A method for calculating the beamforming direction vector of a towed array sonar platform during its steering process
By establishing an array arc trajectory steering model during the steering process of the dragline array sonar platform, decomposing the steering process into 5 stages and giving direction vectors, the problem of array position offset affecting the accuracy of target positioning is solved, and the target position estimation performance is improved.
Patent Information
- Application Number
- CN202111347812.X
- Authority / Receiving Office
- CN · China
- Patent Type
- Patents(China)
- Current Assignee / Owner
- Filing Date
- 2021-11-15
- Publication Date
- 2025-04-11
- Estimated Expiration
- 2041-11-15
Smart Images
Figure CN114063006B_ABST
Abstract
Description
Technical Field
[0001] The invention belongs to the technical field of underwater acoustic detection and sonar signal processing, and relates to a method for measuring a beam forming direction vector during the steering process of a towed array sonar platform. Background Art
[0002] Towed array sonar is an important tool for detecting low-frequency and concealed targets. During the detection process, the towed array sonar needs to use beamforming azimuth spectrum estimation technology, the core of which is the use of the direction vector of the array. During the detection process, the platform of the towed array sonar often needs to change its course, and the flexible towed array also turns accordingly. Affected by the platform's turning, the position of each array element in the array is offset from the position of the ideal linear array element, and the received array data is significantly different from the received array data of the ideal linear array. This results in the use of the direction vector under ideal conditions to estimate the target azimuth for the actual array data with array element errors, and the target positioning result is inaccurate. During the platform turning process, accurately evaluating the real-time direction vector of the array is of great value in improving the target azimuth estimation performance. Summary of the invention
[0003] In order to solve the above problems, the present invention provides a method for calculating the beamforming direction vector during the steering process of a towed array sonar platform, and provides the beamforming direction vector of the towed array sonar in real time when the platform is steering.
[0004] The technical solution of the present invention is:
[0005] The method for calculating the beamforming direction vector during the steering process of the towed array sonar platform is as follows:
[0006] Step 1: Establish a model for the array to turn around a circular trajectory: Assume that the first element of the array is 1, the last element is N, d is the element spacing, θ is the incident direction of the far-field plane wave, λ is the wavelength of the sound wave, and a(θ) is the direction vector.
[0007] Assume that the platform steering angle is α, the turning radius is r, and the array aperture is L = (N-1)d. Define αr < L as a small angle steering, and αr ≥ L as a large angle steering. During the array steering process, the center angle β corresponding to the length of the array on the ring changes with time. According to the value of the center angle β, the array steering is divided into 5 time stages, and the intersection points of each time stage are defined as time t0, t1, t2, and t3 respectively.
[0008] Step 2: Give the direction vectors of the five time stages for small angles, large angles, left turns, and right turns. The details are as follows:
[0009] (I) Small angle steering
[0010] When the angle is small, the value of the central angle β corresponding to the array on the turning ring is given by Figure 1 As shown in Figure 2, the steering process can be decomposed into five stages according to the value of β.
[0011] Case 1: Turn right at a small angle
[0012] When the platform and the array turn right at a small angle, the array formation diagram at each critical moment is as follows: Figure 2 As shown. Figure 2 It can be seen that t0 is the reference time, corresponding to the time when the first array element reaches the turning ring. where v is the platform (array) velocity.
[0013] (1) When 0≤t≤t0, the array has not yet reached the turning ring and is arranged in a standard linear array; the direction vector is:
[0014]
[0015] (2) When t0<t≤t1, part of the array is located on the turning circle, and the other part is tangent to the turning circle in a straight line. represents rounding down, k represents the last number of array elements on the turning circle; m represents the number of array elements between the 1st and kth elements, β m Indicates the center angle of the arc where the array of the 1st to mth array elements is located; the center angle of the array on the ring At this time the direction vector:
[0016]
[0017] (3) When t1<t≤t2, the first array element of the array is out of the ring, and the last array element is not in the ring. k represents the number of the last array element in the front ring of the array; q represents the last number of array elements on the turning circle; k+1≤m≤q, m represents the number of array elements between the first array element k+1 and the qth array element on the turning circle, β m It represents the center angle of the arc from the first tangent point of the array on the turning circle to the array of the mth array element. At this time, the direction vector is:
[0018]
[0019] (4) When t2<t≤t3, the first array element is located on the tangent line of the ring, and the last array element is on the turning ring. k represents the number of the last array element in the front ring of the array; m represents the number of array elements between the first array element k+1 and the Nth array element on the turning circle, β m It represents the center angle of the arc from the tangent point of the array on the turning circle to the array of the mth array element. At this time, the direction vector is:
[0020]
[0021] (5) When t>t3, the last array element has left the ring, and all array elements are located on the tangent line of the ring, forming a uniform linear array. At this time, the direction vector is:
[0022]
[0023] Case 2: Small left turn
[0024] At this time, the five stages of steering are the same as those of a small right turn, such as Figure 1 As shown in the figure, the meaning of the letters in each stage corresponds to that in case 1. When the platform and the array turn left at a small angle, the array formation diagram at the critical moment of each stage is as follows Figure 3 As shown. t0 is the reference time, where v is the platform (array) velocity.
[0025] In the direction vector results of the small right turn at each time stage, the variables r, β, β m All of them are changed to the opposite of the original value (i.e. multiplied by -1), and the direction vector of the small angle left turn at each time stage can be obtained. Specifically:
[0026] (1) When 0≤t≤t0, the array has not yet reached the turning ring and is arranged in a standard linear array; the direction vector is:
[0027]
[0028] (2) When t0<t≤t1, part of the array is located on the turning circle, and the other part is tangent to the turning circle in a straight line. At this time the direction vector is:
[0029]
[0030] (3) When t1<t≤t2, the first array element of the array is out of the ring, and the last array element is not in the ring.
[0031] At this time the direction vector is:
[0032]
[0033] (4) When t2<t≤t3, the first array element is located on the tangent line of the ring, and the last array element is on the turning ring. At this time the direction vector is:
[0034]
[0035] (5) When t>t3, the last array element has moved out of the ring, and all array elements are located on the tangent of the ring, forming a uniform linear array. At this time, the direction vector is:
[0036]
[0037] (II) Large-angle steering
[0038] When the angle is large, the value of the central angle β corresponding to the array on the turning ring is given by Figure 4 As shown in Figure 2, the steering process can be decomposed into five stages according to the value of β.
[0039] Case 3: Large right turn
[0040] When the platform and the array turn right at a large angle, the array formation diagram at each critical moment is as follows: Figure 5 As shown. Figure 5 It can be seen that t0 is the reference time.
[0041] (1) When 0≤t≤t0, the array has not yet reached the turning ring and is arranged in a standard linear array. At this time, the direction vector is:
[0042]
[0043] (2) When t0<t≤t1, part of the array is located on the turning circle, and the other part is tangent to the turning circle in a straight line. k represents the last element number of the array on the turning circle; m represents the number of array elements between the 1st and kth elements, β m Indicates the center angle of the arc where the array from the 1st array element to the mth array element is located; the center angle of the array on the ring At this time the direction vector is:
[0044]
[0045] (3) When t1<t≤t2, the first and last elements of the array are both inside the ring. m represents the number of array elements between the first and Nth array elements, β m Indicates the center angle of the arc corresponding to the array from the 1st element to the mth element. At this time, the direction vector is:
[0046]
[0047] (4) When t2<t≤t3, the first array element is located on the tangent line of the ring, and the last array element is on the turning ring. m represents the number of array elements between the first array element k+1 and the Nth array element on the turning circle, β m It represents the central angle of the arc from the tangent point of the array on the turning circle to the array of the mth array element. At this time, the direction vector is:
[0048]
[0049] (5) When t>t3, the last array element has moved out of the ring, and all array elements are located on the tangent of the ring, forming a uniform linear array. At this time, the direction vector is:
[0050]
[0051] Case 4: Large left turn
[0052] At this time, the five stages of steering are the same as those of a large right turn, such as Figure 4 As shown in the figure, the meaning of the letters in each stage corresponds to that in case 3. When the platform and the array turn left at a large angle, the array formation diagram at the critical moment of each stage is as follows Figure 6 As shown, the critical moments are: t0 is the reference moment,
[0053] In the direction vector results of the large-angle right turn at each time stage, the variables r, β, β m All of them are changed to the opposite of the original value (i.e. multiplied by -1), and the direction vector of the large-angle left turn at each time stage can be obtained. Specifically:
[0054] (1) When 0≤t≤t0, the array has not yet reached the turning ring and is arranged in a standard linear array. At this time, the direction vector is:
[0055]
[0056] (2) When t0<t≤t1, part of the array is located on the turning circle, and the other part is tangent to the turning circle in a straight line. At this time the direction vector is:
[0057]
[0058] (3) When t1<t≤t2, the first and last elements of the array are both inside the ring. At this time the direction vector is:
[0059]
[0060] (4) When t2<t≤t3, the first array element is located on the tangent line of the ring, and the last array element is on the turning ring. At this time the direction vector is:
[0061]
[0062] (5) When t>t3, the last array element has moved out of the ring, and all array elements are located on the tangent of the ring, forming a uniform linear array. At this time, the direction vector is:
[0063]
[0064] Beneficial effects of the present invention: The present invention establishes a steering model of an array around an arc trajectory, and for large-angle steering and small-angle steering, left steering and right steering conditions, uses the center angle corresponding to the array position arc to decompose the steering process into five steering stages, gives the time point values of four adjacent moments in the five stages, and for each steering stage, takes the first array element as a reference, gives the beamforming direction vector at the current moment, which has great application value for improving the target azimuth estimation performance during the steering process. BRIEF DESCRIPTION OF THE DRAWINGS
[0065] Figure 1 It represents the change of the center angle β corresponding to the length of the array position on the circular ring with time during the small-angle steering process of the array; the value of β divides the array steering into five time stages, and the intersection points of each stage are defined as time t0, t1, t2, and t3 respectively. α is the steering angle of the platform or array.
[0066] Figure 2 It represents the positional relationship between the array formation and the turning circle at the four critical moments when the array turns right at a small angle. Among them, O0(0,0) is the position entering the turning circle, and the coordinate system x0y0 is established at this point. The angle that the platform needs to turn is α. O1(0,0), O2(0,0), and O3(0,0) correspond to the current moment coordinate systems x1y1, x2y2, and x3y3 established by the current position of the first array element, respectively. The established coordinate system x0, x1, x2, and x3 directions point to the left vertical direction of the forward direction of the first array element of the array. The y0, y1, y2, and y3 directions point to the tangential direction from the first array element to the second array element. Figure 3 , Figure 5 , Figure 6 The corresponding coordinate origin and parameters in are also set in the same way.
[0067] Figure 3It shows the positional relationship between the array formation and the turning circle at four critical moments when the array turns left at a small angle.
[0068] Figure 4 It represents the change of the center angle β of the array on the ring with time during the large-angle steering process of the array; the value of β divides the array steering into five time stages, and the intersection points of each stage are defined as time t0, t1, t2, and t3 respectively; where β max is the maximum center angle of the array on the turning arc, Figure 5 , Figure 6 Same as in.
[0069] Figure 5 It shows the positional relationship between the array formation and the turning circle at four critical moments when the array turns right at a large angle.
[0070] Figure 6 It shows the positional relationship between the array formation and the turning circle at four critical moments when the array turns left at a large angle. DETAILED DESCRIPTION
[0071] The specific implementation examples of the present invention are described in detail below in conjunction with the invention content and the accompanying drawings.
[0072] The method for calculating the beamforming direction vector during the steering process of the towed array sonar platform is as follows:
[0073] Step 1: Determine the platform turning direction, turning angle α, turning radius r, number of array elements N and array element spacing d. Calculate L = (N-1)d. If αr < L, it is defined as a small angle turn, and αr ≥ L is defined as a large angle turn.
[0074] Step 2: When the platform and array are turning at a small angle, let t0 be the reference time when the first array element enters the turning circle, When the platform and array are turning at a large angle, let t0 be the reference time,
[0075] Step 3: Determine the four possible situations of steering direction and steering angle, and give the direction vectors of the five stages of the steering process in each situation.
[0076] Situation 1. Small right turn
[0077] (1) When 0≤t≤t0,
[0078]
[0079] (2) When t0<t≤t1,
[0080]
[0081] in, means round down,
[0082] (3) When t1<t≤t2,
[0083]
[0084] in,
[0085] (4) When t2<t≤t3,
[0086]
[0087] in,
[0088] (5) When t>t3,
[0089]
[0090] Case 2: Small left turn
[0091] (1) When 0≤t≤t0,
[0092]
[0093] (2) When t0<t≤t1,
[0094]
[0095] in,
[0096] (3) When t1<t≤t2,
[0097]
[0098] in,
[0099] (4) When t2<t≤t3,
[0100]
[0101] in,
[0102] (5) When t>t3,
[0103]
[0104] Case 3: Large right turn
[0105] (1) When 0≤t≤t0,
[0106]
[0107] (2) When t0<t≤t1,
[0108]
[0109] in,
[0110] (3) When t1<t≤t2,
[0111]
[0112] in
[0113] (4) When t2<t≤t3,
[0114]
[0115] in,
[0116] (5) When t>t3,
[0117]
[0118] Situation 4: Large left turn
[0119] (1) When 0≤t≤t0,
[0120]
[0121] (2) When t0<t≤t1,
[0122]
[0123] in,
[0124] (3) When t1<t≤t2,
[0125]
[0126] in,
[0127] (4) When t2<t≤t3,
[0128]
[0129] in,
[0130] (5) When t>t3,
[0131]
Claims
1. A method for calculating the beamforming direction vector during the steering process of a towed array sonar platform, characterized in that: Here are the steps: Step 1: Establish an array turning model around a circular arc trajectory: assume that the first array element of the array is 1, the last array element is N, d is the array element spacing, θ is the incident direction of the far-field plane wave, λ is the wavelength of the sound wave, and a(θ) is the direction vector; Assume that the steering angle of the towed array sonar platform is α, the turning radius is r, and the array aperture is L = (N-1)d; define αr < L as a small angle steering, and αr ≥ L as a large angle steering; During the array turning process, the center angle β corresponding to the length of the array on the ring changes with time; according to the value of the center angle β, the array turning is divided into 5 time stages, and the intersection points of each time stage are defined as time t0, t1, t2, and t3 respectively; Step 2: When the towed array sonar platform and array are turning at a small angle, let t0 be the reference time when the first array element enters the turning circle. When the platform and array are turning at a large angle, let t0 be the reference time when the first array element enters the turning circle. where v is the platform and array velocity; Step 3: Determine the four situations of steering direction and steering angle, and give the direction vectors of the five stages of the steering process in each situation, as follows: Situation 1. Small right turn (1) When 0≤t≤t0, the direction vector is: (2) When t0<t≤t1, the direction vector is: in, Indicates rounding down; (3) When t1<t≤t2, the direction vector is: in, (4) When t2<t≤t3, the direction vector is: in, (5) When t>t3, the direction vector is: Case 2: Small left turn (1) When 0≤t≤t0, the direction vector is: (2) When t0<t≤t1, the direction vector is: in, (3) When t1<t≤t2, the direction vector is: in, (4) When t2<t≤t3, the direction vector is: in, (5) When t>t3, the direction vector is: Case 3: Large right turn (1) When 0≤t≤t0, the direction vector is: (2) When t0<t≤t1, the direction vector is: in, (3) When t1<t≤t2, the direction vector is: in (4) When t2<t≤t3, the direction vector is: in, (5) When t>t3, the direction vector is: Situation 4: Large left turn (1) When 0≤t≤t0, the direction vector is: (2) When t0<t≤t1, the direction vector is: in, (3) When t1<t≤t2, the direction vector is: in, (4) When t2<t≤t3, the direction vector is: in, (5) When t>t3, the direction vector is: