A method for calculating the control reachable set of a redundant drive system under linear constraints

Abstract

This invention proposes a method for calculating the control reachability set of a redundant drive system under linear constraints, belonging to the technical field of dynamic control allocation for redundant drive systems. The method first constructs the control reachability set problem for the redundant drive system under each pair of linearly constrained control components. Then, it divides the boundary surfaces corresponding to the control sets into three types of rectangular boundary surfaces and one type of triangular boundary surface. By grouping these boundary surfaces, key boundary surfaces are determined for each group to form a boundary surface set. After removing duplicate boundary surfaces from the boundary surface set, the control reachability set boundary surfaces are calculated, ultimately yielding the control reachability set. This invention can be used to evaluate the control capabilities of parallel configuration systems with redundant drive characteristics and where each pair of control components is a linearly constrained control component, such as advanced satellites, aircraft, ships, automobiles, and parallel robots. It can provide a foundation for system control allocation and be used for fault-tolerant control of systems after partial actuator failure.

Description

Technical Field

[0001] This invention belongs to the field of dynamic control allocation technology for redundant drive systems, and specifically relates to a method for calculating the reachable set of control for redundant drive systems under linear constraints. Background Technology

[0002] The control reachability set of a redundant drive system can quantitatively characterize the system's control capability, and its calculation is the inverse problem of control assignment. Control assignment is responsible for distributing the desired system control vector to each redundant actuator for execution. The calculation of the control reachability set, given the known variation range of each actuator, determines the boundary of the system control reachability vector that can be achieved by all actuators operating simultaneously, thereby revealing the control capability of the redundant drive system, especially the system control capability after some actuators fail. Control assignment methods based on control reachability sets have become a hot research topic in the field of control assignment.

[0003] The control reachability set of a parallel redundant drive system can be mathematically expressed as:

[0004] Φ={v|v=B·u,u∈Ω} (1-1)

[0005] In the formula, u is the control vector, u = (u1, ..., u2) m ) T , represents the control input of the redundant drive system, where T is the matrix transpose symbol, and the i-th control component u i u represents the control action of the i-th actuator, 1 ≤ i ≤ m, where m is the number of actuators. imin ≤u i ≤u imax ,u imin u is the minimum constraint value of the control action of the i-th actuator. imax u is the maximum constraint value of the control action of the i-th actuator; i There are often linear or nonlinear constraints between them; Ω is the control set, Ω={u}; v is the control reachability vector of the redundant drive system, v=(v1,…,v n ) T , represents the control output of the redundant drive system, where v j Let Φ be the j-th control reachable component, 1≤j≤n, where n is the dimension of the control reachable vector, n<m; Φ is the control reachable set; and B is the control efficiency matrix with n rows and m columns.

[0006] The physical meaning expressed by equation (1-1) above is: given a set Ω of control vectors consisting of m control inputs of a redundant drive system, how to determine the set Φ of control reachable vectors consisting of n control outputs through the control efficiency matrix B. Taking a four-wheel independent drive-independent steering vehicle as an example, the physical meaning of its control reachable set is as follows:

[0007] 1) Given that the four longitudinal forces of the four wheels are F L1 F L2 F L3 F L4 The four lateral forces of the four wheels are F T1 F T2 F T3 F T4 ;

[0008] 2) Let u be the four-wheel independent drive-independent steering vehicle control vector containing 8 control components. V =(F L1 ,F T1 ,F L2 ,F T2 ,F L3 ,F T3 ,F L4 ,F T4 ) T F Li min ≤F Li ≤F Li max F Ti min ≤F Ti ≤F Ti max ,(i=1,···,4),F Li min F Ti min F represents the minimum value of the longitudinal and lateral forces on each wheel. Li max F Ti max The maximum values ​​of longitudinal and lateral forces for each wheel; u V There are often nonlinear constraints or linear constraints after linearization between the components;

[0009] 3) All u V The control set Ω that constitutes a four-wheel independent drive and independent steering vehicle V ={u V};

[0010] 4)Ω V A specific set of data u VS ∈Ω V The efficiency matrix B of wheel force control V Its function is to generate a specific longitudinal force F on the vehicle as a whole. LS Specific overall lateral force F of the vehicle TS And the specific vehicle overall yaw moment M S , denoted as v VS =(F LS ,F TS M S ) T Then we have v VS =B V·u VS ;

[0011] 5) All v VS The control set Φ of a four-wheel independent drive and independent steering vehicle V , that is Φ V ={v VS |v VS =B V ·u VS ,u VS ∈Ω V}

[0012] The literature "Attainable Moments for the Constrained Control Allocation Problem" and the patent "A Control Allocation Method for Overdrive Systems Based on Geometric Intuition to Construct Reachable Sets" (Patent No.: ZL201810131251.1) disclose that the control actions of each actuator are independent of each other, i.e., u i u j The problem of determining the control reachability set when there are no constraints between (1≤i≤m, 1≤j≤m, i≠j) is mathematically represented as follows:

[0013]

[0014] The patent "Method for Determining the Control Reachability Set of an Overdrive System under a Pair of Linear Constraint Control Components" (Patent No.: ZL201911405624.0) solves the problem of determining the control reachability set of a redundant drive system with only one pair of actuators having a linear constraint relationship, but is not applicable to the problem of determining the control reachability set of a redundant drive system with multiple pairs of actuators having linear constraint relationships. The patents "Method for Determining the Control Reachability Set of an Overdrive System under Each Pair of Linear Constraint Control Components" (Patent No.: ZL201911411744.1) and "Method for Determining the Control Reachability Set of an Overdrive System under Multiple Pairs of Linear Constraint Control Components" (Patent No.: ZL20191) are also applicable. 1405939.5) solves the problem of determining the control reachable set in redundant drive systems where there are constraints between each pair of actuators and between multiple pairs of actuators. However, both of these patents require first determining the quasi-critical boundary surface. For each quasi-critical boundary surface, the vertex mapped to the control reachable set is calculated to obtain the plane equation. Then, the vertices of the boundary surface that has been determined as the critical boundary surface are mapped to the vertices of the control reachable set and substituted into the plane equation. Finally, the characteristic that the control reachable set is a convex set is used to determine whether the quasi-critical boundary surface is a critical boundary surface. The calculation steps are complicated and prone to errors, and cannot achieve good application results. Summary of the Invention

[0015] The purpose of this invention is to overcome the shortcomings of existing technologies and propose a method for calculating the control reachability set of redundant drive systems under linear constraints. This invention solves the problem of determining the control reachability set of parallel redundant drive systems where the control reachability set is in three-dimensional space, any three columns of the control efficiency matrix are linearly independent, and each pair of control components are linearly constrained control components. This method can be used to evaluate the control capabilities of parallel systems with redundant drive characteristics and where each pair of control components is a linearly constrained control component, such as advanced satellites, aircraft, ships, automobiles, and parallel robots. It can provide a basis for system control allocation and be used for fault-tolerant control of systems after partial actuator failure.

[0016] This invention proposes a method for calculating the reachability set of a redundant drive system under linear constraints, comprising the following steps:

[0017] 1) The control reachability set expression for the redundant drive system under each pair of linear constraint control components is established as follows:

[0018]

[0019] in, Let m be the integer part of 2, and u be the control vector. Let u be denoted as u ELR u ELR =(u1,...,u m ) T u i min ≤u i ≤u imax ,i=1,…,m,(u k max -u k min )u k+1 +(u k+1 max -u k+1 min )u k ≤u k+1 max u k max -u k+1 min u k min ,

[0020] Wherein, the i-th control component u i u represents the control action of the i-th actuator, 1 ≤ i ≤ m, where m is the number of actuators; imin u is the minimum constraint value of the control action of the i-th actuator. imax Ω represents the maximum constraint value of the control action of the i-th actuator; Ω is the control set, Ω = {u}; v is the control reachability vector of the redundant drive system, v = (v1, v2, v3). T , represents the control output of the redundant drive system; Φ is the control reachability set; B is a 3xm control efficiency matrix;

[0021] Rewrite equation (1) as equation (2):

[0022]

[0023] In the formula, u k u k+1 For a pair of control components with linear constraints, Φ ELR Ω is the control reachability set of a redundant drive system where each pair of control components is a linearly constrained control component; ELR For the control set, u ELR For the control vector, Ω ELR ={u ELR};

[0024] 2) The control set Ω obtained in step 1) ELR All boundary surfaces are divided into four types;

[0025] control vector u ELR =(u1,...,u m ) T Each component u i The control set Ω is the point obtained by taking the maximum or minimum value of its corresponding constraint. ELR The vertex, represented by equation (2), is the m-dimensional control set Ω. ELR have One vertex, It is the remainder when m is divided by 2;

[0026] make Represents Ω ELR The boundary; based on the values ​​of each component of each vertex... The boundary surfaces are divided into four types: Type I rectangular boundary surface, Type II rectangular boundary surface, Type III rectangular boundary surface, and triangular boundary surface.

[0027] Let the four vertices of the rectangular boundary be A, B, D, and C in clockwise order, and the three vertices of the triangular boundary be A, C, and B in clockwise order. Then:

[0028] If any rectangular boundary surface satisfies the following: one component of the components of vertices A and B has a different value and the other components have the same value; another pair of components of vertices A and C has a different value and the other components have the same value; and the above two components of vertices A and D have different values ​​and the other components have the same value, then the rectangular boundary surface is a type I rectangular boundary surface.

[0029] If any rectangular boundary surface satisfies the following: vertex A and vertex B have a pair of linear constraint components with different values ​​and the rest of the components are the same; vertex C and vertex D have a pair of linear constraint components with different values ​​and the rest of the components are the same; and vertex A and vertex C have a pair of linear constraint components with the same values ​​and only one of the other components has a different value and the rest are the same, then the rectangular boundary surface is a type II rectangular boundary surface.

[0030] If any rectangular boundary surface satisfies the following: vertex A and vertex B have a pair of linear constraint components with different values ​​and the rest of the components are the same; vertex A and vertex C have another pair of linear constraint components with different values ​​and the rest of the components are the same; and vertex D and vertex A have each of the above two pairs of linear constraint components with different values ​​and the rest of the components are the same, then the rectangular boundary surface is a type III rectangular boundary surface.

[0031] If any right triangle satisfies the following: vertex A and vertex B have a pair of linear constraint components with different values ​​and the rest of the components have the same value; vertex A and vertex C have a pair of linear constraint components with one component with different value and the rest of the components have the same value; and vertex B and vertex C have a pair of linear constraint components with the other component with different value and the rest of the components have the same value, then the right triangle is a triangular boundary surface.

[0032] 3) Control set Ω ELR All boundary surfaces are divided into One group;

[0033] If u ELR If two components take values ​​between their corresponding minimum and maximum values, and the remaining m-2 components take values ​​that are either their corresponding minimum or maximum values, then these m components form a 2... m-2 Boundary surfaces of control sets;

[0034] remember u ELR Any two components in the control set are the p-th and q-th components, where the values ​​of the p-th and q-th components are between their corresponding minimum and maximum values, 1 ≤ p ≤ m, 1 ≤ q ≤ m, and p < q. The remaining m-2 components take their corresponding minimum or maximum values. The resulting boundary surface is denoted as the pq group. Then, the total number of boundary surfaces in the control set is... One group;

[0035] 4) Regarding step 3): For each of the groups, determine the critical boundary surface;

[0036] Let the mapping be Φ ELR Like at the boundary Ω ELR The boundary surfaces in the equation are called critical boundary surfaces; let Ω be the set of critical boundary surfaces, and initialize Ω as an empty set.

[0037] 4-1) Randomly select a group whose critical boundary surface has not been determined and denote it as the current pq group, then proceed to step 4-2);

[0038] 4-2) Determine whether the p-th control component and the q-th control component are a pair of linear constraint components:

[0039] If so, proceed to step 4-3);

[0040] If not, and when m is even, proceed to step 4-4);

[0041] If not, and when m is odd, if p and q each have a pair of linear constraint components, then proceed to step 4-4);

[0042] If not, and when m is odd, if q has no paired linear constraint component, then proceed to steps 4-5);

[0043] 4-3) When the p-th and q-th control components are a pair of linear constraint components, the specific method for determining the pq-grouped key facets is as follows:

[0044] 4-3-1) When the p-th component and the q-th component are a pair of linearly constrained components, the boundary surface of this group is a triangular boundary surface, and the values ​​of each component at each point on the boundary surface satisfy the following formula:

[0045]

[0046] Each vertex of the boundary surface in this group is represented by an m x 3 matrix Δ, where the three columns are the vectors corresponding to the three vertices. The p-th and q-th components of the three vertices of each boundary surface take values ​​as shown in the p-th and q-th rows of matrix Δ, and the remaining components all take values ​​of u. imax Or both are u imin In the remaining components, a pair of linear constraint components cannot simultaneously take the maximum value of their corresponding constraints; where the p-th row of matrix Δ is (u pmax u pmin u pmin ), the qth line (u qmin u qmin u qmax );

[0047]

[0048] 4-3-2) Constructing coordinate rotation transformations 1 T, so that the transformed coordinate axis v1 is perpendicular to the image of the triangular boundary surface of the group;

[0049] set up in, 1 c ij For matrix 1The element in the i-th row and j-th column of C, 1 t ij For matrix 1 The element in the i-th row and j-th column of T, b ij Let i be the element in the i-th row and j-th column of matrix B;

[0050] make 1 C = 1 T·B, then 1 c 1p =0, 1 c 1q =0;

[0051] Will 1 Substitute T and B 1 C = 1 T·B, we get:

[0052]

[0053] Solving the system of linear equations (4), we obtain... 1 The first row of T ( 1 t 11 , 1 t 12 , 1 t 13 );

[0054] use 1 C = 1 T·B, calculation 1 The first row of matrix C ( 1 c 11 ,..., 1 c 1m );

[0055] 4-3-3) Except for the p-th and q-th rows, process the other rows of matrix Δ according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed in the set Ω, where 1≤i≤m, i is an integer, and i≠p,q;

[0056] The rules are as follows:

[0057] when 1 c 1i · 1 c 1i+1 When <0,

[0058] like 1 c 1i >0 and 1 c 1i+1 <0, let the i-th row of matrix Δ be u imax The (i+1)th row is all u i+1min ;

[0059] like 1 c 1i <0 and 1 c 1i+1 >0, let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1max ;

[0060] when 1 c 1i <0 and 1 c 1i+1 When <0,

[0061] Let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1min ;

[0062] when 1 c 1i >0 and 1 c 1i+1 When >0,

[0063] calculate 1 z 01 = 1 c 1i ·u imax + 1 c 1i+1 ·u i+1min , 1 z 02 = 1 c 1i ·u imin + 1 c 1i+1 ·u i+1max Then determine:

[0064] like 1 z 01 > 1 z 02 Let the i-th row of matrix Δ be all u imax The (i+1)th row is all u i+1min ;like 1 z 01 < 1 z 02 Let the i-th row of matrix Δ be all u imin The (i+1)th row is all u i+1max ;like 1 z 01 = 1 z 02 The i-th and (i+1)-th rows of matrix Δ correspond to two values: one is to let the i-th row of matrix Δ all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Δ is all u.imin The (i+1)th row is all u i+1max ;

[0065] 4-3-4) Except for the p-th and q-th rows, process the other rows of matrix Δ according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into the set Ω, and then proceed to step 4-6), where 1≤i≤m, i is an integer, and i≠p,q;

[0066] The rules are as follows:

[0067] when 1 c 1i · 1 c 1i+1 When <0,

[0068] like 1 c 1i >0 and 1 c 1i+1 <0, let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1max ;

[0069] like 1 c 1i <0 and 1 c 1i+1 >0, let the i-th row of matrix Δ be u imax The (i+1)th row is all u i+1min ;

[0070] when 1 c 1i >0 and 1 c 1i+1 When >0,

[0071] Let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1min ;

[0072] when 1 c 1i <0 and 1 c 1i+1 When <0,

[0073] calculate 2 z 01 = 1 c 1i ·u imax + 1 c 1i+1 ·u i+1min , 2 z 02 = 1 c 1i ·u imin +1 c 1i+1 ·u i+1max Then determine: if 2 z 01 > 2 z 02 Let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1max ;like 2 z 01 < 2 z 02 Let the i-th row of matrix Δ be u imax The (i+1)th row is all u i+1min ;like 1 z 01 = 1 z 02 The i-th and (i+1)-th rows of matrix Δ correspond to two values: one is to let the i-th row of matrix Δ all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Δ is all u. imin The (i+1)th row is all u i+1max ;

[0074] 4-4) When the p-th and q-th control components are not a pair of linear constraint components and both p and q have corresponding linear constraint components, the specific method for determining the pq grouped key facets is as follows:

[0075] 4-4-1) The boundary surface of the pq group contains three types of rectangular boundary surfaces. The values ​​of each component at each point on the boundary surface of the group satisfy the following formula. The component paired with the p-th component is called the p'-th component, and the component paired with the q-th component is called the q'-th component.

[0076]

[0077] When u p' =u p'min u q' =u q'min When the boundary surface formed by the points satisfying equation (5) is a type I rectangular boundary surface;

[0078] when When the boundary surface formed by the points satisfying equation (5) is a type II rectangular boundary surface;

[0079] when When the boundary surface formed by the points satisfying equation (5) is also a type II rectangular boundary surface;

[0080] when

[0081] When the boundary surface formed by the points satisfying equation (5) is a type III rectangular boundary surface;

[0082] The vertices of each rectangular boundary face in this group are represented by an m x 4 matrix, where the four columns are the vectors corresponding to the four vertices; matrix Λ represents the vertices of an I-shaped rectangular boundary face. 1 Π or 2 Π represents a vertex of a type II rectangular boundary surface, and matrix T represents a vertex of a type III rectangular boundary surface; the p-th, p'-th, q-th, and q'-th components of the four vertices of each boundary surface take values ​​as shown in the p-th, p'-th, q-th, and q'-th rows of the matrix, and the remaining components are all u. imax Or both are u imin A pair of linear inequality constraint components cannot simultaneously take the maximum value of their corresponding constraints; where the p-th row of matrix Λ is (u pmax u pmin u pmin u pmax ), the p'th action (u p'min u p'min u p'min u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'min u q'min u q'min u q'min );

[0083] matrix 1 The p-th action of Π (u pmax u pmin u pmin u pmax ), the p'th action (u p'min u p'max u p'max u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'min u q'min u q'min u q'min );

[0084] matrix 2 The p-th action of Π (u pmax u pmin u pmin u pmax ), the p'th action (up'min u p'min u p'min u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'max u q'max u q'min u q'min );

[0085] The p-th row of matrix T (u pmax u pmin u pmin u pmax ), the p'th action (u p'min u p'max u p'max u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'max u q'max u q'min u q'min );

[0086]

[0087] Where p is odd, the p-th row comes before the p'-th row; if p is even, the p-th row comes after the p'-th row; if q is odd, the q-th row comes before the q'-th row; if q is even, the q-th row comes after the q'-th row.

[0088] 4-4-2) Construct a coordinate rotation transformation G such that the transformed coordinate axis v1 is perpendicular to the image of the type III rectangular boundary surface.

[0089] Then the coordinates of the images of all points on the boundary surface on the v1 axis are equal;

[0090] set up Among them, g ij Let e ​​be the element in the i-th row and j-th column of matrix G. ij Let be the element in the i-th row and j-th column of matrix E;

[0091] Let E = G·B, then Substituting G and B into E = G·B, we have:

[0092]

[0093] Solving the linear system of equations (7) yields the first row of G, and using E = G·B, the first row of the E matrix (e 11 ,...,e 1m If e 1p' >0 and e 1q' If e > 0, proceed to step 4-4-3); if e 1p' <0 and e 1q' If e < 0, proceed to step 4-4-4); if e 1p' ·e 1q' If <0, proceed to step 4-4-5);

[0094] 4-4-3) Except for the p, q, p', q' rows, process the other rows of matrix T according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into the set Ω, and then proceed to step 4-4-5), where i = 1, ..., m, i ≠ p, p', q, q';

[0095] The rules are as follows:

[0096] when e 1i ·e 1i+1 When <0,

[0097] If e 1i >0 and e 1i+1 If < 0, then let the i-th row of matrix T be u. imax The (i+1)th row is all u i+1min ;

[0098] If e 1i <0 and e 1i+1 If the value is greater than 0, then let the i-th row of matrix T be all u. imin The (i+1)th row is all u i+1max ;

[0099] when e 1i <0 and e 1i+1 When <0 (i=1,...,m,i≠p,p',q,q')

[0100] Let the i-th row of matrix T be u imin The (i+1)th row is all u i+1min ;

[0101] when e 1i >0 and e 1i+1 When >0,

[0102] calculate 1 d 01 =e 1i ·u imax +e 1i+1 ·u i+1min , 1 d02 =e 1i ·u imin +e 1i+1 ·u i+1max Then determine: if 1 d 01 > 1 d 02 Let the i-th row of matrix T be u imax The (i+1)th row is all u i+1min ;like 1 d 01 < 1 d 02 Let the i-th row of matrix T be u imin The (i+1)th row is all u i+1max ;like 1 d 01 = 1 d 02 Then the i-th row and the (i+1)-th row of matrix T correspond to two values: one is to let the i-th row of matrix T all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix T is all u. imin The (i+1)th row is all u i+1max ;

[0103] 4-4-4) Except for rows p, q, p', q', process the other rows of matrix T according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into set Ω; then proceed to step 4-4-5), where i = 1, ..., m, i ≠ p, p', q, q';

[0104] The rules are as follows:

[0105] when e 1i ·e 1i+1 When <0,

[0106] If e 1i >0 and e 1i+1 <0, let the i-th row of matrix T be u imin The (i+1)th row is all u i+1max ;

[0107] If e 1i <0 and e 1i+1 >0, let the i-th row of matrix T be u imax The (i+1)th row is all u i+1min ;

[0108] when e 1i <0 and e 1i+1 When <0,

[0109] calculate2 d 01 = 2 h 1i ·u imax + 2 h 1i+1 ·u i+1min , 2 d 02 = 2 h 1i ·u imin + 2 h 1i+1 ·u i+1max ;like 2 d 01 > 2 d 02 Let the i-th row of matrix T be u imin The (i+1)th row is all u i+1max ;like 2 d 01 < 2 d 02 Let the i-th row of matrix T be u imax The (i+1)th row is all u i+1min ;like 2 d 01 = 2 d 02 The i-th and (i+1)-th rows of matrix T correspond to two values: one is to let the i-th row of matrix T all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix T is all u. imin The (i+1)th row is all u i+1max ;

[0110] when e 1i >0 and e 1i+1 When >0,

[0111] Let the i-th row of matrix T be u imin The (i+1)th row is all u i+1min ;

[0112] 4-4-5) Construct coordinate rotation transformation 2 T, so that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface of the grouped I-shaped rectangle;

[0113] set up in, 2 c ij For matrix 2 The element in the i-th row and j-th column of C, 2 t ij For matrix 2 The element in the i-th row and j-th column of T;

[0114] make 2 C = 2 T·B, then 2 c 1p =0, 2 c 1q =0; will 2 Substitute T and B 2 C = 2 T·B, we get:

[0115]

[0116] Solving the system of linear equations (8) yields 2 The first row of T ( 2 t 11 , 2 t 12 , 2 t 13 );use 2 C = 2 T·B, calculation 2 The first row of matrix C ( 2 c 11 ,..., 2 c 1m );

[0117] like 2 c 1p' <0 and 2 c 1q' If <0, proceed to step 4-4-6); if 2 c 1p' >0 and 2 c 1q' If > 0, proceed to step 4-4-7); if 2 c 1p' · 2 c 1q' If <0, proceed to step 4-4-8);

[0118] 4-4-6) Except for rows p, q, p', q', process the other rows of matrix Λ according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface, which is placed in set Ω, where i = 1, ..., m, i ≠ p, p', q, q'. Then determine:

[0119] If two critical boundary surfaces have been found in the group, proceed to step 4-6); otherwise, proceed to step 4-4-8.

[0120] The rules are as follows:

[0121] when 2 c 1i · 2 c 1i+1<0 time

[0122] like 2 c 1i >0 and 2 c 1i+1 <0, let the i-th row of matrix Λ be u imax The (i+1)th row is all u i+1min ;

[0123] like 2 c 1i <0 and 2 c 1i+1 >0, let the i-th row of matrix Λ be all u imin The (i+1)th row is all u i+1max ;

[0124] when 2 c 1i <0 and 2 c 1i+1 When <0,

[0125] Let the i-th row of matrix Λ be u imin The (i+1)th row is all u i+1min ;

[0126] when 2 c 1i >0 and 2 c 1i+1 When >0,

[0127] calculate 3 d 01 = 2 c 1i ·u imax + 2 c 1i+1 ·u i+1min , 3 d 02 = 2 c 1i ·u imin + 2 c 1i+1 ·u i+1max .like 3 d 01 > 3 d 02 Let the i-th row of matrix Λ be u imax The (i+1)th row is all u i+1min ;like 3 d 01 < 3 d 02 Let the i-th row of matrix Λ be u imin The (i+1)th row is all u i+1max ;like 3 d01 = 3 d 02 The i-th and (i+1)-th rows of matrix Λ correspond to two values: one is to set all values ​​in the i-th row of matrix Λ to u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Λ is all u. imin The (i+1)th row is all u i+1max ;

[0128] 4-4-7) Except for rows p, q, p', q', process the other rows of matrix Λ according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface, and it is placed into set Ω, where i = 1, ..., m, i ≠ p, p', q, q'; then determine:

[0129] If two critical boundary surfaces have been found in the group, proceed to step 4-6); otherwise, proceed to step 4-4-8.

[0130] The rules are as follows:

[0131] when 2 c 1i · 2 c 1i+1 When <0,

[0132] like 2 c 1i >0 and 2 c 1i+1 <0, let the i-th row of matrix Λ be u imin The (i+1)th row is all u i+1max ;

[0133] like 2 c 1i <0 and 2 c 1i+1 >0, let the i-th row of matrix Λ be all u imax The (i+1)th row is all u i+1min ;

[0134] when 2 c 1i <0 and 2 c 1i+1 When <0,

[0135] calculate 4 d 01 = 2 c 1i ·u imax + 2 c 1i+1 ·u i+1min , 4 d 02 = 2c 1i ·u imin + 2 c 1i+1 ·u i+1max ;like 4 d 01 > 4 d 02 Let the i-th row of matrix Λ be all u imin The (i+1)th row is all u i+1max ;like 4 d 01 < 4 d 02 Let the i-th row of matrix Λ be all u imax The (i+1)th row is all u i+1min ;like 4 d 01 = 4 d 02 The i-th and (i+1)-th rows of matrix Λ correspond to two values: one is to let the i-th row of matrix Λ all be u. imax The (i+1)th row is all u i+1min Another approach is to set all elements in the i-th row of matrix Λ to u. imin The (i+1)th row is all u i+1max ;

[0136] when 2 c 1i >0 and 2 c 1i+1 When >0,

[0137] Let the i-th row of matrix Λ be u imin The (i+1)th row is all u i+1min ;

[0138] 4-4-8) Constructing coordinate rotation transformations 1 F, so that the transformed coordinate axis v1 is perpendicular to the vertex matrix in this group. 1 If the image of the boundary surface of Π is equal to the coordinates of the image of all points on the boundary surface on the v1 axis, then the coordinates of the image of all points on the boundary surface are equal.

[0139] set up

[0140] in, 1 f ij For matrix 1 The element in the i-th row and j-th column of F 1 h ij For matrix 1 The element in the i-th row and j-th column of H;

[0141] make 1 H = 1 F·B, then 1 h 1q =0. 1 Substitute F and B 1 H = 1 F·B, has:

[0142]

[0143] Solving the system of linear equations (9), we obtain... 1 The first line of F, and using 1 H = 1 F·B calculation 1 The first row of the H matrix ( 1 h 11 ,..., 1 h 1m );

[0144] like 1 h 1p' >0 and 1 h 1q' If <0, proceed to step 4-4-9); if 1 h 1p' <0 and 1 h 1q' If > 0, proceed to step 4-4-10); if 1 h 1p' · 1 h 1q' If the value is greater than 0, proceed to step 4-4-11.

[0145] 4-4-9) Except for the p-th, q-th, p'-th, q'-th rows, divide the matrix 1 The other rows of Π are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set Ω, where i = 1, ..., m, i ≠ p, p', q, q'; then determine:

[0146] If two critical boundary surfaces have been found for this group, proceed to step 4-6); otherwise, proceed to step 4-4-11.

[0147] The rules are as follows:

[0148] when 1 h 1i · 1 h 1i+1 When <0,

[0149] like 1 h 1i >0 and 1 h 1i+1 <0, let the matrix 1 The i-th row of Π is all u imax The (i+1)th row is all ui+1min ;

[0150] like 1 h 1i <0 and 1 h 1i+1 >0, let the matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0151] when 1 h 1i <0 and 1 h 1i+1 When <0,

[0152] Let matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1min ;

[0153] when 1 h 1i >0 and 1 h 1i+1 When >0,

[0154] calculate 5 d 01 = 1 h 1i ·u imax + 1 h 1i+1 ·u i+1min , 5 d 02 = 1 h 1i ·u imin + 1 h 1i+1 ·u i+1max ;like 5 d 01 > 5 d 02 Let the matrix 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;like 5 d 01 < 5 d 02 Let the matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;like 5 d 01 = 5 d 02 ,matrix 1The i-th and (i+1)-th rows of Π correspond to two possible values: one is to let the matrix... 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min Another type is a matrix. 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0155] 4-4-10) Except for the p, q, p', q' rows, divide the matrix 1 The other rows of Π are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set Ω, where i = 1,...,m, i ≠ p, p', q, q'. Then, the following determination is made:

[0156] If two critical boundary surfaces have been found for this group, proceed to step 4-6); otherwise, proceed to step 4-4-11.

[0157] The rules are as follows:

[0158] when 1 h 1i · 1 h 1i+1 When <0,

[0159] like 1 h 1i >0 and 1 h 1i+1 <0, let the matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0160] like 1 h 1i <0 and 1 h 1i+1 >0, let the matrix 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;

[0161] when 1 h 1i >0 and 1 h 1i+1 When >0,

[0162] Let matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1min ;

[0163] when 1 h 1i <0 and 1 h1i+1 When <0,

[0164] calculate 6 d 01 = 1 h 1i ·u imax + 1 h 1i+1 ·u i+1min , 6 d 02 = 1 h 1i ·u imin + 1 h 1i+1 ·u i+1max ;like 6 d 01 > 6 d 02 Let the matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;like 6 d 01 < 6 d 02 Let the matrix 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;like 6 d 01 = 6 d 02 ,matrix 1 The i-th and (i+1)-th rows of Π correspond to two possible values: one is to let the matrix... 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min Another type is a matrix. 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0165] 4-4-11) Construct coordinate rotation transformation 2 F, so that the transformed coordinate axis v1 is perpendicular to the vertex matrix in this group. 2 If the image of the boundary surface of Π is equal to the coordinates of the image of all points on the boundary surface on the v1 axis, then the coordinates of the image of all points on the boundary surface are equal.

[0166] set up

[0167] in, 2 f ij For matrix 2 The element in the i-th row and j-th column of F 2 hij For matrix 2 The element in the i-th row and j-th column of H;

[0168] make 2 H = 2 F·B, then 2 h 1p =0. 2 Substitute F and B 2 H = 2 F·B, has:

[0169]

[0170] Solving the system of linear equations (10), we obtain... 2 The first line of F, and using 2 H = 2 F·B calculation 2 The first row of the H matrix ( 2 h 11 ,..., 2 h 1m );like 2 h 1p' <0 and 2 h 1q' If > 0, proceed to step 4-4-12); if 2 h 1p' >0 and 2 h 1q' <0, proceed to step 4-4-13); if 2 h 1p' · 2 h 1q' >0, proceed to steps 4-6);

[0171] 4-4-12) Except for the p-th, q-th, p'-th, q'-th rows, divide the matrix 2 The other rows of Π are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the key boundary surface, which is put into the set Ω, where i = 1, ..., m, i ≠ p, p', q, q', and then proceed to steps 4-6);

[0172] The rules are as follows:

[0173] when 2 h 1i · 2 h 1i+1 When <0,

[0174] like 2 h 1i >0 and 2 h 1i+1 <0, let the matrix 2 The i-th row of Π is all uimax The (i+1)th row is all u i+1min ;

[0175] like 2 h 1i <0 and 2 h 1i+1 >0, let the matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0176] when 2 h 1i <0 and 2 h 1i+1 When <0,

[0177] Let matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1min ;

[0178] when 2 h 1i >0 and 2 h 1i+1 When >0,

[0179] calculate 7 d 01 = 2 h 1i ·u imax + 2 h 1i+1 ·u i+1min , 7 d 02 = 2 h 1i ·u imin + 2 h 1i+1 ·u i+1max .like 7 d 01 > 7 d 02 Let the matrix 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;like 7 d 01 < 7 d 02 Let the matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;like 7 d 01 = 7 d 02 ,matrix2 The i-th and (i+1)-th rows of Π correspond to two possible values: one is to let the matrix... 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min Another type is a matrix. 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0180] 4-4-13) Except for the p, q, p', q' rows, divide the matrix 2 The other rows of Π are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the key boundary surface, which is put into the set Ω, where i = 1, ..., m, i ≠ p, p', q, q', and then proceed to steps 4-6);

[0181] The rules are as follows:

[0182] when 2 h 1i · 2 h 1i+1 When <0,

[0183] like 2 h 1i >0 and 2 h 1i+1 <0, let the matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0184] like 2 h 1i <0 and 2 h 1i+1 >0, let the matrix 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;

[0185] when 2 h 1i <0 and 2 h 1i+1 When <0,

[0186] calculate 8 d 01 = 2 h 1i ·u imax + 2 h 1i+1 ·u i+1min , 8 d 02 = 2 h 1i ·uimin + 2 h 1i+1 ·u i+1max ;like 8 d 01 > 8 d 02 Let the matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;like 8 d 01 < 8 d 02 Let the matrix 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;like 8 d 01 = 8 d 02 ,matrix 2 The i-th and (i+1)-th rows of Π correspond to two possible values: one is to let the matrix... 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min Another type is a matrix. 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0187] when 2 h 1i >0 and 2 h 1i+1 When >0,

[0188] Let matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1min ;

[0189] 4-5) When the p-th and q-th control components are not a pair of linear constraint components and q has no paired linear constraint component, the specific method for determining the pq grouped key facets is as follows:

[0190] 4-5-1) The boundary surface of the pq group contains both type I and type II rectangular boundary surfaces; the values ​​of each component of each point on the boundary surface of the group satisfy equation (11), and the component paired with the p-th component is called the p'-th component.

[0191]

[0192] When u p' =u p'min When, the boundary surface formed by the points satisfying equation (11) is a type I rectangular boundary surface; when When the boundary surface formed by the points satisfying equation (11) is a type II rectangular boundary surface;

[0193] The vertices of each rectangular boundary face in this group are represented by an m x 4 matrix, where the four columns are the vectors corresponding to the four vertices. Matrix Λ' represents the vertex of a type I rectangular boundary face, and matrix Π' represents the vertex of a type II rectangular boundary face. The p-th, p'-th, and q-th components of the four vertices of each boundary face take values ​​as shown in the p-th, p'-th, and q-th rows of the matrix, respectively. The remaining components are all u. imax Or both are u imin A pair of linear constraint components cannot simultaneously take the maximum value of their corresponding constraints; the p-th row of matrix Λ' is (u pmax u pmin u pmin u pmax ), the p'th action (u p'min u p'min u p'min u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'min u q'min u q'min u q'min );

[0194]

[0195] In the matrix shown above, if p is odd, the p-th row is before the p'-th row; if p is even, the p-th row is after the p'-th row.

[0196] 4-5-2) Constructing coordinate rotation transformations 3 T, so that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface of the grouped I-shaped rectangle;

[0197] set up in, 3 t ij For matrix 3 The element in the i-th row and j-th column of T 3 c ij For matrix 3 The element in the i-th row and j-th column of C;

[0198] make 3 C = 3 T·B, then 3 c 1p =0, 3 c 1q =0. 3 Substitute T and B3 C = 3 T·B, we get:

[0199]

[0200] Solving the system of linear equations (12), we obtain... 3 The first row of T ( 3 t 11 , 3 t 12 , 3 t 13 );use 3 C = 3 T·B, calculation 3 The first row of matrix C ( 3 c 11 ,..., 3 c 1m );like 3 c 1p' If <0, proceed to step 4-5-3); if 3 c 1p' If the value is greater than 0, proceed to step 4-5-4.

[0201] 4-5-3) Except for rows p, q, p', process the other rows of matrix Λ' according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into set Ω, and then proceed to step 4-5-5); where i = 1, ..., m, i ≠ p, p', q;

[0202] The rules are as follows:

[0203] when 3 c 1i · 3 c 1i+1 When <0,

[0204] like 3 c 1i >0 and 3 c 1i+1 <0, let the i-th row of matrix Λ' be u imax The (i+1)th row is all u i+1min ;

[0205] like 3 c 1i <0 and 3 c 1i+1 >0, let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1max ;

[0206] when 3 c 1i <0 and 3c 1i+1 When <0,

[0207] Let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1min ;

[0208] when 3 c 1i >0 and 3 c 1i+1 When >0,

[0209] calculate 9 d 01 = 3 c 1i ·u imax + 3 c 1i+1 ·u i+1min , 9 d 02 = 3 c 1i ·u imin + 3 c 1i+1 ·u i+1max ;

[0210] Judgment: If 9 d 01 > 9 d 02 Let the i-th row of matrix Λ' be u imax The (i+1)th row is all u i+1min ;like 9 d 01 < 9 d 02 Let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1max ;like 9 d 01 = 9 d 02 The i-th and (i+1)-th rows of matrix Λ' correspond to two values: one is to let the i-th row of matrix Λ' all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Λ' is all u. imin The (i+1)th row is all u i+1max ;

[0211] 4-5-4) Except for rows p, q, p', process the other rows of matrix Λ' according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into set Ω, and then proceed to step 4-5-5); where i = 1, ..., m, i ≠ p, p', q;

[0212] The rules are as follows:

[0213] when 3 c 1i · 3 c 1i+1 When <0,

[0214] like 3 c 1i >0 and 3 c 1i+1 <0, let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1max ;

[0215] like 3 c 1i <0 and 3 c 1i+1 >0, let the i-th row of matrix Λ' be u imax The (i+1)th row is all u i+1min ;

[0216] when 3 c 1i >0 and 3 c 1i+1 When >0,

[0217] Let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1min ;

[0218] when 3 c 1i <0 and 3 c 1i+1 When <0,

[0219] calculate 10 d 01 = 3 c 1i ·u imax + 3 c 1i+1 ·u i+1min , 10 d 02 = 3 c 1i ·u imin + 3 c 1i+1 ·u i+1max ;like 10 d 01 > 10 d 02 Let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1max ;like 10 d01 < 10 d 02 Let the i-th row of matrix Λ' be u imax The (i+1)th row is all u i+1min ;like 10 d 01 = 10 d 02 The i-th and (i+1)-th rows of matrix Λ' correspond to two values: one is to let the i-th row of matrix Λ' all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Λ' is all u. imin The (i+1)th row is all u i+1max ;

[0220] 4-5-5) Construct coordinate rotation transformation 3 F, so that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface with vertex matrix Π' in the group, then the coordinate values ​​of the images of all points on the boundary surface on the v1 axis are equal;

[0221] set up

[0222] in, 3 h ij For matrix 3 The element in the i-th row and j-th column of H, 3 f ij For matrix 3 The element in the i-th row and j-th column of F;

[0223] make 3 H = 3 F·B, then 3 h 1q =0, will 3 Substitute F and B 3 H = 3 F·B, has:

[0224]

[0225] Solving the system of linear equations (13), we obtain... 3 The first line of F, and using 3 H = 3 F·B calculation 3 The first row of the H matrix ( 3 h 11 ,..., 3 h 1m );like 3 h 1p' If > 0, proceed to steps 4-5-6); if 3 h1p' If <0, proceed to steps 4-5-7);

[0226] 4-5-6) Except for rows p, q, p', process the other rows of matrix Π' according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into set Ω, and then proceed to step 4-6); where i = 1, ..., m, i ≠ p, p', q;

[0227] The rules are as follows:

[0228] when 3 h 1i · 3 h 1i+1 When <0,

[0229] like 3 h 1i >0 and 3 h 1i+1 <0, let the i-th row of matrix Π' be u imax The (i+1)th row is all u i+1min ;

[0230] like 3 h 1i <0 and 3 h 1i+1 >0, let the i-th row of matrix Π' be all u imin The (i+1)th row is all u i+1max ;

[0231] when 3 h 1i <0 and 3 h 1i+1 When <0,

[0232] Let the i-th row of matrix Π' be u imin The (i+1)th row is all u i+1min ;

[0233] when 3 h 1i >0 and 3 h 1i+1 When >0,

[0234] calculate 11 d 01 = 3 h 1i ·u imax + 3 h 1i+1 ·u i+1min , 11 d 02 = 3 h 1i ·u imin +3 h 1i+1 ·u i+1max ;like 11 d 01 > 11 d 02 Let the i-th row of matrix Π' be all u imax The (i+1)th row is all u i+1min ;like 11 d 01 < 11 d 02 Let the i-th row of matrix Π' be all u imin The (i+1)th row is all u i+1max ;like 11 d 01 = 11 d 02 The i-th and (i+1)-th rows of matrix Π' have two possible values: one is to set all values ​​in the i-th row of matrix Π' to u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Π' is all u. imin The (i+1)th row is all u i+1max ;

[0235] 4-5-7) Except for rows p, q, p', process the other rows of matrix Π' according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into set Ω, and then proceed to step 4-6); where i = 1, ..., m, i ≠ p, p', q;

[0236] The rules are as follows:

[0237] when 3 h 1i · 3 h 1i+1 When <0,

[0238] like 3 h 1i >0 and 3 h 1i+1 <0, let the i-th row of matrix Π' be u imin The (i+1)th row is all u i+1max ;

[0239] like 3 h 1i <0 and 3 h 1i+1 >0, let the i-th row of matrix Π' be all u imax The (i+1)th row is all u i+1min ;

[0240] when 3 h 1i <0 and3 h 1i+1 When <0,

[0241] calculate 12 d 01 = 3 h 1i ·u imax + 3 h 1i+1 ·u i+1min , 12 d 02 = 3 h 1i ·u imin + 3 h 1i+1 ·u i+1max ;like 12 d 01 > 12 d 02 Let the i-th row of matrix Π' be all u imin The (i+1)th row is all u i+1max ;like 12 d 01 < 12 d 02 Let the i-th row of matrix Π' be all u imax The (i+1)th row is all u i+1min ;like 12 d 01 = 12 d 02 The i-th and (i+1)-th rows of matrix Π' have two possible values: one is to set all values ​​in the i-th row of matrix Π' to u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Π' is all u. imin The (i+1)th row is all u i+1max ;

[0242] when 3 h 1i >0 and 3 h 1i+1 When >0,

[0243] Let the i-th row of matrix Π' be u imin The (i+1)th row is all u i+1min ;

[0244] 4-6) Return to step 4-1), select the next group whose critical boundary surface has not been determined, until all groups have determined their critical boundary surfaces, and then proceed to step 5);

[0245] 5) Check all critical boundary surfaces in Ω and determine: If there are identical critical boundary surfaces, keep only one of them and remove the rest that are identical to the kept critical boundary surface; after processing all critical boundary surfaces, a new set of critical boundary surfaces is formed, denoted as Ω'.

[0246] Then proceed to step 6);

[0247] 6) All critical boundary surfaces in Ω', i.e., all critical boundary surfaces in the control set; map all vertices of each critical boundary surface in Ω' using v = B·u. ELR This process obtains all vertices of the corresponding control reachable set boundary surface, thus determining a control reachable set boundary surface, which can be either quadrilateral or triangular. The control reachable set boundary surfaces determined by all critical boundary surfaces in Ω' constitute the control reachable set boundary.

[0248] Features and beneficial effects of the present invention:

[0249] This invention overcomes the limitation of existing technologies that all control components must be independent of each other. It solves the problem of determining the control reachable set of a parallel configuration redundant drive system where the control reachable set is in three-dimensional space, any three columns of the control efficiency matrix are linearly independent, and each pair of control components are linearly constrained control components. It can eliminate the step of determining the quasi-critical boundary surface and directly find the critical boundary surface of each group. The method is simple, computationally efficient, and has high application value.

[0250] This method can be used to evaluate the control capabilities of parallel configuration systems with redundant drive characteristics and linearly constrained control components in each pair of control components, such as advanced satellites, aircraft, ships, automobiles, and parallel robots. It can provide a basis for system control allocation and be used for fault-tolerant control of systems after partial actuator failure. Attached Figure Description

[0251] Figure 1 This is an overall flowchart of a method for calculating the reachable set of a redundant drive system under linear constraints, according to the present invention. Detailed Implementation

[0252] This invention proposes a method for calculating the reachable set of control in a redundant drive system under linear constraints, which is further described in detail below with reference to the accompanying drawings and specific embodiments.

[0253] This invention proposes a method for calculating the reachable set of a redundant drive system under linear constraints. The overall process is as follows: Figure 1 As shown, it includes the following steps:

[0254] 1) The control reachability set expression for the redundant drive system under each pair of linear constraint control components is as follows:

[0255]

[0256] in, Let m be the integer part of 2, and u be the control vector. Let u be denoted as u ELR u ELR =(u1,...,u m ) T u imin ≤u i ≤u imax ,i=1,…,m,(u kmax -u kmin )u k+1 +(u k+1max -u k+1min )u k ≤u k+1max u kmax -u k+1min u kmin ,

[0257] Wherein, the i-th control component u i u represents the control action of the i-th actuator, 1 ≤ i ≤ m, where m is the number of actuators; imin u is the minimum constraint value of the control action of the i-th actuator. imax Let Ω be the maximum constraint value of the control action of the i-th actuator. Ω is the control set, Ω = {u}; v is the control reachability vector of the redundant drive system, v = (v1, v2, v3). T , represents the control output of the redundant drive system; Φ is the control reachability set; B is the 3x3 m control efficiency matrix.

[0258] For ease of expression later, equation (1) is rewritten as equation (2):

[0259]

[0260] In the formula, u k u k+1 A pair of control components with linear constraints is called a pair of linearly constrained control components, Φ ELR Ω is the control reachability set of a redundant drive system where each pair of control components is a linearly constrained control component. ELR For the control set, u ELR For the control vector, Ω ELR ={u ELR}.use Represents Ω ELR The boundary, Φ ELRThe boundary. Thus, the problem of determining the reachability set of a redundant drive system where each pair of control components is a linearly constrained control component is: given Ω ELR And B, how to determine

[0261] 2) The control set Ω obtained in step 1) ELR All boundary surfaces are divided into four types.

[0262] In this embodiment, the control set Ω ELR ={u ELR}, control vector u ELR =(u1,...,u m ) T Each component u i The control set Ω is obtained by taking the maximum or minimum value of the corresponding constraint for (i = 1, ..., m). ELR The vertices. m is the number of actuators, m > 3. An m-dimensional control set has at most 2... m There are vertices. Since paired linear constraint control components cannot simultaneously reach their maximum values, the m-dimensional control set Ω represented by equation (2) is... ELR have One vertex. The integer part of m divided by 2, It is the remainder when m is divided by 2.

[0263] Ω ELR Ω is the control set of a redundant drive system where each pair of control components is a linearly constrained control component. ELR The boundary is The values ​​of each component at each vertex determine the composition. There are four types of boundary surfaces: Type I rectangular boundary surface, Type II rectangular boundary surface, Type III rectangular boundary surface, and triangular boundary surface, defined as follows:

[0264] Let the four vertices of the rectangular boundary be A, B, D, and C in clockwise order, and the three vertices of the triangular boundary be A, C, and B in clockwise order.

[0265] A rectangle that simultaneously satisfies the following three conditions is called an I-shaped rectangular boundary surface: ① One component of vertices A and B has a different value, while the other components have the same value; ② Another pair of components of vertices A and C has one component with a different value, while the other components have the same value; ③ The above two components of vertices A and D have different values, while the other components have the same value.

[0266] A rectangle that satisfies the following three conditions is called a Type II rectangular boundary surface: ① Vertex A and vertex B have one pair of linear constraint components with different values, and the other components are the same; ② Vertex C and vertex D have one pair of linear constraint components with different values, and the other components are the same; ③ Vertex A and vertex C have one pair of linear constraint components with the same values, and only one of the other components has a different value, while the rest are the same.

[0267] A rectangle that satisfies the following three conditions is called a Type III rectangular boundary surface: ① Vertex A and vertex B have one pair of linear constraint components with different values, while the other components are the same; ② Vertex A and vertex C have another pair of linear constraint components with different values, while the other components are the same; ③ Vertex D and vertex A have each of the above two pairs of linear constraint components with different values, while the other components are the same.

[0268] A right triangle that satisfies the following three conditions is called a triangular boundary surface: ① Vertex A and vertex B have one pair of linear constraint components with different values, while the other components have the same value; ② Vertex A and vertex C have one linear constraint component with a different value, while the other components have the same value; ③ Vertex B and vertex C have another linear constraint component with a different value, while the other components have the same value.

[0269] 3) Control set Ω ELR All boundary surfaces are divided into One group;

[0270] If u ELR Two components take values ​​between their corresponding minimum and maximum values, and the remaining m-2 components take values ​​that are either their corresponding minimum or maximum values. Then, these m components form a 2... m-2 Boundary surfaces of control sets;

[0271] remember u ELR Any two components in the control set are the p-th and q-th components. The p-th and q-th components take values ​​between their corresponding minimum and maximum values, where 1 ≤ p ≤ m, 1 ≤ q ≤ m, and p < q. The remaining m-2 components take values ​​of their corresponding minimum or maximum values. These boundary surfaces are grouped together and called the pq group. Therefore, the control set contains a total of [number missing] boundary surfaces. One group;

[0272] 4) Regarding step 3): For each of the groups, determine the critical boundary surface;

[0273] Let the mapping be Φ ELR Like at the boundary Ω ELR The boundary surfaces in the equation are called critical boundary surfaces; let Ω be the set of critical boundary surfaces, and initialize Ω as an empty set.

[0274] 4-1) Randomly select a group whose critical boundary surface has not been determined and denote it as the current pq group, then proceed to step 4-2);

[0275] 4-2) Determine whether the p-th control component and the q-th control component are a pair of linear constraint components:

[0276] If so, proceed to step 4-3);

[0277] If not, and when m is even, proceed to step 4-4);

[0278] If not, and when m is odd, if p and q each have a pair of linear constraint components, then proceed to step 4-4);

[0279] If not, and when m is odd, if q has no paired linear constraint component, then proceed to steps 4-5.

[0280] It should be noted that only when m is odd does the m-th component not have a paired linear constraint component. Since p < q has been specified earlier, p must have a paired linear constraint component.

[0281] 4-3) When the p-th and q-th control components are a pair of linear constraint components, the specific method for determining the pq-grouped key facets is as follows:

[0282] 4-3-1) When the p-th component and the q-th component are a pair of linearly constrained components, the boundary surface of this group is a triangular boundary surface, and the values ​​of each component at each point on the boundary surface satisfy the following formula:

[0283]

[0284] Each vertex of the boundary surface in this group can be represented by an m x 3 matrix Δ, where the three columns are the vectors corresponding to the three vertices. The p-th and q-th components of the three vertices of each boundary surface take values ​​as shown in the p-th and q-th rows of matrix Δ, respectively. The remaining i-th (i = 1, ..., m, i ≠ p, q) component takes the value u. imax Or both are u imin The i-th row (i = 1, ..., m, i ≠ p, q) of the corresponding matrix Δ is u imax Or both are u imin However, among the remaining components, a pair of linear constraint components cannot simultaneously take the maximum value of their corresponding constraints; this is uniformly represented by an ellipsis in matrix Δ. The p-th row of matrix Δ is (u pmax u pmin u pmin ), the qth line (u qmin u qmin u qmax ).

[0285]

[0286] 4-3-2) Constructing coordinate rotation transformations 1 T, so that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface of the grouped triangle. Since only the v1 axis is considered, it is only necessary to... 1 The first line of T.

[0287] set up in, 1 c ij For matrix 1 The element in the i-th row and j-th column of C, 1 t ij For matrix 1 The element in the i-th row and j-th column of T, b ij Let be the element in the i-th row and j-th column of matrix B.

[0288] make 1 C = 1 T·B, must have 1 c 1p =0, 1 c 1q =0. (This likely refers to a specific value or quantity.) 1 Substitute T and B 1 C = 1 T·B, we get:

[0289]

[0290] Solving the system of linear equations (4), we obtain... 1 The first row of T ( 1 t 11 , 1 t 12 , 1 t 13 Reuse 1 C = 1 T·B can be calculated 1 The first row of matrix C ( 1 c 11 ,..., 1 c 1m );

[0291] 4-3-3) Except for the p-th and q-th rows, process the other rows of matrix Δ according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed in the set Ω, where 1≤i≤m, i is an integer, and i≠p,q;

[0292] The rules are as follows:

[0293] when 1 c 1i · 1 c1i+1 When <0,

[0294] like 1 c 1i >0 and 1 c 1i+1 <0, let the i-th row of matrix Δ be u imax The (i+1)th row is all u i+1min ;

[0295] like 1 c 1i <0 and 1 c 1i+1 >0, let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1max .

[0296] when 1 c 1i <0 and 1 c 1i+1 When <0,

[0297] Let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1min .

[0298] when 1 c 1i >0 and 1 c 1i+1 When >0,

[0299] calculate 1 z 01 = 1 c 1i ·u imax + 1 c 1i+1 ·u i+1min , 1 z 02 = 1 c 1i ·u imin + 1 c 1i+1 ·u i+1max Then determine:

[0300] like 1 z 01 > 1 z 02 Let the i-th row of matrix Δ be all u imax The (i+1)th row is all u i+1min ;like 1 z 01 < 1 z 02 Let the i-th row of matrix Δ be all uimin The (i+1)th row is all u i+1max ;like 1 z 01 = 1 z 02 The i-th and (i+1)-th rows of matrix Δ correspond to two values: one is to let the i-th row of matrix Δ all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Δ is all u. imin The (i+1)th row is all u i+1max .

[0301] 4-3-4) Except for the p-th and q-th rows, process the other rows of matrix Δ according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into the set Ω, and then proceed to step 4-6), where 1≤i≤m, i is an integer, and i≠p,q;

[0302] The rules are as follows:

[0303] when 1 c 1i · 1 c 1i+1 When <0,

[0304] like 1 c 1i >0 and 1 c 1i+1 <0, let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1max ;

[0305] like 1 c 1i <0 and 1 c 1i+1 >0, let the i-th row of matrix Δ be u imax The (i+1)th row is all u i+1min .

[0306] when 1 c 1i >0 and 1 c 1i+1 When >0,

[0307] Let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1min .

[0308] when 1 c 1i <0 and 1 c 1i+1 When <0,

[0309] calculate 2z 01 = 1 c 1i ·u imax + 1 c 1i+1 ·u i+1min , 2 z 02 = 1 c 1i ·u imin + 1 c 1i+1 ·u i+1max Then determine: if 2 z 01 > 2 z 02 Let the i-th row of matrix Δ be u imin The (i+1)th row is all u i+1max ;like 2 z 01 < 2 z 02 Let the i-th row of matrix Δ be u imax The (i+1)th row is all u i+1min ;like 1 z 01 = 1 z 02 The i-th and (i+1)-th rows of matrix Δ correspond to two values: one is to let the i-th row of matrix Δ all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Δ is all u. imin The (i+1)th row is all u i+1max .

[0310] 4-4) When the p-th and q-th control components are not a pair of linear constraint components and both p and q have corresponding linear constraint components, the specific method for determining the pq grouped key facets is as follows:

[0311] 4-4-1) The boundary surface of the pq group contains three types of rectangular boundary surfaces. The values ​​of each component at each point on the boundary surface of the group satisfy the following formula, and the component paired with the p-th component is called the p'-th component, and the component paired with the q-th component is called the q'-th component.

[0312]

[0313] When u p' =u p'min u q' =u q'min When the boundary surface formed by the points satisfying equation (5) is a type I rectangular boundary surface;

[0314] When u p'=u p'min , When the boundary surface formed by the points satisfying equation (5) is a type II rectangular boundary surface;

[0315] when u q' =u q'min When the boundary surface formed by the points satisfying equation (5) is also a type II rectangular boundary surface;

[0316] when

[0317] When the boundary surface formed by the points satisfying equation (5) is a type III rectangular boundary surface.

[0318] The vertices of each rectangular boundary face in this group can be represented by an m x 4 matrix, where the four columns are the vectors corresponding to the four vertices. Matrix Λ represents the vertices of an I-shaped rectangular boundary face. 1 Π or 2 Let Π represent a vertex of a type II rectangular boundary surface, and matrix T represent a vertex of a type III rectangular boundary surface. The p-th, p'-th, q-th, and q'-th components of the four vertices of each boundary surface take values ​​as shown in the p-th, p'-th, q-th, and q'-th rows of the matrix, respectively. The remaining i-th component is u. imax Or both are u imin The i-th row of the corresponding matrix is ​​u imax Or both are u imin However, a pair of linear inequality constraint components cannot simultaneously take the maximum value of their corresponding constraints; this is uniformly represented by an ellipsis in the matrix. The p-th row of matrix Λ is (u pmax u pmin u pmin u pmax ), the p'th action (u p'min u p'min u p'min u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'min u q'min u q'min u q'min );

[0319] matrix 1 The p-th action of Π (u pmax u pmin u pmin u pmax ), the p'th action (u p'min up'max u p'max u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'min u q'min u q'min u q'min );

[0320] matrix 2 The p-th action of Π (u pmax u pmin u pmin u pmax ), the p'th action (u p'min u p'min u p'min u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'max u q'max u q'min u q'min );

[0321] The p-th row of matrix T (u pmax u pmin u pmin u pmax ), the p'th action (u p'min u p'max u p'max u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'max u q'max u q'min u q'min ).

[0322]

[0323] It should be noted that in the expressions of the four matrices shown above, if p is odd, then the p-th row comes before the p'-th row; if p is even, then the p-th row comes after the p'-th row. Similarly, if q is odd, then the q-th row comes before the q'-th row; if q is even, then the q-th row comes after the q'-th row.

[0324] 4-4-2) Construct a coordinate rotation transformation G such that the transformed coordinate axis v1 is perpendicular to the image of the type III rectangular boundary surface. Then, the coordinate values ​​of the images of all points on the boundary surface on the v1 axis are equal.

[0325] set up

[0326] Among them, g ij Let e ​​be the element in the i-th row and j-th column of matrix G. ij Let be the element in the i-th row and j-th column of matrix E.

[0327] Let E = G·B, then we must have Substituting G and B into E = G·B, we have:

[0328]

[0329] Solving the linear system of equations (7) yields the first row of G, and using E = G·B, the first row of the E matrix (e 11 ,...,e 1m If e 1p' >0 and e 1q' If e > 0, proceed to step 4-4-3); if e 1p' <0 and e 1q' If e < 0, proceed to step 4-4-4); if e 1p' ·e 1q' If <0, proceed to step 4-4-5);

[0330] 4-4-3) Except for the p, q, p', q' rows, process the other rows of matrix T according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into the set Ω, and then proceed to step 4-4-5), where i = 1, ..., m, i ≠ p, p', q, q';

[0331] The rules are as follows:

[0332] when e 1i ·e 1i+1 When <0,

[0333] If e 1i >0 and e 1i+1 If < 0, then let the i-th row of matrix T be u. imax The (i+1)th row is all u i+1min ;

[0334] If e 1i <0 and e 1i+1 If the value is greater than 0, then let the i-th row of matrix T be all u. imin The (i+1)th row is all u i+1max .

[0335] when e 1i <0 and e 1i+1 When <0,

[0336] Let the i-th row of matrix T be u imin The (i+1)th row is all u i+1min ;

[0337] when e 1i >0 and e 1i+1 When >0,

[0338] calculate 1 d 01 =e 1i ·u imax +e 1i+1 ·u i+1min , 1 d 02 =e 1i ·u imin +e 1i+1 ·u i+1max Then determine: if 1 d 01 > 1 d 02 Let the i-th row of matrix T be u imax The (i+1)th row is all u i+1min ;like 1 d 01 < 1 d 02 Let the i-th row of matrix T be u imin The (i+1)th row is all u i+1max ;like 1 d 01 = 1 d 02 Then the i-th row and the (i+1)-th row of matrix T correspond to two values: one is to let the i-th row of matrix T all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix T is all u. imin The (i+1)th row is all u i+1max .

[0339] 4-4-4) Except for rows p, q, p', q', process the other rows of matrix T according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into set Ω. Then proceed to step 4-4-5), where i = 1, ..., m, i ≠ p, p', q, q';

[0340] The rules are as follows:

[0341] when e 1i ·e 1i+1 When <0,

[0342] If e 1i >0 and e 1i+1 <0, let the i-th row of matrix T be u imin The (i+1)th row is all u i+1max ;

[0343] If e 1i <0 and e 1i+1 >0, let the i-th row of matrix T be u imax The (i+1)th row is all u i+1min .

[0344] when e 1i <0 and e 1i+1 When <0,

[0345] calculate 2 d 01 = 2 h 1i ·u imax + 2 h 1i+1 ·u i+1min , 2 d 02 = 2 h 1i ·u imin + 2 h 1i+1 ·u i+1max .like 2 d 01 > 2 d 02 Let the i-th row of matrix T be u imin The (i+1)th row is all u i+1max ;like 2 d 01 < 2 d 02 Let the i-th row of matrix T be u imax The (i+1)th row is all u i+1min ;like 2 d 01 = 2 d 02 The i-th and (i+1)-th rows of matrix T correspond to two values: one is to let the i-th row of matrix T all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix T is all u. imin The (i+1)th row is all u i+1max .

[0346] when e 1i >0 and e 1i+1 When >0,

[0347] Let the i-th row of matrix T be u imin The (i+1)th row is all u i+1min .

[0348] 4-4-5) Construct coordinate rotation transformation 2 T, so that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface of the grouped I-shaped rectangle. Since only the v1 axis is considered, it is only necessary to... 2 The first line of T.

[0349] set up

[0350] in, 2 c ij For matrix 2 The element in the i-th row and j-th column of C, 2 t ij For matrix 2 The element in row i and column j of T.

[0351] make 2 C = 2 T·B, must have 2 c 1p =0, 2 c 1q =0. (This likely refers to a specific value or quantity.) 2 Substitute T and B 2 C = 2 T·B, we get:

[0352]

[0353] Solving the system of linear equations (8) yields 2 The first row of T ( 2 t 11 , 2 t 12 , 2 t 13 Reuse 2 C = 2 T·B can be calculated 2 The first row of matrix C ( 2 c 11 ,..., 2 c 1m ).

[0354] like 2 c 1p' <0 and 2 c 1q' If <0, proceed to step 4-4-6); if 2 c 1p' >0 and 2 c 1q' If > 0, proceed to step 4-4-7); if 2 c1p' · 2 c 1q' If <0, proceed to step 4-4-8);

[0355] 4-4-6) Except for rows p, q, p', q', process the other rows of matrix Λ according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface, which is placed in set Ω, where i = 1, ..., m, i ≠ p, p', q, q'. Then determine:

[0356] If two critical boundary surfaces have been found in the group, proceed to step 4-6); otherwise, proceed to step 4-4-8.

[0357] The rules are as follows:

[0358] when 2 c 1i · 2 c 1i+1 <0 time

[0359] like 2 c 1i >0 and 2 c 1i+1 <0, let the i-th row of matrix Λ be u imax The (i+1)th row is all u i+1min ;

[0360] like 2 c 1i <0 and 2 c 1i+1 >0, let the i-th row of matrix Λ be all u imin The (i+1)th row is all u i+1max .

[0361] when 2 c 1i <0 and 2 c 1i+1 When <0,

[0362] Let the i-th row of matrix Λ be u imin The (i+1)th row is all u i+1min .

[0363] when 2 c 1i >0 and 2 c 1i+1 When >0,

[0364] calculate 3 d 01 = 2 c 1i ·u imax + 2 c 1i+1 ·ui+1min , 3 d 02 = 2 c 1i ·u imin + 2 c 1i+1 ·u i+1max .like 3 d 01 > 3 d 02 Let the i-th row of matrix Λ be u imax The (i+1)th row is all u i+1min ;like 3 d 01 < 3 d 02 Let the i-th row of matrix Λ be u imin The (i+1)th row is all u i+1max ;like 3 d 01 = 3 d 02 The i-th and (i+1)-th rows of matrix Λ correspond to two values: one is to set all values ​​in the i-th row of matrix Λ to u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Λ is all u. imin The (i+1)th row is all u i+1max .

[0365] 4-4-7) Except for rows p, q, p', q', process the other rows of matrix Λ according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface, and it is placed into set Ω, where i = 1, ..., m, i ≠ p, p', q, q'; then determine:

[0366] If two critical boundary surfaces have been found in the group, proceed to step 4-6); otherwise, proceed to step 4-4-8.

[0367] The rules are as follows:

[0368] when 2 c 1i · 2 c 1i+1 When <0,

[0369] like 2 c 1i >0 and 2 c 1i+1 <0, let the i-th row of matrix Λ be u imin The (i+1)th row is all u i+1max ;

[0370] like 2 c 1i<0 and 2 c 1i+1 >0, let the i-th row of matrix Λ be all u imax The (i+1)th row is all u i+1min .

[0371] when 2 c 1i <0 and 2 c 1i+1 When <0,

[0372] calculate 4 d 01 = 2 c 1i ·u imax + 2 c 1i+1 ·u i+1min , 4 d 02 = 2 c 1i ·u imin + 2 c 1i+1 ·u i+1max .like 4 d 01 > 4 d 02 Let the i-th row of matrix Λ be all u imin The (i+1)th row is all u i+1max ;like 4 d 01 < 4 d 02 Let the i-th row of matrix Λ be all u imax The (i+1)th row is all u i+1min ;like 4 d 01 = 4 d 02 The i-th and (i+1)-th rows of matrix Λ correspond to two values: one is to let the i-th row of matrix Λ all be u. imax The (i+1)th row is all u i+1min Another approach is to set all elements in the i-th row of matrix Λ to u. imin The (i+1)th row is all u i+1max .

[0373] when 2 c 1i >0 and 2 c 1i+1 When >0,

[0374] Let the i-th row of matrix Λ be u imin The (i+1)th row is all u i+1min .

[0375] 4-4-8) Constructing coordinate rotation transformations 1 F, so that the transformed coordinate axis v1 is perpendicular to the vertex matrix in this group. 1 If the image of the boundary surface of Π is such that the coordinates of the images of all points on the boundary surface are equal along the v1 axis, then since we only consider the v1 axis, we only need to... 1 The first line of F.

[0376] set up

[0377] in, 2 f ij For matrix 2 The element in the i-th row and j-th column of F 2 h ij For matrix 2 The element in the i-th row and j-th column of H.

[0378] make 1 H = 1 F·B, must have 1 h 1q =0. (This likely refers to a specific value or quantity.) 1 Substitute F and B 1 H = 1 F·B, has:

[0379]

[0380] Solving the system of linear equations (9), we obtain... 1 The first line of F, and using 1 H = 1 F·B calculation 1 The first row of the H matrix ( 1 h 11 ,..., 1 h 1m ).

[0381] like 1 h 1p' >0 and 1 h 1q' If <0, proceed to step 4-4-9); if 1 h 1p' <0 and 1 h 1q' If > 0, proceed to step 4-4-10); if 1 h 1p' · 1 h 1q' If the value is greater than 0, proceed to step 4-4-11.

[0382] 4-4-9) Except for the p-th, q-th, p'-th, q'-th rows, divide the matrix 1The other rows of Π are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set Ω, where i = 1, ..., m, i ≠ p, p', q, q'. Then determine:

[0383] If two critical boundary surfaces have been found for this group, proceed to step 4-6); otherwise, proceed to step 4-4-11.

[0384] The rules are as follows:

[0385] when 1 h 1i · 1 h 1i+1 When <0,

[0386] like 1 h 1i >0 and 1 h 1i+1 <0, let the matrix 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;

[0387] like 1 h 1i <0 and 1 h 1i+1 >0, let the matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max .

[0388] when 1 h 1i <0 and 1 h 1i+1 When <0,

[0389] Let matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1min .

[0390] when 1 h 1i >0 and 1 h 1i+1 When >0,

[0391] calculate 5 d 01 = 1 h 1i ·u imax + 1 h 1i+1 ·u i+1min , 5 d 02 = 1 h1i ·u imin + 1 h 1i+1 ·u i+1max .like 5 d 01 > 5 d 02 Let the matrix 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;like 5 d 01 < 5 d 02 Let the matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;like 5 d 01 = 5 d 02 ,matrix 1 The i-th and (i+1)-th rows of Π correspond to two possible values: one is to let the matrix... 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min Another type is a matrix. 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max .

[0392] 4-4-10) Except for the p, q, p', q' rows, divide the matrix 1 The other rows of Π are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set Ω, where i = 1,...,m, i ≠ p, p', q, q'. Then, the following determination is made:

[0393] If two critical boundary surfaces have been found for this group, proceed to step 4-6); otherwise, proceed to step 4-4-11.

[0394] The rules are as follows:

[0395] when 1 h 1i · 1 h 1i+1 When <0,

[0396] like 1 h 1i >0 and 1 h 1i+1 <0, let the matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0397] like 1 h 1i <0 and 1 h 1i+1 >0, let the matrix 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min .

[0398] when 1 h 1i >0 and 1 h 1i+1 When >0,

[0399] Let matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1min .

[0400] when 1 h 1i <0 and 1 h 1i+1 When <0,

[0401] calculate 6 d 01 = 1 h 1i ·u imax + 1 h 1i+1 ·u i+1min , 6 d 02 = 1 h 1i ·u imin + 1 h 1i+1 ·u i+1max .like 6 d 01 > 6 d 02 Let the matrix 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;like 6 d 01 < 6 d 02 Let the matrix 1 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;like 6 d 01 = 6 d 02 ,matrix 1 The i-th and (i+1)-th rows of Π correspond to two possible values: one is to let the matrix... 1 The i-th row of Π is all uimax The (i+1)th row is all u i+1min Another type is a matrix. 1 The i-th row of Π is all u imin The (i+1)th row is all u i+1max .

[0402] 4-4-11) Construct coordinate rotation transformation 2 F, so that the transformed coordinate axis v1 is perpendicular to the vertex matrix in this group. 2 If the image of the boundary surface of Π is such that the coordinates of the images of all points on the boundary surface are equal along the v1 axis, then since we only consider the v1 axis, we only need to... 2 The first line of F.

[0403] set up

[0404] in, 2 f ij For matrix 2 The element in the i-th row and j-th column of F 2 h ij For matrix 2 The element in the i-th row and j-th column of H;

[0405] make 2 H = 2 F·B, must have 2 h 1p =0. 2 Substitute F and B 2 H = 2 F·B, has:

[0406]

[0407] Solving the system of linear equations (10), we obtain... 2 The first line of F, and using 2 H = 2 F·B calculation 2 The first row of the H matrix ( 2 h 11 ,..., 2 h 1m ).like 2 h 1p' <0 and 2 h 1q' If > 0, proceed to step 4-4-12); if 2 h 1p' >0 and 2 h 1q' If <0, proceed to step 4-4-13); if 2 h 1p' · 2h 1q' If the value is greater than 0, proceed to steps 4-6.

[0408] 4-4-12) Except for the p-th, q-th, p'-th, q'-th rows, divide the matrix 2 The other rows of Π are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the key boundary surface, which is put into the set Ω, where i = 1, ..., m, i ≠ p, p', q, q', and then proceed to steps 4-6);

[0409] The rules are as follows:

[0410] when 2 h 1i · 2 h 1i+1 When <0,

[0411] like 2 h 1i >0 and 2 h 1i+1 <0, let the matrix 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;

[0412] like 2 h 1i <0 and 2 h 1i+1 >0, let the matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max .

[0413] when 2 h 1i <0 and 2 h 1i+1 When <0,

[0414] Let matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1min .

[0415] when 2 h 1i >0 and 2 h 1i+1 When >0,

[0416] calculate 7 d 01 = 2 h 1i ·u imax + 2 h 1i+1 ·u i+1min , 7 d02 = 2 h 1i ·u imin + 2 h 1i+1 ·u i+1max .like 7 d 01 > 7 d 02 Let the matrix 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;like 7 d 01 < 7 d 02 Let the matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;like 7 d 01 = 7 d 02 ,matrix 2 The i-th and (i+1)-th rows of Π correspond to two possible values: one is to let the matrix... 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min Another type is a matrix. 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max .

[0417] 4-4-13) Except for the p, q, p', q' rows, divide the matrix 2 The other rows of Π are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the key boundary surface, which is put into the set Ω, where i = 1, ..., m, i ≠ p, p', q, q', and then proceed to steps 4-6);

[0418] The rules are as follows:

[0419] when 2 h 1i · 2 h 1i+1 When <0,

[0420] like 2 h 1i >0 and 2 h 1i+1 <0, let the matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;

[0421] like 2 h1i <0 and 2 h 1i+1 >0, let the matrix 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min .

[0422] when 2 h 1i <0 and 2 h 1i+1 When <0,

[0423] calculate 8 d 01 = 2 h 1i ·u imax + 2 h 1i+1 ·u i+1min , 8 d 02 = 2 h 1i ·u imin + 2 h 1i+1 ·u i+1max ;like 8 d 01 > 8 d 02 Let the matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max ;like 8 d 01 < 8 d 02 Let the matrix 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min ;like 8 d 01 = 8 d 02 ,matrix 2 The i-th and (i+1)-th rows of Π correspond to two possible values: one is to let the matrix... 2 The i-th row of Π is all u imax The (i+1)th row is all u i+1min Another type is a matrix. 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1max .

[0424] when 2 h 1i >0 and 2 h 1i+1 When >0,

[0425] Let matrix 2 The i-th row of Π is all u imin The (i+1)th row is all u i+1min .

[0426] 4-5) When the p-th and q-th control components are not a pair of linear constraint components and q has no paired linear constraint component, the specific method for determining the pq grouped key facets is as follows:

[0427] 4-5-1) This situation only occurs when m is odd and q = m. The boundary surface of the pq group contains both type I and type II rectangular boundary surfaces. The values ​​of each component at each point on the boundary surface of this group satisfy equation (11). The component paired with the p-th component is called the p'-th component.

[0428]

[0429] When u p' =u p'min When, the boundary surface formed by the points satisfying equation (11) is a type I rectangular boundary surface; when When the boundary surface formed by the points satisfying equation (11) is a type II rectangular boundary surface.

[0430] The vertices of each rectangular boundary face in this group can be represented by an m x 4 matrix, where the four columns are the vectors corresponding to the four vertices. Let matrix Λ' represent the vertices of a type I rectangular boundary face, and matrix Π' represent the vertices of a type II rectangular boundary face. The p-th, p'-th, and q-th components of the four vertices of each boundary face are as shown in the p-th, p'-th, and q-th rows of the matrix, respectively. The remaining i-th (i = 1, ..., m, i ≠ p, p', q) component is u. imax Or both are u imin The i-th row (i = 1, ..., m, i ≠ p, p', q) of the corresponding matrix is ​​u imax Or both are u imin However, a pair of linear constraint components cannot simultaneously take the maximum value of their corresponding constraints; this is uniformly represented by an ellipsis in the matrix. The p-th row of matrix Λ' is (u pmax u pmin u pmin u pmax ), the p'th action (u p'min u p'min u p'min u p'min ), the qth line (u qmin u qmin u qmax u qmax ), the q'th action (u q'min u q'min uq'min u q'min ).

[0431]

[0432] In the matrix shown above, if p is odd, the p-th row is before the p'-th row; if p is even, the p-th row is after the p'-th row.

[0433] 4-5-2) Constructing coordinate rotation transformations 3 T, so that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface of the grouped I-shaped rectangle. Since only the v1 axis is considered, it is only necessary to... 3 The first line of T.

[0434] set up in, 3 t ij For matrix 3 The element in the i-th row and j-th column of T 3 c ij For matrix 3 The element in row i and column j of C.

[0435] make 3 C = 3 T·B, must have 3 c 1p =0, 3 c 1q =0. 3 Substitute T and B 3 C = 3 T·B, we get:

[0436]

[0437] Solving the system of linear equations (12), we obtain... 3 The first row of T ( 3 t 11 , 3 t 12 , 3 t 13 Reuse 3 C = 3 T·B can be calculated 3 The first row of matrix C ( 3 c 11 ,..., 3 c 1m ).like 3 c 1p' If <0, proceed to step 4-5-3); if 3 c 1p' If the value is greater than 0, proceed to step 4-5-4.

[0438] 4-5-3) Except for rows p, q, p', process the other rows of matrix Λ' according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into set Ω, and then proceed to step 4-5-5); where i = 1, ..., m, i ≠ p, p', q;

[0439] The rules are as follows:

[0440] when 3 c 1i · 3 c 1i+1 When <0,

[0441] like 3 c 1i >0 and 3 c 1i+1 <0, let the i-th row of matrix Λ' be u imax The (i+1)th row is all u i+1min ;

[0442] like 3 c 1i <0 and 3 c 1i+1 >0, let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1max .

[0443] when 3 c 1i <0 and 3 c 1i+1 When <0,

[0444] Let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1min .

[0445] when 3 c 1i >0 and 3 c 1i+1 When >0,

[0446] calculate 9 d 01 = 3 c 1i ·u imax + 3 c 1i+1 ·u i+1min , 9 d 02 = 3 c 1i ·u imin + 3 c 1i+1 ·u i+1max .

[0447] Judgment: If 9 d 01 > 9 d 02 Let the i-th row of matrix Λ' be u imax The (i+1)th row is all u i+1min ;like 9 d 01 < 9 d 02 Let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1max ;like 9 d 01 = 9 d 02 The i-th and (i+1)-th rows of matrix Λ' correspond to two values: one is to let the i-th row of matrix Λ' all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Λ' is all u. imin The (i+1)th row is all u i+1max .

[0448] 4-5-4) Except for rows p, q, p', process the other rows of matrix Λ' according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into set Ω, and then proceed to step 4-5-5); where i = 1, ..., m, i ≠ p, p', q;

[0449] The rules are as follows:

[0450] when 3 c 1i · 3 c 1i+1 When <0,

[0451] like 3 c 1i >0 and 3 c 1i+1 <0, let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1max ;

[0452] like 3 c 1i <0 and 3 c 1i+1 >0, let the i-th row of matrix Λ' be u imax The (i+1)th row is all u i+1min .

[0453] when 3 c 1i >0 and 3 c1i+1 When >0,

[0454] Let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1min .

[0455] when 3 c 1i <0 and 3 c 1i+1 When <0,

[0456] calculate 10 d 01 = 3 c 1i ·u imax + 3 c 1i+1 ·u i+1min , 10 d 02 = 3 c 1i ·u imin + 3 c 1i+1 ·u i+1max .like 10 d 01 > 10 d 02 Let the i-th row of matrix Λ' be u imin The (i+1)th row is all u i+1max ;like 10 d 01 < 10 d 02 Let the i-th row of matrix Λ' be u imax The (i+1)th row is all u i+1min ;like 10 d 01 = 10 d 02 The i-th and (i+1)-th rows of matrix Λ' correspond to two values: one is to let the i-th row of matrix Λ' all be u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Λ' is all u. imin The (i+1)th row is all u i+1max .

[0457] 4-5-5) Construct coordinate rotation transformation 3 F is used to make the transformed coordinate axis v1 perpendicular to the image of the boundary surface with vertex matrix Π' in this group. Then, the coordinate values ​​of the images of all points on the boundary surface on the v1 axis are equal. Since only the v1 axis is considered, it is only necessary to... 3 The first line of F.

[0458] set up

[0459] in, 3 h ij For matrix 3 The element in the i-th row and j-th column of H, 3 f ij For matrix 3 The element in the i-th row and j-th column of F.

[0460] make 3 H = 3 F·B, must have 3 h 1q =0; will 3 Substitute F and B 3 H = 3 F·B, has:

[0461]

[0462] Solving the system of linear equations (13), we obtain... 3 The first line of F, and using 3 H = 3 F·B calculation 3 The first row of the H matrix ( 3 h 11 ,..., 3 h 1m ).like 3 h 1p' If > 0, proceed to steps 4-5-6); if 3 h 1p' If <0, proceed to steps 4-5-7);

[0463] 4-5-6) Except for rows p, q, p', process the other rows of matrix Π' according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into set Ω, and then proceed to step 4-6); where i = 1, ..., m, i ≠ p, p', q;

[0464] The rules are as follows:

[0465] when 3 h 1i · 3 h 1i+1 When <0,

[0466] like 3 h 1i >0 and 3 h 1i+1 <0, let the i-th row of matrix Π' be u imax The (i+1)th row is all u i+1min ;

[0467] like 3 h 1i <0 and 3 h 1i+1 >0, let the i-th row of matrix Π' be all u imin The (i+1)th row is all u i+1max .

[0468] when 3 h 1i <0 and 3 h 1i+1 When <0,

[0469] Let the i-th row of matrix Π' be u imin The (i+1)th row is all u i+1min .

[0470] when 3 h 1i >0 and 3 h 1i+1 When >0,

[0471] calculate 11 d 01 = 3 h 1i ·u imax + 3 h 1i+1 ·u i+1min , 11 d 02 = 3 h 1i ·u imin + 3 h 1i+1 ·u i+1max .like 11 d 01 > 11 d 02 Let the i-th row of matrix Π' be all u imax The (i+1)th row is all u i+1min ;like 11 d 01 < 11 d 02 Let the i-th row of matrix Π' be all u imin The (i+1)th row is all u i+1max ;like 11 d 01 = 11 d 02 The i-th and (i+1)-th rows of matrix Π' have two possible values: one is to set all values ​​in the i-th row of matrix Π' to u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Π' is all u. iminThe (i+1)th row is all u i+1max .

[0472] 4-5-7) Except for rows p, q, p', process the other rows of matrix Π' according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface. Put it into set Ω, and then proceed to step 4-6); where i = 1, ..., m, i ≠ p, p', q;

[0473] The rules are as follows:

[0474] when 3 h 1i · 3 h 1i+1 When <0,

[0475] like 3 h 1i >0 and 3 h 1i+1 <0, let the i-th row of matrix Π' be u imin The (i+1)th row is all u i+1max ;

[0476] like 3 h 1i <0 and 3 h 1i+1 >0, let the i-th row of matrix Π' be all u imax The (i+1)th row is all u i+1min .

[0477] when 3 h 1i <0 and 3 h 1i+1 When <0,

[0478] calculate 12 d 01 = 3 h 1i ·u imax + 3 h 1i+1 ·u i+1min , 12 d 02 = 3 h 1i ·u imin + 3 h 1i+1 ·u i+1max .like 12 d 01 > 12 d 02 Let the i-th row of matrix Π' be all u imin The (i+1)th row is all u i+1max ;like 12 d01 < 12 d 02 Let the i-th row of matrix Π' be all u imax The (i+1)th row is all u i+1min ;like 12 d 01 = 12 d 02 The i-th and (i+1)-th rows of matrix Π' have two possible values: one is to set all values ​​in the i-th row of matrix Π' to u. imax The (i+1)th row is all u i+1min Another type is where the i-th row of matrix Π' is all u. imin The (i+1)th row is all u i+1max .

[0479] when 3 h 1i >0 and 3 h 1i+1 When >0,

[0480] Let the i-th row of matrix Π' be u imin The (i+1)th row is all u i+1min .

[0481] 4-6) Return to step 4-1), select the next group whose critical boundary surface has not been determined, until all groups have determined their critical boundary surfaces, and then proceed to step 5).

[0482] 5) Check all critical boundary surfaces in Ω and determine: if there are identical critical boundary surfaces, keep only one of them and remove the other critical boundary surfaces that are the same as the kept critical boundary surface; after processing all critical boundary surfaces, form a new set of critical boundary surfaces denoted as Ω', and then proceed to step 6).

[0483] 6) All critical boundary surfaces in Ω' are the same as all critical boundary surfaces in the control set. Map all vertices of each critical boundary surface in Ω' using the mapping v = B·u. ELR This yields all vertices of the corresponding control reachable set boundary surface, thus determining a control reachable set boundary surface, which can be either quadrilateral or triangular. The control reachable set boundary surfaces determined by all critical boundary surfaces in Ω' constitute the control reachable set boundary θ(Φ). ELR ).

[0484] The method of the present invention will be further described in detail below with reference to a specific embodiment.

[0485] Example

[0486] This embodiment calculates the control reachability set of a four-wheel independent drive, independent steering vehicle.

[0487] Given that the four longitudinal forces of the four wheels are F L1 F L2 F L3 F L4 The four lateral forces are F T1 F T2 F T3 F T4 , let u V =(F L1 ,F T1 ,F L2 ,F T2 ,F L3 ,F T3 ,F L4 ,F T4 ) T F Li min ≤F Li ≤F Li max F Ti min ≤F Ti ≤F Ti max , i = 1,...,4, F Li min F Ti min F represents the minimum value of the longitudinal and lateral forces on each wheel. Li max F Ti max The maximum values ​​of longitudinal and lateral forces for each wheel; [F] L1 min ,...,F L4 min ]=[-16,-16,-19.3,-19.3];[F T1min ,...,F T4min ] = [-5, -5, -9.21, -9.21]; [F L1max ,...,F L4max ] = [16,16,9.3,9.3]; [F T1max ,...,F T4max ]=[5,5,9.21,9.21]; The i-th longitudinal force and the i-th lateral force have the following linear relationship: (F Limax -F Limin )F Ti +(F Ti max -F Ti min )F Li ≤F Timax F Limax -F Ti min F iLmin ; All u V The control set Ω that constitutes a four-wheel independent drive and independent steering vehicle V ={u V Meanwhile, the wheel force control efficiency matrix B is known. V ,

[0488]

[0489] Calculate the control reachable set Φ of a four-wheel independent drive-independent steering vehicle V Φ V ={v VS |v VS =B V ·u VS =(F LS ,F TS M S ) T u VS ∈Ω V}, where F LS For a specific vehicle overall longitudinal force, F TS For a specific vehicle overall lateral force, M S The overall yaw moment of a specific vehicle.

[0490] This embodiment proposes a method for calculating the reachable set of a redundant drive system under linear constraints, including the following steps:

[0491] 1) Construct the control reachable set of the redundant drive system under each pair of linear constraint control components.

[0492] The correspondences between the physical quantities in the embodiments and the terms used above are as follows: Ω V Corresponding Ω ELR B V For the corresponding B, Φ V Corresponding Φ ELR u V =(F L1 ,F T1 ,F L2 ,F T2 ,F L3 ,F T3 ,F L4 ,F T4 ) T Corresponding to u ELR =(u1,...,u8) T The physics problem is:

[0493] Determine an 8-dimensional control vector u ELR =(u1,...,u8) T The control reachable set, where u1 and u2 are a pair of control components, u3 and u4 are a pair of control components, u5 and u6 are a pair of control components, and u7 and u8 are a pair of control components; each control component u i There is a range of values, namely u. imin ≤u i ≤u imax i = 1, ..., 8

[0494] [u 1min ,...,u 8min ] = [-16, -5, -16, -5, -19.3, -9.21, -19.3, -9.21],

[0495] [u 1max ,...,u 8max ] = [16,5,16,5,19.3,9.21,19.3,9.21];

[0496] Each pair of control components has the following linear relationship:

[0497] (u kmax -u kmin )u k+1 +(u k+1max -u k+1min )u k ≤u k+1max u kmax -u k+1min u kmin k = 1, 3, 5, 7;

[0498] Let Ω ELR ={u ELR},in,

[0499] u ELR =(u1,...,u8) T

[0500] st

[0501] u imin ≤u i ≤u imax i = 1, ..., 8;

[0502] (u kmax -u kmin )u k+1 +(u k+1max -u k+1min )u k ≤u k+1max u kmax -u k+1min u kmin k = 1, 3, 5, 7; Ω ELR Each control vector u ELR The control reachability vector v, v = B·u, is generated through the following mapping. ELR , where B:R 8 →R 3 ,

[0503]

[0504] Requirement: Determine the control reachable set

[0505] 2) Control set Ω ELR All boundary surfaces are divided into four types;

[0506] control vector u ELR =(u1,...,u8) T Each component u i The control set Ω is obtained by taking the maximum or minimum value of the corresponding constraint for (i = 1, ..., 8). ELR The vertex. The 8-dimensional control set Ω represented in this embodiment. ELR There are 3 4 Each vertex constitutes a... There are four types of boundary surfaces: Type I rectangular boundary surface, Type II rectangular boundary surface, Type III rectangular boundary surface, and triangular boundary surface.

[0507] 3) Divide all boundary surfaces of the control set into Grouped.

[0508] The p-th and q-th control components u p u q (1≤p≤8, 1≤q≤8, p<q) takes values ​​between the corresponding minimum and maximum values, while the other 6 components u i (1≤i≤8,i≠p,i≠q) takes the corresponding minimum or maximum value, i.e., u i =u imax or u imin This combination yields a total of 2 6 The boundary surfaces are grouped into a single group, called the pq group. All boundary surfaces in the control set are divided into groups 1-2, 1-3, ..., 7-8, for a total of 28 groups.

[0509] 4) Perform the steps in step 3) sequentially. Each group is divided into groups, and the critical boundary surfaces are determined in each group.

[0510] There are 28 groups in total. Due to space limitations, only groups 1-2 and 3-5 will be used as examples to illustrate the process of determining the critical boundary surfaces. Let Ω be the set of critical boundary surfaces, and Ω is initialized to an empty set.

[0511] 4-1) Record group 1-2 as the current pq group:

[0512] 4-2) The first and second control components are a pair of linear constraint components, therefore, proceed to step 4-3);

[0513] 4-3)

[0514] 4-3-1) Should point Group All boundary surfaces are triangular boundary surfaces ；

[0515] 4-3-2)

[0516] Construct coordinate rotation transformation 1 T, so that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface of the grouped triangle. Since only the v1 axis is considered, it is only necessary to... 1 The first row of T. Given

[0517] set up make 1 C = 1 T·B, must have 1 c 11 =0, 1 c 12 =0. (This likely refers to a specific value or quantity.) 1 Substitute T and B 1 C = 1 T·B, obtained

[0518]

[0519] Solve the system of linear equations (14), let 1 t 13 =1, calculation yields 1 t 11 =-1.4, 1 t 12 = -0.7.

[0520] Reuse 1 C = 1 T·B can be calculated 1 The first row of matrix C ( 1 c 11 ,..., 1 c 18 = (0, 0, 0.4093, -1.3388, -2.8738, -2.0252, -2.5334, -0.5849). Proceed to step 4-3-3);

[0521] 4-3-3) Due to 1 c 13 · 1 c 14 <0, and 1 c 13 >0, 1 c 14 <0, so let the third row of matrix Δ all be u. 3max The fourth row is all u 4min ;because 1 c 15 <0,1 c 16 <0, so let the 5th row of matrix Δ be all u. 5min The 6th line is all u 6min ;because 1 c 17 <0, 1 c 18 <0, so let the 7th row of matrix Δ be all u. 7min The 8th line is all u 8min The resulting matrix is ​​denoted by Δ1. The boundary surface represented by matrix Δ1 is the critical boundary surface and is placed into set Ω. Proceed to step 4-3-4).

[0522]

[0523] 4-3-4) Due to 1 c 13 · 1 c 14 <0, and 1 c 13 >0, 1 c 14 <0, so let the third row of matrix Δ all be u. 3min The fourth row is all u 4max ;because 1 c 15 <0, 1 c 16 <0, and 1 c 15 ·u 5max + 1 c 16 ·u 6min < 1 c 15 ·u 5min + 1 c 16 ·u 6max Therefore, let the 5th row of matrix Δ be all u. 5max The 6th line is all u 6min ;because 1 c 17 <0, 1 c 18 <0, 1 c 17 ·u 7max + 1 c 18 ·u 8min < 1 c 17 ·u 7min + 1 c 18 ·u 8max Therefore, let the 7th row of matrix Δ be all u.7max The 8th line is all u 8min The resulting matrix is ​​represented by Δ2. The boundary surface represented by matrix Δ2 is the critical boundary surface and is placed into set Ω. Proceed to steps 4-6).

[0524]

[0525] 4-1) Record group 3-5 as the current pq group:

[0526] 4-2) If the 3rd and 5th control components are not a pair of linear constraint components and m = 8 is an even number, proceed to step 4-4);

[0527] 4-4)

[0528] 4-4-1) The boundary surfaces of this pq group contain three types of rectangular boundary surfaces. Proceed to step 4-4-2);

[0529] 4-4-2) Construct a coordinate rotation transformation G such that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface of the type III rectangle in this group. Since only the v1 axis is considered, only the first row of G is needed. Given...

[0530]

[0531] set up Let E = G·B, then we must have Substituting G and B into E = G·B, we have

[0532]

[0533] Solving the linear system of equations (15) yields the first row of G, and the first row of the E matrix (e) is calculated using E = G·B. 11 ,...,e 18 ) = (0.2621 1.8566 0.5041 1.6130 1.0 2.0955 1.0152 2.4431). Since e 14 >0 and e 16 >0, proceed to step 4-4-3);

[0534] 4-4-3) Except for rows 3, 4, 5, and 6, process the other rows of matrix T according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface, and it is placed into set Ω. Then proceed to step 4-4-5);

[0535] Because of e 11 >0 and e 12 >0, and e 11 ·u 1max +e 12 ·u 2min <e11 ·u 1min +e 12 ·u 2max Therefore, let the first row of matrix T all be u. 1min The second row is all u 2max ; due to e 17 >0 and e 18 >0, and e 17 ·u 7max +e 18 ·u 8min <e 17 ·u 7min +e 18 ·u 8max Therefore, let the 7th row of matrix T all be u. 7min The 8th line is all u 8max The resulting matrix is ​​denoted by T1. The boundary surface represented by matrix T1 is the critical boundary surface and is placed into the set Ω.

[0536]

[0537] 4-4-5) Construct coordinate rotation transformation 2 T, so that the transformed coordinate axis v1 is perpendicular to the image of the boundary surface of the grouped I-shaped rectangle. Since only the v1 axis is considered, it is only necessary to... 2 The first row of T. Given

[0538]

[0539] set up make 2 C = 2 T·B, must have 2 c 13 =0, 2 c 18 =0. (This likely refers to a specific value or quantity.) 2 Substitute T and B 2 C = 2 T·B, obtained

[0540]

[0541] Solve the system of linear equations (16), let 2 t 13 =1, calculation yields 2 t 11 =-0.0143, 2 t 12 =5.2321.

[0542] Reuse 2 C = 2 T·B can be calculated 2 The first line of C is ( 2c 11 ,..., 2 c 18 ) = (-0.7266, 5.1005, 0, 3.9289, 0, 4.0633, 0.151, 5.507). Since 2 c 14 >0 and 2 c 16 >0, proceed to step 4-4-7);

[0543] 4-4-7) Except for rows 3, 4, 5, and 6, process the other rows of matrix Λ according to the following rules. The boundary surfaces represented by the resulting matrix are the critical boundary surfaces, and are placed into set Ω. This group has found two critical boundary surfaces, proceed to step 4-6);

[0544] because 2 c 11 · 2 c 12 <0, and 2 c 11 <0, 2 c 12 >0, so let the first row of matrix Λ be all u. 1max The second row is all u 2min ;because 2 c 17 >0, 2 c 18 >0, so let the 7th row of matrix Λ be all u. 7min The 8th line is all u 8min The resulting matrix is ​​represented by Λ1. The boundary surface represented by matrix Λ1 is the critical boundary surface and is placed into set Ω.

[0545]

[0546] 4-6) According to steps 4-1) to 4-5-7) of the present invention, the key boundary surface can be determined for each group.

[0547] Due to space limitations, the key boundary surfaces of all groups will not be determined. Therefore, since the above steps do not provide all elements of set Ω, step 5 is omitted, and we proceed to step 6.

[0548] Step 6) The control reachable set boundary surfaces determined by all critical boundary surfaces in Ω' constitute the control reachable set boundary. According to the implementation steps of this invention, all 36 boundary surfaces of the control reachable set in this embodiment can be calculated. Since step 5) is omitted, all control reachable set boundary surfaces are no longer calculated. Only the control reachable set boundary surfaces of groups 1-2 and 3-5 are calculated below.

[0549] The three vertices of the critical boundary surface Δ1 of group 1-2 are mapped using v = B·u. ELR This yields the three vertices of the corresponding control reachable set boundary surface, thus defining a control reachable set boundary surface that is triangular, which is then determined using a matrix. This means that each column represents a vertex.

[0550]

[0551] The three vertices of the other critical boundary surface Δ2 of group 1-2 are mapped using v = B·u. ELR This yields the three vertices of the corresponding control reachable set boundary surface, thus determining a control reachable set boundary surface, which is a triangle, represented by matrix θ(Δ2), with each column representing a vertex.

[0552]

[0553] The four vertices of the critical boundary surface T1 in the 3-5 group are mapped using v = B·u. ELR This yields the four vertices of the corresponding control reachable set boundary surface, thus defining a control reachable set boundary surface that is quadrilateral, determined using a matrix. This means that each column represents a vertex.

[0554]

[0555] The four vertices of the other critical boundary surface Λ1 in group 3-5 are mapped using v = B·u. ELR This yields the four vertices of the corresponding control reachable set boundary surface, thus defining a control reachable set boundary surface that is quadrilateral, determined using a matrix. This means that each column represents a vertex.

[0556]

[0557] In summary, the above are merely preferred embodiments of the present invention and are not intended to limit the scope of protection of the present invention. Any modifications, equivalent substitutions, improvements, etc., made within the spirit and principles of the present invention should be included within the scope of protection of the present invention.

Claims

1. A method for computing the control reachable set of a redundant drive system under linear constraints, characterized in that, Includes the following steps: 1) The control reachability set expression for the redundant drive system under each pair of linear constraint control components is established as follows: （1） wherein is the integer part of is a control vector, and denoted by , , , , ; Among them, the One control component For the corresponding number The control action of each actuator , The number of actuators; For the first Minimum constraint value of each actuator control action. For the first The maximum constraint value of the control action of each actuator; For control set, ; For the control reachability vector of the redundant drive system, , representing the control output of the redundant drive system; To control reachability sets; for OK The control efficiency matrix of the column; Rewrite equation (1) as equation (2): （2） In the formula, , For a pair of control components with linear constraints, The control reachable set of a redundant drive system where each pair of control components is a linearly constrained control component; For control set, For control vectors, ; 2) The control set obtained in step 1) All boundary surfaces are divided into four types; Control Vector Each component The control set is obtained by taking the maximum or minimum value of the corresponding constraint. The vertex, represented by equation (2) Dimensional control set have One vertex, for The remainder when divided by 2; make express The boundary; based on the values ​​of each component of each vertex... The boundary surfaces are divided into four types: Type I rectangular boundary surface, Type II rectangular boundary surface, Type III rectangular boundary surface, and triangular boundary surface. Here, the four vertices of the rectangular boundary surface are denoted as follows, in clockwise order: , , , The three vertices of the triangular boundary face are arranged clockwise as follows: , , ,but: If any rectangular boundary surface satisfies: Vertex and One component in the set has a different value, while the rest of the components have the same value, and the vertex... and In one pair of components, one component has a different value, while the other components are the same, and the vertex... and If the two components mentioned above have different values ​​and the other components are the same, then the rectangular boundary surface is a type I rectangular boundary surface. If any rectangular boundary surface satisfies: Vertex With vertex There is a pair of linear constraint components whose values ​​are all different, while the other components are the same, and the vertex... With vertex The values ​​of these two linear constraint components are all different, while the values ​​of the other components are the same, and the vertex... With vertex If the linear constraint components have the same value, and only one of the other components has a different value while the rest are the same, then the rectangular boundary surface is a type II rectangular boundary surface. If any rectangular boundary surface satisfies: Vertex With vertex There is a pair of linear constraint components whose values ​​are all different, while the other components are the same, and the vertex... With vertex There is another pair of linear constraints where all component values ​​are different, the remaining components are the same, and the vertex... With vertex If each component of the two pairs of linear constraint components mentioned above has a different value, and the remaining components have the same value, then the rectangular boundary surface is a type III rectangular boundary surface. If any right triangle satisfies: vertex With vertex There is a pair of linear constraint components with all different values, while the other components have the same value, and the vertex... With vertex The linear constraint components have one component with a different value, while the other components have the same value, and the vertex... With vertex If the other component of this pair of linear constraint components has a different value, and the other components have the same value, then the right triangle is a triangular boundary surface. 3) Control set All boundary surfaces are divided into One group; like The two components take values ​​between their corresponding minimum and maximum values, and the rest... If each component takes the corresponding minimum or maximum value, then the Each component forms Boundary surfaces of control sets; remember Any two components in the middle are the first The component and the first The component, the first The component and the first Each component takes a value between its corresponding minimum and maximum values. , ,the remaining Each component takes the corresponding minimum or maximum value, and the resulting boundary surface is denoted as . Grouping; then the control set obtains a total of all boundary surfaces. One group; 4) Regarding step 3): For each of the groups, determine the critical boundary surface; Mapping to Like at the boundary of The boundary surface in the middle is the critical boundary surface; let it be... For the set of critical boundary surfaces, Initialize to an empty set; 4-1) Randomly select a group whose critical boundary surface has not been determined and denote it as the current group. Grouping, proceed to step 4-2). 4-2) Determine the first The first control component and the second Are the control components a pair of linear constraint components? If so, proceed to step 4-3). If not, and when If the number is even, proceed to step 4-4). If not, and when When it is an odd number, if If each component has a paired linear constraint component, then proceed to step 4-4). If not, and when When it is an odd number, if If there is no paired linear constraint component, proceed to steps 4-5). 4-3) When the first The and the first When each control component is a pair of linear constraint components, determine The specific method for grouping key facets is as follows: 4-3-1) When the first The component and the first When each component is a pair of linearly constrained components, the boundary surface of this group is a triangular boundary surface, and the value of each component at each point on the boundary surface satisfies the following formula: （3） Each vertex of the boundary face in this group is represented by a... OK Column matrix express, Each column is The vector corresponding to the 3 vertices; the 3 vertices of each boundary surface The, the Each component takes values ​​as shown in the matrix. The line, number As shown in the row, the values ​​of the remaining components are all or all In the remaining components, a pair of linear constraint components cannot simultaneously take the maximum value of their corresponding constraints; where, the matrix The Behavior , No. Behavior ; ； 4-3-2) Constructing coordinate rotation transformations The transformed coordinate axes The image perpendicular to the boundary plane of the triangular group; set up , , ,in, For matrix The Line number Column elements, For matrix The Line number Column elements, For matrix The Line number Column elements; make ,but , ; Will , Substitution ,get: （4） Solving the system of linear equations (4), we obtain... The first line ; use ,calculate The first row of the matrix ; 4-3-3) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. ,in, ; The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , Then determine: like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-3-4) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. Then proceed to steps 4-6), where, ; The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , Then determine: if Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-4) When the first The and the first The control components are not a pair of linear constraint components and When each has a paired linear constraint component, determine Grouping key small faces; 4-5) When the first The and the first The control components are not a pair of linear constraint components and When there are no paired linear constraint components, determine Grouping key small faces; 4-6) Return to step 4-1), select the next group whose critical boundary surfaces have not been determined, until all groups have determined their critical boundary surfaces, and then proceed to step 5). 5) Inspection We process all critical boundary surfaces and determine: if duplicate critical boundary surfaces exist, we retain only one of them and remove the rest that are identical to the retained critical boundary surface; after processing all critical boundary surfaces, a new set of critical boundary surfaces is formed, denoted as . Then proceed to step 6). 6) All critical boundary surfaces in the control set; All vertices of each critical boundary surface are mapped This yields all the vertices of the corresponding control reachable set boundary surface, thereby determining a control reachable set boundary surface, which can be a quadrilateral or a triangle. The control reachable set boundary surfaces determined by all critical boundary surfaces constitute the control reachable set boundary. .

2. The method according to claim 1, characterized in that, Also includes: 4-4-1) This The grouped boundary surface contains three types of rectangular boundary surfaces. The values ​​of each component at every point on this grouped boundary surface satisfy the following formula, denoted as the... The components that are paired together are the first component. The component, and the first component The components that are paired together are the first component. One component; （5） when , When the boundary surface formed by the points satisfying equation (5) is a type I rectangular boundary surface; when , When the boundary surface formed by the points satisfying equation (5) is a type II rectangular boundary surface; when , When the boundary surface formed by the points satisfying equation (5) is also a type II rectangular boundary surface; when , When the boundary surface formed by the points satisfying equation (5) is a type III rectangular boundary surface; The vertices of each rectangular boundary face in this group are represented by a single vertex. OK Column matrix representation, The columns are respectively The vectors corresponding to each vertex; the matrix The matrix represents the vertices of a type I rectangular boundary surface. or The matrix represents the vertices of a type II rectangular boundary surface. Represents a vertex of a type III rectangular boundary surface; the fourth vertex of each boundary surface is... The, the The, the The, the The value of the nth component is as shown in the matrix. line, number line, number line, number As shown in the row, the remaining components are or all A pair of linear inequality constraint components cannot simultaneously take the maximum value of their corresponding constraints; where, matrix The Behavior , No. Behavior , No. Behavior , No. Behavior ; matrix The Behavior , No. Behavior , No. Behavior , No. Behavior ; matrix The Behavior , No. Behavior , No. Behavior , No. Behavior ; matrix The Behavior , No. Behavior , No. Behavior , No. Behavior ; ， ， ， （6） in, like If it is an odd number, then the first... line in Before going; if If it is even, then the first... line in After the action; if If it is an odd number, then the first... line in Before going; if If it is even, then the first... line in After the action; 4-4-2) Constructing coordinate rotation transformations The transformed coordinate axes If the image is perpendicular to the boundary surface of the type III rectangle, then the images of all points on the boundary surface are in... The coordinate values ​​of the axes are equal; set up , , ,in, For matrix The Line number Column elements, For matrix The Line number Column elements; make ,but ;Will , Substitution ,have: （7） Solving the system of linear equations (7) yields The first line, and using calculate The first row of the matrix ;like and Then proceed to step 4-4-3); if and Then proceed to step 4-4-4); if Then proceed to step 4-4-5). 4-4-3) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. Then proceed to steps 4-4-5), where, ； in, The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , Then determine: if Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like Then the matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-4-4) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. Then proceed to steps 4-4-5), where, ； in, The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, calculate , ;like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; 4-4-5) Constructing coordinate rotation transformations The transformed coordinate axes Image perpendicular to the boundary surface of the type I rectangle in this group; set up , , ,in, For matrix The Line number Column elements, For matrix The Line number Column elements; make ,but , ;Will , Substitution ,get: （8） Solving the system of linear equations (8) yields The first line ; use ,calculate The first row of the matrix ; like and Then proceed to step 4-4-6); if and Then proceed to step 4-4-7); if Then proceed to step 4-4-8). 4-4-6) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. ,in, Then determine: If two critical boundary surfaces have been found in the group, proceed to step 4-6; otherwise, proceed to step 4-4-8. in, The rules are as follows: when hour like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , .like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-4-7) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. ,in, ； Then determine: If two critical boundary surfaces have been found in the group, proceed to step 4-6; otherwise, proceed to step 4-4-8. in, The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, calculate , ;like Let the matrix The All actions , No. All actions ;like Let the matrix The All actions , No. All actions ;like ,matrix The line, number The row corresponds to two values: one is to let the matrix... The All rows are , No. All rows are Another approach is to let the matrix... The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; 4-4-8) Constructing coordinate rotation transformations The transformed coordinate axes The matrix perpendicular to the vertices in this group is The image of the boundary surface, then the image of all points on the boundary surface is in The coordinate values ​​of the axes are equal; set up , , ,in, For matrix The Line number Column elements, For matrix The Line number Column elements; make ,but , .Will , Substitution ,have: （9） Solving the system of linear equations (9), we obtain... The first line, and using calculate The first row of the matrix ; like and Then proceed to step 4-4-9); if and Then proceed to step 4-4-10); if Then proceed to step 4-4-11). 4-4-9) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. ,in, ； Then determine: If two critical boundary surfaces have been found in the group, proceed to step 4-6); otherwise, proceed to step 4-4-11). in, The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , ;like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-4-10) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. ,in, Then determine: If two critical boundary surfaces have been found in the group, proceed to step 4-6); otherwise, proceed to step 4-4-11). in, The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , ;like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-4-11) Constructing coordinate rotation transformations The transformed coordinate axes The matrix perpendicular to the vertices in this group is The image of the boundary surface, then the image of all points on the boundary surface is in The coordinate values ​​of the axes are equal; set up , , ,in, For matrix The Line number Column elements, For matrix The Line number Column elements; make ,but , .Will , Substitution ,have: （10） Solving the system of linear equations (10), we obtain... The first line, and using calculate The first row of the matrix ;like and Then proceed to step 4-4-12); if and Proceed to step 4-4-13); if Proceed to steps 4-6). 4-4-12) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. ,in, Then proceed to steps 4-6). in, The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , .like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-4-13) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. ,in, Then proceed to steps 4-6). in, The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, calculate , ;like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are .

3. The method according to claim 1, characterized in that, Also includes: 4-5-1) This The boundary surface of the group contains both type I and type II rectangular boundary surfaces; the values ​​of each component at each point on the boundary surface of the group satisfy equation (11), denoted as the first... The components that are paired together are the first component. One component; （11） when When, the boundary surface formed by the points satisfying equation (11) is a type I rectangular boundary surface; when When the boundary surface formed by the points satisfying equation (11) is a type II rectangular boundary surface; The vertices of each rectangular boundary face in this group are represented by a single vertex. OK Column matrix representation, The columns are respectively The vectors corresponding to each vertex; using a matrix The matrix represents the vertices of a type I rectangular boundary surface. Let each vertex of a type II rectangular boundary face be a vertex of a type II rectangular boundary face. The, the The, the The value of the nth component is as shown in the matrix. line, number line, number As shown in the row, the remaining components are or all A pair of linear constraint components cannot simultaneously take the maximum value of their corresponding constraints; matrix The first Behavior , No. Behavior , No. Behavior , No. Behavior ; ， As shown in the above formula, if the matrix is... If it is an odd number, the first line in Before going, if If it is even, the first line in After the action; 4-5-2) Constructing coordinate rotation transformations The transformed coordinate axes Image perpendicular to the boundary surface of the type I rectangle in this group; set up , , ,in, For matrix The Line number Column elements, For matrix The Line number Column elements; make ,but , .Will , Substitution ,get: （12） Solving the system of linear equations (12), we obtain... The first line ;use ,calculate The first row of the matrix ;like Then proceed to step 4-5-3); if Then proceed to step 4-5-4). 4-5-3) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. Then proceed to steps 4-5-5); where, ; The rules are as follows: when hour like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , ; Judgment: If Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-5-4) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. Then proceed to steps 4-5-5); where, ; The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , ;like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-5-5) Construct coordinate rotation transformation The transformed coordinate axes The matrix perpendicular to the vertices in this group is The image of the boundary surface, then the image of all points on the boundary surface is in The coordinate values ​​of the axes are equal; set up , , ,in, For matrix The Line number Column elements, For matrix The Line number Column elements; make ,but , ,Will , Substitution ,have: （13） Solving the system of linear equations (13), we obtain... The first line, and using calculate The first row of the matrix ;like Then proceed to steps 4-5-6); if Then proceed to steps 4-5-7). 4-5-6) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. Then proceed to steps 4-6); where, ; The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are ; when hour, calculate , ;like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; 4-5-7) Except for the first Outside the row, the matrix The other rows are processed according to the following rules. The boundary surface represented by the resulting matrix is ​​the critical boundary surface and is placed into the set. Then proceed to steps 4-6); where, ; The rules are as follows: when hour, like and Let the matrix The All rows are , No. All rows are ; like and Let the matrix The All rows are , No. All rows are ; when hour, calculate , ;like Let the matrix The All rows are , No. All rows are ;like Let the matrix The All rows are , No. All rows are ;like ,matrix The line, number Rows correspond to two values: one is to let the matrix... The All rows are , No. All rows are Another type is a matrix. The All rows are , No. All rows are ; when hour, Let matrix The All rows are , No. All rows are .

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