CN101827452B - Rate splitting based method for allocating user rates in wireless multiple access channels - Google Patents

Abstract

The invention provides a rate splitting based method for allocating user rates in wireless multiple access channels, which is characterized by firstly computing the parameters of a rate splitting based multiple access communication system, including computing the splitting sequences and the splitting parameters; obtaining the splitting sequences of the user by merging binary trees; splitting the rates of a plurality of users into a series of two-user rate splitting based on the splitting sequences; alternately using the splitting formulas of two kinds of father nodes according to the splitting sequences to compute the splitting coefficients of all the users and the interference noise sum corresponding to the virtual users; and adjusting the code rate at the transmitting end of the user and controlling the successive decoding sequences of the virtual users at the receiving end according to the obtained splitting parameters so as to further realize the preset rate allocation strategy. The method can realize any rate allocation on the boundary of the multiple access channel capacity region, farthest utilize the wireless communication resources, not only reduce coding complexity but also dispense with strict synchronization among the users. In addition, the method is simple and effective and is used for allocating the wireless communication resources.

Description

Based on user rate distribution method in the wireless multiple access channels of code check division

Technical field

The invention belongs to wireless communication resources distribution technique field, relate to a kind of method of utilizing code check division multi-access technology to realize multiple access communication system rate-allocation, specifically is a kind of method of calculating code check division multi-access systems parameter.

Background technology

The wireless multiple access channels model is applicable to many practical application scenes, the for example up link of mobile cellular network.A plurality of users occupy radio channel resource simultaneously in the multiple access channel, and radio channel resource is very precious non-renewable resources, and this just makes the rational and effective allocation of channel resources may become very important, even are determining the performance of whole network.The existing resources distribution method mainly comprises three kinds of time division multiplexing, frequency division multiplexing and code division multiplexings, and these three kinds of methods all belong to the quadrature distribution method.The quadrature distribution method is a kind of channel resource allocation method of suboptimum, does not make full use of radio channel resource, has caused the waste of radio channel resource.A kind of new technology that code check division multiple access (RSMA) technology produces in order to make full use of radio channel resource just.This method not only requires low to system synchronization, and can reduce the coding expense, only combines the decoding technique one by one of receiving terminal just can realize on the multiple access channel capacity region border rate-allocation arbitrarily with single subscriber-coded transmission.

Decoding technique is existing a kind of ripe multi-user's decoding technique one by one, is widely used in the fields such as interference eliminated, relay transmission.If receiving terminal receives N user's transmission signal.At first with the 1st user's reception signal as data message, remaining N-1 user's signal is deciphered out the 1st user's information as interference.For the 1st user, its channel conditions is relatively poor, and transmission rate is low.Then, first user's the decoding information transmission information as first user is deducted from the signal that receives, with remaining signal as total reception signal.Further, as data message, remaining N-2 user's signal is deciphered out the 2nd user's information as interference with the 2nd user's reception signal.According to this method repetitive operation, until the information that translates all users.Decipher the communication channel that order is equivalent to different user has been distributed different quality one by one for different.Through regulating the order of decoding one by one, can realize distribution to radio channel resource to the user.

In N user's multiple access transmitted, code check splitting technique N-1 user at the most respectively was split into two Virtual User, and remaining users remains unchanged.For the user who is divided, the power division of its Virtual User is determined by bundle factor.Confirm that a component splits coefficient and to the order of decoding one by one of 2N-1 Virtual User at the most, just can confirm one group of unique user's transfer rate.Sending user side, according to bundle factor its power is divided into two parts, carry out coding transmission respectively.At receiving terminal, the Virtual User information that receives is deciphered one by one according to the order of deciphering one by one of gained.By this method, just can realize that the user rate of presetting distributes.Utilize the code check splitting technique to realize the multiple access channel resource allocation, at first will calculate the code check splitting system parameter value of realizing preset rate-allocation strategy, comprise user's bundle factor and the order of decoding one by one of Virtual User.Yet, also there is not a kind of method of practicality to calculate the system parameters of code check division so far, this just feasible research to the code check splitting technique rests on the theory analysis stage always.To code check division birth, this technology just is applied to various theory analysis scenes, and in practical application, is only limited to multi-user's code check division of N=2.For number of users is the code check division multiple access of N (N＞2), does not have feasible method to calculate the system parameters of code check division.

Summary of the invention

The purpose of this invention is to provide a kind ofly based on user rate distribution method in the wireless multiple access channels of code check division, this method can realize the borderline rate-allocation of multiple access channel capacity, makes wireless communication resources effectively utilized.

Provided by the invention based on user rate distribution method in the wireless multiple access channels of code check division, comprise that calculating code check division multi-access systems parameter and configuration-system transmit two parts.Calculate code check division multi-access systems parameter and comprise that again the division of calculating the user is in proper order with based on two parts of the order of decoding one by one of user's division order computation bundle factor and Virtual User.User's division order can be calculated by merging binary tree.After calculating user's division order, the division of multi-user's code check just turns to a series of two users' code check divisions.The code check splitting system parameter of calculating two users further can be divided into two kinds of situation.According to the division order,, just can calculate the system parameters of multi-user's code check division multiple access according to the computing formula that needs two users' code check splitting parameter under two kinds of situation of use alternately.Use the gained parameter code check division multi-access systems is configured, just can realize the preset distribution of multiple access channel resource.

Concrete steps are following:

(1) calculate the code check division multi-access systems parameter that realizes preset rate-allocation:

A) calculating the dried of user makes an uproar and array (Δ ₁..., Δ _{| I|}), wherein, I={1,2 ..., | I|} is the user label set, | I| is a number of users, $\∀i\∈I,$ ${\mathrm{\Δ}}_i=\frac{P_i}{(2^{2R_i}-1)}$ For user i dried made an uproar and, P _iBe the transmitted power of user i, R _iBe the speed that user i realizes in advance, according to the user dried make an uproar with from small to large order to user's ordering and renumber, make

Δ ₁≤...≤Δ _|I|；

B) seek user i and user j, i, j ∈ I satisfies Δ _j≤Δ _i＜Δ _j+ P _j, merge into a ultra user S to user i and user j, the power of ultra user S is P _s=P _i+ P _j, code check is R _s=R _i+ R _j, it is dried makes an uproar and does

${\mathrm{\Δ}}_s=\frac{P_i+P_j}{(2^{2(R_i+R_j)}-1)};$

C) collect among the I by ultra user's alternate user i and user j the user, number of users reduces 1, and user's collection is I={1, and 2 ..., | I|};

D) repeat a), b), c) three steps are until obtaining a series of nonoverlapping users or ultra user, for i, j ∈ I satisfies Δ _j+ P _j≤Δ _i, write down above-mentioned merging process, obtain a series of merging binary trees;

E) from merging the root node of binary tree; Each division is split into two node corresponding to father node; Until the leaf node that merges binary tree; This is the division order of being asked in proper order; Father node comprises two types: one type comprises two nonoverlapping power block, another kind ofly comprises a complete power block;

F) calculating user's bundle factor (ε ₁, ε ₂..., ε _{| I|}):

For comprising two not father nodes of overlapping power block, satisfy

$P_s={\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}$

$R_s={\mathrm{<figure-callout\; id="1"\; label="R\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">R</figure-callout>}}_s^{}+{\mathrm{<figure-callout\; id="2"\; label="R\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">R</figure-callout>}}_s^{}=\frac{1}{2}\log (1+\frac{{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}}{{\mathrm{\&Delta;}}_s^1})+\frac{1}{2}\log (1+\frac{{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}}{{\mathrm{\&Delta;}}_s^2})$

${\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2<{\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="1"\; label="s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">s</figure-callout>}}^1$

In the formula, P _s ¹And P _s ²Be respectively the performance number of two power block, Δ _s ¹And Δ _s ²Be that corresponding dried of two power block made an uproar and, R _s ¹And R _s ²Be the rate value of two power block, father node is split into two node, difference respective user i and user j, and the bundle factor of user j can calculate according to formula (3)-(20):

① $P_j\&le;P_s^1,$ $P_j\&le;P_s^2:$

${\mathrm{\&epsiv;}}_j=\frac{({\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="1"\; label="s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">s</figure-callout>}}^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2\&CenterDot;2^{2R_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(3\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(4\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(5\right)$

② $P_j\&GreaterEqual;{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},$ $P_j\&GreaterEqual;P_s^2:$

${\mathrm{\&epsiv;}}_j=1-\frac{({\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^2)({\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="1"\; label="s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">s</figure-callout>}}^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^1)}{P_j({\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="1"\; label="s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">s</figure-callout>}}^1\&CenterDot;2^{2R_j}-P_s^2-{\mathrm{\&Delta;}}_s^2)}---\left(6\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{\&Delta;}}_s^1---\left(7\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^2-(1-{\mathrm{\&epsiv;}}_j)P_j---\left(8\right)$

③ $P_j\&le;{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},$ $P_j\&GreaterEqual;P_s^2:$

When $R_j\&GreaterEqual;g({\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{\&Delta;}}_s^2)+g(P_j-{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},P_s^1+{\mathrm{\&Delta;}}_s^1-(P_j-P_s^2)),$ Then

${\mathrm{\&epsiv;}}_j=(1-{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}/P_j)---\left(9\right)$

${\mathrm{\&Delta;}}_{j,1}=f(R_j-g((1-{\mathrm{\&epsiv;}}_j)P_j,{\mathrm{\&Delta;}}_s^2),{\mathrm{\&epsiv;}}_j{P}_j)---\left(10\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(11\right)$

When $R_j\&le;g({\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{\&Delta;}}_s^2)+g(P_j-{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^1-(P_j-P_s^2)),$ Then

${\mathrm{\&epsiv;}}_j=\frac{({\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="1"\; label="s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">s</figure-callout>}}^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2\&CenterDot;2^{2R_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(12\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(13\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(14\right)$

④ $P_j\&GreaterEqual;{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},$ $P_j\&le;P_s^2:$

When $R_j\&GreaterEqual;g({\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{\&Delta;}}_s^1)+g(P_j-{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{\&Delta;}}_s^2),$ Then

${\mathrm{\&epsiv;}}_j=\frac{({\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="1"\; label="s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">s</figure-callout>}}^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2\&CenterDot;2^{2R_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(15\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(16\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(17\right)$

When $R_j\&le;g({\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{\&Delta;}}_s^1)+g(P_j-{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{\&Delta;}}_s^2),$ Then

${\mathrm{\&epsiv;}}_j={\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}/P_j---\left(18\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{\&Delta;}}_s^1---\left(19\right)$

${\mathrm{\&Delta;}}_{j,2}=f(R_j-g({\mathrm{\&epsiv;}}_j{P}_j,{\mathrm{\&Delta;}}_s^1),(1-{\mathrm{\&epsiv;}}_j)P_j)---\left(20\right)$

In the formula, ε _jBe the bundle factor of user j, Δ _{J, 1}And Δ _{J, 2}For user j split into corresponding respectively dried of two power block make an uproar with, according to two child nodes equal the power of father node with power, two child nodes equal the speed of father node with speed, the bundle factor of user i can correspondingly be tried to achieve;

For the father node that comprises a complete power block, satisfy

$P_s={\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}$

$R_s={\mathrm{<figure-callout\; id="1"\; label="R\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">R</figure-callout>}}_s^{}+{\mathrm{<figure-callout\; id="2"\; label="R\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">R</figure-callout>}}_s^{}=\frac{1}{2}\log (1+\frac{{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}}{{\mathrm{\&Delta;}}_s^1})+\frac{1}{2}\log (1+\frac{{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}}{{\mathrm{\&Delta;}}_s^2})$

${\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}+{\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2={\mathrm{\&Delta;}}_{\mathrm{<figure-callout\; id="1"\; label="s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">s</figure-callout>}}^1$

Wherein, P _s ¹And P _s ²Be respectively the performance number of two power block, Δ _s ¹And Δ _s ²Be corresponding with two power block dried making an uproar and, R _s ¹And R _s ²Be the rate value of two power block, the bundle factor of user j can calculate according to formula (21)-(23):

${\mathrm{\&epsiv;}}_j=\frac{P_s+{\mathrm{\&Delta;}}_s-(P_i+{\mathrm{\&Delta;}}_i)}{P_j}---\left(21\right)$

${\mathrm{\&Delta;}}_{j,1}=P_i+{\mathrm{\&Delta;}}_i---\left(22\right)$

Δ _j，2＝Δ _s (23)

In the formula, ε _jBe the bundle factor of user j, Δ _{J, 1}And Δ _{J, 2}For user j split into corresponding respectively dried of two power block make an uproar with, user i does not divide, its bundle factor ε _i=1, dried making an uproar with constant;

G) order of decoding one by one of calculating Virtual User:

According to e) in the division order of gained; Be used alternatingly f according to situation) in the computing formula of two types of father nodes user or ultra user are divided; Calculate corresponding dried of user's bundle factor and division user two power block make an uproar with; Splitting parameter until the user who obtains the leaf node representative that merges binary tree; Two power block after user's division are two Virtual User of respective user respectively, and Virtual User is deciphered order according to dried making an uproar with sorting from big to small one by one;

(2) configuration communication system parameters:

Send user side: for user i, i ∈ I is according to its bundle factor ε _i, power is divided into two parts ε _iP _i(1-ε _i) P _i, respectively according to speed g (ε _iP _i, Δ _{I, 1}) and g ((1-ε _iP _i), Δ _{I, 2}) transmission of encoding;

Receiving terminal: according to g) decoding one by one of gained time ordered pair Virtual User is deciphered in.

In the above-mentioned steps, user dried made an uproar and represented under the power constraint in the additive white gaussian channel, when the user realizes preset transfer rate patient maximum interference noise power with.User j dried make an uproar with ${\mathrm{\&Delta;}}_j=\frac{P_j}{(2^{2R_j}-1)}={\mathrm{\&delta;}}_j+{\mathrm{\&sigma;}}^2,$ σ ²Noise power in the expression channel, δ _jExpression is from the power of other Gauss's interference sources.What the Virtual User that obtains after the user division was corresponding is two power block that realize different rates, its power and be original subscriber's power, speed and be original subscriber's speed.

This method provides a kind of method of calculating code check division multi-access systems parameter, according to the system parameters configuration communication system that obtains, effectively distributes wireless communication resources with regard to code check division multi-access technology capable of using, realizes preset rate-allocation strategy.This method can be utilized radio channel resource to greatest extent, not only reduces the coding expense, only needs the decoding one by one of single subscriber-coded combination receiving terminal just can realize that the user rate of presetting distributes, and need not send between the user strict synchronous.

Description of drawings

Fig. 1 first kind father node division sketch map.

Second type of father node division of Fig. 2 sketch map.

Fig. 3 calculates and merges the binary tree algorithm flow chart.

The code check splitting algorithm operation sketch map of Fig. 4 multi-access network.

Fig. 5 calculates the bundle factor flow chart.

Fig. 6 multi-access network transmitting scene.

Embodiment

The present invention is applicable under given transmission number of users in Gauss's multiple access channel, interchannel noise and the transmitted power array situation, utilizes code check division multi-access technology to realize on the capacity region border rate-allocation strategy arbitrarily.

Utilize code check division multi-access technology to realize that the channel resource allocation strategy is divided into three parts.At first to calculate the system parameters of code check division; Then according to the code check bundle factor of gained and gained Virtual User dried make an uproar and the transmission parameter of size configure user side, make an uproar and from big to small sequential control receiving terminal the order of decoding one by one according to dried again to Virtual User.According to above method, just can realize rate-allocation strategy preset on the multiple access channel.

Concrete steps are following:

I={1,2 ..., | I|} representes the user label set in the system, | I| is a number of users.User's transmitted power array is (P ₁..., P _{| I|}), the noise power σ of receiving terminal ²The user rate array that realizes in advance is (R ₁..., R _{| I|}).

∑=(ε ₁, ε ₂..., ε _{| I|}) be that user's bundle factor array is with ⊿=(Δ _1,1, Δ _1,2..., Δ _{| I|, 1}, Δ _{| I|, 2}) make an uproar and array for the dried of Virtual User.For user j, j ∈ I, ε _jThe power of expression user j is split into two parts, respectively corresponding Virtual User u ₁ ^jAnd u ₂ ^j, its power is respectively ε _jP _j(1-ε _j) P _j, Δ _{J, 1}And Δ _{J, 2}Two Virtual User u of representative of consumer j ₁ ^jAnd u ₂ ^jDivide other dried make an uproar with.

(1) calculate the code check division multi-access systems parameter that realizes preset rate-allocation:

Calculate the division order:, in the user, seek overlapping user based on user's power array and preset speed array.User i and user j are overlapping, then Δ _j≤Δ _i＜Δ _j+ P _jWherein, ${\mathrm{\&Delta;}}_i=\frac{P_i}{(2^{2R_i}-1)},$ ${\mathrm{\&Delta;}}_j=\frac{P_j}{(2^{2R_j}-1)}$ Corresponding respectively dried of expression user i and user j make an uproar and.Overlapping user is merged into a ultra user, and replace overlapping user with ultra user, number of users subtracts 1.Repetitive operation is listed as nonoverlapping user or ultra user until obtaining some.Corresponding one of each nonoverlapping user or ultra user merge binary tree.Have only corresponding nonoverlapping user of merging binary tree of a root node, this user does not need division.The corresponding nonoverlapping ultra user of the merging binary tree of a plurality of points; From merging the root node of binary tree, each division is split into two node corresponding to a father node, until the leaf node that merges binary tree; This is the division order of being asked in proper order, specifically can be divided into following steps:

A) calculating the dried of user makes an uproar and array (Δ ₁..., Δ _{| I|}).Wherein, ${\mathrm{\&Delta;}}_i=\frac{P_i}{(2^{2R_i}-1)}$ For user i dried make an uproar with.According to the user dried make an uproar with from small to large order to user's ordering and renumber, make

Δ ₁≤...≤Δ _|I|。

B) seek user i and user j, i, j ∈ I satisfies Δ _j≤Δ _i＜Δ _j+ P _jMerge into a ultra user S to user i and user j, the power of ultra user S is P _s=P _i+ P _j, code check is R _s=R _i+ R _j, it is dried makes an uproar and does

${\mathrm{\&Delta;}}_s=\frac{P_i+P_j}{(2^{2(R_i+R_j)}-1)}.$

C) collect among the I by ultra user's alternate user i and user j the user, number of users reduces 1, and user's collection is I={1, and 2 ..., | I|}.

D) repeat a), b), c) three steps are until obtaining a series of nonoverlapping users or ultra user.For i, j ∈ I satisfies Δ _j+ P _j≤Δ _iWrite down above-mentioned merging process, obtain a series of merging binary trees.

E) from merging the root node of binary tree, each division is split into two node corresponding to father node, and until the leaf node that merges binary tree, this is the division order of being asked in proper order.Father node comprises two types: one type comprises two nonoverlapping power block, another kind ofly comprises a complete power block.

Calculate bundle factor and Virtual User dried make an uproar and: the user for having neither part nor lot in merging, it is not divided, its pairing Virtual User is exactly itself, bundle factor can be made as 1, driedly makes an uproar and remains unchanged.For the user who participate in to merge, according to e) in the division order that obtains, the code check fission process turns to some row two users code check fission processes.The code check splitting system parameter of calculating two users further can be divided into two kinds of situation, respectively corresponding two types of different father nodes.According to 1) in the division order that obtains, iteration is used the computing formula of code check bundle factor under two kinds of situation, until the leaf node that merges binary tree, bundle factor and corresponding Virtual User dried that calculates all users make an uproar with.

F) calculating user's bundle factor (ε ₁, ε ₂..., ε _{| I|}):

For comprising two not father nodes of overlapping power block, satisfy

$P_s=P_s^1+P_s^2$

$R_s=R_s^1+R_s^2=\frac{1}{2}\log (1+\frac{P_s^1}{{\mathrm{\&Delta;}}_s^1})+\frac{1}{2}\log (1+\frac{P_s^2}{{\mathrm{\&Delta;}}_s^2})$

$P_s^2+{\mathrm{\&Delta;}}_s^2<{\mathrm{\&Delta;}}_s^1$

Wherein, P _s ¹And P _s ²Be respectively the performance number of two power block, Δ _s ¹And Δ _s ²Be that corresponding dried of two power block made an uproar and, R _s ¹And R _s ²It is the rate value of two power block.Father node is split into two node, and its division sketch map is as shown in Figure 1, difference respective user (ultra user) i and user (ultra user) j, and its bundle factor can calculate according to formula (3)-(20):

① $P_j\&le;P_s^1,$ $P_j\&le;P_s^2:$

${\mathrm{\&epsiv;}}_j=\frac{(P_s^1+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_s^2\&CenterDot;2^{2R_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(3\right)$

${\mathrm{\&Delta;}}_{j,1}=P_s^1+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(4\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(5\right)$

② $P_j\&le;P_s^1,$ $P_j\&GreaterEqual;P_s^2:$

${\mathrm{\&epsiv;}}_j=1-\frac{(P_s^2+{\mathrm{\&Delta;}}_s^2)({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^1)}{P_j({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_s^2-{\mathrm{\&Delta;}}_s^2)}---\left(6\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{\&Delta;}}_s^1---\left(7\right)$

${\mathrm{\&Delta;}}_{j,2}=P_s^2+{\mathrm{\&Delta;}}_s^2-(1-{\mathrm{\&epsiv;}}_j)P_j---\left(8\right)$

③ $P_j\&le;P_s^1,$ $P_j\&GreaterEqual;P_s^2:$

When $R_j\&GreaterEqual;g(P_s^2,{\mathrm{\&Delta;}}_s^2)+g(P_j-{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},P_s^1+{\mathrm{\&Delta;}}_s^1-(P_j-P_s^2)),$ Then

${\mathrm{\&epsiv;}}_j=(1-{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{}/P_j)---\left(9\right)$

${\mathrm{\&Delta;}}_{j,1}=f(R_j-g((1-{\mathrm{\&epsiv;}}_j)P_j,{\mathrm{\&Delta;}}_s^2),{\mathrm{\&epsiv;}}_j{P}_j)---\left(10\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(11\right)$

When $R_j\&le;g(P_s^2,{\mathrm{\&Delta;}}_s^2)+g(P_j-{\mathrm{<figure-callout\; id="2"\; label="P\; s"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},P_s^1+{\mathrm{\&Delta;}}_s^1-(P_j-P_s^2)),$ Then

${\mathrm{\&epsiv;}}_j=\frac{(P_s^1+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{{2R}_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_s^2\&CenterDot;2^{{2R}_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(12\right)$

${\mathrm{\&Delta;}}_{j,1}=P_s^1+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(13\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(14\right)$

④ $P_j\&GreaterEqual;P_s^1,$ $P_j\&le;P_s^2:$

When $R_j\&GreaterEqual;g(P_s^1,{\mathrm{\&Delta;}}_s^1)+g(P_j-{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{\&Delta;}}_s^2),$ Then

${\mathrm{\&epsiv;}}_j=\frac{(P_s^1+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_s^2\&CenterDot;2^{2R_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(15\right)$

${\mathrm{\&Delta;}}_{j,1}=P_s^1+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(16\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(17\right)$

When $R_j\&le;g(P_s^1,{\mathrm{\&Delta;}}_s^1)+g(P_j-{\mathrm{<figure-callout\; id="1"\; label="P\; s"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_s^{},{\mathrm{\&Delta;}}_s^2),$ Then

${\mathrm{\&epsiv;}}_j=P_s^1/P_j---\left(18\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{\&Delta;}}_s^1---\left(19\right)$

${\mathrm{\&Delta;}}_{j,2}=f(R_j-g({\mathrm{\&epsiv;}}_j{P}_j,{\mathrm{\&Delta;}}_s^1),(1-{\mathrm{\&epsiv;}}_j)P_j)---\left(20\right)$

Wherein, ε _jBe the bundle factor of user j, Δ _{J, 1}And Δ _{J, 2}For user j split into corresponding respectively dried of two power block make an uproar with.The bundle factor of user i can correspondingly be tried to achieve.

For the father node that comprises a complete power block, satisfy

$P_s=P_s^1+P_s^2$

$R_s=R_s^1+R_s^2=\frac{1}{2}\log (1+\frac{P_s^1}{{\mathrm{\&Delta;}}_s^1})+\frac{1}{2}\log (1+\frac{P_s^2}{{\mathrm{\&Delta;}}_s^2})$

$P_s^2+{\mathrm{\&Delta;}}_s^2={\mathrm{\&Delta;}}_s^1$

Wherein, P _s ¹And P _s ²Be respectively the performance number of two power block, Δ _s ¹And Δ _s ²Be corresponding with two power block dried making an uproar and, R _s ¹And R _s ²It is the rate value of two power block.Father node is split into two node, and its division sketch map is as shown in Figure 2, and respective user (ultra user) i and user (ultra user) j bundle factor can calculate according to formula (21)-(23) respectively:

${\mathrm{\&epsiv;}}_j=\frac{P_s+{\mathrm{\&Delta;}}_s-(P_i+{\mathrm{\&Delta;}}_i)}{P_j}---\left(21\right)$

Δ _j，1＝P _i+Δ _i (22)

Δ _j，2＝Δ _s (23)

Wherein, ε _jBe the bundle factor of user j, Δ _{J, 1}And Δ _{J, 2}For user j split into corresponding respectively dried of two power block make an uproar with.User i does not divide, its bundle factor ε _i=1, dried making an uproar with constant.

G) order of decoding one by one of calculating Virtual User:

According to e) in the division order of gained; Be used alternatingly f according to situation) in the computing formula of two types of father nodes user or ultra user are divided; And calculate corresponding dried of bundle factor and Virtual User make an uproar with, until the user's who obtains the leaf node representative that merges binary tree splitting parameter.Virtual User is deciphered order according to dried making an uproar with sorting from big to small one by one.

(2) configuration communication system parameters:

Send user side: for user i, i ∈ I is according to its bundle factor ε _i, power is divided into two parts ε _iP _i(1-ε _i) P _i, respectively according to speed g (ε _iP _i, Δ _{I, 1}) and g ((1-ε _iP _i), Δ _{I, 2}) transmission of encoding.

Receiving terminal: according to g) decoding one by one of gained time ordered pair user deciphers in.

Be that the code check division is transmitted as example in 3 the multi-access network with a number of users below, implementation method of the present invention is described.If in 3 users' the multiple access channel, receiving terminal white Gaussian noise power is σ ², user's transmitted power array is (P ₁, P ₂, P ₃), certain rate-allocation array (R on the multiple access channel capacity region border ₁, R ₂, R ₃) be the rate-allocation strategy of realizing in advance.

1. calculate code check splitting system parameter:

At first calculate the merging binary tree, its algorithm flow chart is as shown in Figure 3.With reference to the division binary tree among Fig. 4, the order of user's division is:

1) ultra user 1,2, and 3} be split into user 1 and ultra user 2,3}.

2) { 2,3} is split into two users, user 2 and user 3 to surpass the user.

Calculate bundle factor and Virtual User dried make an uproar and algorithm flow chart as shown in Figure 5.Calculate bundle factor and each Virtual User dried make an uproar and the time, ultra user 1,2, meet the corresponding situation of the first kind father node during 3} division.Ultra user 2, meet the corresponding situation of second type of father node during the 3} division.Utilize formula (3)-(23) to calculate ∑=(ε ₁, ε ₂, ε ₃) He ⊿=(Δ _1,1, Δ _1,2, Δ _2,1, Δ _2,2, Δ _3,1, Δ _3,2).User 1 is not by division only corresponding Virtual User, then a ε ₁=1, dried making an uproar and Δ _1,2Can be made as Δ _1,2=0.

2. user side:

Bundle factor (ε according to gained ₁, ε ₂, ε ₃), each user splits into two parts with its power, and every part is corresponding to a Virtual User.According to the dried Zao of gained with ⊿=(Δ _1,1, Δ _1,2, Δ _2,1, Δ _2,2, Δ _3,1, Δ _3,2), calculate each Virtual User corresponding codes speed, the transmission of encoding then.The transmission scene is as shown in Figure 6.

1) not division of user 1, only corresponding Virtual User u ₁ ¹, its power is P ₁, driedly make an uproar and be Δ _1,1=Δ ₁, then its code rate does

${\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^1=g({\mathrm{<figure-callout\; id="1"\; label="P"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_1,{\mathrm{\&Delta;}}_{\mathrm{1,1}})=\frac{1}{2}\log (1+\frac{{\mathrm{<figure-callout\; id="1"\; label="P"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_1}{{\mathrm{\&Delta;}}_{\mathrm{1,1}}})$

2) user 2 power is split into two parts, corresponds respectively to two Virtual User u ₁ ²And u ₂ ²According to the bundle factor of gained, its power is respectively ε ₂P ₂(1-ε ₂) P ₂Corresponding dried of these two Virtual User made an uproar and is Δ _2,1And Δ _2,2, then code rate should be set at respectively

${\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2=g({\mathrm{\&epsiv;}}_2{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_2,{\mathrm{\&Delta;}}_{\mathrm{2,1}})=\frac{1}{2}\log (1+\frac{{\mathrm{\&epsiv;}}_2{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_2}{{\mathrm{\&Delta;}}_{\mathrm{2,1}}})$

${\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2=g((1-{\mathrm{\&epsiv;}}_2){\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_2,{\mathrm{\&Delta;}}_{\mathrm{2,2}})=\frac{1}{2}\log (1+\frac{(1-{\mathrm{\&epsiv;}}_2){\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_2}{{\mathrm{\&Delta;}}_{\mathrm{2,2}}})$

3) similar with user 2, two Virtual User u of user 3 ₁ ³And u ₂ ³Power be respectively ε ₃P ₃(1-ε ₃) P ₃, corresponding driedly make an uproar and be respectively Δ _3,1And Δ _3,2Its code rate should be set at respectively

${\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^3=g({\mathrm{\&epsiv;}}_3{\mathrm{<figure-callout\; id="3"\; label="P"\; filenames="HSA00000030377400051.png"\; state="\{\{state\}\}">P</figure-callout>}}_3,{\mathrm{\&Delta;}}_{\mathrm{3,1}})=\frac{1}{2}\log (1+\frac{{\mathrm{\&epsiv;}}_3{\mathrm{<figure-callout\; id="3"\; label="P"\; filenames="HSA00000030377400051.png"\; state="\{\{state\}\}">P</figure-callout>}}_3}{{\mathrm{\&Delta;}}_{\mathrm{3,1}}})$

${\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^3=g((1-{\mathrm{\&epsiv;}}_3){\mathrm{<figure-callout\; id="3"\; label="P"\; filenames="HSA00000030377400051.png"\; state="\{\{state\}\}">P</figure-callout>}}_3,{\mathrm{\&Delta;}}_{\mathrm{3,2}})=\frac{1}{2}\log (1+\frac{(1-{\mathrm{\&epsiv;}}_3){\mathrm{<figure-callout\; id="3"\; label="P"\; filenames="HSA00000030377400051.png"\; state="\{\{state\}\}">P</figure-callout>}}_3}{{\mathrm{\&Delta;}}_{\mathrm{3,2}}})$

3. receiving terminal

After receiving the information that all Virtual User transmit, receiving terminal according to Virtual User dried make an uproar and size Virtual User is deciphered one by one, dry and big Virtual User is benefitsed from one's ancestor, dried make an uproar and little after translate.Can get Δ by Fig. 4 _3,2≤Δ _2,2≤Δ _1,1≤Δ _2,1≤Δ _3,1, decoding is u in proper order so ₁ ³→ u ₁ ²→ u ₁ ¹→ u ₂ ²→ u ₂ ³

So far, the speed of each user's realization is respectively:

The user 1: ${\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">R</figure-callout>}}_1={\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^1=\frac{1}{2}\mathrm{Log}(1+\frac{{\mathrm{<figure-callout\; id="1"\; label="P"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">P</figure-callout>}}_1}{{\mathrm{\&Delta;}}_{\mathrm{1,1}}})$

The user 2: ${\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">R</figure-callout>}}_2={\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2+{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2=\frac{1}{2}\mathrm{Log}(1+\frac{{\mathrm{\&epsiv;}}_2{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_2}{{\mathrm{\&Delta;}}_{\mathrm{2,1}}})+\frac{1}{2}\mathrm{Log}(1+\frac{(1-{\mathrm{\&epsiv;}}_2){\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">P</figure-callout>}}_2}{{\mathrm{\&Delta;}}_{\mathrm{2,2}}})$

The user 3: ${\mathrm{<figure-callout\; id="3"\; label="R"\; filenames="HSA00000030377400051.png"\; state="\{\{state\}\}">R</figure-callout>}}_3={\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HSA00000030377400011.png,HSA00000030377400021.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^3+{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HSA00000030377400011.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^3=\frac{1}{2}\mathrm{Log}(1+\frac{{\mathrm{\&epsiv;}}_3{\mathrm{<figure-callout\; id="3"\; label="P"\; filenames="HSA00000030377400051.png"\; state="\{\{state\}\}">P</figure-callout>}}_3}{{\mathrm{\&Delta;}}_{\mathrm{3,1}}})+\frac{1}{2}\mathrm{Log}(1+\frac{(1-{\mathrm{\&epsiv;}}_3){\mathrm{<figure-callout\; id="3"\; label="P"\; filenames="HSA00000030377400051.png"\; state="\{\{state\}\}">P</figure-callout>}}_3}{{\mathrm{\&Delta;}}_{\mathrm{3,2}}})$

It is the rate-allocation strategy that we preset.

Claims (1)

1. based on user rate distribution method in the wireless multiple access channels of code check division, this method may further comprise the steps:

(1) calculate the code check division multi-access systems parameter that realizes preset rate-allocation:

A) calculating the dried of user makes an uproar and array (Δ ₁..., Δ _{| I|}), wherein, I={1,2 ..., | I|} is the user label set, | I| is a number of users, $\&ForAll;i\&Element;I,$ ${\mathrm{\&Delta;}}_i=\frac{P_i}{(2^{2R_i}-1)}$ For user i dried made an uproar and, P _iBe the transmitted power of user i, R _iBe the speed that user i realizes in advance, according to the user dried make an uproar with from small to large order to user's ordering and renumber, make

Δ ₁≤…≤Δ _|I|；

B) seek user i and user j, i, j ∈ I satisfies Δ _j≤Δ _i＜Δ _j+ P _j, merge into a ultra user S to user i and user j, the power of ultra user S is P _s=P _i+ P _j, code check is R _s=R _i+ R _j, it is dried makes an uproar and does

${\mathrm{\&Delta;}}_s=\frac{P_i+P_j}{(2^{2(R_i+R_j)}-1)};$

C) collect among the I by ultra user's alternate user i and user j the user, number of users reduces 1, and user's collection is I={1, and 2 ..., | I|};

D) repeat a), b), c) three steps are until obtaining a series of nonoverlapping users or ultra user, for i, j ∈ I satisfies Δ _j+ P _j≤Δ _i, write down above-mentioned merging process, obtain a series of merging binary trees;

E) from merging the root node of binary tree; Each division is split into two node corresponding to father node; Until the leaf node that merges binary tree; This is the division order of being asked in proper order; Father node comprises two types: one type comprises two nonoverlapping power block, another kind ofly comprises a complete power block;

F) calculating user's bundle factor (ε ₁, ε ₂..., ε _{| I|}):

For comprising two not father nodes of overlapping power block, satisfy

$P_s=P_s^1+P_s^2$

$R_s=R_s^1+R_s^2=\frac{1}{2}\log (1+\frac{P_s^1}{{\mathrm{\&Delta;}}_s^1})+\frac{1}{2}\log (1+\frac{P_s^2}{{\mathrm{\&Delta;}}_s^2})$

$P_s^2+{\mathrm{\&Delta;}}_s^2<{\mathrm{\&Delta;}}_s^1$

In the formula, P _s ¹And P _s ²Be respectively the performance number of two power block, Δ _s ¹And Δ _s ²Be that corresponding dried of two power block made an uproar and, R _s ¹And R _s ²Be the rate value of two power block, father node is split into two node, difference respective user i and user j, and the bundle factor of user j can calculate according to formula (3)-(20):

① $P_j\&le;P_s^1,$ $P_j\&le;P_s^2:$

${\mathrm{\&epsiv;}}_j=\frac{(P_s^1+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_s^2\&CenterDot;2^{2R_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(3\right)$

${\mathrm{\&Delta;}}_{j,1}=P_s^1+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(4\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(5\right)$

② $P_j\&GreaterEqual;P_s^1,$ $P_j\&GreaterEqual;P_s^2:$

${\mathrm{\&epsiv;}}_j=1-\frac{(P_s^2+{\mathrm{\&Delta;}}_s^2)({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^1)}{P_j({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_s^2-{\mathrm{\&Delta;}}_s^2)}---\left(6\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{\&Delta;}}_s^1---\left(7\right)$

${\mathrm{\&Delta;}}_{j,2}=P_s^2+{\mathrm{\&Delta;}}_s^2-(1-{\mathrm{\&epsiv;}}_j)P_j---\left(8\right)$

③ $P_j\&le;P_s^1,$ $P_j\&GreaterEqual;P_s^2:$

When $P_j\&GreaterEqual;g(P_s^2,{\mathrm{\&Delta;}}_s^2)+g(P_j-P_s^2,P_s^1+{\mathrm{\&Delta;}}_s^1-(P_j-P_s^2)),$ Then

${\mathrm{\&epsiv;}}_j=(1-P_s^2/P_j)---\left(9\right)$

${\mathrm{\&Delta;}}_{j,1}=f(R_j-g((1-{\mathrm{\&epsiv;}}_j)P_j,{\mathrm{\&Delta;}}_s^2),{\mathrm{\&epsiv;}}_j{P}_j)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(11\right)$

When $P_j\&le;g(P_s^2,{\mathrm{\&Delta;}}_s^2)+g(P_j-P_s^2,P_s^1+{\mathrm{\&Delta;}}_s^1-(P_j-P_s^2)),$ Then

${\mathrm{\&epsiv;}}_j=\frac{(P_s^1+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_s^2\&CenterDot;2^{2R_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(12\right)$

${\mathrm{\&Delta;}}_{j,1}=P_s^1+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(13\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(14\right)$

④ $P_j\&GreaterEqual;P_s^1,$ $P_j\&le;P_s^2:$

When $P_j\&GreaterEqual;g(P_s^1,{\mathrm{\&Delta;}}_s^1)+g(P_j-P_s^1,{\mathrm{\&Delta;}}_s^2),$ Then

${\mathrm{\&epsiv;}}_j=\frac{(P_s^1+{\mathrm{\&Delta;}}_s^1)({\mathrm{\&Delta;}}_s^1\&CenterDot;2^{2R_j}-P_j-{\mathrm{\&Delta;}}_s^2)}{P_j({\mathrm{\&Delta;}}_s^2\&CenterDot;2^{2R_j}-P_s^1-{\mathrm{\&Delta;}}_s^1)}---\left(15\right)$

${\mathrm{\&Delta;}}_{j,1}=P_s^1+{\mathrm{\&Delta;}}_s^1-{\mathrm{\&epsiv;}}_j{P}_j---\left(16\right)$

${\mathrm{\&Delta;}}_{j,2}={\mathrm{\&Delta;}}_s^2---\left(17\right)$

When $P_j\&le;g(P_s^1,{\mathrm{\&Delta;}}_s^1)+g(P_j-P_s^1,{\mathrm{\&Delta;}}_s^2),$ Then

${\mathrm{\&epsiv;}}_j=P_s^1/P_j---\left(18\right)$

${\mathrm{\&Delta;}}_{j,1}={\mathrm{\&Delta;}}_s^1---\left(19\right)$

${\mathrm{\&Delta;}}_{j,2}=f(R_j-g({\mathrm{\&epsiv;}}_j{P}_j,{\mathrm{\&Delta;}}_s^1),(1-{\mathrm{\&epsiv;}}_j)P_j)---\left(20\right)$

In the formula, ε _jBe the bundle factor of user j, Δ _{J, 1}And Δ _{J, 2}For user j split into corresponding respectively dried of two power block make an uproar with, according to two child nodes equal the power of father node with power, two child nodes equal the speed of father node with speed, the bundle factor of user i can correspondingly be tried to achieve;

For the father node that comprises a complete power block, satisfy

$P_s=P_s^1+P_s^2$

$R_s=R_s^1+R_s^2=\frac{1}{2}\log (1+\frac{P_s^1}{{\mathrm{\&Delta;}}_s^1})+\frac{1}{2}\log (1+\frac{P_s^2}{{\mathrm{\&Delta;}}_s^2})$

$P_s^2+{\mathrm{\&Delta;}}_s^2={\mathrm{\&Delta;}}_s^1$

Wherein, P _s ¹And P _s ²Be respectively the performance number of two power block, Δ _s ¹And Δ _s ²Be corresponding with two power block dried making an uproar and, R _s ¹And R _s ²Be the rate value of two power block, the bundle factor of user j can calculate according to formula (21)-(23):

${\mathrm{\&epsiv;}}_j=\frac{P_s+{\mathrm{\&Delta;}}_s-(P_i+{\mathrm{\&Delta;}}_i)}{P_j}---\left(21\right)$

Δ _j，1＝P _i+Δ _i (22)

Δ _j，2＝Δ _s (23)

In the formula, ε _jBe the bundle factor of user j, Δ _{J, 1}And Δ _{J, 2}For user j split into corresponding respectively dried of two power block make an uproar with, user i does not divide, its bundle factor ε _i=1, dried making an uproar with constant;

G) order of decoding one by one of calculating Virtual User:

According to e) in the division order of gained; Be used alternatingly f according to situation) in the computing formula of two types of father nodes user or ultra user are divided; Calculate corresponding dried of user's bundle factor and division user two power block make an uproar with; Splitting parameter until the user who obtains the leaf node representative that merges binary tree; Two power block after user's division are two Virtual User of respective user respectively, and Virtual User is deciphered order according to dried making an uproar with sorting from big to small one by one;

(2) configuration communication system parameters:

Send user side: for user i, i ∈ I is according to its bundle factor ε _i, power is divided into two parts ε _iP _i(1-ε _i) P _i, respectively according to speed g (ε _iP _i, Δ _{I, 1}) and g ((1-ε _iP _i), Δ _{I, 2}) transmission of encoding; Receiving terminal: according to g) decoding one by one of gained time ordered pair Virtual User is deciphered in.

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