CN110807248A - Grouping judgment method for spacecraft cluster - Google Patents

Grouping judgment method for spacecraft cluster Download PDF

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CN110807248A
CN110807248A CN201910979035.7A CN201910979035A CN110807248A CN 110807248 A CN110807248 A CN 110807248A CN 201910979035 A CN201910979035 A CN 201910979035A CN 110807248 A CN110807248 A CN 110807248A
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topology
subgraph
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党朝辉
罗建军
周昊
代洪华
王明明
马卫华
孙冲
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Northwestern Polytechnical University
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Abstract

The invention discloses a method for judging the clustering of spacecraft clusters, which comprises the following steps: 1, inputting an inter-satellite communication relation, and establishing a communication topology subgraph among multiple spacecrafts; 2, inputting relative navigation relations and establishing relative navigation topological subgraphs among the multiple spacecrafts; 3, combining the communication topological subgraph and the relative navigation topological subgraph to construct an information topological graph among the multiple spacecrafts; and 4, constructing an adjacent matrix corresponding to the information topological graph, and judging whether the multiple spacecrafts are clustered or not through existence of the directed spanning tree. The judgment method established by the invention considers the essential equivalence of inter-satellite communication and relative navigation in the problem of cluster formation of multiple spacecrafts, thereby being more suitable for the practical application of spacecraft clusters.

Description

一种航天器集群的成群判定方法A method for swarm determination of spacecraft swarms

技术领域technical field

本发明涉及航空航天技术领域,尤其涉及一种航天器集群的成群条件判定方法。The invention relates to the technical field of aerospace, in particular to a method for judging group conditions of spacecraft clusters.

背景技术Background technique

航天器集群是近年来航天领域新出现的一种分布式卫星系统概念。典型的航天器集群飞行计划有美国的F6计划、以色列的SAMSON计划等。尽管航天器集群的概念已产生了近十年,但集群的定义仍很模糊。现有航天器集群的描述体系中,通常将集群定义为“近距离伴飞”的一群航天器。然而,在工程实践中,经常会出现尽管距离很近但实际上不属于同一个集群的航天器。相反,即便一些航天器彼此之间离得很远,但它们能够通过航天器间通信的“多跳”保持信息的沟通,因而在本质上仍是同一个集群。这就说明,采用距离来界定多个航天器是否属于同一个集群是不严谨的。为此,有文献提出将航天器集群定义为“通过星间无线通信网络互联形成的多航天器分布式系统”。但这样的定义仍存在问题,因为星间通信并非构成集群的充要条件。已有文献指出,采用相对导航替代部分或全部星间通信也能实现集群飞行。对当前研究现状深入分析可知,造成集群定义模糊、不准确的主要原因在于没有给出多航天器成群的判定方法。所谓成群,即多个航天器构成同一个集群。由于缺乏这样的成群条件判断方法,这使得航天器集群的在轨应用存在诸多困难。Spacecraft cluster is a new concept of distributed satellite system in the aerospace field in recent years. Typical spacecraft cluster flight plans include the F6 plan of the United States and the SAMSON plan of Israel. Although the concept of a spacecraft swarm has been around for nearly a decade, the definition of a swarm remains vague. In the description system of existing spacecraft clusters, clusters are usually defined as a group of spacecraft that "fly at close range". In engineering practice, however, there are often spacecraft that, despite their close proximity, do not actually belong to the same cluster. Conversely, even though some spacecraft are far apart from each other, they are able to maintain information communication through the "multi-hop" of inter-spacecraft communication and thus remain essentially the same cluster. This shows that it is not rigorous to use distance to define whether multiple spacecraft belong to the same cluster. For this reason, some literature proposes to define a spacecraft cluster as "a multi-spacecraft distributed system formed by the interconnection of inter-satellite wireless communication networks". But such a definition is still problematic, because inter-satellite communication is not a necessary and sufficient condition to form a cluster. Existing literature points out that the use of relative navigation to replace part or all of the inter-satellite communication can also achieve swarm flight. In-depth analysis of the current research status shows that the main reason for the vague and inaccurate definition of clusters is that there is no method for judging clusters of multiple spacecraft. The so-called swarm, that is, multiple spacecraft form the same cluster. Due to the lack of such a swarm condition judgment method, there are many difficulties in the on-orbit application of spacecraft swarms.

为克服上述问题,有必要建立一种有效的多航天器成群判定方法,通过对多航天器信息交互拓扑在数学上的定量计算实现多航天器是否成群的量化判定,从而为航天器集群飞行的在轨应用奠定基础。In order to overcome the above problems, it is necessary to establish an effective multi-spacecraft group determination method, through the mathematical quantitative calculation of the multi-spacecraft information interaction topology to realize the quantitative determination of whether the multi-spacecraft is in a group or not, so as to provide a basis for the spacecraft cluster. Lay the groundwork for on-orbit applications of flight.

发明内容SUMMARY OF THE INVENTION

针对现有技术中无法有效完成多航天器成群判定的问题,本发明的目的在于提供一种航天器集群的成群判定方法,本发明的方法从而为航天器集群的在轨应用提供必要的理论基础,从而解决现有技术中不能量化确定多个航天器是否属于同一集群的困难。Aiming at the problem that the multi-spacecraft cluster determination cannot be effectively completed in the prior art, the purpose of the present invention is to provide a cluster determination method for a spacecraft cluster. Theoretical basis, so as to solve the difficulty of quantitatively determining whether multiple spacecraft belong to the same cluster in the existing technology.

为了实现上述目的,本发明采用的技术方案如下:In order to achieve the above object, the technical scheme adopted in the present invention is as follows:

一种航天器集群的成群判定方法,包括步骤如下:A swarm determination method for a spacecraft swarm, comprising the following steps:

S1,获取星间通信关系,建立多航天器间的通信拓扑子图;S1, obtain the inter-satellite communication relationship, and establish a communication topology subgraph between multiple spacecraft;

S2,获取相对导航关系,建立多航天器间的相对导航拓扑子图;S2, obtain a relative navigation relationship, and establish a relative navigation topology subgraph among multiple spacecraft;

S3,联合通信拓扑子图和相对导航拓扑子图,构建多航天器间的信息拓扑图;S3, combine the communication topology sub-map and the relative navigation topology sub-map to construct an information topology map between multiple spacecraft;

S4,构建信息拓扑图对应的邻接矩阵,并通过有向生成树的存在性判定多航天器是否成群。S4, construct the adjacency matrix corresponding to the information topology graph, and determine whether the multi-spacecraft is in a group through the existence of the directed spanning tree.

所述S1中的通信拓扑子图是由代表航天器序号的节点构成的节点集合V与具有星间通信关系的边集合E1联合形成的二元组G1(V,E1);The communication topology subgraph in the S1 is a binary group G 1 (V, E 1 ) that is jointly formed by a node set V composed of nodes representing spacecraft serial numbers and an edge set E 1 with an inter-satellite communication relationship;

其中,通信拓扑子图的节点集合V={vi|i=1,…,N},其中N为航天器的数量,vi表示节点,为1到N的自然数;通信拓扑子图的边集合E1={eij|(vi∈V;vj∈V;i≠j)},其中边eij表示节点vi是否指向vj进行了单向通信,若是则边eij=1,否则eij=0。Among them, the node set V={v i |i=1,...,N} of the communication topology subgraph, where N is the number of spacecraft, and v i represents the node, which is a natural number from 1 to N; the edge of the communication topology subgraph Set E 1 ={e ij |(vi ∈V; v j ∈ V; i≠ j )}, where edge e ij indicates whether node v i points to v j for unidirectional communication, and if so, edge e ij =1 , otherwise e ij =0.

所述S2中的相对导航拓扑子图是由代表航天器序号的节点构成的节点集合V与具有相对导航关系的边集合E2联合形成的二元组G2(V,E2);The relative navigation topology subgraph in S2 is a binary group G 2 (V, E 2 ) formed by the joint formation of a node set V composed of nodes representing the spacecraft serial numbers and an edge set E 2 with a relative navigation relationship;

其中,相对导航拓扑子图的节点集合与通信拓扑子图的节点集合相同,相对导航拓扑子图的边集合E2={eij|(vi∈V;vj∈V;i≠j)},其中边eij表示节点vj是否对节点vi进行了相对导航测量,若是则边eij=1,否则eij=0。Among them, the node set of the relative navigation topology subgraph is the same as the node set of the communication topology subgraph, and the edge set of the relative navigation topology subgraph E 2 ={e ij |(vi ∈V; v j ∈V; i≠ j ) }, where edge e ij represents whether node v j has performed relative navigation measurement to node v i , if so, edge e ij =1, otherwise e ij =0.

所述S3中的信息拓扑图G(V,E)是通信拓扑子图与相对导航拓扑子图的联合,即G=G1∪G2The information topology graph G(V, E) in S3 is the combination of the communication topology subgraph and the relative navigation topology subgraph, that is, G=G 1 ∪ G 2 .

信息拓扑图的节点集合等价于通信拓扑子图的节点集合,也等价于相对导航拓扑子图的节点集合。The node set of the information topology graph is equivalent to the node set of the communication topology subgraph, and it is also equivalent to the node set of the relative navigation topology subgraph.

通信拓扑图的边集合E1与相对导航图的边集合E2形成的联合构成了信息拓扑图的边集合,即E=E1∪E2The union formed by the edge set E 1 of the communication topology graph and the edge set E 2 of the relative navigation graph constitutes the edge set of the information topology graph, that is, E=E 1 ∪ E 2 .

所述S4中的信息拓扑图的邻接矩阵A,是与信息拓扑图G(V,E)一一对应的N阶方阵;其非对角线上的第i行第j列的元素aij=eij(i≠j),对角线上的元素aii=0(vi∈V)。The adjacency matrix A of the information topology map in S4 is an N-order square matrix corresponding to the information topology map G(V, E) one-to-one; the elements a ij in the i-th row and the j-th column on the off-diagonal =e ij (i≠j), the elements on the diagonal a ii =0(vi ∈V ).

所述多航天器是否属于同一个集群是由信息拓扑图G是否包含有向生成树决定的:Whether the multiple spacecraft belong to the same cluster is determined by whether the information topology graph G contains a directed spanning tree:

当G包含有向生成树时N个航天器属于同一个集群;当G不包含有向生成树时N个航天器不属于同一个集群。When G contains a directed spanning tree, N spacecraft belong to the same cluster; when G does not contain a directed spanning tree, N spacecraft do not belong to the same cluster.

所述信息拓扑图G是否包含有向生成树是通过邻接矩阵A来判定的,具体判定的步骤如下:Whether the information topology graph G contains a directed spanning tree is determined by the adjacency matrix A, and the specific determination steps are as follows:

S401,计算矩阵A的1~N-1阶幂:A、A2、A3、……、AN-1S401, calculate the 1-N-1 order power of the matrix A: A, A 2 , A 3 , ..., A N-1 ;

S402,将所有N-1阶幂求和得到幂和矩阵

Figure BDA0002234581850000031
S402, sum all N-1 order powers to obtain a power sum matrix
Figure BDA0002234581850000031

S403,检查和矩阵A的所有元素:S403, check all elements of the sum matrix A∑:

若存在某一行(记作第j行,1≤j≤N)除了对角线外所有元素均大于零,即

Figure BDA0002234581850000032
则信息拓扑图G包含有向生成树;若对于任意一行,总是存在某个非对角线元素为零,则信息拓扑图G不包含有向生成树。If there is a row (denoted as the jth row, 1≤j≤N), all elements except the diagonal are greater than zero, that is
Figure BDA0002234581850000032
Then the information topology graph G contains a directed spanning tree; if for any row, there is always a non-diagonal element that is zero, then the information topology graph G does not contain a directed spanning tree.

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

本发明先通过联合通信拓扑子图和相对导航拓扑子图,构建多航天器间的信息拓扑图,然后构建信息拓扑图对应的邻接矩阵,进一步通过邻接矩阵计算有向生成树的存在性并据此判定多航天器是否成群。该方法给出了定量计算的步骤,因此克服了现有文献无法对多航天器是否属于同一集群进行量化判断的困难。此外,该方法在具体计算过程中充分利用了邻接矩阵与拓扑关系的等价性,因而将判断拓扑是否存在生成树的较难的问题转化为了寻找幂和矩阵是否存在非对角线零元素的简单问题。故该方法能够快速且量化确定具有复杂信息交互关系的多个航天器是否属于同一集群,进而为清晰定义集群和设计集群协同控制方法奠定理论基础。本发明所建立的判定方法,同时考虑了星间通信与相对导航在多航天器构成集群问题中的本质等价性,因而更适合航天器集群的实际应用。The invention firstly constructs the information topology map among multiple spacecraft by combining the communication topology submap and the relative navigation topology submap, then constructs the adjacency matrix corresponding to the information topology map, and further calculates the existence of the directed spanning tree through the adjacency matrix and calculates the existence of the directed spanning tree according to the data. This determines whether multiple spacecraft are swarmed. This method gives the steps of quantitative calculation, so it overcomes the difficulty that the existing literature cannot quantitatively judge whether multiple spacecraft belong to the same cluster. In addition, this method makes full use of the equivalence between the adjacency matrix and the topological relationship in the specific calculation process, so the difficult problem of judging whether there is a spanning tree in the topology is transformed into the problem of finding whether there are off-diagonal zero elements in the power sum matrix. Simple question. Therefore, this method can quickly and quantitatively determine whether multiple spacecraft with complex information interaction relationships belong to the same cluster, thereby laying a theoretical foundation for clearly defining clusters and designing cluster collaborative control methods. The determination method established by the present invention also considers the essential equivalence of inter-satellite communication and relative navigation in the problem of multi-spacecraft cluster formation, so it is more suitable for the practical application of spacecraft clusters.

附图说明Description of drawings

图1是本发明的一种航天器集群的成群判定方法的流程示意图;1 is a schematic flowchart of a method for determining a group of a spacecraft cluster according to the present invention;

图2是具体实施例中的三个航天器的星间通信和相对导航关系;Fig. 2 is the inter-satellite communication and relative navigation relationship of three spacecraft in the specific embodiment;

图3是具体实施例中的三个航天器的星间通信拓扑子图;Fig. 3 is the inter-satellite communication topology subgraph of three spacecraft in the specific embodiment;

图4是具体实施例中的三个航天器的相对导航拓扑子图;Fig. 4 is the relative navigation topology subgraph of three spacecraft in the specific embodiment;

图5是具体实施例中的三个航天器的信息拓扑图。FIG. 5 is an information topology diagram of three spacecraft in a specific embodiment.

具体实施方式Detailed ways

为了使本技术领域的人员更好地理解本发明中的技术方案,下面将结合本发明实施例中的附图,对本发明实施例中的技术方案进行清楚、完整地描述,显然,所描述的实施例仅仅是本发明一部分实施例,而不是全部的实施例。基于本发明中的实施例,本领域普通技术人员在没有做出创造性劳动的前提下所获得的所有其他实施例,都应当属于本发明保护的范围。In order to make those skilled in the art better understand the technical solutions of the present invention, the technical solutions in the embodiments of the present invention will be clearly and completely described below with reference to the accompanying drawings in the embodiments of the present invention. Obviously, the described The embodiments are only some of the embodiments of the present invention, but not all of the embodiments. Based on the embodiments of the present invention, all other embodiments obtained by persons of ordinary skill in the art without creative work shall fall within the protection scope of the present invention.

除非另有定义,本文所使用的所有的技术和科学术语与属于本发明的技术领域的技术人员通常理解的含义相同。本文中在本发明的说明书中所使用的术语只是为了描述具体的实施例的目的,不是旨在于限制本发明。本文所使用的术语“和/或”包括一个或多个相关的所列项目的任意的和所有的组合。Unless otherwise defined, all technical and scientific terms used herein have the same meaning as commonly understood by one of ordinary skill in the art to which this invention belongs. The terms used herein in the description of the present invention are for the purpose of describing specific embodiments only, and are not intended to limit the present invention. As used herein, the term "and/or" includes any and all combinations of one or more of the associated listed items.

如图1所示,本发明一种航天器集群的成群判定方法,包括以下步骤:As shown in FIG. 1 , a method for judging groups of spacecraft clusters of the present invention includes the following steps:

步骤1,输入星间通信关系,建立多航天器间的通信拓扑子图;Step 1, enter the inter-satellite communication relationship, and establish a communication topology subgraph between multiple spacecraft;

步骤2,输入相对导航关系,建立多航天器间的相对导航拓扑子图;Step 2, input the relative navigation relationship, and establish the relative navigation topology subgraph between multiple spacecraft;

步骤3,联合通信拓扑子图和相对导航拓扑子图,构建多航天器间的信息拓扑图;Step 3, combine the communication topology sub-map and the relative navigation topology sub-map to construct an information topology map between multiple spacecraft;

步骤4,构建信息拓扑图对应的邻接矩阵,并通过有向生成树的存在性判定多航天器是否成群。Step 4, construct an adjacency matrix corresponding to the information topology graph, and determine whether multiple spacecraft are grouped by the existence of a directed spanning tree.

作为优选地实施例,所述步骤1中的通信拓扑子图是由代表航天器序号的节点构成的节点集合V与具有星间通信关系的边集合E1联合形成的二元组G1(V,E1)。As a preferred embodiment, the communication topology subgraph in the step 1 is a binary group G 1 ( V ,E1 ) .

其中,通信拓扑子图的节点集合V={vi|i=1,…,N},其中N为航天器的数量,vi表示节点,为1到N的自然数。通信拓扑子图的边集合E1={eij|(vi∈V;vj∈V;i≠j)},其中边eij表示节点vi是否指向vj进行了单向通信,若是则边eij=1,否则eij=0。Among them, the node set V={v i |i=1,...,N} of the communication topology subgraph, where N is the number of spacecraft, and v i represents nodes, which are natural numbers from 1 to N. The edge set E 1 ={e ij |(vi ∈V; v j ∈ V; i≠ j )} of the communication topology subgraph, where the edge e ij indicates whether the node v i points to v j for one-way communication, and if so Then edge e ij =1, otherwise e ij =0.

作为优选地实施例,所述步骤2中的相对导航拓扑子图是由代表航天器序号的节点构成的节点集合V与具有相对导航关系的边集合E2联合形成的二元组G2(V,E2)。As a preferred embodiment, the relative navigation topology subgraph in step 2 is a binary group G 2 ( V , E2 ).

其中,相对导航拓扑子图的节点集合与通信拓扑子图的节点集合相同。The node set of the relative navigation topology subgraph is the same as the node set of the communication topology subgraph.

相对导航拓扑子图的边集合E2={eij|(vi∈V;vj∈V;i≠j)},其中边eij表示节点vj是否对节点vi进行了相对导航测量,若是则边eij=1,否则eij=0。The edge set E 2 ={e ij |(vi ∈V; v j ∈ V; i≠ j )} of the relative navigation topology subgraph, where the edge e ij represents whether the node v j has performed relative navigation measurements to the node v i , if the edge e ij =1, otherwise e ij =0.

作为优选的实施例,所述步骤3中的信息拓扑图G(V,E)是通信拓扑子图与相对导航拓扑子图的联合,即G=G1∪G2As a preferred embodiment, the information topology graph G(V, E) in the step 3 is a combination of the communication topology subgraph and the relative navigation topology subgraph, that is, G=G 1 ∪ G 2 .

优选的,信息拓扑图的节点集合等价于通信拓扑子图的节点集合,也等价于相对导航拓扑子图的节点集合。通信图的边集合E1与相对导航图的边集合E2形成的联合构成了信息拓扑图的边集合,即E=E1∪E2Preferably, the node set of the information topology graph is equivalent to the node set of the communication topology subgraph, and is also equivalent to the node set of the relative navigation topology subgraph. The union formed by the edge set E 1 of the communication graph and the edge set E 2 of the relative navigation graph constitutes the edge set of the information topology graph, that is, E=E 1 ∪ E 2 .

作为优选的实施例,所述步骤4中的信息拓扑图的邻接矩阵A,是与信息拓扑图G(V,E)一一对应的N阶方阵;其非对角线上的第i行第j列的元素aij=eij(i≠j),对角线上的元素aii=0(vi∈V)。As a preferred embodiment, the adjacency matrix A of the information topology map in the step 4 is an N-order square matrix corresponding to the information topology map G(V, E) one-to-one; The element in the jth column a ij =e ij (i≠j), the element on the diagonal a ii =0 (vi ∈V ).

优选的,所述N个航天器是否属于同一个集群是由信息拓扑图G是否包含有向生成树决定的;当G包含有向生成树时N个航天器属于同一个集群;当G不包含有向生成树时N个航天器不属于同一个集群。Preferably, whether the N spacecraft belong to the same cluster is determined by whether the information topology graph G includes a directed spanning tree; when G includes a directed spanning tree, N spacecraft belong to the same cluster; when G does not include a directed spanning tree In a directed spanning tree, N spacecraft do not belong to the same cluster.

优选的,所述信息拓扑图G是否包含有向生成树是通过邻接矩阵A来判定的;其判定的步骤如下:Preferably, whether the information topology graph G contains a directed spanning tree is determined by the adjacency matrix A; the steps of its determination are as follows:

S1,计算矩阵A的1~N-1阶幂:A、A2、A3、……、AN-1S1, calculate the 1-N-1 order power of the matrix A: A, A 2 , A 3 , ..., A N-1 ;

S2,将所有N-1阶幂求和得到幂和矩阵 S2, sum all N-1 order powers to get the power sum matrix

S3,检查和矩阵A的所有元素;若存在某一行(记作第j行,1≤j≤N)除了对角线外所有元素均大于零,即

Figure BDA0002234581850000062
则信息拓扑图G包含有向生成树;若对于任意一行,总是存在某个非对角线元素为零,则信息拓扑图G不包含有向生成树。S3, check all elements of the sum matrix A ; if there is a row (denoted as the jth row, 1≤j≤N), all elements except the diagonal are greater than zero, that is
Figure BDA0002234581850000062
Then the information topology graph G contains a directed spanning tree; if for any row, there is always a non-diagonal element that is zero, then the information topology graph G does not contain a directed spanning tree.

实施例1Example 1

如图2所示,3个航天器的星间通信关系为:航天器1和3之间能够进行双向通信,但航天器2不与其它航天器进行任何通信;3个航天器的相对导航关系为:航天器1和3分别可对航天器2进行相对导航测量。下面采用本发明的方法判定这三个航天器是否构成同一个集群。As shown in Figure 2, the inter-satellite communication relationship of the three spacecraft is: two-way communication is possible between spacecraft 1 and 3, but spacecraft 2 does not communicate with other spacecraft; the relative navigation relationship of the three spacecraft For: spacecraft 1 and 3 can respectively perform relative navigation measurements on spacecraft 2. The method of the present invention is used below to determine whether the three spacecraft form the same cluster.

步骤1,根据星间通信关系,可建立通信拓扑子图G1(V,E1)如图3所示。其中,V={1,2,3},E1={e13,e31}。Step 1, according to the inter-satellite communication relationship, a communication topology subgraph G 1 (V, E 1 ) can be established as shown in FIG. 3 . Wherein, V={1,2,3}, E1 = { e13 , e31 }.

步骤2,根据相对导航关系,可建立相对导航拓扑子图如图4所示。其中,V={1,2,3},E2={e21,e23}Step 2, according to the relative navigation relationship, a relative navigation topology subgraph can be established as shown in FIG. 4 . where, V={1,2,3}, E 2 ={e 21 ,e 23 }

步骤3,将通信拓扑子图与相对导航拓扑子图进行联合,得到信息拓扑图如图5所示。其中,V={1,2,3},E=E1∪E2={e13,e31,e21,e23}。In step 3, the communication topology subgraph and the relative navigation topology subgraph are combined to obtain the information topology graph as shown in FIG. 5 . Wherein, V={1, 2, 3}, E=E 1 ∪ E 2 ={e 13 , e 31 , e 21 , e 23 }.

步骤4,根据图5中的信息拓扑图具体形式,得邻接矩阵为:Step 4, according to the specific form of the information topology diagram in Figure 5, the adjacency matrix is obtained as:

Figure BDA0002234581850000071
Figure BDA0002234581850000071

进一步,按照如下方法判断信息拓扑图G是否包含有向生成树:Further, determine whether the information topology graph G contains a directed spanning tree according to the following method:

S1,计算矩阵A的1~2阶幂:S1, calculate the 1st to 2nd order power of matrix A:

Figure BDA0002234581850000072
Figure BDA0002234581850000072

S2,将所有2阶幂求和得到和矩阵

Figure BDA0002234581850000073
S2, sum all 2nd order powers to get the sum matrix
Figure BDA0002234581850000073

Figure BDA0002234581850000074
Figure BDA0002234581850000074

S3,检查矩阵A的除对角线外的所有元素a∑,ij(i∈V;j∈V;i≠j);由于第2行除对角线元素外其它元素均大于零,因此信息拓扑图G包含有向生成树。S3, check all elements a ∑,ij (i∈V; j∈V; i≠j) of the matrix A except the diagonal elements; since all elements in the second row except the diagonal elements are greater than zero, so The information topology graph G contains a directed spanning tree.

由于信息拓扑图G包含有向生成树,因此上述三个航天器在给定的星间通信和相对导航交互关系的基础上构成集群。Since the information topology graph G contains a directed spanning tree, the above three spacecraft form clusters based on the given inter-satellite communication and relative navigation interactions.

为了更进一步说明本发明方法的有益效果,在步骤4中对另外两种假设情形进行成群判断。In order to further illustrate the beneficial effect of the method of the present invention, in step 4, group judgment is performed on the other two hypothetical situations.

第一种假设情形为上述三个航天器仅有星间通信(如图2实线箭头所示),The first assumption is that the above three spacecraft only have inter-satellite communication (as shown by the solid arrows in Figure 2),

第二种假设情形为上述三个航天器仅有相对导航(如图2虚线箭头所示)。The second hypothetical situation is that the above three spacecraft only have relative navigation (as shown by the dashed arrows in Figure 2).

采用本发明的判定方法,得到结果如表1所示。由表中结果可知,两种假设情形下都无法成群。由该结果可得下述结论:采用信息拓扑图描述集群是否成群更为准确;且在单纯星间通信或单纯相对导航无法成群时,可将两类信息传递方式结合,从而完成集群成群。这就看出,本发明的成群判定方法更接近工程实际应用,具有良好的使用效益。Using the judgment method of the present invention, the obtained results are shown in Table 1. From the results in the table, it can be seen that the two hypothetical situations cannot be grouped. From this result, the following conclusions can be drawn: it is more accurate to use the information topology map to describe whether clusters are clustered; and when pure inter-satellite communication or pure relative navigation cannot be clustered, the two types of information transmission methods can be combined to complete cluster formation. group. It can be seen that the group determination method of the present invention is closer to the practical application of engineering, and has good use benefits.

表1采用本发明方法计算得到的两种假设情形下的成群结果Table 1. Clustering results under two hypothetical situations calculated by the method of the present invention

Figure BDA0002234581850000081
Figure BDA0002234581850000081

通过上述实施例,可以得出本发明所建立的判定方法,同时考虑了星间通信与相对导航在多航天器构成集群问题中的本质等价性,因而更适合航天器集群的实际应用。Through the above embodiments, the determination method established by the present invention can be obtained, and the essential equivalence of inter-satellite communication and relative navigation in the multi-spacecraft cluster formation problem can be considered, so it is more suitable for the practical application of spacecraft clusters.

以上所述仅是本发明的优选实施方式,应当指出,对于本技术领域的普通技术人员来说,在不脱离本发明原理的前提下,还可以作出若干改进和润饰,这些改进和润饰也应视为本发明的保护范围。The above are only the preferred embodiments of the present invention. It should be pointed out that for those skilled in the art, without departing from the principles of the present invention, several improvements and modifications can be made, and these improvements and modifications should also be It is regarded as the protection scope of the present invention.

Claims (9)

1.一种航天器集群的成群判定方法,其特征在于,包括步骤如下:1. a group determination method of spacecraft cluster, is characterized in that, comprises the steps as follows: S1,获取星间通信关系,建立多航天器间的通信拓扑子图;S1, obtain the inter-satellite communication relationship, and establish a communication topology subgraph between multiple spacecraft; S2,获取相对导航关系,建立多航天器间的相对导航拓扑子图;S2, obtain a relative navigation relationship, and establish a relative navigation topology subgraph among multiple spacecraft; S3,联合通信拓扑子图和相对导航拓扑子图,构建多航天器间的信息拓扑图;S3, combine the communication topology sub-map and the relative navigation topology sub-map to construct an information topology map between multiple spacecraft; S4,构建信息拓扑图对应的邻接矩阵,并通过有向生成树的存在性判定多航天器是否成群。S4, construct the adjacency matrix corresponding to the information topology graph, and determine whether the multi-spacecraft is in a group through the existence of the directed spanning tree. 2.根据权利要求1所述的一种航天器集群的成群判定方法,其特征在于,所述S1中的通信拓扑子图是由代表航天器序号的节点构成的节点集合V与具有星间通信关系的边集合E1联合形成的二元组G1(V,E1);2. The method for group determination of a spacecraft cluster according to claim 1, wherein the communication topology subgraph in the S1 is a node set V composed of nodes representing the spacecraft serial numbers and a set of nodes with inter-satellite numbers. A binary group G 1 (V, E 1 ) formed by the joint set of edges E 1 of the communication relationship; 其中,通信拓扑子图的节点集合V={vi|i=1,…,N},其中N为航天器的数量,vi表示节点,为1到N的自然数;通信拓扑子图的边集合E1={eij|(vi∈V;vj∈V;i≠j)},其中边eij表示节点vi是否指向vj进行了单向通信,若是则边eij=1,否则eij=0。Among them, the node set V={v i |i=1,...,N} of the communication topology subgraph, where N is the number of spacecraft, and v i represents the node, which is a natural number from 1 to N; the edge of the communication topology subgraph Set E 1 ={e ij |(vi ∈V; v j ∈ V; i≠ j )}, where edge e ij indicates whether node v i points to v j for unidirectional communication, and if so, edge e ij =1 , otherwise e ij =0. 3.根据权利要求1所述的一种航天器集群的成群判定方法,其特征在于,所述S2中的相对导航拓扑子图是由代表航天器序号的节点构成的节点集合V与具有相对导航关系的边集合E2联合形成的二元组G2(V,E2);3. The method for group determination of a spacecraft cluster according to claim 1, wherein the relative navigation topology subgraph in S2 is a node set V composed of nodes representing spacecraft serial numbers and a relative navigation topology sub-graph. A binary group G 2 (V, E 2 ) formed by the joint set of edges E 2 of the navigation relation; 其中,相对导航拓扑子图的节点集合与通信拓扑子图的节点集合相同,相对导航拓扑子图的边集合E2={eij|(vi∈V;vj∈V;i≠j)},其中边eij表示节点vj是否对节点vi进行了相对导航测量,若是则边eij=1,否则eij=0。Among them, the node set of the relative navigation topology subgraph is the same as the node set of the communication topology subgraph, and the edge set of the relative navigation topology subgraph E 2 ={e ij |(vi ∈V; v j ∈V; i≠ j ) }, where edge e ij represents whether node v j has performed relative navigation measurement to node v i , if so, edge e ij =1, otherwise e ij =0. 4.根据权利要求1所述的一种航天器集群的成群判定方法,其特征在于,所述S3中的信息拓扑图G(V,E)是通信拓扑子图与相对导航拓扑子图的联合,即G=G1∪G24. The method for grouping determination of a spacecraft cluster according to claim 1, wherein the information topology graph G(V, E) in the S3 is the communication topology subgraph and the relative navigation topology subgraph. Union, ie G=G 1 ∪ G 2 . 5.根据权利要求4所述的一种航天器集群的成群判定方法,其特征在于,信息拓扑图的节点集合等价于通信拓扑子图的节点集合,也等价于相对导航拓扑子图的节点集合。5. the method for grouping determination of a kind of spacecraft cluster according to claim 4 is characterized in that, the node set of the information topology graph is equivalent to the node set of the communication topology subgraph, and is also equivalent to the relative navigation topology subgraph set of nodes. 6.根据权利要求4所述的一种航天器集群的成群判定方法,其特征在于,通信拓扑图的边集合E1与相对导航图的边集合E2形成的联合构成了信息拓扑图的边集合,即E=E1∪E26. The method for grouping determination of a spacecraft cluster according to claim 4, wherein the combination formed by the edge set E 1 of the communication topology graph and the edge set E 2 of the relative navigation graph constitutes the information topology graph. The set of edges, that is, E=E 1 ∪ E 2 . 7.根据权利要求1所述的一种航天器集群的成群判定方法,其特征在于,所述S4中的信息拓扑图的邻接矩阵A,是与信息拓扑图G(V,E)一一对应的N阶方阵;其非对角线上的第i行第j列的元素aij=eij(i≠j),对角线上的元素aii=0(vi∈V)。7. The method for grouping determination of a spacecraft cluster according to claim 1, wherein the adjacency matrix A of the information topology graph in the S4 is one-to-one with the information topology graph G(V, E). The corresponding N-order square matrix; its off-diagonal element a ij = e ij (i≠j) in the i-th row and j-th column, and the diagonal element a ii =0 (vi ∈V ). 8.根据权利要求1所述的一种航天器集群的成群判定方法,其特征在于,所述多航天器是否属于同一个集群是由信息拓扑图G是否包含有向生成树决定的:8. The method for group determination of a spacecraft cluster according to claim 1, wherein whether the multiple spacecraft belong to the same cluster is determined by whether the information topology graph G contains a directed spanning tree: 当G包含有向生成树时N个航天器属于同一个集群;当G不包含有向生成树时N个航天器不属于同一个集群。When G contains a directed spanning tree, N spacecraft belong to the same cluster; when G does not contain a directed spanning tree, N spacecraft do not belong to the same cluster. 9.根据权利要求8所述的一种航天器集群的成群判定方法,其特征在于,所述信息拓扑图G是否包含有向生成树是通过邻接矩阵A来判定的,具体判定的步骤如下:9. The method for group determination of a spacecraft cluster according to claim 8, wherein whether the information topology graph G contains a directed spanning tree is determined by an adjacency matrix A, and the concrete determination steps are as follows : S401,计算矩阵A的1~N-1阶幂:A、A2、A3、……、AN-1S401, calculate the 1-N-1 order power of the matrix A: A, A 2 , A 3 , ..., A N-1 ; S402,将所有N-1阶幂求和得到幂和矩阵
Figure FDA0002234581840000021
S402, sum all N-1 order powers to obtain a power sum matrix
Figure FDA0002234581840000021
S403,检查和矩阵A的所有元素:S403, check all elements of the sum matrix A∑: 若存在某一行(记作第j行,1≤j≤N)除了对角线外所有元素均大于零,即
Figure FDA0002234581840000022
则信息拓扑图G包含有向生成树;若对于任意一行,总是存在某个非对角线元素为零,则信息拓扑图G不包含有向生成树。
If there is a row (denoted as the jth row, 1≤j≤N), all elements except the diagonal are greater than zero, that is
Figure FDA0002234581840000022
Then the information topology graph G contains a directed spanning tree; if for any row, there is always a non-diagonal element that is zero, then the information topology graph G does not contain a directed spanning tree.
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