CN103308913A

CN103308913A - Foresight SAR ambiguity resolving algorithm of double antennas carried by high-speed aircraft

Info

Publication number: CN103308913A
Application number: CN2013102947124A
Authority: CN
Inventors: 包敏; 周松; 周鹏; 史林
Original assignee: Xidian University
Current assignee: Xidian University
Priority date: 2013-07-12
Filing date: 2013-07-12
Publication date: 2013-09-18

Abstract

A high-speed aircraft-mounted dual-antenna forward-looking SAR defuzzification algorithm, which includes, step 1): establishing a dual-antenna forward-looking SAR imaging model; step 2): performing range FFT processing on echoes, and then correcting linear walking and range pulse pressure ; Step 3): Azimuth FFT processing, and then carry out secondary range compression and range bending correction; Step 4): range IFFT processing and azimuth compression processing; Step 5): phase compensation; Step 6): use the distance between the two antennas The phase difference caused by the wave path difference establishes the steering vector, and the left and right blur is resolved by beamforming, and the images on the left and right sides are respectively obtained. This method is simple and easy, and is suitable for the processing requirements of high-speed aircraft platforms.

Description

A Defuzzification Algorithm for Forward-Looking SAR with Dual Antennas on High-Speed Aircraft

【技术领域】【Technical field】

本发明涉及一种高速飞行器载双天线前视SAR解模糊算法。The invention relates to a high-speed aircraft-mounted dual-antenna forward-looking SAR defuzzification algorithm.

【背景技术】【Background technique】

正侧视与斜视式SAR可以对载体（飞机、卫星、导弹等）运动方向两侧的目标（场景）进行高分辨成像，其良好的成像性能已经得到了广泛的研究与应用。然而，在实际雷达成像应用中，有很大一部分需求要求雷达能够具有对平台运动方向正前方场景进行成像的功能，即对前视SAR的需求。例如，在民用方面，前视SAR可用于全天候全天时的飞机盲降系统；在军事上，装配有前视SAR的战斗机或侦察机可以提高发现前方目标的能力；特别的，结合前视SAR成像制导的导弹攻击精确度和攻击方式的灵活性将大大地提高。由SAR成像的原理可知，在前视情况下，距离分辨率梯度与方位分辨率梯度的夹角接近于零，在与载体运动方向相垂直的方向甚至会出现等于零的情况，此时传统的SAR成像方式已经无法正常成像，并且在航向两侧对称的目标具有相同的多普勒历程并因此造成了SAR图像的模糊，即左右模糊问题。Side-looking and squint-looking SAR can perform high-resolution imaging of targets (scenes) on both sides of the moving direction of carriers (aircraft, satellites, missiles, etc.), and its good imaging performance has been widely studied and applied. However, in actual radar imaging applications, a large part of the requirements require the radar to have the function of imaging the scene directly in front of the platform movement direction, that is, the requirement for forward-looking SAR. For example, in civilian use, forward-looking SAR can be used for all-weather and all-weather aircraft blind landing systems; in military affairs, fighter jets or reconnaissance aircraft equipped with forward-looking SAR can improve the ability to find targets ahead; in particular, combined with forward-looking SAR The attack accuracy and attack mode flexibility of imaging-guided missiles will be greatly improved. According to the principle of SAR imaging, in the case of forward-looking, the angle between the range resolution gradient and the azimuth resolution gradient is close to zero, and it may even be equal to zero in the direction perpendicular to the moving direction of the carrier. At this time, the traditional SAR The imaging method can no longer be used for normal imaging, and the symmetrical targets on both sides of the course have the same Doppler history, which causes the blurring of the SAR image, that is, the problem of left and right blurring.

由于与线性阵列相比，双天线前视系统结构比较简单且便于研制与安装，代胜利，刘光炎等对双天线前视成像系统特性和成像算法进行了研究，提出了一种解左右模糊的方法，取得了一定的效果，但是该方法需要对场景进行区域划分，对不同区域的双通道数据进行重排列，然后对重排列后的数据构造不同的匹配函数进行成像，其算法复杂度比较高，难以满足高速飞行器载前视SAR实时成像的要求。Compared with the linear array, the structure of the dual-antenna forward-looking system is relatively simple and easy to develop and install. Dai Shengli, Liu Guangyan, etc. studied the characteristics and imaging algorithms of the dual-antenna forward-looking imaging system, and proposed a method to solve the left-right blur , and achieved certain effects, but this method needs to divide the scene into regions, rearrange the dual-channel data in different regions, and then construct different matching functions for the rearranged data for imaging, and the algorithm complexity is relatively high. It is difficult to meet the real-time imaging requirements of forward-looking SAR carried by high-speed aircraft.

【发明内容】【Content of invention】

本发明的目的在于提供一种高速飞行器载双天线前视SAR解模糊算法,其特征在于：其包括，The object of the present invention is to provide a kind of forward-looking SAR defuzzification algorithm with dual antennas carried by high-speed aircraft, it is characterized in that: it comprises,

步骤1）:建立双天线前视SAR成像模型；Step 1): Establish a dual-antenna forward-looking SAR imaging model;

步骤2）:对回波进行距离FFT处理，然后校正线性走动和距离脉压；Step 2): Perform distance FFT processing on the echo, and then correct linear walking and distance pulse pressure;

步骤3）:方位向FFT处理，然后进行二次距离压缩和距离弯曲校正；Step 3): Azimuth FFT processing, and then perform secondary distance compression and distance bending correction;

步骤4）:距离IFFT处理并进行方位压缩处理；Step 4): distance IFFT processing and azimuth compression processing;

步骤5）:相位补偿；Step 5): phase compensation;

步骤6）:利用双天线间的波程差所导致的相位差建立导向矢量，通过波束形成的方式解左右模糊，分别得到左右两侧的图像。Step 6): Using the phase difference caused by the wave path difference between the two antennas to establish a steering vector, the left and right blur is resolved by beamforming, and the images on the left and right sides are respectively obtained.

在上述技术方案的基础上，所述步骤1包括，On the basis of the above technical solution, the step 1 includes,

设所述双天线为天线A及天线B,设飞行器正前方的波束照射区域内有P₁和P₂两个点目标，P₁和P₂以Y轴左右对称，坐标分别为(x0,y0,0)与(-x0,y0,0)；Assume that the dual antennas are antenna A and antenna B, and there are two point targets P ₁ and P ₂ in the beam irradiation area directly in front of the aircraft, P ₁ and P ₂ are left and right symmetrical about the Y axis, and the coordinates are (x0, y0 ,0) and (-x0,y0,0);

目标P₁、P₂与天线A之间的双程斜距方程为The two-way slant distance equation between targets P ₁ , P ₂ and antenna A is

$\begin{matrix} {R R}_{11__A A} (({t t}_{m m})) = = 22 \sqrt{{x x}_{00}^{22} + + {((v v {t t}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \\ {R R}_{22__A A} (({t t}_{m m})) = = 22 \sqrt{{((- - {x x}_{00}))}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \end{matrix}\} - - - - - - ((11)),,$

P₁、P₂与天线B之间的双程斜距方程为The two-way slant distance equation between P ₁ , P ₂ and antenna B is

$\begin{matrix} {R R}_{11__B B} (({t t}_{m m})) = = \sqrt{{x x}_{00}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} - - d d))}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \\ {R R}_{22__B B} (({t t}_{m m})) = = \sqrt{{x x}_{00}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} + + d d))}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \end{matrix}\} - - - - - - ((22))$

其中，H为载机高度，v为载机速度，t_m为慢时间。Among them, H is the height of the carrier, v is the speed of the carrier, and t _m is the slow time.

在上述技术方案的基础上，所述步骤2包括，On the basis of the above technical solution, the step 2 includes,

设点P(x,y,0)为场景中一个点目标，经过t_m时间后，飞行器移动到A'点，其Y轴坐标为vt_m，设点目标P与C点之间的距离为y_n，则从⊿A'PC'可得P与天线A间的双程斜距RP_A(t_m)为Let point P(x, y, 0) be a point target in the scene. After t _m time, the aircraft moves to point A', its Y-axis coordinate is vt _m , and the distance between point P and point C is y _n , then the two-way slant distance RP_A(t _m ) between P and antenna A can be obtained from ⊿A'PC' as

${R R}_{P P__A A} (({t t}_{m m})) = = 22 \sqrt{{x x}^{22} + + {(({vt vt}_{m m} - - y the y))}^{22} + + {H h}^{22}}$

$= = 22 \sqrt{{(({vt vt}_{m m} - - {y the y}_{n no} - - {y the y}_{c c}))}^{22} + + {R R}_{B B}^{22}} - - - - - - ((33))$

$= = 22 \sqrt{{(({vt vt}_{m m} - - {y the y}_{n no}))}^{22} + + {R R}_{00}^{22} - - 22 {R R}_{00} (({vt vt}_{m m} - - {y the y}_{n no})) sin sin θ θ}$

其中 $R_{B} = \sqrt{x^{2} + H^{2}},$ $R_{0} = \sqrt{x^{2} + y_{c}^{2} + H^{2}},$ θ是斜视角。in $R_{B} = \sqrt{x^{2} + h^{2}},$ $R_{0} = \sqrt{x^{2} + {the y}_{c}^{2} + h^{2}},$ θ is the oblique angle.

将式(3)在y_n处作泰勒级数展开，保留(vt_m-y_n)的二次以下项，得Expand the formula (3) by Taylor series at y _n , keep the quadratic term of (vt _m -y _n ), get

${R R}_{P P__A A} (({t t}_{m m})) \approx \approx 22 [[{R R}_{00} - - (({vt vt}_{m m} - - {y the y}_{n no})) sin sin θ θ + + \frac{{v v}^{22} {cos cos}^{22} θ θ}{22 {R R}_{00}} {(({vt vt}_{m m} - - {y the y}_{n no}))}^{22}]] - - - - - - ((44))$

将式(4)改写为精度相同的另一形式Rewrite formula (4) into another form with the same accuracy

${R R}_{P P__A A} (({t t}_{m m})) = = 22 \sqrt{{R R}_{00}^{22} + + {cos cos}^{22} θ θ {(({vt vt}_{m m} - - {y the y}_{n no}))}^{22}} - - 22 (({vt vt}_{m m} - - {y the y}_{n no})) sin sin θ θ - - - - - - ((55))$

则天线A接收到的目标P的回波信号写为Then the echo signal of target P received by antenna A is written as

${S S}_{A A} ((\overset{^^}{t t},, {t t}_{m m})) = = {σ σ}_{P P} {w w}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{P P__A A} (({t t}_{m m}))}{c c})) {w w}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{v v})) exp exp [[jπγ jπγ {((\overset{^^}{t t} - - \frac{{R R}_{P P__A A} (({t t}_{m m}))}{c c}))}^{22}]] exp exp [[- - j j \frac{22 π π}{λ λ} {R R}_{P P__A A} (({t t}_{m m}))]] - - - - - - ((66))$

其中，

是快时间，λ是信号波长，c是光速，γ是信号调频率，w_r(·)和w_a(·)分别是距离和方位窗函数，σ_P为目标P的反射系数，in,

is the fast time, λ is the signal wavelength, c is the speed of light, γ is the signal modulation frequency, w _r (·) and w _a (·) are the distance and azimuth window functions respectively, σ _P is the reflection coefficient of the target P,

将式(6)的回波基频信号从二维时域变换到距离频域-方位时域，Transform the echo fundamental frequency signal of formula (6) from the two-dimensional time domain to the range frequency domain-azimuth time domain,

${S S}_{A A} (({f f}_{r r},, {t t}_{m m})) = = {σ σ}_{P P} {w w}_{r r} (({f f}_{r r})) {w w}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{v v})) exp exp [[- - jπ jπ \frac{{f f}_{r r}^{22}}{γ γ}]] exp exp [[- - j j \frac{22 π π}{c c} {R R}_{P P__A A} (({t t}_{m m})) (({f f}_{r r} + + {f f}_{c c}))]] - - - - - - ((77))$

其中f_c为载频，f_r为距离频率，根据式(5)，对式(7)乘以式(8)相位项对数据进行统一的线性走动校正并进行距离向脉压处理：Where f _c is the carrier frequency, and f _r is the range frequency. According to formula (5), multiply formula (7) by the phase item of formula (8) to perform uniform linear motion correction on the data and perform distance pulse pressure processing:

${H h}_{LRMC LRMC} (({f f}_{r r},, {t t}_{m m})) = = exp exp [[- - j j 22 π π \frac{v v sin sin {θt θt}_{m m}}{c c} (({f f}_{r r} {+ + f f}_{c c}))]] exp exp [[jπ jπ \frac{{f f}_{r r}^{22}}{γ γ}]] . . - - - - - - ((88))$

在上述技术方案的基础上，所述步骤3包括，On the basis of the above technical solution, the step 3 includes,

将距离走动校正和距离向脉压后的信号变换到二维频域，并使用以下相位项进行距离徙动校正和二次距离脉压Transform the range-walking-corrected and range-pulse signals into the two-dimensional frequency domain, and use the following phase terms for range-migration correction and quadratic range-pulse pressure

${H h}_{rg r g} (({f f}_{r r},, {f f}_{a a})) = = exp exp [[- - jπ jπ \frac{22 λ λ {R R}_{s the s} {((\frac{λ λ {f f}_{a a}}{22 v v cos cos θ θ}))}^{22}}{{c c}^{22} {((\sqrt{11 - - {((\frac{λ λ {f f}_{a a}}{22 v v cos cos θ θ}))}^{22}}))}^{33}} {f f}_{r r}^{22}]] exp exp [[j j \frac{22 π π {R R}_{s the s}}{c c} {((\frac{λ λ {f f}_{a a}}{22 v v cos cos θ θ}))}^{22} {f f}_{r r}]] - - - - - - ((99))$

对信号进行距离向逆傅里叶变换，得到距离-多普勒域的信号表达式如下The range-to-inverse Fourier transform is performed on the signal, and the signal expression in the range-Doppler domain is obtained as follows

${S S}_{A A} ((\overset{^^}{t t},, {f f}_{a a})) = = {σ σ}_{P P} sin sin c c [[Δ Δ {f f}_{r r} ((\overset{^^}{t t} - - \frac{22 (({R R}_{00} + + {y the y}_{n no} sin sin θ θ))}{c c}))]] {a a}_{a a} ((\frac{{R R}_{00} λ λ {f f}_{a a}}{{22 v v}^{22} \sqrt{11 - - {(({f f}_{a a} / / {f f}_{aM a M}))}^{22}}})) - - - - - - ((1010))$

$\cdot \cdot exp exp [[- - j j \frac{22 π π}{v v cos cos θ θ} {R R}_{00} \sqrt{{f f}_{aM a M}^{22} - - {f f}_{a a}^{22}}]] exp exp [[- - j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ]] exp exp [[- - j j 22 π π {f f}_{a a} \frac{{y the y}_{n no}}{v v}]]$

其中，Δf_r为发射信号带宽，f_a为方位频率，

R_s为场景中心线处的射线斜距，完成回波信号距离向的处理。Among them, Δf _r is the transmission signal bandwidth, f _a is the azimuth frequency,

R _s is the slant distance of the ray at the center line of the scene, and completes the processing of the distance direction of the echo signal.

在上述技术方案的基础上，所述步骤4包括，On the basis of the above technical solution, the step 4 includes,

将式(10)改写为能同时表示天线A、B接收的目标P1、P2的信号的形式Rewrite formula (10) into a form that can simultaneously represent the signals of targets P1 and P2 received by antennas A and B

${S S}_{M m} ((\overset{^^}{t t},, {f f}_{a a})) = = {σ σ}_{N N} sin sin c c [[Δ Δ {f f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__M m__00} + + 22 {y the y}_{n no} sin sin θ θ}{c c}))]] {w w}_{a a} ((\frac{{R R}_{N N__M m__00} λ λ {f f}_{a a}}{44 {v v}^{22} \sqrt{11 - - {(({f f}_{a a} / / {f f}_{aM a M}))}^{22}}})) - - - - - - ((1111))$

$\cdot \cdot exp exp [[- - j j \frac{22 π π}{v v cos cos θ θ} {R R}_{N N__M m__00} \sqrt{{f f}_{aM a M}^{22} - - {f f}_{a a}^{22}}]] exp exp [[- - j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ]] exp exp [[- - j j 22 π π {f f}_{a a} \frac{{y the y}_{n no}}{v v}]]$

其中N=1,2表示目标P₁与P₂，而M=A，B，表示天线A和B。R_{N_M}为式(1)与(2)中的四个斜距，而R_{N_M_0}代表t_m=0时的斜距，即Among them, N=1,2 represent the targets P ₁ and P ₂ , and M=A, B, represent the antennas A and B. R _{N_M} is the four slant distances in formulas (1) and (2), and R _{N_M_0} represents the slant distance when t _m =0, namely

${R R}_{N N__M m__00} \{\begin{matrix} {R R}_{11__A A__00} = = 22 \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} \\ {R R}_{22__A A__00} = = 22 \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} \\ {R R}_{11__B B__00} = = \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} - - d d))}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} \\ {R R}_{22__B B__00} = = \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} + + d d))}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} \end{matrix} - - - - - - ((1212))$

式(11)中的第一个相位项是方位调制项，将它展开为f_a的泰勒级数得The first phase term in Eq. (11) is the azimuth modulation term, and expanding it into the Taylor series of f _a gives

$exp exp [[- - j j \frac{π π}{V V cos cos θ θ} {R R}_{N N__M m__00} \sqrt{{f f}_{aM a M}^{22} - - {f f}_{a a}^{22}}]] \approx \approx exp exp [[- - j j \frac{22 π π {R R}_{N N__M m__00}}{λ λ} ((11 - - \frac{11}{22} \frac{{f f}_{a a}^{22}}{{f f}_{aM a M}^{22}} - - \frac{11}{88} \frac{{f f}_{a a}^{44}}{{f f}_{aM a M}^{44}}))]] - - - - - - ((1313))$

构造一个没有常数项的方位匹配滤波器Construct an orientation matched filter with no constant term

${H h}_{az az} ((\overset{^^}{t t},, {f f}_{a a})) = = exp exp [[j j \frac{22 π π {R R}_{N N__M m__00}}{λ λ} ((- - \frac{11}{22} \frac{{f f}_{a a}^{22}}{{f f}_{aM a M}^{22}} - - \frac{11}{88} \frac{{f f}_{a a}^{44}}{{f f}_{aM a M}^{44}}))]] - - - - - - ((1414))$

得到方位匹配滤波后的聚焦图像为The focused image obtained after azimuth matched filtering is

${S S}_{M m} ((\overset{^^}{t t},, {t t}_{m m})) = = {σ σ}_{N N} sin sin c c [[Δ Δ {f f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__M m__00} + + 22 {y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[Δ Δ {f f}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{v v}))]] - - - - - - ((1515))$

$exp exp [[- - j j \frac{22 π π {R R}_{N N__M m__00}}{λ λ}]] exp exp [[- - j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ]]$

其中Δf_a为方位向多普勒带宽。where Δf _a is the Doppler bandwidth in azimuth.

在上述技术方案的基础上，所述步骤5包括，On the basis of the above technical solution, the step 5 includes,

式(12)中四个斜距公式拥有一个共同项

式(15)表示的四个聚集图像中包含有一个公共相位

用式(16)相位项消掉公共相位和剩余相位

\exp (- j \frac{4 π}{λ} y_{n} \sin θ)

The four slant distance formulas in formula (12) have a common term

The four aggregated images represented by Equation (15) contain a common phase

Eliminate the common phase and the residual phase by using the phase term of formula (16)

\exp (- j \frac{4 π}{λ} {the y}_{no} \sin θ)

${H h}_{phase phase} = = exp exp [[j j \frac{22 π π \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}}{λ λ}]] exp exp ((j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ)) - - - - - - ((1616))$

得到信号get the signal

${S S}_{M m} ((\overset{^^}{t t},, {t t}_{m m})) = = {σ σ}_{N N} sin sin c c [[Δ Δ {f f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__M m__00} + + 22 {y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[Δ Δ {f f}_{a a} (({t t}_{m m} - - \frac{{Y Y}_{n no}}{V V}))]]$

$exp exp [[- - j j \frac{22 π π (({R R}_{N N__M m__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] . . - - - - - - ((1717))$

在上述技术方案的基础上，所述步骤6包括，On the basis of the above technical solution, the step 6 includes,

由(17)分别写出两个天线中混叠信号的表达式From (17), write the expressions of the aliased signals in the two antennas respectively

$\{\begin{matrix} {S S}_{A A} ((\overset{^^}{t t},, {t t}_{m m})) = = {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} sin sin c c [[Δ Δ {f f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__A A__00} + + 22 {y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[Δ Δ {f f}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{V V}))]] \\ \cdot \cdot exp exp [[- - j j \frac{22 π π (({R R}_{N N__A A__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \\ {S S}_{B B} ((\overset{^^}{t t},, {t t}_{m m})) = = {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} sin sin c c [[Δ Δ {f f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__B B__00} + + 22 {y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[Δ Δ {f f}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{V V}))]] \\ \cdot &Center Dot; exp exp [[- - j j \frac{22 π π (({R R}_{N N__B B__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \end{matrix} - - - - - - ((1818))$

取出(18)中脉冲响应的峰值Peak_A和Peak_B，由于sinc函数的峰值为1，可以得到二元线阵接收的观测信号向量S为Take out the peak values Peak _A and Peak _B of the impulse response in (18), since the peak value of the sinc function is 1, the observed signal vector S received by the binary linear array can be obtained as

$S S = = [\begin{matrix} {Peak Peak}_{A A} \\ {Peak Peak}_{B B} \end{matrix}]$

$= = [\begin{matrix} {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} exp exp [[- - j j \frac{22 π π (({R R}_{N N__A A__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \\ {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} exp exp [[- - j j \frac{22 π π (({R R}_{N N__B B__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \end{matrix}]$

$= = [\begin{matrix} {σ σ}_{11} {e e}^{- - j j \frac{22 π π}{λ λ} \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}} + + {σ σ}_{22} {e e}^{- - j j \frac{22 π π}{λ λ} \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}} \\ {σ σ}_{11} {e e}^{- - j j \frac{22 π π}{λ λ} ((\sqrt{{(({x x}_{00} - - d d))}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))} + + {σ σ}_{22} {e e}^{- - j j \frac{22 π π}{λ λ} ((\sqrt{{(({x x}_{00} + + d d))}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))} \end{matrix}] - - - - - - ((1919))$

$= = [\begin{matrix} {σ σ}_{11} {a a}_{11__A A} + + {σ σ}_{22} {a a}_{22__A A} \\ {σ σ}_{11} {a a}_{11__B B} + + {σ σ}_{22} {a a}_{22__B B} \end{matrix}]$

$= = [\begin{matrix} {a a}_{11__A A},, {a a}_{22__A A} \\ {a a}_{11__B B},, {a a}_{22__B B} \end{matrix}] [\begin{matrix} {σ σ}_{11} \\ {σ σ}_{22} \end{matrix}]$

其中[a_{1_A},a_{2_A}]^T和[a_{1_B},a_{2_B}]^T分别为P₁与P₂的导向矢量，Where [a _{1_A} , a _{2_A} ] ^T and [a _{1_B} , a _{2_B} ] ^T are the steering vectors of P ₁ and P ₂ respectively,

通过波束形成的方法分别抑制来自Y轴一侧信号，而得到Y轴另一侧的信号，若令阵列波束指向P₁的加权向量为W₁，指向P₂的加权向量为W₂，则加权矩阵为W＝[W₁,W₂]，The signal from one side of the Y-axis is suppressed by the beamforming method, and the signal on the other side of the Y-axis is obtained. If the weighted vector of the array beam pointing to P ₁ is W ₁ , and the weighting vector pointing to P ₂ is W ₂ , then the weighted The matrix is W=[W ₁ , W ₂ ],

求解W的表达式，从式(19)中得到方程组Solve the expression of W, and get the equation system from formula (19)

$\{\begin{matrix} {Peak Peak}_{A A} = = {σ σ}_{11} {a a}_{11__A A} + + {σ σ}_{22} {a a}_{22__A A} \\ {Peak Peak}_{B B} = = {σ σ}_{11} {a a}_{11__B B} + + {σ σ}_{22} {a a}_{22__B B} \end{matrix} - - - - - - ((2020))$

解方程（20）组得到Solve equation (20) to get

$\{\begin{matrix} {σ σ}_{11} = = \frac{11}{b b} [[{Peak Peak}_{A A} {a a}_{22__B B} - - {Peak Peak}_{B B} {a a}_{22__A A}]] \\ {σ σ}_{22} = = \frac{11}{b b} [[- - {Peak Peak}_{A A} {a a}_{11__B B} + + {Peak Peak}_{B B} {a a}_{11__A A}]] \end{matrix} - - - - - - ((21 twenty one))$

其中b＝a_{1_A}a_{2_B}-a_{2_A}a_{1_B}，where b=a _{1_A} a _{2_B} -a _{2_A} a _{1_B} ,

由(21)写出W₁为From (21) write W ₁ as

${W W}_{11} = = \frac{11}{b b} [\begin{matrix} {a a}_{22__B B} \\ - - {a a}_{22__A A} \end{matrix}] - - - - - - ((22 twenty two))$

而W₂为while _W2 is

${W W}_{22} = = \frac{11}{b b} [\begin{matrix} - - {a a}_{11__B B} \\ {a a}_{11__A A} \end{matrix}] - - - - - - ((23 twenty three))$

则加权矩阵W为Then the weight matrix W is

$W W = = [[{W W}_{11},, {W W}_{22}]] = = \frac{11}{b b} [\begin{matrix} {a a}_{22__B B},, - - {a a}_{11__B B} \\ - - {a a}_{22__A A},, {a a}_{11__A A} \end{matrix}] - - - - - - ((24 twenty four))$

对混叠信号解左右模糊的过程为The process of deblurring the left and right aliased signals is

${W W}^{H h} S S = = {W W}^{H h} [\begin{matrix} {a a}_{11__A A},, {a a}_{22__A A} \\ {a a}_{11__B B},, {a a}_{22__B B} \end{matrix}] [\begin{matrix} {σ σ}_{11} \\ {σ σ}_{22} \end{matrix}] = = [\begin{matrix} {σ σ}_{11} \\ {σ σ}_{22} \end{matrix}] - - - - - - ((2525)) . .$

与现有技术相比，本发明首先对双天线接收的雷达回波信号分别进行斜视成像，得到左右模糊的SAR图像，然后将两副图像中的对应像素取出，利用双天线间的波程差所导致的相位差建立导向矢量，通过波束形成的方式解开左右模糊，分别得到左右两侧的图像。仿真实验证明，此方法简便易行，适用于高速飞行器平台的处理要求。Compared with the prior art, the present invention first performs squint imaging on the radar echo signals received by the dual antennas to obtain left and right blurred SAR images, then takes out the corresponding pixels in the two images, and utilizes the wave path difference between the dual antennas The resulting phase difference establishes a steering vector, and the left and right blurs are resolved by beamforming to obtain images on the left and right sides, respectively. The simulation experiment proves that this method is simple and feasible, and is suitable for the processing requirements of the high-speed aircraft platform.

【附图说明】【Description of drawings】

图1为本发明飞行器载双天线前视SAR解模糊算法SAR几何模型图；Fig. 1 is the SAR geometric model diagram of the aircraft carrying dual antenna forward-looking SAR defuzzification algorithm of the present invention;

图2为本发明Y轴右侧场景斜视SAR几何关系图；Fig. 2 is the SAR geometric relationship diagram of the scene squint SAR on the right side of the Y axis of the present invention;

图3为本发明飞行器载双天线前视SAR解模糊算法处理流程图；Fig. 3 is the processing flowchart of the forward-looking SAR defuzzification algorithm with dual antennas carried by the aircraft of the present invention;

图4为本发明点目标设置方式图；Fig. 4 is a point target setting mode diagram of the present invention;

图5为天线A成像结果图；Fig. 5 is an imaging result diagram of antenna A;

图6为天线B成像结果图；Fig. 6 is an imaging result diagram of antenna B;

图7为天线A对中心点成像结果图；Fig. 7 is the imaging result diagram of antenna A to the central point;

图8为天线B对中心点成像结果图；Fig. 8 is a picture of the imaging result of antenna B to the central point;

图9为Y轴左侧图像；Figure 9 is the image on the left side of the Y axis;

图10为Y轴右侧图像；Figure 10 is the image on the right side of the Y axis;

图11为Y轴左侧场景中心点图；Figure 11 is a diagram of the center point of the scene on the left side of the Y axis;

图12为Y轴右侧场景中心点图。Fig. 12 is a diagram of the center point of the scene on the right side of the Y axis.

【具体实施方式】【Detailed ways】

请参考图1至图3，一种算法简单的飞行器载双天线前视SAR解模糊算法，其包括，Please refer to Fig. 1 to Fig. 3, a kind of algorithm is simple to carry dual-antenna forward-looking SAR defuzzification algorithm, which includes,

步骤5）:相位补偿；Step 5): phase compensation;

请参考图1，图中雷达平台以速度v沿O′Y′匀速直线飞行。平台上装有两个天线，分别是天线A与天线B。当雷达工作时，天线A发射信号，A与B同时接收回波信号。天线间距为d。若λ为发射信号的波长，为避免栅瓣，则有d≤λ/2。令t_m为慢时间，当t_m=0时，雷达平台位于O′，高度为H，此时天线A位于(0,0,H)，天线B位于(d,0,H)。步骤1包括，设所述双天线为天线A及天线B,设飞行器正前方的波束照射区域内有P₁和P₂两个点目标，P₁和P₂以Y轴左右对称，坐标分别为(x0,y0,0)与(-x0,y0,0)；Please refer to Figure 1, in which the radar platform is flying along O'Y' in a straight line with speed v. There are two antennas installed on the platform, namely Antenna A and Antenna B. When the radar is working, antenna A transmits signals, and A and B receive echo signals at the same time. The distance between the antennas is d. If λ is the wavelength of the transmitted signal, in order to avoid grating lobes, there is d≤λ/2. Let t _m be the slow time. When t _m =0, the radar platform is located at O′ and the height is H. At this time, antenna A is located at (0,0,H), and antenna B is located at (d,0,H). Step 1 includes, assuming that the dual antennas are antenna A and antenna B, and assuming that there are two point targets _P1 and _P2 in the beam irradiation area directly in front of the aircraft, _P1 and _P2 are left and right symmetrical about the Y axis, and the coordinates are respectively (x0,y0,0) and (-x0,y0,0);

$\begin{matrix} {R R}_{11__B B} (({t t}_{m m})) = = \sqrt{{x x}_{00}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} - - d d))}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \\ {R R}_{22__B B} (({t t}_{m m})) = = \sqrt{{x x}_{00}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} + + d d))}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \end{matrix}\} - - - - - - ((22)) . .$

请参考图2，步骤2包括，飞行器沿Y轴以速度v匀速直线运动。设点P(x,y,0)为场景中一个点目标，经过t_m时间后，飞行器移动到A'点，其Y轴坐标为vt_m，设点目标P与C点之间的距离为y_n，则从⊿A'PC'可得P与天线A间的双程斜距RP_A(t_m)为Please refer to FIG. 2 , step 2 includes, the aircraft moves in a straight line with a speed v along the Y axis. Let point P(x, y, 0) be a point target in the scene. After t _m time, the aircraft moves to point A', its Y-axis coordinate is vt _m , and the distance between point P and point C is y _n , then the two-way slant distance RP_A(t _m ) between P and antenna A can be obtained from ⊿A'PC' as

其中，

其中f_c为载频，f_r为距离频率，根据式(5)，对式(7)乘以以下相位项可以对数据进行统一的线性走动校正并进行距离向脉压处理：Where f _c is the carrier frequency, and f _r is the range frequency. According to formula (5), formula (7) can be multiplied by the following phase items to perform uniform linear motion correction on the data and perform range pulse pressure processing:

${H h}_{LRMC LRMC} (({f f}_{r r},, {t t}_{m m})) = = exp exp [[- - j j 22 π π \frac{v v sin sin {θt θt}_{m m}}{c c} (({f f}_{r r} {+ + f f}_{c c}))]] exp exp [[jπ jπ \frac{{f f}_{r r}^{22}}{γ γ}]] - - - - - - ((88))$

将距离走动校正和距离向脉压后的信号变换到二维频域，接着使用以下相位项进行距离徙动校正和二次距离脉压Transform the signal after range walking correction and range pulse pressure into the two-dimensional frequency domain, and then use the following phase terms for range migration correction and secondary range pulse pressure

$\cdot &Center Dot; exp exp [[- - j j \frac{22 π π}{v v cos cos θ θ} {R R}_{00} \sqrt{{f f}_{aM a M}^{22} - - {f f}_{a a}^{22}}]] exp exp [[- - j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ]] exp exp [[- - j j 22 π π {f f}_{a a} \frac{{y the y}_{n no}}{v v}]]$

其中，Δf_r为发射信号带宽，f_a为方位频率，

$\cdot \cdot exp exp [[- - j j \frac{π π}{v v cos cos θ θ} {R R}_{N N__M m__00} \sqrt{{f f}_{aM a M}^{22} - - {f f}_{a a}^{22}}]] exp exp [[- - j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ]] exp exp [[- - j j 22 π π {f f}_{a a} \frac{{y the y}_{n no}}{v v}]]$

可以注意到，式(13)中泰勒展开后的第一项是常数项，对应目标到天线的距离。对于不同的天线和不同目标来说，这一常数项各不相同，即具有不同的波程差。因此我们接下来考虑利用这一项的差别对P₁和P₂解模糊。It can be noticed that the first term after Taylor expansion in formula (13) is a constant term, corresponding to the distance from the target to the antenna. For different antennas and different targets, this constant term is different, that is, they have different wave path differences. So we next consider using the difference of this item to defuzzify _P1 and _P2 .

由于常数项对方位聚焦效果并没有影响，因此可以构造一个没有常数项的方位匹配滤波器Since the constant term has no effect on the azimuth focusing effect, an azimuth matched filter without a constant term can be constructed

可以得到方位匹配滤波后的聚焦图像为The focused image after azimuth matching filtering can be obtained as

可以发现，在式(12)中四个斜距公式拥有一个共同项因此式(15)表示的四个聚集图像中包含有一个公共相位

用以下相位项将此公共相位和剩余相位

\exp (- j \frac{4 π}{λ} y_{n} \sin θ)

一起消掉It can be found that the four slant distance formulas in formula (12) have a common term Therefore, the four aggregated images represented by equation (15) contain a common phase

Combine this common phase and the remaining phase with the following phase term

\exp (- j \frac{4 π}{λ} {the y}_{no} \sin θ)

disappear together

得到信号get the signal

$exp exp [[- - j j \frac{22 π π (({R R}_{N N__M m__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] - - - - - - ((1717))$

其中sinc函数的峰值与点目标的反射系数相对应，而sinc函数在图像中的位置和点目标的实际位置相对应，由于天线A，B中信号是混叠的，可以由(17)分别写出两个天线中混叠信号的表达式The peak value of the sinc function corresponds to the reflection coefficient of the point target, and the position of the sinc function in the image corresponds to the actual position of the point target. Since the signals in antennas A and B are aliased, they can be written by (17) respectively Find the expression of the aliased signal in the two antennas

$\{\begin{matrix} {S S}_{A A} ((\overset{^^}{t t},, {t t}_{m m})) = = {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} sin sin c c [[Δ Δ {f f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__A A__00} + + 22 {y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[Δ Δ {f f}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{V V}))]] \\ \cdot \cdot exp exp [[- - j j \frac{22 π π (({R R}_{N N__A A__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \\ {S S}_{B B} ((\overset{^^}{t t},, {t t}_{m m})) = = {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} sin sin c c [[Δ Δ {f f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__B B__00} + + 22 {y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[Δ Δ {f f}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{V V}))]] \\ \cdot \cdot exp exp [[- - j j \frac{22 π π (({R R}_{N N__B B__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \end{matrix} - - - - - - ((1818))$

$S S = = [\begin{matrix} {Peak Peak}_{A A} \\ {Peak Peak}_{B B} \end{matrix}]$

显然，对于式(19)，可以通过波束形成的方法分别抑制来自Y轴一侧信号，而得到另一侧的信号，若令阵列波束指向P₁的加权向量为W₁，指向P₂的加权向量为W₂，则加权矩阵为W＝[W₁,W₂]，Obviously, for equation (19), the beamforming method can be used to suppress the signal from one side of the Y axis and obtain the signal on the other side. If the weighted vector of the array beam pointing to P ₁ is W ₁ , the weighted vector pointing to P ₂ The vector is W ₂ , then the weighting matrix is W=[W ₁ ,W ₂ ],

求解W的表达式，从式(19)中得到一个方程组Solve the expression for W and get a system of equations from equation (19)

解这个方程组得到Solving this system of equations gives

由(21)写出W₁为From (21) write W ₁ as

而W₂为while _W2 is

则加权矩阵W为Then the weight matrix W is

于是，对混叠信号解左右模糊的过程即为Therefore, the process of solving the left and right ambiguity of the aliased signal is

${W W}^{H h} S S = = {W W}^{H h} [\begin{matrix} {a a}_{11__A A},, {a a}_{22__A A} \\ {a a}_{11__B B,,} {a a}_{22__B B} \end{matrix}] [\begin{matrix} {σ σ}_{11} \\ {σ σ}_{22} \end{matrix}] = = [\begin{matrix} {σ σ}_{11} \\ {σ σ}_{22} \end{matrix}] - - - - - - ((2525)) . .$

方程(25)说明我们可以通过波束形成由混叠的脉冲响应峰值中分别解得P₁和P₂两个对称点目标各自的反射系数σ₁与σ₂。同理，在对混叠图像中的每一个像素进行波束形成后，我们将分别得到Y轴两侧的图像。Equation (25) shows that we can obtain the respective reflection coefficients σ ₁ and σ ₂ of the two symmetrical point targets P ₁ and P ₂ from the aliased impulse response peaks through beamforming. Similarly, after performing beamforming on each pixel in the aliased image, we will obtain images on both sides of the Y axis respectively.

仿真实验Simulation

下面用计算机仿真来验证所提的高速飞行器载双天线前视SAR解模糊算法的有效性。仿真参数如表1所示。Next, computer simulation is used to verify the effectiveness of the proposed high-speed aircraft-mounted dual-antenna forward-looking SAR defuzzification algorithm. The simulation parameters are shown in Table 1.

表1仿真参数Table 1 Simulation parameters

雷达工作在X波段，波长0.03m。天线A，B的间距d为波长λ的一半。弹体在1km的高度以500m/s的速度沿Y轴向前飞行。Y轴右侧的场景中心位于(176,2500,0)，Y轴左侧的场景中心位于(-176,2500,0)，左右场景中心斜距均为2698.33m。The radar works in the X-band with a wavelength of 0.03m. The distance d between the antennas A and B is half of the wavelength λ. The projectile flies forward along the Y axis at a speed of 500m/s at a height of 1km. The scene center on the right side of the Y axis is at (176,2500,0), the scene center on the left side of the Y axis is at (-176,2500,0), and the slant distance between the left and right scene centers is 2698.33m.

仿真所用的点目标设置方式如图4所示。以Y轴为对称轴，在波束照射范围内放置两两对称的10个点目标。其中，在Y轴右侧放置成十字型的五个点目标，Y轴左侧放置与它们对称的五个点目标。十个点目标分别标记为P₁,P₂,P₃,…P₁₀，它们的反射系数分别设为σ_n=n，n=1,2,3…10。The point target setting method used in the simulation is shown in Figure 4. Take the Y axis as the axis of symmetry, and place 10 point targets symmetrically in pairs within the beam irradiation range. Among them, on the right side of the Y axis, place five point targets in a cross shape, and on the left side of the Y axis, place five point targets symmetrical to them. The ten point targets are respectively marked as P ₁ , P ₂ , P ₃ ,...P ₁₀ , and their reflection coefficients are respectively set to σ _n =n, n=1,2,3...10.

需要注意的是，为了便于对成像后的点目标二维脉冲响应进行观测，我们将天线X方向的波束宽度设置的较大，相应的点目标之间的距离也设置的较大，以保证成像后各个二维脉冲响应之间保持一定距离，不影响我们对实验结果的分析。It should be noted that, in order to facilitate the observation of the two-dimensional impulse response of the point target after imaging, we set the beam width in the X direction of the antenna to be relatively large, and the distance between the corresponding point targets is also set to be relatively large to ensure that the imaging A certain distance is kept between each two-dimensional impulse response, which does not affect our analysis of the experimental results.

下面利用所提算法对点目标进行成像和解模糊处理。Next, the proposed algorithm is used to image and deblur the point target.

首先用所提算法对A、B两天线所接收的回波信号分别进行成像，结果如图5与图6所示。由于在天线A、B所接收的信号中，关于Y轴对称的点目标的信号相互混叠，因此我们设置的十个点目标在图5与图6中只产生了五个混叠的脉冲响应。图5与图6中脉冲响应旁标注的P_n+P_m（n=1,2,···,5，m=6,7,···,10）表示该脉冲响应是由哪两个点目标的脉冲响应混叠而成的。Firstly, the proposed algorithm is used to image the echo signals received by the antennas A and B respectively, and the results are shown in Figure 5 and Figure 6. Since in the signals received by antennas A and B, the signals of the point targets that are symmetrical about the Y axis are aliased with each other, the ten point targets we set only produce five aliased impulse responses in Figures 5 and 6 . P _n +P _m (n=1,2,···,5, m=6,7,···,10) next to the impulse response in Figure 5 and Figure 6 indicates which two The impulse response of the point target is aliased.

这里需要注意的是，SAR成像后所得的成像结果是斜距平面图像，因此图5与图6中脉冲响应之间的位置关系与图4的中点目标在XOY平面的位置关系有所不同。It should be noted here that the imaging result obtained after SAR imaging is an oblique-range plane image, so the positional relationship between the impulse responses in Figures 5 and 6 is different from the positional relationship of the midpoint target in the XOY plane in Figure 4 .

为了测量脉冲响应的峰值，我们从图5与图6中取出场景中心点的脉冲响应进行细致观测，如图7与图8所示。从图7与图8中可以看出，混叠后的脉冲响应的包络形状仍为二维sinc函数，其峰值分别为373.12和358.89。考虑到SAR成像过程对信号幅度进行了线性变换，这两个峰值的具体数值并不能直接反应点目标反射系数之间的关系，因此我们对两个峰值幅度取其比值，得373.12:358.89=1.04:1。由之前的点目标设置可知，由于两个场景中心点目标混叠，因此它们的反射系数互相叠加后的比值应该为7:7=1:1。显然，仿真实验得到的脉冲响应峰值幅度间的比值与预设的反射系数间的比值相一致。In order to measure the peak value of the impulse response, we take out the impulse response of the center point of the scene from Figure 5 and Figure 6 for careful observation, as shown in Figure 7 and Figure 8. It can be seen from Fig. 7 and Fig. 8 that the envelope shape of the aliased impulse response is still a two-dimensional sinc function, and its peak values are 373.12 and 358.89 respectively. Considering that the SAR imaging process linearly transforms the signal amplitude, the specific values of these two peaks cannot directly reflect the relationship between the reflection coefficients of point targets, so we take the ratio of the two peak amplitudes to get 373.12:358.89=1.04 :1. It can be seen from the previous point target settings that since the two scene center point targets are aliased, the ratio of their reflection coefficients after superimposing each other should be 7:7=1:1. Obviously, the ratio between the peak amplitudes of the impulse response obtained from the simulation experiment is consistent with the preset ratio of the reflection coefficients.

接着，对图5与图6中的各个脉冲响应的进行峰值测量并求出它们之间的比值，其结果如表2所示。将比值与预设的点目标反射系数间的比值进行比较，可以看出脉冲响应的混叠符合理论上的分析。Next, measure the peak value of each impulse response in Fig. 5 and Fig. 6 and calculate the ratio between them, and the results are shown in Table 2. Comparing the ratio with the ratio between the preset point target reflection coefficients, it can be seen that the aliasing of the impulse response conforms to the theoretical analysis.

表2混叠的脉冲响应测量结果Table 2 Impulse Response Measurement Results for Aliasing

然后，使用所提的解模糊方法对图5与图6进行解模糊处理，解模糊后所得的图像如图9与图10所示。其中，图9为Y轴左侧图像，图10为Y轴右侧图像。图中脉冲响应旁标注的P_n（n=1,2,···10）表示该脉冲响应是由哪个点目标产生。Then, the proposed deblurring method is used to deblurh Figures 5 and 6, and the resulting images after deblurring are shown in Figures 9 and 10. Among them, FIG. 9 is the image on the left side of the Y axis, and FIG. 10 is the image on the right side of the Y axis. The P _n (n=1, 2,...10) marked next to the impulse response in the figure indicates which point target the impulse response is generated from.

同样地，从图9与图10中取出场景中心点的脉冲响应进行细致观测，得到图11与图12。Similarly, the impulse response of the center point of the scene is taken out from Figures 9 and 10 for careful observation, and Figures 11 and 12 are obtained.

由图11与图12可以看出，解模糊后的场景中心点的包络形状依然保持为二维sinc函数，但其峰值变化为320.16与53.56，其比值320.16:53.56=5.96:1与预设的反射率比值6:1相比，其比值几乎完全一致。It can be seen from Figure 11 and Figure 12 that the envelope shape of the center point of the scene after deblurring remains as a two-dimensional sinc function, but its peak value changes to 320.16 and 53.56, and its ratio is 320.16:53.56=5.96:1 and the preset Compared with the reflectance ratio of 6:1, the ratio is almost exactly the same.

再测量图9与图10中各个脉冲响应的峰值，其结果如表3所示。对测量值求其比值后可以看出，其比值在四舍五入后和预设值几乎完全一致。可见，经过解模糊处理，Y轴左右两侧混叠的点目标被成功解开了。Then measure the peak value of each impulse response in Figure 9 and Figure 10, and the results are shown in Table 3. After calculating the ratio of the measured values, it can be seen that the ratio is almost exactly the same as the preset value after rounding. It can be seen that after the defuzzification process, the aliased point targets on the left and right sides of the Y axis have been successfully solved.

表3解模糊后的脉冲响应峰值测量结果Table 3. Impulse response peak measurement results after defuzzification

需要注意的是，以上成像结果均为具有一定几何形变的斜距面图像，各点的坐标与相对位置和各点在地面的相对位置和坐标有所不同。如果需要得到在地距平面的成像结果可以进行几何形变校正。It should be noted that the above imaging results are oblique range surface images with certain geometric deformation, and the coordinates and relative positions of each point and the relative position and coordinates of each point on the ground are different. If it is necessary to obtain the imaging results at the ground distance plane, geometric deformation correction can be performed.

由以上仿真结果我们可以看出，通过本文发明的基于波束形成的解模糊方法，可以在双天线前视模式下方便地解开左右模糊。From the above simulation results, we can see that through the beamforming-based deblurring method invented in this paper, the left and right ambiguities can be easily resolved in the dual-antenna forward-looking mode.

Claims

1. A high-speed aircraft carrying dual-antenna forward-looking SAR defuzzification algorithm is characterized in that: it comprises,

Step 1): Establish a dual-antenna forward-looking SAR imaging model;

Step 2): Perform distance FFT processing on the echo, and then correct linear walking and distance pulse pressure;

Step 3): Azimuth FFT processing, and then perform secondary distance compression and distance bending correction;

Step 4): distance IFFT processing and azimuth compression processing;

Step 5): phase compensation;

Step 6): Using the phase difference caused by the wave path difference between the two antennas to establish a steering vector, the left and right blur is resolved by beamforming, and the images on the left and right sides are respectively obtained.

2. A kind of high-speed aircraft carrying dual-antenna forward-looking SAR defuzzification algorithm as claimed in claim 1, it is characterized in that: described step 1 comprises,

Assume that the dual antennas are antenna A and antenna B, and there are two point targets P ₁ and P ₂ in the beam irradiation area directly in front of the aircraft, P ₁ and P ₂ are left and right symmetrical about the Y axis, and the coordinates are (x0, y0 ,0) and (-x0,y0,0);

The two-way slant distance equation between targets P ₁ , P ₂ and antenna A is

\begin{matrix} {R R}_{11__A A} (({t t}_{m m})) = = 22 \sqrt{{x x}_{00}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \\ {R R}_{22__A A} (({t t}_{m m})) = = 22 \sqrt{{((- - {x x}_{00}))}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \end{matrix}\} - - - - - - ((11)),,

The two-way slant distance equation between P ₁ , P ₂ and antenna B is

\begin{matrix} {R R}_{11__B B} (({t t}_{m m})) = = \sqrt{{x x}_{00}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} - - d d))}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \\ {R R}_{22__B B} (({t t}_{m m})) = = \sqrt{{x x}_{00}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} + + d d))}^{22} + + {(({vt vt}_{m m} - - {y the y}_{00}))}^{22} + + {H h}^{22}} \end{matrix}\} - - - - - - ((22))

Among them, H is the height of the carrier, v is the speed of the carrier, and t _m is the slow time.

3. A kind of high-speed aircraft carrying dual-antenna forward-looking SAR defuzzification algorithm as claimed in claim 1, it is characterized in that: described step 2 comprises,

Let point P(x, y, 0) be a point target in the scene. After t _m time, the aircraft moves to point A', its Y-axis coordinate is vt _m , and the distance between point P and point C is y _n , then the two-way slant distance RP_A(t _m ) between P and antenna A can be obtained from ⊿A'PC' as

{R R}_{P P__A A} (({t t}_{m m})) = = 22 \sqrt{{x x}^{22} + + {(({vt vt}_{m m} - - y the y))}^{22} + + {H h}^{22}}

= = 22 \sqrt{{(({vt vt}_{m m} - - {y the y}_{n no} - - {y the y}_{c c}))}^{22} + + {R R}_{B B}^{22}} - - - - - - ((33))

= = 22 \sqrt{{(({vt vt}_{m m} - - {y the y}_{n no}))}^{22} + + {R R}_{00}^{22} - - {22 R R}_{00} (({vt vt}_{m m} - - {y the y}_{n no})) sin sin θ θ}

in

R_{B} = \sqrt{x^{2} + h^{2}},

R_{0} = \sqrt{x^{2} + {the y}_{c}^{2} + h^{2}},

θ is the oblique angle.

Expand the formula (3) by Taylor series at y _n , keep the quadratic term of (vt _m -y _n ), get

{R R}_{P P__A A} (({t t}_{m m})) \approx \approx 22 [[{R R}_{00} - - (({vt vt}_{m m} - - {y the y}_{n no})) sin sin θ θ + + \frac{{v v}^{22} {cos cos}^{22} θ θ}{{22 R R}_{00}} {(({vt vt}_{m m} - - {y the y}_{n no}))}^{22}]] - - - - - - ((44))

Rewrite formula (4) into another form with the same accuracy

{R R}_{P P__A A} (({t t}_{m m})) = = 22 \sqrt{{R R}_{00}^{22} + + {cos cos}^{22} θ θ {(({vt vt}_{m m} - - {y the y}_{n no}))}^{22}} - - 22 (({vt vt}_{m m} - - {y the y}_{n no})) sin sin θ θ - - - - - - ((55))

Then the echo signal of target P received by antenna A is written as

{S S}_{A A} ((\overset{^^}{t t},, {t t}_{m m})) = = {σ σ}_{P P} {w w}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{P P__A A} (({t t}_{m m}))}{c c})) {w w}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{v v})) exp exp [[jπγ jπγ {((\overset{^^}{t t} - - \frac{{R R}_{P P__A A} (({t t}_{m m}))}{c c}))}^{22}]] exp exp [[- - j j \frac{22 π π}{λ λ} {R R}_{P P__A A} (({t t}_{m m}))]]

(6)

in,

is the fast time, λ is the signal wavelength, c is the speed of light, γ is the signal modulation frequency, w _r (.) and w _a (.) are the distance and azimuth window functions respectively, σ _P is the reflection coefficient of the target P,

Transform the echo fundamental frequency signal of formula (6) from the two-dimensional time domain to the range frequency domain-azimuth time domain,

{S S}_{A A} (({f f}_{r r},, {t t}_{m m})) = = {σ σ}_{P P} {w w}_{r r} (({f f}_{r r})) {w w}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{v v})) exp exp [[jπ jπ \frac{{f f}_{r r}^{22}}{γ γ}]] exp exp [[- - j j \frac{22 π π}{c c} {R R}_{P P__A A} (({t t}_{m m})) (({f f}_{r r} + + {f f}_{c c})) - - - - - - ((77))

Where f _c is the carrier frequency, and f _r is the range frequency. According to formula (5), multiply formula (7) by the phase item of formula (8) to perform uniform linear motion correction on the data and perform distance pulse pressure processing:

{H h}_{LRMC LRMC} (({f f}_{r r},, {t t}_{m m})) = = exp exp [[- - j j 22 π π \frac{v v sin sin {θt θt}_{m m}}{c c} (({f f}_{r r} + + {f f}_{c c}))]] exp exp [[jπ jπ \frac{{f f}_{r r}^{22}}{γ γ}]] . . - - - - - - ((88))

4. A kind of high-speed aircraft carrying dual-antenna forward-looking SAR defuzzification algorithm as claimed in claim 3, it is characterized in that: described step 3 comprises,

Transform the range-walking-corrected and range-pulse signals into the two-dimensional frequency domain, and use the following phase terms for range-migration correction and quadratic range-pulse pressure

{H h}_{rg r g} (({f f}_{r r},, {f f}_{a a})) = = exp exp [[- - jπ jπ \frac{{22 λR λR}_{s the s} {((\frac{{λf λf}_{a a}}{22 v v cos cos θ θ}))}^{22}}{{c c}^{22} {((\sqrt{11 - - {((\frac{{λf λf}_{a a}}{22 v v cos cos θ θ}))}^{22}}))}^{33}} {f f}_{11}^{22}]] exp exp [[j j \frac{{22 πR πR}_{s the s}}{c c} {((\frac{{λf λf}_{a a}}{22 v v cos cos θ θ}))}^{22} {f f}_{r r}]] - - - - - - ((99))

The range-to-inverse Fourier transform is performed on the signal, and the signal expression in the range-Doppler domain is obtained as follows

{S S}_{A A} ((\overset{^^}{t t},, {f f}_{a a})) = = {σ σ}_{P P} sin sin cp cp [[{Δf Δf}_{r r} ((\overset{^^}{t t} - - \frac{22 (({R R}_{00} + + {y the y}_{n no} sin sin θ θ))}{c c}))]] {a a}_{a a} ((\frac{{R R}_{00} λ λ {f f}_{a a}}{{22 v v}^{22} \sqrt{11 - - {(({f f}_{a a} / / {f f}_{aM a M}))}^{22}}}))

\cdot &Center Dot; exp exp [[- - j j \frac{22 π π}{v v cos cos θ θ} {R R}_{00} \sqrt{{f f}_{aM a M}^{22} - - {f f}_{a a}^{22}}]] exp exp [[- - j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ]] exp exp [[- - j j 22 {πf πf}_{a a} \frac{{y the y}_{n no}}{v v}]]

(10)

Among them, Δf _r is the transmission signal bandwidth, f _a is the azimuth frequency, R _s is the slant distance of the ray at the center line of the scene, and completes the processing of the distance direction of the echo signal.

5. A kind of high-speed aircraft carrying dual-antenna forward-looking SAR defuzzification algorithm as claimed in claim 4, it is characterized in that: described step 4 comprises,

Rewrite formula (10) into a form that can simultaneously represent the signals of targets P1 and P2 received by antennas A and B

{S S}_{M m} ((\overset{^^}{t t},, {f f}_{a a})) = = {σ σ}_{N N} sin sin c c [[{Δf Δ f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__M m__00} + + {22 y the y}_{n no} sin sin θ θ}{c c}))]] {w w}_{a a} ((\frac{{R R}_{N N__M m__00} {λf λ f}_{a a}}{{44 v v}^{22} \sqrt{11 - - {(({f f}_{a a} / / {f f}_{aM a M}))}^{22}}}))

(11)

. . exp exp [[- - j j \frac{π π}{v v cos cos θ θ} {R R}_{N N__M m__00} \sqrt{{f f}_{aM a M}^{22} - - {f f}_{a a}^{22}}]] exp exp [[- - j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ]] exp exp [[- - j j {22 πf πf}_{a a} \frac{{y the y}_{n no}}{v v}]]

Among them, N=1,2 represent the targets P ₁ and P ₂ , and M=A, B, represent the antennas A and B. R _{N_M} is the four slant distances in formulas (1) and (2), and R _{N_M_0} represents the slant distance when t _m =0, namely

{R R}_{N N__M m__00} \{\begin{matrix} {R R}_{11__A A__00} = = 22 \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} \\ {R R}_{22__A A__00} = = 22 \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} \\ {R R}_{11__B B__00} = = \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} - - d d))}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} \\ {R R}_{22__B B__00} = = \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} + + \sqrt{{(({x x}_{00} + + d d))}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}} \end{matrix} - - - - - - ((1212))

The first phase term in Eq. (11) is the azimuth modulation term, and expanding it into the Taylor series of f _a gives

exp exp [[- - j j \frac{π π}{V V cos cos θ θ} {R R}_{N N__M m__00} \sqrt{{f f}_{aM a M}^{22} - - {f f}_{a a}^{22}}]] \approx \approx exp exp [[- - j j \frac{{22 πR πR}_{N N__M m__00}}{λ λ} ((11 - - \frac{11}{22} \frac{{f f}_{a a}^{22}}{{f f}_{aM a M}^{22}} - - \frac{11}{88} \frac{{f f}_{a a}^{44}}{{f f}_{aM a M}^{44}}))]] - - - - - - ((1313))

Constructs an orientation matched filter with no constant term

{H h}_{az az} ((\overset{^^}{t t},, {f f}_{a a})) = = exp exp [[j j \frac{{22 πR πR}_{N N__M m__00}}{λ λ} ((- - \frac{11}{22} \frac{{f f}_{a a}^{22}}{{f f}_{aM a M}^{22}} - - \frac{11}{88} \frac{{f f}_{a a}^{44}}{{f f}_{aM a M}^{44}}))]] - - - - - - ((1414))

The focused image obtained after azimuth matched filtering is

{S S}_{M m} ((\overset{^^}{t t},, {t t}_{m m})) = = {σ σ}_{N N} sin sin c c [[{Δf Δ f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__M m__00} + + {22 y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[{Δf Δ f}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{v v}))]]

(15)

exp exp [[- - j j \frac{{22 πR πR}_{N N__M m__00}}{λ λ}]] exp exp [[- - j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ]]

where Δf _a is the Doppler bandwidth in azimuth.

6. A kind of high-speed aircraft carrying dual-antenna forward-looking SAR defuzzification algorithm as claimed in claim 5, it is characterized in that: described step 5 comprises,

The four slant distance formulas in formula (12) have a common term

The four aggregated images represented by Equation (15) contain a common phase Eliminate the common phase and the residual phase by using the phase term of formula (16)

\exp (- j \frac{4 π}{λ} {the y}_{no} \sin θ)

{H h}_{phase phase} = = exp exp [[j j \frac{22 π π \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}}{λ λ}]] exp exp ((j j \frac{44 π π}{λ λ} {y the y}_{n no} sin sin θ θ)) - - - - - - ((1616))

get the signal

{S S}_{M m} ((\overset{^^}{t t},, {t t}_{m m})) = = {σ σ}_{N N} sin sin c c [[{Δf Δf}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__M m__00} + + {22 y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[{Δf Δf}_{a a} (({t t}_{m m} - - \frac{{Y Y}_{n no}}{V V}))]]

(17)

exp exp [[- - j j \frac{22 π π (({R R}_{N N__M m__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]]

7. A kind of high-speed aircraft carrying dual-antenna forward-looking SAR defuzzification algorithm as claimed in claim 6, it is characterized in that: described step 6 comprises,

From (17), write the expressions of the aliased signals in the two antennas respectively

\{\begin{matrix} {S S}_{A A} ((\overset{^^}{t t},, {t t}_{m m})) = = {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} sin sin c c [[{Δf Δ f}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__A A__00} + + {22 y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[{Δf Δf}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{V V}))]] \\ \cdot \cdot exp exp [[- - j j \frac{22 π π (({R R}_{N N__A A__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \\ {S S}_{B B} ((\overset{^^}{t t},, {t t}_{m m})) = = {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} sin sin c c [[{Δf Δf}_{r r} ((\overset{^^}{t t} - - \frac{{R R}_{N N__B B__00} + + {22 y the y}_{n no} sin sin θ θ}{c c}))]] sin sin c c [[{Δf Δf}_{a a} (({t t}_{m m} - - \frac{{y the y}_{n no}}{V V}))]] \\ \cdot \cdot exp exp [[- - j j \frac{22 π π (({R R}_{N N__B B__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} {H h}^{22}}))}{λ λ} \end{matrix} - - - - - - ((1818))

Take out the peak values Peak _A and Peak _B of the impulse response in (18), since the peak value of the sinc function is 1, the observed signal vector S received by the binary linear array can be obtained as

S S = = [[\begin{matrix} {Peak Peak}_{A A} \\ {Peak Peak}_{B B} \end{matrix}]

= = [\begin{matrix} {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} exp exp [[- - j j \frac{22 π π (({R R}_{N N__A A__00} + + \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \\ {Σ Σ}_{N N}^{N N = = 1,2 1,2} {σ σ}_{N N} exp exp [[- - j j \frac{22 π π (({R R}_{N N__B B__00} \sqrt{{x x}_{00}^{22} + + {y the y}_{00}^{22} + + {H h}^{22}}))}{λ λ}]] \end{matrix}]

= [\begin{matrix} {σ_{1} e}^{- j \frac{2 π}{λ} \sqrt{x_{0}^{2} + {the y}_{0}^{2} + h^{2}}} + {σ_{2} e}^{- j \frac{2 π}{λ} \sqrt{x_{0}^{2} + {the y}_{0}^{2} + h^{2}}} \\ {σ_{1} e}^{- j \frac{2 π}{λ} (\sqrt{{(x_{0} - d)}^{2} + {the y}_{0}^{2} + h^{2}})} + {σ_{2} e}^{- j \frac{2 π}{λ} (\sqrt{{(x_{0} + d)}^{2} + {the y}_{0}^{2} + h^{2}})} \end{matrix}]

(19)

= = [\begin{matrix} {σ σ}_{11} {a a}_{11__A A} + + {σ σ}_{22} {a a}_{22__A A} \\ {σ σ}_{11} {a a}_{11__B B} + + {σ σ}_{22} {a a}_{22 - - B B} \end{matrix}]

= = [\begin{matrix} {a a}_{11__A A},, {a a}_{22 - - A A} \\ {a a}_{11__B B},, {a a}_{22 - - B B} \end{matrix}] [\begin{matrix} {σ σ}_{11} \\ {σ σ}_{22} \end{matrix}]

Where [a _{1_A} , a _{2_A} ] ^T and [a _{1_B} , a _{2_B} ] ^T are the steering vectors of P ₁ and P ₂ respectively,

The signal from one side of the Y-axis is suppressed by the beamforming method, and the signal on the other side of the Y-axis is obtained. If the weighted vector of the array beam pointing to P ₁ is W ₁ , and the weighting vector pointing to P ₂ is W ₂ , then the weighted The matrix is W=[W ₁ , W ₂ ],

Solve the expression of W, and get the equation system from formula (19)

\{\begin{matrix} {Peak Peak}_{A A} = = {σ σ}_{11} {a a}_{11__A A} + + {σ σ}_{22} {a a}_{22__A A} \\ {Peak Peak}_{B B} = = {σ σ}_{11} {a a}_{11__B B} + + {σ σ}_{22} {a a}_{22__B B} \end{matrix} - - - - - - ((2020))

Solving the system of equations (20) gives

\{\begin{matrix} {σ σ}_{11} = = \frac{11}{b b} [[{Peak Peak}_{A A} {a a}_{22__B B} - - {Peak Peak}_{B B} {a a}_{22__A A}]] \\ {σ σ}_{22} = = \frac{11}{b b} [[- - {Peak Peak}_{A A} {a a}_{11__B B} + + {Peak Peak}_{B B} {a a}_{11__A A}]] \end{matrix} - - - - - - ((21 twenty one))

where b=a _{1_A} a _{2_B} -a _{2_A} a _{1_B} ,

From (21) write W ₁ as

{W W}_{11} = = \frac{11}{b b} [\begin{matrix} {a a}_{22__B B} \\ - - {a a}_{22__A A} \end{matrix}] - - - - - - ((22 twenty two))

while _W2 is

{W W}_{22} = = \frac{11}{b b} [\begin{matrix} - - {a a}_{11__B B} \\ {a a}_{11__A A} \end{matrix}] - - - - - - ((23 twenty three))

Then the weight matrix W is

W W = = [[{W W}_{11},, {W W}_{22}]] = = \frac{11}{b b} [\begin{matrix} {a a}_{22__B B},, - - {a a}_{11__B B} \\ - - {a a}_{22__A A,,} {a a}_{11__A A} \end{matrix}] - - - - - - ((24 twenty four))

The process of unambiguous left and right for aliased signals is

{W W}^{H h} S S = = {W W}^{H h} [\begin{matrix} {a a}_{11__A A},, {a a}_{22__A A} \\ {a a}_{11__B B},, {a a}_{22__B B} \end{matrix}] [\begin{matrix} {σ σ}_{11} \\ {σ σ}_{22} \end{matrix}] = = [\begin{matrix} {σ σ}_{11} \\ {σ σ}_{22} \end{matrix}] - - - - - - ((2525)) . .