最大所对应的发送速率作为基站的发送速率进行发送数据。3) Select the transmission rate corresponding to the maximum equivalent transmission rate as the transmission rate of the base station to transmit data.

进一步，上述步骤1)的具体方法如下：Further, the specific method of above-mentioned step 1) is as follows:

单基站的无线通信系统一共有M个用户，用户的信道分布状态已知，第i个用户信噪比的概率密度函数为f_i(x)，单基站的发送速率是离散的，且总共有K个可供选择的离散速率，构成集合A＝{r₁,r₂,…,r_K}，每次都以相同的速率r_k发送数据包，其中k＝1～K为优化参数，直到所有用户都收到，才进行下一数据包的发送；每次传输的时间长度为T，将其定义为一个时间块；信道系数在一个时间块内保持不变，在不同的时间块上保持独立，定义snr_k满足下式：A wireless communication system with a single base station has a total of M users, and the channel distribution state of the users is known. The probability density function of the i-th user's SNR is f _i (x), the transmission rate of a single base station is discrete, and there are a total of K optional discrete rates form a set A={r ₁ ,r ₂ ,…,r _K }, and send data packets at the same rate r _k every time, where k=1～K is the optimization parameter, until All users have received it before sending the next data packet; the time length of each transmission is T, which is defined as a time block; the channel coefficient remains unchanged in a time block, and remains constant in different time blocks Independently, define snr _k to satisfy the following formula:

r_k＝log₂(1+snr_k)(k＝1～K) (1)r _k ＝log ₂ (1+snr _k )(k＝1～K) (1)

可以得到can get

${snr snr}_{k k} = = 22^{{r r}_{k k}} - - 11 - - - - - - ((22))$

P_n ^k的计算可以分为以下几种不同的情况：The calculation of P _n ^k can be divided into the following different situations:

(1)同构网络(1) Isomorphic network

这种场景下所有用户的信噪比独立同分布，第i个用户的接收信号表示为：In this scenario, the signal-to-noise ratios of all users are independently and identically distributed, and the received signal of the i-th user is expressed as:

${y the y}_{i i} = = \sqrt{P P} {h h}_{i i} x x + + {n no}_{i i} - - - - - - ((33))$

h_i表示信道系数，对应的信噪比为h _i represents the channel coefficient, and the corresponding signal-to-noise ratio is

${SNR SNR}_{i i} = = \frac{P P {| | {h h}_{i i} | |}^{22}}{{N N}_{00}} - - - - - - ((44))$

所有用户信噪比概率密度函数相同且为f(x),在这种网络场景下，以下考虑三种方案中P_n ^k的求解：The SNR probability density function of all users is the same and is f(x). In this network scenario, the solution of P _n ^k in the three schemes is considered as follows:

a)普通情况a) Ordinary situation

P_n ^k由下式计算得到：P _n ^k is calculated by the following formula:

P_n ^k＝[1-(1-p)ⁿ]^M-[1-(1-p)^n-1]^M (5)P _n ^k ＝[1-(1-p) ⁿ ] ^M -[1-(1-p) ^n-1 ] ^M (5)

其中in

$p p = = {&Integral; &Integral;}_{{snr snr}_{k k}}^{\infty \infty} f f ((x x)) dx dx - - - - - - ((66))$

b)采用最大比合并技术b) Using the maximum ratio combining technique

将来自不同时间块的信号进行最大比合并，在这种情况下，The signals from different time blocks are combined by maximum ratio, in this case,

${P P}_{n no}^{k k} = = {[[11 - - {Π Π}_{i i = = 11}^{n no} ((11 - - {p p}_{i i}^{MRC MRC}))]]}^{M m} - - {[[11 - - {Π Π}_{i i = = 11}^{n no - - 11} ((11 - - {p p}_{i i}^{MRC MRC}))]]}^{M m} - - - - - - ((77))$

其中in

p₁ ^MRC＝p₁ (8)p ₁ ^MRC = p ₁ (8)

${p p}_{i i}^{MRC MRC} = = \frac{{p p}_{i i} - - {p p}_{i i - - 11}}{11 - - {p p}_{i i - - 11}} ((i i &GreaterEqual; &Greater Equal; 22)) - - - - - - ((99))$

p_i的计算如下p _i is calculated as follows

p₁＝p (10)p ₁ =p (10)

${p p}_{i i} = = {&Integral; &Integral;}_{{snr snr}_{k k}}^{\infty \infty} {f f}_{i i} ((x x)) dx dx ((i i &GreaterEqual; &Greater Equal; 22)) - - - - - - ((1111))$

f_i(y)的计算如下：f _i (y) is calculated as follows:

f₁(x)＝f(x) (12)f ₁ (x)=f(x) (12)

${f f}_{i i} ((y the y)) {&Integral; &Integral;}_{00}^{\infty \infty} f f ((x x)) {f f}_{i i - - 11} ((y the y - - x x)) dx dx ((i i &GreaterEqual; &Greater Equal; 22)) - - - - - - ((1313))$

c)使用喷泉码c) using fountain code

每个用户至少需要正确接收数据大小为S_m bit的信息才能恢复原始数据，此时：Each user needs to receive at least information with a data size of S _m bits correctly to restore the original data. At this time:

${P P}_{n no}^{k k} = = {((11 - - {Σ Σ}_{i i = = 00}^{U u - - 11} {C C}_{n no}^{i i} {p p}^{i i} {((11 - - p p))}^{n no - - i i}))}^{M m} - - {((11 - - {Σ Σ}_{i i = = 00}^{U u - - 11} {C C}_{n no - - 11}^{i i} {p p}^{i i} {((11 - - p p))}^{n no - - 11 - - i i}))}^{M m} - - - - - - ((1414))$

其中p由(6)式计算所得；Where p is calculated by formula (6);

(2)异构网络(2) Heterogeneous network

这种场景下考虑用户的大尺度衰落，所有用户的信噪比独立但是服从不同的分布，第i个用户的接收信号如下：Considering the large-scale fading of users in this scenario, the signal-to-noise ratios of all users are independent but obey different distributions. The received signal of the i-th user is as follows:

${y the y}_{i i} = = \sqrt{{Pd PD}_{i i}^{η η}} {h h}_{i i} x x + + {n no}_{i i} - - - - - - ((1616))$

d_i表示用户i到基站的距离。η大尺度衰落系数，对应的信噪比为d _i represents the distance from user i to the base station. η large-scale fading coefficient, the corresponding signal-to-noise ratio is

${SNR SNR}_{i i} = = \frac{{Pd PD}_{i i}^{- - η η} {| | {h h}_{i i} | |}^{22}}{{N N}_{00}} - - - - - - ((1717))$

每个用户的信道系数h_i依然是独立同分布，但是由于考虑大尺度衰落，所以每个用户的信噪比分布并不相同。The channel coefficient _hi of each user is still independent and identically distributed, but due to the consideration of large-scale fading, the SNR distribution of each user is not the same.

在异构网络场景下，在已知用户到基站距离d的情况下，信噪比的概率密度函数为f(x|d)是可以知道的，假设用户均匀分布在小区内，小区的半径为R_d，所以用户到基站的概率密度函数也是可以求得的；因此所有用户信噪比的概率密度函数可以统一表示为：In a heterogeneous network scenario, when the distance d between the user and the base station is known, the probability density function of the signal-to-noise ratio is f(x|d), which can be known. Assuming that the users are evenly distributed in the cell, the radius of the cell is R _d , so the probability density function of the user to the base station can also be obtained; therefore, the probability density function of all user signal-to-noise ratios can be uniformly expressed as:

$f f ((x x)) = = {&Integral; &Integral;}_{00}^{{R R}_{d d}} f f ((x x | | y the y)) {f f}_{{d d}_{i i}} ((y the y)) dy dy - - - - - - ((1818))$

经过这样的处理后，所有用户的信噪比分布就又可以看成是一样的，也就是说异构网络就可以看成是同构网络，接下来按照P_n ^k求解的三种方案。After such processing, the SNR distribution of all users can be regarded as the same again, that is to say, the heterogeneous network can be regarded as a homogeneous network, and the following three solutions are solved according to _Pn ^k .

在异构网络场景下，已知用户与基站的距离后，在异构网络中分三种情况：In the heterogeneous network scenario, after the distance between the user and the base station is known, there are three situations in the heterogeneous network:

d)普通情况d) Ordinary situation

此时，发送速率r_k所对应的P_n ^k At this time, P _n ^k corresponding to the sending rate r _k

${P P}_{n no}^{k k} = = {Π Π}_{i i = = 11}^{M m} [[11 - - {(({11 - - p p}_{i i}))}^{n no}]] - - {Π Π}_{i i = = 11}^{M m} [[11 - - {(({11 - - p p}_{i i}))}^{n no - - 11}]] - - - - - - ((1919))$

其中in

${p p}_{i i} = = {&Integral; &Integral;}_{{snr snr}_{k k}}^{\infty \infty} {f f}_{i i} ((x x)) dx dx - - - - - - ((2020))$

e)采用最大比合并技术e) Using maximum ratio combining technology

考虑最大比合并后的P_n ^k可以表示为Considering the maximum ratio combined P _n ^k can be expressed as

${P P}_{n no}^{k k} = = {Π Π}_{i i = = 11}^{M m} [[11 - - {Π Π}_{j j = = 11}^{n no} ((11 - - {P P}_{ij ij}^{MRC MRC}))]] - - {Π Π}_{i i = = 11}^{M m} [[11 - - {Π Π}_{j j = = 11}^{n no - - 11} ((11 - - {p p}_{ij ij}^{MRC MRC}))]] - - - - - - ((21 twenty one))$

其中，in,

p_i1 ^MRC＝p_i1 (21)p _i1 ^MRC = p _i1 (21)

${p p}_{ij ij}^{MRC MRC} = = \frac{{p p}_{ij ij} - - {p p}_{i i ((j j - - 11))}}{11 - - {p p}_{i i ((j j - - 11))}} ((j j &GreaterEqual; &Greater Equal; 22)) - - - - - - ((22 twenty two))$

${p p}_{ij ij} = = {&Integral; &Integral;}_{{snr snr}_{T T}}^{\infty \infty} {f f}_{ij ij} ((x x)) dx dx - - - - - - ((23 twenty three))$

f_i1(y)＝f_i(y) (24)f _i1 (y) = f _i (y) (24)

${f f}_{ij ij} ((y the y)) = = {&Integral; &Integral;}_{00}^{\infty \infty} {f f}_{i i 11} ((x x)) {f f}_{i i ((j j - - 11))} ((y the y - - x x)) dx dx ((j j &GreaterEqual; &Greater Equal; 22)) - - - - - - ((2525))$

f)使用喷泉码f) Using Fountain Code

使用喷泉码时：When using fountain codes:

${P P}_{n no}^{k k} = = {Π Π}_{i i = = 11}^{M m} ((11 - - {Σ Σ}_{j j = = 00}^{U u - - 11} {C C}_{n no}^{j j} {p p}_{i i}^{j j} {(({11 - - p p}_{i i}))}^{n no - - j j})) - - {Π Π}_{i i = = 11}^{M m} ((11 - - {Σ Σ}_{j j = = 00}^{U u - - 11} {C C}_{n no - - 11}^{j j} {p p}_{i i}^{j j} {((11 - - p p))}^{n no - - 11 - - j j})) . .$

当基站与用户间的距离未知，假设用户每次都将信噪比能够反馈会基站，这时相当于知道信噪比分布，此时根据反馈回来的数据预测分布的参数，用数据拟合的方法来求得近似参数，求得每个用户的信噪比分布参数。When the distance between the base station and the user is unknown, assuming that the user can feed back the signal-to-noise ratio to the base station every time, it is equivalent to knowing the distribution of the signal-to-noise ratio. At this time, the parameters of the distribution are predicted according to the feedback data, and the data fitting is used. method to obtain the approximate parameters, and obtain the SNR distribution parameters of each user.

进一步，以上步骤2)的具体方法如下：Further, the specific method of the above step 2) is as follows:

A、求得P_n ^k后，当系统采用普通情况或者使用最大比合并技术时，发送速率r_k所对应的等效传输速率为：A. After obtaining P _n ^k , when the system adopts the normal situation or uses the maximum ratio combination technology, the equivalent transmission rate corresponding to the sending rate r _k is:

${R R}_{eq eq}^{k k} = = E E. {{\frac{{r r}_{k k} \times \times M m}{n no}}} = = {r r}_{k k} \times \times M m \times \times E E. {{\frac{11}{n no}}}_{k k} - - - - - - ((2626))$

其中in

$E E. {{\frac{11}{n no}}}_{k k} = = {Σ Σ}_{n no = = 11}^{\infty \infty} \frac{11}{n no} {P P}_{n no}^{k k} - - - - - - ((2727))$

B、当系统使用喷泉码时，发送速率r_k所对应的等效传输速率为：B. When the system uses the fountain code, the equivalent transmission rate corresponding to the sending rate r _k is:

${R R}_{eq eq}^{k k} = = E E. {{\frac{{MS MS}_{m m}}{nT n}}}_{k k} = = \frac{{MS MS}_{m m}}{T T} E E. {{\frac{11}{n no}}}_{k k} - - - - - - ((2828))$

$E E. {{\frac{11}{n no}}}_{k k} = = {Σ Σ}_{n no = = 11}^{\infty \infty} \frac{11}{n no} {P P}_{n no}^{k k} - - - - - - ((2929))$

进一步，以上步骤3)的具体方法如下：Further, the specific method of above step 3) is as follows:

最优的的发送速率索引可以表示为：The optimal sending rate index can be expressed as:

${k k}_{opt opt} = = \underset{k k}{arg arg max max {{{R R}_{eq eq}^{k k}}}} ((11 \leq \leq k k \leq \leq K K)) - - - - - - ((3030))$

然后选择作为基站的发送速率进行数据的传输。then choose Data transmission is performed at the transmission rate of the base station.

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

本发明在已知信道分布的条件下，几乎不需要反馈数据的情况下，给出了等发送速率机会多播机制，并且根据概率论与随机过程相关知识，提出了如何获得最优的发送速率R的方案。该方案极大的降低了反馈的同时有效的提高了系统的容量，使其更加适用于实际系统，并且本发明考虑了多种场景，多种情况下的最优方案，使得方案更加具有通用性。Under the condition of known channel distribution and almost no feedback data, the present invention provides an equal transmission rate opportunity multicast mechanism, and proposes how to obtain the optimal transmission rate according to the knowledge of probability theory and random process R's scheme. This solution greatly reduces the feedback and effectively improves the system capacity, making it more suitable for practical systems, and the present invention considers various scenarios and optimal solutions in various situations, making the solution more versatile .

附图说明Description of drawings

图1多媒体多播网络示意图；Fig. 1 is a schematic diagram of a multimedia multicast network;

图2所提出方案的流程图；The flowchart of the proposed scheme in Fig. 2;

图3所讨论的各种情形示意图；Figure 3 is a schematic diagram of the various situations discussed;

具体实施方式Detailed ways

下面结合附图和具体实施实例对本发明做进一步的详细说明。The present invention will be further described in detail below in conjunction with the accompanying drawings and specific implementation examples.

如图(1)所示，考虑一个单小区单天线的下行无线通信网，小区半径为R_d基站(BS)同时向M个用户发送同样的数据，信道为块衰落无线信道，每个时间块的长度为T，近似等于信道相关时间。用户的信道分布已知，h_i(k)表示第i个用户在第k个时间块上的信道系数，假设h_i(k)服从瑞利分布，这里用零均值的复高斯随机变量表示，且方差为1。假设信道系数在一个时间块内保持不变，在不同的时间块上独立变化。假设基站的发送速率是离散的，且总共有K个可供选择的离散速率，构成集合A＝{r₁,r₂,…,r_K}。每次都以相同的速率r_k(k＝1～K为优化参数)发送数据包，直到所有用户都收到，才进行下一数据包的发送。每次传输的时间长度为T，将其定义为一个时间块。定义snr_k满足下式：As shown in Figure (1), consider a downlink wireless communication network with single cell and single antenna. The cell radius is R _d . The base station (BS) sends the same data to M users at the same time. The channel is a block fading wireless channel. Each time block The length of T is approximately equal to the channel correlation time. The channel distribution of the user is known, h _i (k) represents the channel coefficient of the i-th user in the k-th time block, assuming that h _i (k) obeys the Rayleigh distribution, here it is represented by a complex Gaussian random variable with zero mean value, And the variance is 1. It is assumed that the channel coefficients remain constant within a time block and vary independently over different time blocks. Assuming that the sending rate of the base station is discrete, and there are K discrete rates available for selection, a set A={r ₁ , r ₂ ,...,r _K } is formed. The data packets are sent at the same rate r _k (k=1~K is an optimization parameter) each time, and the next data packet is not sent until all users have received it. The time length of each transmission is T, which is defined as a time block. Define snr _k to satisfy the following formula:

r_k＝log₂(1+snr_k)(k＝1～K) (1)r _k ＝log ₂ (1+snr _k )(k＝1～K) (1)

可以得到：can get:

${snr snr}_{k k} = = 22^{{r r}_{k k}} - - 11 - - - - - - ((22))$

用户i在第j个时间块上收到的信号可以表示为：The signal received by user i at the jth time block can be expressed as:

${y the y}_{i i} ((j j)) = = \sqrt{{Pd PD}_{i i}^{η η}} {h h}_{i i} ((j j)) x x + + {n no}_{i i} ((j j))$

式中：In the formula:

P——发送功率；P - transmit power;

n_i(j)——高斯白噪声，方差为N₀；n _i (j)——Gaussian white noise, variance is N ₀ ;

d_i——用户i与基站之间的距离；d _i ——the distance between user i and the base station;

η——大尺度衰落系数；η——large-scale fading coefficient;

对应的信噪比为：The corresponding signal-to-noise ratio is:

${SNR SNR}_{i i} = = \frac{P P {d d}_{i i}^{- - η η} {| | {h h}_{i i} | |}^{22}}{{N N}_{00}}$

服从参数为的指数分布。如果无线通信网络为同构网络，也就是所有用户的信道分布相同，此时，令d_i＝1。定义用户i所能达到的速率:Obedience parameters are exponential distribution of . If the wireless communication network is a homogeneous network, that is, all users have the same channel distribution, then d _i =1 is set. Define the rate that user i can achieve:

R_i＝log(1+SNR_i)R _i =log(1+SNR _i )

如果r_i大于基站的发送速率，则用户i能够正确接收数据，否则不能。对于传统多播方案，也就是基站每次都以最差用户所能正确接收数据的速率来发送数据。此时，发送速率为If r _i is greater than the sending rate of the base station, user i can receive data correctly, otherwise it cannot. For the traditional multicast scheme, that is, the base station sends data every time at the rate that the worst user can correctly receive the data. At this time, the sending rate is

R_t＝min{R_i}(i＝1,2,...,M)R _t =min{R _i }(i=1,2,...,M)

进行等发送速率最优机会多播方案调度时，如图(2)所示，包括以下步骤：When scheduling the optimal opportunity multicast scheme with equal transmission rate, as shown in Figure (2), the following steps are included:

1)在每个固定的发送速率rk情况下，计算各种场景各种方案下的恰好需要n次所有用户都能正确接收数据的概率P_n ^k；在P_n ^k的计算过程中，如图(3)所示，可分为以下几种情况分别讨论：1) In the case of each fixed transmission rate rk, calculate the probability P _n ^k that exactly n times all users can receive data correctly under various scenarios and schemes; during the calculation of P _n ^k , as shown in As shown in (3), it can be divided into the following situations and discussed separately:

(1)同构网络(1) Isomorphic network

这种场景下所有用户的信噪比独立同分布，第i个用户的接收信号可以表示为：In this scenario, the signal-to-noise ratios of all users are independently and identically distributed, and the received signal of the i-th user can be expressed as:

在假设瑞丽分布的信道条件下，服从参数为N₀/P的指数分布。其概率密度函数为Under the channel condition of assuming Rayleigh distribution, it obeys the exponential distribution with parameter N ₀ /P. Its probability density function is

$f f ((x x)) = = {λe λ e}^{- - λx λx} ((λ λ = = \frac{{N N}_{00}}{P P})) - - - - - - ((55))$

在这种网络场景下，以下考虑三种方案中P_n ^k的求解：a)普通情况，b)采用最大比合并技术，c)使用喷泉码。In this network scenario, the solution of P _n ^k in three schemes is considered as follows: a) general case, b) maximum ratio combining technique, and c) fountain code.

a)普通情况a) Ordinary situation

普通情况下，P_n ^k的计算可由下式计算得到：In general, the calculation of P _n ^k can be calculated by the following formula:

P_n ^k＝[1-(1-p)ⁿ]^M-[1-(1-p)^n-1]^M (6)P _n ^k ＝[1-(1-p) ⁿ ] ^M -[1-(1-p) ^n-1 ] ^M (6)

其中in

$p p = = {&Integral; &Integral;}_{{snr snr}_{k k}}^{\infty \infty} f f ((x x)) dx dx - - - - - - ((77))$

其中f(x)由(5)式给出。where f(x) is given by (5).

b)采用最大比合并技术b) Using the maximum ratio combining technique

用户可以将来自不同时间块的信号进行最大笔合并，在这种情况下，The user can combine signals from different time blocks for a maximum sum, in this case,

${P P}_{n no}^{k k} = = {[[11 - - {Π Π}_{i i = = 11}^{n no} ((11 - - {p p}_{i i}^{MRC MRC}))]]}^{M m} - - {[[11 - - {Π Π}_{i i = = 11}^{n no - - 11} ((11 - - {p p}_{i i}^{MRC MRC}))]]}^{M m} - - - - - - ((88))$

其中in

p₁ ^MRC＝p₁ (9)p ₁ ^MRC = p ₁ (9)

${p p}_{i i}^{MRC MRC} = = \frac{{p p}_{i i} - - {p p}_{i i - - 11}}{11 - - {p p}_{i i - - 11}} ((i i &GreaterEqual; &Greater Equal; 22)) - - - - - - ((1010))$

p_i的计算如下p _i is calculated as follows

p₁＝p (11)p ₁ =p (11)

${p p}_{i i} = = {&Integral; &Integral;}_{{snr snr}_{k k}}^{\infty \infty} {f f}_{i i} ((x x)) dx dx ((i i &GreaterEqual; &Greater Equal; 22)) - - - - - - ((1212))$

f_i(y)的计算如下：f _i (y) is calculated as follows:

f₁(x)＝f(x) (13)f ₁ (x)=f(x) (13)

${f f}_{i i} ((y the y)) {&Integral; &Integral;}_{00}^{\infty \infty} f f ((x x)) {f f}_{i i - - 11} ((y the y - - x x)) dx dx ((i i &GreaterEqual; &Greater Equal; 22)) - - - - - - ((1414))$

c)使用喷泉码c) using fountain code

当数据经过喷泉码编码后，就会与a)和b)两种情况有所不同。在使用喷泉码时，假设每个用户至少需要正确接收S_mbit数据的信息才能恢复原始数据。此时：When the data is encoded by the fountain code, it will be different from a) and b). When using fountain codes, it is assumed that each user needs at least the information of correctly receiving S _m bit data in order to restore the original data. at this time:

${P P}_{n no}^{k k} = = {((11 - - {Σ Σ}_{i i = = 00}^{U u - - 11} {C C}_{n no}^{i i} {p p}^{i i} {((11 - - p p))}^{n no - - i i}))}^{M m} - - {((11 - - {Σ Σ}_{i i = = 00}^{U u - - 11} {C C}_{n no - - 11}^{i i} {p p}^{i i} {((11 - - p p))}^{n no - - 11 - - i i}))}^{M m} - - - - - - ((1515))$

其中p由(7)式计算所得。Where p is calculated by (7).

(2)异构网络(2) Heterogeneous network

${y the y}_{i i} = = \sqrt{{Pd PD}_{i i}^{η η}} {h h}_{i i} x x + + {n no}_{i i} - - - - - - ((1717))$

${SNR SNR}_{i i} = = \frac{{Pd PD}_{i i}^{- - η η} {| | {h h}_{i i} | |}^{22}}{{N N}_{00}} - - - - - - ((1818))$

每个用户的信道系数h_i依然是独立同分布，但是由于考虑大尺度衰落，所以每个用户的信噪比分布并不相同。下面给出三种解决方案：The channel coefficient _hi of each user is still independent and identically distributed, but due to the consideration of large-scale fading, the SNR distribution of each user is not the same. Three solutions are given below:

1)similar_homo方案1) similar_homo scheme

在已知用户到基站距离d的情况下，信噪比概率密度函数为When the distance d from the user to the base station is known, the SNR probability density function is

$f f ((x x | | d d)) = = {λe λ e}^{- - λx λx} ((λ λ = = \frac{{N N}_{00}}{{Pd PD}^{- - η η}})) - - - - - - ((1919))$

对于圆形小区，用户到基站距离的概率密度函数为For a circular cell, the probability density function of the distance from the user to the base station is

${f f}_{{d d}_{i i}} ((d d)) = = \frac{22 d d}{{R R}_{d d}^{22}} - - - - - - ((2020))$

因此所有用户信噪比的概率密度函数可以统一表示为：Therefore, the probability density function of all user signal-to-noise ratios can be uniformly expressed as:

$f f ((x x)) = = {&Integral; &Integral;}_{00}^{{R R}_{d d}} f f ((x x | | y the y)) {f f}_{{d d}_{i i}} ((y the y)) dy dy = = {&Integral; &Integral;}_{00}^{{R R}_{d d}} {λe λ e}^{- - λx λx} \frac{22 y the y}{{R R}_{d d}^{22}} dy dy ((λ λ = = \frac{{N N}_{00}}{{Py Python}^{- - η η}})) - - - - - - ((21 twenty one))$

经过这样的处理后，所有用户的信噪比分布就又可以看成是一样的，也就是说异构网络就可以看成是同构网络，方案的其它步骤也和同构网络一样，可以参考a)～c)的处理过程。After such processing, the signal-to-noise ratio distribution of all users can be regarded as the same again, that is to say, the heterogeneous network can be regarded as a homogeneous network, and other steps of the scheme are the same as the homogeneous network. a) ~ c) process.

2)最优算法(假设已知用户与基站间的距离)2) Optimal algorithm (assuming that the distance between the user and the base station is known)

知道用户与基站的距离后，每个用户信噪比的概率密度函数f_i(x)是可以表示为：After knowing the distance between the user and the base station, the probability density function f _i (x) of the signal-to-noise ratio of each user can be expressed as:

${f f}_{i i} ((x x)) = = {λ λ}_{i i} {e e}^{- - {λ λ}_{i i} x x} (({λ λ}_{i i} = = \frac{{N N}_{00}}{{Pd PD}_{i i}^{- - η η}})) - - - - - - ((22 twenty two))$

这样我们在异构网络中也可以分三种情况讨论：d)普通情况，e)考虑最大比合并，f)使用喷泉码。In this way, we can also discuss three cases in the heterogeneous network: d) common case, e) consider maximum ratio combination, f) use fountain code.

d)普通情况d) Ordinary situation

${P P}_{n no}^{k k} = = {Π Π}_{i i = = 11}^{M m} [[11 - - {(({11 - - p p}_{i i}))}^{n no}]] - - {Π Π}_{i i = = 11}^{M m} [[11 - - {(({11 - - p p}_{i i}))}^{n no - - 11}]] - - - - - - ((23 twenty three))$

其中in

${p p}_{i i} = = {&Integral; &Integral;}_{{snr snr}_{k k}}^{\infty \infty} {f f}_{i i} ((x x)) dx dx - - - - - - ((24 twenty four))$

e)采用最大比合并技术e) Using maximum ratio combining technology

${P P}_{n no}^{k k} = = {Π Π}_{i i = = 11}^{M m} [[11 - - {Π Π}_{j j = = 11}^{n no} ((11 - - {p p}_{ij ij}^{MRC MRC}))]] - - {Π Π}_{i i = = 11}^{M m} [[11 - - {Π Π}_{j j = = 11}^{n no - - 11} ((11 - - {p p}_{ij ij}^{MRC MRC}))]] - - - - - - ((2525))$

其中，in,

p_i1 ^MRC＝p_i1 (26)p _i1 ^MRC = p _i1 (26)

${p p}_{ij ij}^{MRC MRC} = = \frac{{p p}_{ij ij} - - {p p}_{i i ((j j - - 11))}}{11 - - {p p}_{i i ((j j - - 11))}} ((j j &GreaterEqual; &Greater Equal; 22)) - - - - - - ((2727))$

${p p}_{ij ij} = = {&Integral; &Integral;}_{{snr snr}_{T T}}^{\infty \infty} {f f}_{ij ij} ((x x)) dx dx - - - - - - ((2828))$

f_i1(y)＝f_i(y) (29)f _i1 (y) = f _i (y) (29)

${f f}_{ij ij} ((y the y)) = = {&Integral; &Integral;}_{00}^{\infty \infty} {f f}_{i i 11} ((x x)) {f f}_{i i ((j j - - 11))} ((y the y - - x x)) dx dx ((j j &GreaterEqual; &Greater Equal; 22)) - - - - - - ((3030))$

f)使用喷泉码f) Using Fountain Code

使用喷泉码时：When using fountain codes:

${P P}_{n no}^{k k} = = {Π Π}_{i i = = 11}^{M m} ((11 - - {Σ Σ}_{j j = = 00}^{U u - - 11} {C C}_{n no}^{j j} {p p}_{i i}^{j j} {(({11 - - p p}_{i i}))}^{n no - - j j})) - - {Π Π}_{i i = = 11}^{M m} ((11 - - {Σ Σ}_{j j = = 00}^{U u - - 11} {C C}_{n no - - 11}^{j j} {p p}_{i i}^{j j} {((11 - - p p))}^{n no - - 11 - - j j})) - - - - - - ((3131))$

相关定义见c)See c) for related definitions

3)次优算法(基站与用户间的距离未知)3) Suboptimal algorithm (the distance between the base station and the user is unknown)

在这种情况下，假设用户每次都将信噪比能够反馈会基站，此时我们可以根据反馈回来的数据预测分布的参数，服从参数为λ的指数分布变量x的数学期望E[x]＝1/λ。In this case, assuming that the user can feed back the signal-to-noise ratio to the base station every time, we can predict the parameters of the distribution according to the feedback data, and obey the mathematical expectation E[x] of the exponential distribution variable x with the parameter λ =1/λ.

Step1：假设M个用户信噪比的概率密度函数由(21)式获得，然后根据同构网络相关理论求出最优速率R_opt；Step1: Assume that the probability density function of the SNR of M users is obtained by formula (21), and then calculate the optimal rate R _opt according to the related theory of isomorphic network;

Step2：每经过K次传输后，对每个用户的信噪比概率密度函数进行修正，修正后每个用户的分布参数为Step2: After every K transmissions, the SNR probability density function of each user is corrected, and the distribution parameter of each user after correction is

${λ λ}_{i i} \approx \approx \frac{11}{\frac{11}{K K} {Σ Σ}_{j j = = 11}^{K K} {SNR SNR}_{ij ij}} ((i i = = 1,2 1,2,, . . . . . .,, M m)) - - - - - - ((3232))$

SNR_ij表示第i个用户在第j次传输时的信噪比。修正后，根据异构网络的相关理论计算得到最优的速率R_opt(参见最优算法的步骤)。值得注意的是，以上的算法还适用于每个用户缓慢移动的情况。SNR _ij represents the signal-to-noise ratio of the i-th user at the j-th transmission. After correction, the optimal rate R _opt is calculated according to the related theory of the heterogeneous network (refer to the steps of the optimal algorithm). It is worth noting that the above algorithm is also applicable to the case where each user moves slowly.

2)求得P_n ^k后，计算每个发送速率rk所对应的等效传输速率，分两种情况讨论；2) After obtaining P _n ^k , calculate the equivalent transmission rate corresponding to each sending rate rk, and discuss in two cases;

A、求得各种场景各种方案下的P_n ^k后，当系统采用普通情况或者使用最大比合并技术时，发送速率r_k所对应的等效传输速率为：A. After obtaining P _n ^k under various schemes in various scenarios, when the system adopts the normal situation or uses the maximum ratio combination technology, the equivalent transmission rate corresponding to the sending rate r _k is:

${R R}_{eq eq}^{k k} = = E E. {{\frac{{r r}_{k k} \times \times M m}{n no}}} = = {r r}_{k k} \times \times M m \times \times E E. {{\frac{11}{n no}}}_{k k} - - - - - - ((3333))$

其中in

$E E. {{\frac{11}{n no}}}_{k k} = = {Σ Σ}_{n no = = 11}^{\infty \infty} \frac{11}{n no} {P P}_{n no}^{k k} - - - - - - ((3434))$

${R R}_{eq eq}^{k k} = = E E. {{\frac{{MS MS}_{m m}}{nT n}}}_{k k} = = \frac{{MS MS}_{m m}}{T T} E E. {{\frac{11}{n no}}}_{k k} - - - - - - ((3535))$

$E E. {{\frac{11}{n no}}}_{k k} = = {Σ Σ}_{n no = = 11}^{\infty \infty} \frac{11}{n no} {P P}_{n no}^{k k} - - - - - - ((3636))$

3)在各种场景各种方案下，选择使得等效传输速率最大所对应的发送速率作为基站的发送速率进行发送数据。具体操作如下：3) Under various scenarios and schemes, the transmission rate corresponding to the maximum equivalent transmission rate is selected as the transmission rate of the base station to transmit data. The specific operation is as follows:

${k k}_{opt opt} = = \underset{k k}{arg arg max max {{{R R}_{eq eq}^{k k}}}} ((11 \leq \leq k k \leq \leq K K)) - - - - - - ((3737))$

Claims

1. An optimal opportunity multicast method under a medium sending rate mechanism in a multimedia multicast technology is characterized by comprising the following steps:

1) at each fixed transmission rate r_kIn this case, after the data is transmitted n times, the probability P that all users can correctly receive the data is calculated_n ^k；

The specific method comprises the following steps:

the wireless communication system of a single base station has M users in total, the channel distribution state of the users is known, and the probability density function of the signal-to-noise ratio of the ith user isf_i(x) The transmission rate of a single base station is discrete and there are K total discrete rates available for selection, forming a set a ═ r₁,r₂,…,r_KEvery time at the same rate r_kSending a data packet, wherein K is 1-K which is an optimization parameter, and sending the next data packet until all users receive the data packet; the time length of each transmission is T, and the transmission is defined as a time block; the channel coefficients remain unchanged in one time block and independent in different time blocks, defining snr_kSatisfies the following formula:

r_k＝log₂(1+snr_k) (k＝1～K) (1)

can obtain

{snr}_{k} = 2^{r_{k}} - 1 - - - (2)

P_n ^kThe calculation of (a) is divided into the following different cases:

(1) homogeneous network

In this scenario, the snr of all users is independently and equally distributed, and the received signal of the ith user is represented as:

y_{i} = \sqrt{P} h_{i} x + n_{i} - - - (3)

p is the transmission power; n is_iIs white gaussian noise; x is the user signal-to-noise ratio;

h_irepresenting the channel coefficient, corresponding to a signal-to-noise ratio of

{SNR}_{i} = \frac{P {| h_{i} |}^{2}}{N_{0}} - - - (4)

N₀For variance, the snr probability density function of all users is the same and is f (x), and in this network scenario, P in the three schemes is considered below_n ^kSolving:

a) general conditions

P_n ^kCalculated from the following formula:

P_n ^k＝[1-(1-p)ⁿ]^M-[1-(1-p)^n-1]^M (5)

wherein

<math> <mrow> <mi>p</mi> <mo>=</mo> <munderover> <mo>&Integral;</mo> <msub> <mi>snr</mi> <mi>k</mi> </msub> <mo>∞</mo> </munderover> <mi>f</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mi>dx</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> </math>

b) Using maximal ratio combining techniques

The signals from different time blocks are maximal ratio combined, in this case,

wherein

p₁ ^MRC＝p₁ (8)

<math> <mrow> <msup> <msub> <mi>p</mi> <mi>i</mi> </msub> <mi>MRC</mi> </msup> <mo>=</mo> <mfrac> <mrow> <msub> <mi>p</mi> <mi>i</mi> </msub> <mo>-</mo> <msub> <mi>p</mi> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <msub> <mi>p</mi> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>,</mo> <mrow> <mo>(</mo> <mi>i</mi> <mo>&GreaterEqual;</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> </math>

p_iIs calculated as follows

p₁＝p (10)

<math> <mrow> <msub> <mi>p</mi> <mi>i</mi> </msub> <mo>=</mo> <munderover> <mo>&Integral;</mo> <msub> <mi>snr</mi> <mi>k</mi> </msub> <mo>∞</mo> </munderover> <msub> <mi>f</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mi>dx</mi> <mo>,</mo> <mrow> <mo>(</mo> <mi>i</mi> <mo>&GreaterEqual;</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> </math>

f_iThe calculation of (y) is as follows:

f₁(x)＝f(x) (12)

<math> <mrow> <msub> <mi>f</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>∞</mo> </munderover> <mi>f</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <msub> <mi>f</mi> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>-</mo> <mi>x</mi> <mo>)</mo> </mrow> <mi>dx</mi> <mo>,</mo> <mrow> <mo>(</mo> <mi>i</mi> <mo>&GreaterEqual;</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow> </math>

c) use fountain sign indicating number

Each user needs to receive data of at least size S correctly_mThe bit information can restore the original data, at this time:

wherein p is calculated from the formula (6);

(2) heterogeneous network

In this scenario, large-scale fading of users is considered, the signal-to-noise ratios of all users are independent but obey different distributions, and the received signal of the ith user is as follows:

d_irepresents the distance from user i to the base station; eta large scale fading coefficient corresponding to a signal-to-noise ratio of

Channel coefficient h for each user_iThe same distribution is still independent, but the signal-to-noise ratio distribution of each user is different due to the consideration of large-scale fading;

2) find P_n ^kThen, each transmission rate r is calculated_kThe corresponding equivalent transmission rate;

3) and selecting the sending rate corresponding to the maximum equivalent transmission rate as the sending rate of the base station to send data.

2. The method of claim 1, wherein the distance d between the known user and the base station is determined in a heterogeneous network scenarioIn the case of (1), the probability density function of the snr is f (x | d), and it is known that the users are uniformly distributed in the cell with radius R_dSo probability density function of user to base stationCan also be obtained; the probability density function of the signal-to-noise ratio of all users can be uniformly expressed as:

<math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&Integral;</mo> <mn>0</mn> <msub> <mi>R</mi> <mi>d</mi> </msub> </msubsup> <mi>f</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>|</mo> <mi>y</mi> <mo>)</mo> </mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>i</mi> </msub> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> <mi>dy</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>18</mn> <mo>)</mo> </mrow> </mrow> </math>

after such processing, the snr distributions of all users can be considered to be the same, that is, the heterogeneous network can be considered to be a homogeneous network, and then according to P_n ^kAnd (5) solving three schemes.

3. The method of claim 1, wherein in the heterogeneous network scenario, after knowing the distance between the user and the base station, there are three cases in the heterogeneous network:

d) general conditions

At this time, the transmission rate r_kCorresponding P_n ^k

Wherein

<math> <mrow> <msub> <mi>p</mi> <mi>i</mi> </msub> <mo>=</mo> <munderover> <mo>&Integral;</mo> <msub> <mi>snr</mi> <mi>k</mi> </msub> <mo>∞</mo> </munderover> <msub> <mi>f</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mi>dx</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>20</mn> <mo>)</mo> </mrow> </mrow> </math>

e) Using maximal ratio combining techniques

Considering P after maximum ratio combining_n ^kCan be expressed as

Wherein,

p_i1 ^MRC＝p_i1 (21)

<math> <mrow> <msup> <msub> <mi>p</mi> <mi>ij</mi> </msub> <mi>MRC</mi> </msup> <mo>=</mo> <mfrac> <mrow> <msub> <mi>p</mi> <mi>ij</mi> </msub> <mo>-</mo> <msub> <mi>p</mi> <mrow> <mi>i</mi> <mrow> <mo>(</mo> <mi>j</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </msub> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <msub> <mi>p</mi> <mrow> <mi>i</mi> <mrow> <mo>(</mo> <mi>j</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </msub> </mrow> </mfrac> <mo>,</mo> <mrow> <mo>(</mo> <mi>i</mi> <mo>&GreaterEqual;</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>22</mn> <mo>)</mo> </mrow> </mrow> </math>

<math> <mrow> <msub> <mi>p</mi> <mi>ij</mi> </msub> <mo>=</mo> <msubsup> <mo>&Integral;</mo> <msub> <mi>snr</mi> <mi>T</mi> </msub> <mo>∞</mo> </msubsup> <msub> <mi>f</mi> <mi>ij</mi> </msub> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mi>dx</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>23</mn> <mo>)</mo> </mrow> </mrow> </math>

f_i1(y)＝f_i(y) (24)

<math> <mrow> <msub> <mi>f</mi> <mi>ij</mi> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>∞</mo> </munderover> <msub> <mi>f</mi> <mrow> <mi>i</mi> <mn>1</mn> </mrow> </msub> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <msub> <mi>f</mi> <mrow> <mi>i</mi> <mrow> <mo>(</mo> <mi>j</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>-</mo> <mi>x</mi> <mo>)</mo> </mrow> <mi>dx</mi> <mo>,</mo> <mrow> <mo>(</mo> <mi>j</mi> <mo>&GreaterEqual;</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>25</mn> <mo>)</mo> </mrow> </mrow> </math>

f) use fountain sign indicating number

When the fountain code is used:

<math> <mrow> <mi>p</mi> <mo>=</mo> <munderover> <mo>&Integral;</mo> <msub> <mi>snr</mi> <mi>k</mi> </msub> <mo>∞</mo> </munderover> <mi>f</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mi>dx</mi> <mo>;</mo> </mrow> </math>

p_iis calculated as follows:

p₁＝p (11)

<math> <mrow> <msub> <mi>p</mi> <mi>i</mi> </msub> <mo>=</mo> <munderover> <mo>&Integral;</mo> <msub> <mi>snr</mi> <mi>k</mi> </msub> <mo>∞</mo> </munderover> <msub> <mi>f</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mi>dx</mi> <mo>,</mo> <mrow> <mo>(</mo> <mi>i</mi> <mo>&GreaterEqual;</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> </math>

4. the method according to claim 1, wherein when the distance between the base station and the user is unknown, the user can feed back the snr to the base station each time, which is equivalent to knowing the snr distribution, and then predict the distribution parameters according to the data fed back, and obtain the snr distribution parameters of each user by using a data fitting method.

5. The method for optimal opportunistic multicasting under the medium sending rate mechanism in multimedia multicasting technology according to any one of claims 1-4, wherein the specific method in step 2) is as follows:

A. find P_n ^kThen, when the system adopts the normal condition or uses the maximum ratio combining technology, the sending rate r_kThe corresponding equivalent transmission rates are:

wherein

B. When the system uses fountain codes, the sending rate r_kThe corresponding equivalent transmission rates are:

{R_{eq}}^{k} = E {\frac{{MS}_{m}}{nT}}_{k} = \frac{{MS}_{m}}{T} E {\frac{1}{n}}_{k} - - - (28)

6. the method for optimal opportunistic multicasting under the medium sending rate mechanism in multimedia multicasting technology according to any one of claims 1-4, wherein the specific method in step 3) is as follows:

the optimal transmission rate index can be expressed as:

then selectThe transmission of data is performed as the transmission rate of the base station.