CN110392378B - Compromise power distribution method in downlink multi-cluster NOMA system - Google Patents

Compromise power distribution method in downlink multi-cluster NOMA system Download PDF

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CN110392378B
CN110392378B CN201910777897.1A CN201910777897A CN110392378B CN 110392378 B CN110392378 B CN 110392378B CN 201910777897 A CN201910777897 A CN 201910777897A CN 110392378 B CN110392378 B CN 110392378B
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CN110392378A (en
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李兴旺
李晓静
张丹青
彭成章
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Henan University of Technology
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    • H—ELECTRICITY
    • H04—ELECTRIC COMMUNICATION TECHNIQUE
    • H04W—WIRELESS COMMUNICATION NETWORKS
    • H04W24/00—Supervisory, monitoring or testing arrangements
    • H04W24/02—Arrangements for optimising operational condition
    • H—ELECTRICITY
    • H04—ELECTRIC COMMUNICATION TECHNIQUE
    • H04W—WIRELESS COMMUNICATION NETWORKS
    • H04W72/00—Local resource management
    • H04W72/04—Wireless resource allocation
    • H04W72/044—Wireless resource allocation based on the type of the allocated resource
    • H04W72/0473—Wireless resource allocation based on the type of the allocated resource the resource being transmission power
    • H—ELECTRICITY
    • H04—ELECTRIC COMMUNICATION TECHNIQUE
    • H04W—WIRELESS COMMUNICATION NETWORKS
    • H04W72/00—Local resource management
    • H04W72/50—Allocation or scheduling criteria for wireless resources
    • H04W72/54—Allocation or scheduling criteria for wireless resources based on quality criteria
    • H04W72/542—Allocation or scheduling criteria for wireless resources based on quality criteria using measured or perceived quality
    • Y—GENERAL TAGGING OF NEW TECHNOLOGICAL DEVELOPMENTS; GENERAL TAGGING OF CROSS-SECTIONAL TECHNOLOGIES SPANNING OVER SEVERAL SECTIONS OF THE IPC; TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
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Abstract

The invention discloses a compromise power distribution method in a downlink multi-cluster NOMA system, which is suitable for a system comprising 1 base station andMKthe method comprises the steps that a downlink NOMA system of each user is provided, a base station and the users are both provided with a single antenna base station, the lowest power required by a single cluster and the maximum power which can be achieved by the single cluster are calculated according to channel conditions and the lowest speed requirement of each user, a power distribution optimization problem of compromise between the maximized speed and the energy efficiency is established, the power distribution of the compromise between the maximized speed and the energy efficiency in the single cluster is solved firstly, the relation between the maximum value of the compromise between the speed and the energy efficiency in the single cluster and the total power of the cluster is obtained, based on the result, the power distribution among the users in the optimization problem is converted into the power distribution among the clusters, the power distribution among the clusters is solved, and the power is distributed to each user according to the result.

Description

Compromise power distribution method in downlink multi-cluster NOMA system
Technical Field
The invention relates to the field of communication, in particular to a compromise power allocation method in a downlink multi-cluster NOMA system.
Background
With the rapid development of mobile communications, it has been difficult for conventional multiple access techniques to meet the explosive growth of wireless data traffic. Therefore, the fifth generation mobile communication employs a Non-Orthogonal Multiple Access (NOMA) technology with higher system throughput and higher spectral efficiency. Compared with the research of the traditional multiple access technology in time domain, frequency domain and code domain, the NOMA technology introduces a new dimension, namely power domain, distributes different powers for a plurality of users at a base station end, then superposes the signals of the users on the same time-frequency resource, and after receiving the signals, the users adopt the Successive Interference Cancellation (SIC) technology to detect the expected received signals. Power allocation not only relates to the detection order of each user signal, but also affects the reliability and effectiveness of the system, and therefore, power allocation in NOMA is one of the research hotspots in recent years.
Many documents have studied power allocation in single cell downlink NOMA systems, where the targets of power allocation are of three types: maximize sum rate, maximize energy efficiency, and maximize fairness. The power distribution scheme for maximizing the sum rate takes the total power or the rate of a single user as a constraint condition, solves the power distributed to each user by taking the sum rate of the maximized user as a target, and also solves the power distributed to each user by taking the ratio of the sum rate of the maximized user to the total power as a target by taking the sum rate of the maximized user or the rate of the single user as a constraint condition. In recent years, many documents have studied the relationship between energy efficiency and sum rate. The document "Energy efficiency and spectral-efficiency trade off in downlink NOMA systems" studies a compromised power allocation scheme in a multi-cluster NOMA system, however each cluster contains only two users.
Disclosure of Invention
The invention provides a compromise power allocation method in a downlink multi-cluster NOMA system, which is suitable for the downlink NOMA system comprising 1 base station and MK users, wherein the base station and the users are both provided with a single antenna.
The method comprises the steps of calculating the minimum power required by a single cluster and the maximum power which can be achieved by the single cluster by a base station according to channel conditions and the minimum speed requirement of each user, establishing an optimization problem of compromise between the maximized speed and the energy efficiency by taking the total power of the base station and the minimum speed requirement of the user as constraint conditions, solving power distribution of the maximized user in the single cluster and the compromise between the speed and the energy efficiency to obtain the relation between the maximum value of the compromise between the single cluster speed and the energy efficiency and the total power of the cluster, converting power distribution among the users in the optimization problem into power distribution among the clusters based on the result, solving power distribution among the clusters and distributing power for each user according to the result.
In summary, the compromised power allocation method in the downlink multi-cluster NOMA systems proposed in the present invention is applicable to downlink NOMA systems including 1 base station and MK users, and the base station and the users are both configured with a single antenna, and includes the following steps:
a, the base station clusters the users according to the channels from the base station to the MK users,each cluster comprises M users, which are divided into K clusters and ukmDenotes the mth user in the kth cluster, K1, 2kmIs hkm，|hk1|2≥|hk2|2≥…≥|hkM|2The base station allocates a sub-band for each cluster, and the sub-bands among the clusters are orthogonal;
b, the base station calculates the minimum power required by a single cluster and the maximum power which can be reached by the single cluster;
c, the base station establishes an optimization problem of compromise between the maximum rate and the energy efficiency, and simplifies the optimization problem;
d, the base station solves the simplified optimization problem in the step C to obtain the power distributed to the kth cluster;
e, the base station is uk1Distributing power
Figure BDA0002175115190000031
Base station is ukmDistribution power pkm0K1, 2,., K, M2, 3., M, K is the total number of clusters, M is the total number of users in each cluster.
Further, the step B specifically includes:
b1, with pkmIndicating a base station as ukmAllocated power, pk1≤pk2≤…≤pkM，
Figure BDA0002175115190000032
pkIs the total power allocated by the base station for the kth cluster, using rkmRepresents ukmThe minimum rate requirement of (a) is,
Figure BDA0002175115190000033
represents ukmMinimum rate requirement rkmThe corresponding signal-to-interference-and-noise ratio is calculated by the base station to obtain pkmIs taken to satisfy
Figure BDA0002175115190000034
Figure BDA0002175115190000035
σ2Is the variance of the noise received by the user, and therefore ukmThe minimum power required is
Figure BDA0002175115190000036
Figure BDA0002175115190000041
K is the total number of clusters, M is the total number of users in each cluster;
b2, the base station obtains p according to the step B1kmSatisfied condition and
Figure BDA0002175115190000042
calculating to obtain the lowest power p required by the kth clusterkmin，
Figure BDA0002175115190000043
Figure BDA0002175115190000044
Figure BDA0002175115190000045
PmaxIs the maximum transmit power of the base station, K is the total number of clusters, M is the total number of users in each cluster;
b3, the base station calculates the maximum power that the kth cluster can reach
Figure BDA0002175115190000046
K is the total number of clusters.
Further, the step C specifically includes:
c1, the base station sets up an optimization problem that maximizes the compromise of rate and energy efficiency,
Figure BDA0002175115190000047
wherein beta is a weighting factor for rate, 1-beta is a weighting factor for energy efficiency, 0 < beta < 1,
Figure BDA0002175115190000051
constraint conditions
Figure BDA0002175115190000052
Means that the sum of the power of all users cannot exceed PmaxConstraint pkm≥pkm0A rate requirement for guaranteeing users, K1, 2,., K, M1, 2., M, K being the total number of clusters, M being the total number of users in each cluster;
c2, the base station reduces the optimization problem in the formula (1) to,
Figure BDA0002175115190000053
wherein the content of the first and second substances,
Figure BDA0002175115190000054
Figure BDA0002175115190000055
constraint conditions
Figure BDA0002175115190000056
Means that the sum of the powers of all clusters cannot exceed PmaxConstraint pk≥pkminFor guaranteeing the user rate requirements of the kth cluster.
Further, the step D specifically includes:
d1, using
Figure BDA0002175115190000057
Expressed as the power allocated for the kth cluster, order
Figure BDA0002175115190000058
Figure BDA0002175115190000059
Figure BDA00021751151900000510
Wherein the content of the first and second substances,
Figure BDA0002175115190000061
Figure BDA0002175115190000062
for the kth cluster, the base station calculates
Figure BDA0002175115190000063
And
Figure BDA0002175115190000064
k is the total number of clusters;
d2, if ηk(pkmin) Is greater than 0 and
Figure BDA0002175115190000065
power is allocated to the kth cluster
Figure BDA0002175115190000066
If etak(pkmax) Is < 0 and
Figure BDA0002175115190000067
power is allocated to the kth cluster
Figure BDA0002175115190000068
K is the total number of clusters;
d3, if ηk(pkmin) Is greater than 0 and
Figure BDA0002175115190000069
then calculate λk(pkmax) And λk(pkmin)，λk(pkmin)＞λk(pkmax) Then, power is allocated to the cluster
Figure BDA00021751151900000610
Otherwise, not allocating power to the cluster, where K is 1,2, …, and K is the total number of clusters;
d4, if ηk(pkmax) Is < 0 and
Figure BDA00021751151900000611
calculating lambdak(pk) Maximum time pkBy taking the value of (a), pk' represents, let pkmax＝pk', K-1, 2, K being the total number of clusters;
d5, placing the clusters without power distribution into the set A;
d6, for any cluster a in the set A, if etaa(pamax) Greater than 0 and etaa(pamin) If the cluster is less than 0, the cluster is placed in a set B;
d7, for any cluster B in the set B, the base station solves etab(pb) When p is 0bBy taking the value of (a), pb *That is, if equation (3) is satisfied, power p is allocated to cluster bb＝pbminIf the cluster is deleted from the set A and the formula (4) is satisfied, the cluster is deleted in the section [ p ]bmin,pb *]To find out lambdab(pb) P corresponding to the maximum value ofbBy pb' to, compare λb(pbmax) And λb(pb'), if λb(pbmax)＜λb(pb') let pbmax＝pb′；
Figure BDA0002175115190000071
Figure BDA0002175115190000072
D8, calculating for the clusters in the set A
Figure BDA0002175115190000073
a belongs to A, and selects
Figure BDA0002175115190000074
Is represented by cluster m, and power is allocated to the cluster corresponding to the maximum value in the group
Figure BDA0002175115190000075
And deleting the cluster from the set A;
d9, if the sum of the power allocated to the clusters with allocated power is lower than PmaxThe remaining power is denoted by p ', if p' > p for any cluster a in set AamaxCalculating
Figure BDA0002175115190000076
Otherwise calculate
Figure BDA0002175115190000077
Select out
Figure BDA0002175115190000078
Is represented by cluster n, if p' > pnmaxA 1 is to pnmaxAllocating p 'to the cluster n, otherwise, allocating p' to the cluster n and deleting the cluster from the set A;
d10, repeating the step D9 until all power is allocated or the set A is empty.
Advantageous effects
The scheme disclosed by the invention expands the power distribution scheme with compromise between the maximum rate and the energy efficiency in the NOMA system to the scene that each cluster comprises a plurality of users, and under the condition of meeting the minimum rate requirement of all the users, the compromise between the maximum system and the rate and the energy efficiency is suitable for more scenes than the existing scheme.
Drawings
FIG. 1 is a system model of an implementation of the present invention;
fig. 2 is a flow chart of the present invention.
Detailed Description
An embodiment of the present invention is given below, and the present invention will be described in further detail. As shown in fig. 1, consider a single-cell downlink NOMA system including 1 base station and MK users,both the base station and the user configure a single antenna. The users are divided into K clusters, each cluster containing M users, ukmDenotes the mth user in the kth cluster, K being 1,2, …, K, M being 1,2, …, M. Base station to ukmIs hkm，|hk1|2≥|hk2|2≥…≥|hkM|2. The base station allocates the total power p for the kth clusterkWherein u iskmHas a power of pkm，pk1≤pk2≤…≤pkM，
Figure BDA0002175115190000081
The base station allocates a sub-band for each cluster, and the sub-bands among the clusters are orthogonal.
By ykmRepresents ukmOf the received signal, ykmIs expressed in the form of
Figure BDA0002175115190000082
Wherein x iskmIs ukmDesired received signal of nkmIs ukmReceived white Gaussian noise with mean value of zero and variance of sigma2。
uk1First detecting xkMAnd eliminating the signal pair yk1The interference caused, and then x is detectedk(M-1)And eliminating the signal pair yk1The interference caused by this, in turn, detects other signals and cancels these signal pairs yk1The interference caused until x is detectedk1。
ukmDetecting xkiThe Signal to Interference and Noise Ratio (SINR) is expressed as
Figure BDA0002175115190000083
ukmPer unit bandwidth rate Rkm(pkm) Is expressed in the form of
Figure BDA0002175115190000091
The sum of the unit bandwidth rates of all users in the kth cluster is
Figure BDA0002175115190000092
The sum of the unit bandwidth rates of MK users in the system is
Figure BDA0002175115190000093
Let r be0Is the minimum requirement for SINR when correctly detecting signals, rkmIs ukmSINR r corresponding to the lowest unit bandwidth rate requirementkm≥r0Therefore, the following equation is required to be satisfied
Figure BDA0002175115190000094
Figure BDA0002175115190000095
In formula (6), j is 1,2, …, M is 1,2, …, M. Thus can be derived, pkmIs taken to satisfy
Figure BDA0002175115190000096
Order to
Figure BDA0002175115190000097
j≤m，l(|hkj|2) Is | hkj|2A monotonically decreasing function. Due to | hk1|2≥|hk2|2≥…≥|hkM|2When j is m, l (| h)kj|2) A maximum value is reached. And because rkm≥r0Therefore, the formula (9) holds
Figure BDA0002175115190000098
In this case, the formula (8) can be represented as
Figure BDA0002175115190000101
Let the equality sign in equation (10) be true and m be 1 to obtain uk1Minimum power required pk10Is composed of
Figure BDA0002175115190000102
Let the equality sign in equation (10) be true and m be 2, yielding uk2Minimum power required pk20And pk1In a relationship of
Figure BDA0002175115190000103
Let the equality sign in equation (10) be true and m be 3 to obtain uk3Minimum power required pk30And pk1In a relationship of
Figure BDA0002175115190000104
Let the equality sign in equation (10) be true and m be 4, yielding uk4Minimum power required pk40And pk1In a relationship of
Figure BDA0002175115190000105
Let the equality sign in equation (10) be true and m be 5 to obtain uk5Minimum power required pk50And pk1In a relationship of
Figure BDA0002175115190000106
Obtained by induction method, M is 2,3, …, when M is ukmMinimum power required pkm0And pk1In a relationship of
Figure BDA0002175115190000111
When equation (16) is satisfied, ukmJust to the required minimum rate. U is obtained by bringing formula (11) into formula (16)kmMinimum power required pkm0Is composed of
Figure BDA0002175115190000112
So that the lowest total power p required for the kth clusterk0Is composed of
Figure BDA0002175115190000113
By PminRepresenting the minimum total power required to meet the minimum rate requirements of all users,
Figure BDA0002175115190000114
the rate vs. energy efficiency trade-off is formulated as
Figure BDA0002175115190000115
Where β is a weighting factor for rate, 1- β is a weighting factor for energy efficiency, 0 < β < 1.
From the document "On the optimization of power allocaFor normal downlink with induced qos constraints ", it is known that for a single cluster, the second to mth users in the cluster just meet the minimum rate requirement, while the remaining power is allocated to the first user, so as to maximize the sum rate of all users in the cluster. Thus, the total power of the kth cluster is pkThe base station is uk1Distributing power
Figure BDA0002175115190000116
Is ukmDistribution power pkm0K is 1,2, …, K, M is 2,3
Figure BDA0002175115190000121
Wherein the content of the first and second substances,
Figure BDA0002175115190000122
Figure BDA0002175115190000123
when the compromise between the speed and the energy efficiency in the cluster is maximum
Figure BDA0002175115190000124
The targets of the power allocation are: at a given total power Pmax(Pmax≥Pmin) And under the condition of meeting the speed requirement of each user, the inter-cluster power p is adjustedkK1, 2, …, K, a trade-off that maximizes the rate and energy efficiency of the system, formulated as
Figure BDA0002175115190000125
Wherein C1 represents that the total power of the base station is not higher than PmaxC2 is used to guarantee the minimum rate requirement of the user.
The optimal solution of the optimization problem in the formula (23) cannot be directly given, the optimization problem is decomposed into a plurality of subproblems, the power distribution which maximizes compromise between single cluster rate and energy efficiency is firstly solved, and the power distribution is expressed as a formula
Figure BDA0002175115190000131
The above optimization problem is to find λk(pk) Maximum time corresponding pkThe value of (a). Followed by analysis of lambdak(pk) The increase or decrease of (2). Lambda is foundk(pk) With respect to pkThe partial derivative of (a) of (b),
Figure BDA0002175115190000132
wherein, Deltak6＝Δk4-Δk3If, if
Figure BDA0002175115190000133
Then
Figure BDA0002175115190000134
Order to
Figure BDA0002175115190000135
And is
Figure BDA0002175115190000136
Then formula (26) can be written as
βηk(pk)＞θk(pk) (27)
Namely, it is
Figure BDA0002175115190000137
Equivalent to β ηk(pk)＞θk(pk). From the above analysis, λ can be derivedk(pk) The increase and decrease of (A) are as follows:
if etakIs greater than 0 and
Figure BDA0002175115190000138
λk(pk) Is pkIs a monotonically increasing function of.
If etakIs greater than 0 and
Figure BDA0002175115190000139
λk(pk) Is pkIs a monotonically decreasing function of (a).
If etakIs < 0 and
Figure BDA0002175115190000141
λk(pk) Is pkIs a monotonically increasing function of.
If etakIs < 0 and
Figure BDA0002175115190000142
λk(pk) Is pkIs a monotonically decreasing function of (a).
Separately calculate etak(pk) And thetak(pk) With respect to pkThe partial derivative of (a) of (b),
Figure BDA0002175115190000143
Figure BDA0002175115190000144
because of the fact that
Figure BDA0002175115190000145
And
Figure BDA0002175115190000146
is always true, so ηk(pk) And thetak(pk) Is pkIs a monotonically increasing function of. The derivation can be obtained by the following steps,
Figure BDA0002175115190000147
when the temperature of the water is higher than the set temperature,
Figure BDA0002175115190000148
is pkIs a monotonically decreasing function of (a).
To simplify the analysis, let
Figure BDA0002175115190000149
By pkmaxRepresents the maximum power of the k-th cluster,
Figure BDA00021751151900001410
next, p is analyzedk∈[pkmin,pkmax]When is lambdak(pk) The increase or decrease of (2).
If etak(pkmin) Is greater than 0 and
Figure BDA00021751151900001411
then p isk∈[pkmin,pkmax]When is lambdak(pk) Is pkIs a monotonically decreasing function of (a).
If etak(pkmin) Is greater than 0 and
Figure BDA00021751151900001412
then p isk∈[pkmin,pkmax]When is lambdak(pk) Is pkIs a monotonically increasing function of.
If etak(pkmin) Is greater than 0 and
Figure BDA00021751151900001413
then p isk∈[pkmin,pkmax]When, with pkIncrease of (a)k(pk) Is pkIs decreasing and then increasing.
If etak(pkmax) Is < 0 and
Figure BDA0002175115190000151
then p isk∈[pkmin,pkmax]When is lambdak(pk) Is pkIs a monotonically decreasing function of (a).
If etak(pkmax) Is < 0 and
Figure BDA0002175115190000152
then p isk∈[pkmin,pkmax]When is lambdak(pk) Is pkIs a monotonically increasing function of.
If etak(pkmax) Is < 0 and
Figure BDA0002175115190000153
then p isk∈[pkmin,pkmax]When, with pkIncrease of (a)k(pk) Is pkIs incremented and then decremented.
If etak(pkmax) Greater than 0 and etak(pkmin) < 0, there must be a unique pk *∈[pkmin,pkmax]So that ηk(pk *)＝0。
Due to ηk(pk) Is a monotonically increasing function, pk∈[pkmin,pk *]Then ηk(pk) Is less than 0. If it is
Figure BDA0002175115190000154
Then p isk∈[pkmin,pk *]When is lambdak(pk) Is pkIs a monotonically decreasing function of (a). If it is
Figure BDA0002175115190000155
Then p isk∈[pkmin,pk *]When is lambdak(pk) Is pkIs a monotonically increasing function of. If it is
Figure BDA0002175115190000156
Then p isk∈[pkmin,pk *]When, with pkIncrease of (a)k(pk) Is pkIs incremented and then decremented.
Due to ηk(pk) Is a monotonically increasing function, pk∈[pk *,pkmax]Then ηk(pk) Is greater than 0. If it is
Figure BDA0002175115190000157
Then p isk∈[pk *,pkmax]When is lambdak(pk) Is pkIs a monotonically increasing function of. If it is
Figure BDA0002175115190000158
Then p isk∈[pk *,pkmax]When is lambdak(pk) Is pkIs a monotonically decreasing function of (a). If it is
Figure BDA0002175115190000159
Then p isk∈[pk *,pkmax]When, with pkIncrease of (a)k(pk) Is pkIs decreasing and then increasing.
Table 1 shows pk∈[pkmin,pkmax]Time lambdak(pk) The increase or decrease of (2). Referring to table 1, the following conclusion holds:
pk∈[pkmin,pkmax]when condition 1 or condition 4 in table 1 is satisfied, i.e., λk(pk) Is pkIs assigned the lowest power p for the clusterkminThe trade-off between rate and energy efficiency of the cluster can be maximized.
pk∈[pkmin,pkmax]When condition 2 or condition 5 is satisfied, i.e. λk(pk) Is pkIs assigned the highest power p for the clusterkmaxThe trade-off between rate and energy efficiency of the cluster can be maximized.
pk∈[pkmin,pkmax]And when condition 3 is satisfied, the ratio is higherComparison of lambdak(pkmax) And λk(pkmin) If λk(pkmax)＞λk(pkmin) Then power p is allocated to the clusterkmaxOtherwise, allocating power pkmin。
pk∈[pkmin,pkmax]When condition 6 is satisfied, λ is foundk(pk) Maximum time pkBy taking the value of (a), pk' to indicate, the power p is allocated to the clusterk′。
If conditions 7 and 11 are satisfied simultaneously, λ isk(pk) Is pkIs a monotonically decreasing function of, assigns pkmin。
If conditions 8 and 10 are satisfied simultaneously, λ isk(pk) Is pkIs assigned p as a monotonically increasing function ofkmax。
If conditions 9 and 12 are satisfied simultaneously, λk(pk) Increment, decrement, and then increment, at pkmin,pk *]Finding lambda internallyk(pk) Maximum time pkBy taking the value of (a), pk' to, compare λk(pkmax) And λk(pk'), if λk(pkmax)＞λk(pk') then allocate power p for the clusterkmaxOtherwise, allocating power pk′。
Of the 12 conditions in table 1, only the above three possible combinations are simultaneously true, and no other combinations exist.
TABLE 1pk∈[pkmin,pkmax]Time lambdak(pk) The increase or decrease of (2).
Figure BDA0002175115190000171
Figure BDA0002175115190000181
The power allocation method for maximizing the compromise of rate and energy efficiency of a single cluster is given above. Next, a power allocation method between clusters is given to maximize compromise, and the specific steps are as follows:
step 1, first calculate p for each clusterkminAnd pkmaxCalculating
Figure BDA0002175115190000182
And
Figure BDA0002175115190000183
if etak(pkmax) Greater than 0 and etak(pkmin) If < 0, find out etak(pk) When p is 0kBy taking the value of (a), pk *And (4) showing.
Step 2, if the condition 1 or the condition 4 is met, directly distributing power p to the clusterkmin(ii) a If condition 3 is satisfied, λ is calculatedk(pkmax) And λk(pkmin) If λk(pkmin)＞λk(pkmax) Then the power p is allocated directly to the clusterkminOtherwise, putting the cluster into the set A and turning to the step 3; if class 6 is satisfied, then the calculation is such that λk(pk) Maximum time pkBy taking the value of (a), pk' represents, let pkmax＝pk', put the cluster into set A and go to step 3; if the classification 2 and the classification 5 are met, putting the cluster into the set A and turning to the step 3; if both condition 7 and condition 11 are satisfied, then power p is allocated directly to the clusterkmin(ii) a If the condition 8 and the condition 10 are met simultaneously, putting the cluster into the set A and turning to the step 3; if both condition 9 and condition 12 are satisfied, [ p ] iskmin,pk *]Finding lambda internallyk(pk) Maximum time pkBy taking the value of (a), pk' to, compare λk(pkmax) And λk(pk'), if λk(pkmax)＞λk(pk') put the cluster into set A and go to step 3 if lambdak(pkmax)＜λk(pk') let pkmax＝pk', willThe cluster is placed in set a and proceeds to step 3.
Step 3, for the clusters in the set A, calculating
Figure BDA0002175115190000191
a belongs to A, and selects
Figure BDA0002175115190000192
Is represented by cluster m, and power p is allocated to the clustermmaxAnd the cluster is deleted from set a.
Step 4, if the sum of the power distributed to the clusters with distributed power is lower than PmaxThe remaining power is denoted by p ', if p' > p for any cluster a in set AamaxCalculating
Figure BDA0002175115190000193
Otherwise calculate
Figure BDA0002175115190000194
Select out
Figure BDA0002175115190000195
Is represented by cluster n, if p' > pnmaxA 1 is to pnmaxAnd allocating p 'to the cluster n, otherwise, allocating p' to the cluster n, deleting the cluster from the set A, and repeating the steps until all power is allocated or the set A is an empty set.
The four steps are a power distribution method among clusters, and
Figure BDA0002175115190000196
represents the power allocated by the base station for the kth cluster, the base station is uk1Distributing power
Figure BDA0002175115190000197
Base station is ukmDistribution power pkm0K1, 2,., K, M2, 3., M, K is the total number of clusters, M is the total number of users in each cluster.
With reference to the flowchart of the present invention, that is, fig. 2, the specific steps of the compromise power allocation method in the downlink multi-cluster NOMA system are as follows:
a, a base station clusters users according to channels from the base station to MK users, each cluster comprises M users, the M users are divided into K clusters, and u is usedkmDenotes the mth user in the kth cluster, K1, 2, …, K, M1, 2, …, M, base station to ukmIs hkm，|hk1|2≤|hk2|2≤…≤|hkM|2The base station allocates a sub-band for each cluster, and the sub-bands among the clusters are orthogonal;
b, the base station calculates the minimum power required by a single cluster and the maximum power which can be reached by the single cluster;
c, the base station establishes an optimization problem of compromise between the maximum rate and the energy efficiency, and simplifies the optimization problem;
d, the base station solves the simplified optimization problem in the step C to obtain the power distributed to the kth cluster;
e, the base station is uk1Distributing power
Figure BDA0002175115190000201
Base station is ukmDistribution power pkm0K1, 2,., K, M2, 3., M, K is the total number of clusters, M is the total number of users in each cluster.
The above embodiments are merely illustrative of the present invention, and those skilled in the art can make various changes and modifications to the present invention without departing from the spirit and scope of the present invention. Thus, if such modifications and variations of the present invention fall within the scope of the claims of the present invention and their equivalents, the present invention is also intended to include such modifications and variations.

Claims (1)

1. The compromise power allocation method in the downlink multi-cluster NOMA system is characterized in that: the method is suitable for a downlink NOMA system comprising 1 base station and MK users, wherein the base station and the users are both provided with a single antenna, and the method comprises the following steps:
a, a base station clusters users according to channels from the base station to MK users, each cluster comprises M users and is divided into K clusters,by ukmDenotes the mth user in the kth cluster, K1, 2kmIs hkm，|hk1|2≥|hk2|2≥…≥|hkM|2The base station allocates a sub-band for each cluster, and the sub-bands among the clusters are orthogonal;
b, the base station calculates the minimum power required by a single cluster and the maximum power which can be reached by the single cluster, and the specific process is as follows:
b1, with pkmIndicating a base station as ukmAllocated power, pk1≤pk2≤…≤pkM，
Figure FDA0002175115180000011
pkIs the total power allocated by the base station for the kth cluster, using rkmRepresents ukmThe base station calculates p to obtain the Signal to Interference and Noise Ratio (SINR) corresponding to the minimum rate requirementkmIs taken to satisfy
Figure FDA0002175115180000012
σ2Is the variance of the noise received by the user, and therefore ukmThe minimum power required is
Figure FDA0002175115180000013
Figure FDA0002175115180000014
K is the total number of clusters, M is the total number of users in each cluster;
b2, the base station obtains p according to the step B1kmSatisfied condition and
Figure FDA0002175115180000015
calculating to obtain the lowest power p required by the kth clusterkmin，
Figure FDA0002175115180000021
Figure FDA0002175115180000022
PmaxIs the maximum transmit power of the base station, K is the total number of clusters, M is the total number of users in each cluster;
b3, the base station calculates the maximum power that the kth cluster can reach
Figure FDA0002175115180000023
K is the total number of clusters;
c, the base station establishes an optimization problem of compromise between the maximum rate and the energy efficiency, and simplifies the optimization problem, and the specific process is as follows:
c1, the base station sets up an optimization problem that maximizes the compromise of rate and energy efficiency,
Figure FDA0002175115180000024
wherein beta is a weighting factor for rate, 1-beta is a weighting factor for energy efficiency, 0 < beta < 1,
Figure FDA0002175115180000025
constraint conditions
Figure FDA0002175115180000026
Means that the sum of the power of all users cannot exceed PmaxConstraint pkm≥pkm0For guaranteeing the rate requirements of the user;
c2, the base station reduces the optimization problem in the formula (1) to,
Figure FDA0002175115180000031
wherein the content of the first and second substances,
Figure FDA0002175115180000032
Figure FDA0002175115180000033
constraint conditions
Figure FDA0002175115180000034
Means that the sum of the powers of all clusters cannot exceed PmaxConstraint pk≥pkminUser rate requirements for guaranteeing a kth cluster;
d, the base station solves the simplified optimization problem in the step C formula (2) to obtain the power distributed to the kth cluster
Figure FDA0002175115180000035
K is the total number of clusters, and the specific process is as follows:
d1, using
Figure FDA0002175115180000036
Expressed as the power allocated for the kth cluster, order
Figure FDA0002175115180000037
Figure FDA0002175115180000038
Figure FDA0002175115180000039
Wherein, Deltak6＝Δk4-Δk3，
Figure FDA00021751151800000310
Figure FDA00021751151800000311
For the kth cluster, the base station calculates
Figure FDA00021751151800000312
And
Figure FDA0002175115180000041
k is the total number of clusters;
d2, if ηk(pkmin) Is greater than 0 and
Figure FDA0002175115180000042
power is allocated to the kth cluster
Figure FDA0002175115180000043
If etak(pkmax) Is < 0 and
Figure FDA0002175115180000044
power is allocated to the kth cluster
Figure FDA0002175115180000045
K is the total number of clusters;
d3, if ηk(pkmin) Is greater than 0 and
Figure FDA0002175115180000046
then calculate λk(pkmax) And λk(pkmin)，λk(pkmin)＞λk(pkmax) Then, power is allocated to the cluster
Figure FDA0002175115180000047
Otherwise, no power is allocated for the cluster, K being 1, 2., K being the total number of clusters;
d4, if ηk(pkmax) Is < 0 and
Figure FDA0002175115180000048
calculating lambdak(pk) Maximum time pkBy taking the value of (a), pk' represents, let pkmax＝pk', K-1, 2, K being the total of the clusterCounting;
d5, placing the clusters without power distribution into the set A;
d6, for any cluster a in the set A, if etaa(pamax) Greater than 0 and etaa(pamin) If the cluster is less than 0, the cluster is placed in a set B;
d7, for any cluster B in the set B, the base station solves etab(pb) When p is 0bBy taking the value of (a), pb *That is, if equation (3) is satisfied, power is allocated to cluster b
Figure FDA0002175115180000049
Deleting the cluster from the set A, and if the formula (4) is satisfied, determining that the cluster is in the interval [ p ]bmin,pb *]To find out lambdab(pb) P corresponding to the maximum value ofbBy pb' to, compare λb(pbmax) And λb(pb'), if λb(pbmax)＜λb(pb') let pbmax＝pb′；
Figure FDA00021751151800000410
Figure FDA0002175115180000051
D8, calculating for the clusters in the set A
Figure FDA0002175115180000052
Select out
Figure FDA0002175115180000053
Is represented by cluster m, and power is allocated to the cluster corresponding to the maximum value in the group
Figure FDA0002175115180000054
And the cluster is selected from the set ADeleting;
d9, if the sum of the power allocated to the clusters with allocated power is lower than PmaxThe remaining power is denoted by p ', if p' > p for any cluster a in set AamaxCalculating
Figure FDA0002175115180000055
Otherwise calculate
Figure FDA0002175115180000056
Select out
Figure FDA0002175115180000057
Is represented by cluster n, if p' > pnmaxA 1 is to pnmaxTo a cluster n, i.e.
Figure FDA0002175115180000058
Otherwise, allocating p' to the cluster n and deleting the cluster from the set A;
d10, repeating the step D9 until all power is allocated or the set A is an empty set;
e, the base station is uk1Distributing power
Figure FDA0002175115180000059
Base station is ukmDistribution power pkm0K1, 2,., K, M2, 3., M, K is the total number of clusters, M is the total number of users in each cluster.
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