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

Abstract

The invention provides a foresight SAR ambiguity resolving algorithm of double antennas carried by a high-speed aircraft. The foresight SAR ambiguity resolving algorithm includes: step 1), building a double-antenna foresight SAR imaging model; step 2), performing distance FFT processing on echo and then correcting linear motion and distance pulse pressure; step 3), performing azimuth FFT processing and then performing secondary distance compression and range curvature correction; step 4), performing distance IFFT processing and orientation compression processing; step 5), performing phase compensation; step 6), utilizing phase difference caused by wave path difference between the double antennas to build an orientation vector, and resolving left and right ambiguity through a wave beam forming mode to obtain an image on the left side and an image on the right side respectively. The foresight SAR ambiguity resolving algorithm is simple and feasible and meets processing requirements of high-speed aircraft platforms.

Description

A kind of high-speed aircraft carries double antenna Forward-looking SAR ambiguity solution algorithm

[technical field]

The present invention relates to a kind of high-speed aircraft and carry double antenna Forward-looking SAR ambiguity solution algorithm.

[background technology]

Positive side-looking can be carried out high-resolution imaging to the target (scene) of carrier (aircraft, satellite, guided missile etc.) direction of motion both sides with stravismus formula SAR, and its good imaging performance has obtained widely research and used.Yet, in actual radar imagery is used, have greatly that demand requires radar can have the function that platform direction of motion dead ahead scene is carried out imaging, namely to the demand of Forward-looking SAR.For example, aspect civilian, Forward-looking SAR can be used for the aircraft blind landing system of round-the-clock round-the-clock; Militarily, the fighter plane or the reconnaissance plane that are equipped with Forward-looking SAR can improve the ability of finding the place ahead target; Especially, will improve widely in conjunction with the missile attack degree of accuracy of Forward-looking SAR Imaging Guidance and the dirigibility of attack pattern.By the principle of SAR imaging as can be known, front lower according to circumstances, the mitre joint of range resolution gradient and azimuthal resolution gradient is bordering on zero, with the perpendicular direction of carrier movement direction even null situation can occur, this moment, traditional SAR imaging mode can't normal imaging, and monosymmetric target has identical Doppler history and has therefore caused the fuzzy of SAR image, i.e. left and right sides fuzzy problem in the course.

Owing to comparing with linear array, double antenna forward sight system architecture is fairly simple and be convenient to development and install, the generation triumph, Liu Guangyan etc. are studied double antenna forward sight imaging system characteristic and imaging algorithm, method fuzzy about a kind of solution has been proposed, obtained certain effect, but the method need to be carried out the zone to scene and be divided, binary channels data to zones of different are carried out permutatation, then the different adaptation function of the data configuration after the permutatation is carried out imaging, its algorithm complex is higher, is difficult to satisfy the requirement that high-speed aircraft carries the Forward-looking SAR real time imagery.

[summary of the invention]

The object of the present invention is to provide a kind of high-speed aircraft to carry double antenna Forward-looking SAR ambiguity solution algorithm, it is characterized in that: it comprises,

Step 1): set up double antenna Forward-looking SAR imaging model;

Step 2): echo is carried out processing apart from FFT, proofread and correct then that linearity is walked about and apart from pulse pressure;

Step 3): the orientation is processed to FFT, then carries out secondary range compression and range curvature and proofreaies and correct;

Step 4): process and carry out Azimuth Compression and process apart from IFFT;

Step 5): phase compensation;

Step 6): the phase differential that utilizes the wave path-difference between double antenna to cause is set up steering vector, by fuzzy about the mode solution of wave beam formation, obtains respectively the image of the left and right sides.

On the basis of technique scheme, described step 1 comprises,

If described double antenna is antenna A and antenna B, establishing in the wave beam irradiation area in aircraft dead ahead has P ₁And P ₂Two point targets, P ₁And P ₂With the Y-axis left-right symmetric, coordinate is respectively (x0, y0,0) and (x0, y0,0);

Target P ₁, P ₂And the round trip oblique distance equation between the antenna A is

$\left.\begin{array}{c}{R}_{1\_A}\left(t_m\right)=2\sqrt{x_0^2+{(v{t}_m-y_0)}^2+H^2}\\ R_{2\_A}\left(t_m\right)=2\sqrt{{(-x_0)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\end{array}\right\}---\left(1\right),$

P ₁, P ₂And the round trip oblique distance equation between the antenna B is

$\left.\begin{array}{c}{R}_{1\_B}\left(t_m\right)=\sqrt{x_0^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}+\sqrt{{(x_0-d)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\\ R_{2\_B}\left(t_m\right)=\sqrt{x_0^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}+\sqrt{{(x_0+d)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\end{array}\right\}---\left(2\right)$

Wherein, H is the carrier aircraft height, and v is carrier aircraft speed, t _mBe the slow time.

On the basis of technique scheme, described step 2 comprises,

The P (x, y, 0) that sets up an office is point target in the scene, through t _mAfter time, aircraft moves to the A' point, and its Y-axis coordinate is vt _m, the distance between the target P of setting up an office and the C point is y _n, Ze Cong ⊿ A'PC' can get the round trip oblique distance RP_A (t between P and antenna A _m) be

$R_{P\_A}\left(t_m\right)=2\sqrt{x^2+{({\mathrm{vt}}_m-y)}^2+H^2}$

$=2\sqrt{{({\mathrm{vt}}_m-y_n-y_c)}^2+{R_B}^2}---\left(3\right)$

$=2\sqrt{{({\mathrm{vt}}_m-y_n)}^2+R_0^2-2R_0({\mathrm{vt}}_m-y_n)\sin \mathrm{\θ}}$

Wherein $R_B=\sqrt{x^2+H^2},$ $R_0=\sqrt{x^2+y_c^2+H^2},$ θ is the angle of squint.

With formula (3) at y _nThe place makes Taylor series expansion, keeps (vt _m-y _n) following of secondary,

$R_{P\_A}\left(t_m\right)\≈2[R_0-({\mathrm{vt}}_m-y_n)\sin \mathrm{\θ}+\frac{v^2{\cos }^2\mathrm{\θ}}{2R_0}{({\mathrm{vt}}_m-y_n)}^2]---\left(4\right)$

Formula (4) is rewritten as another identical form of precision

$R_{P\_A}\left(t_m\right)=2\sqrt{R_0^2+{\cos }^2\mathrm{\θ}{({\mathrm{vt}}_m-y_n)}^2}-2({\mathrm{vt}}_m-y_n)\sin \mathrm{\θ}---\left(5\right)$

Then the echoed signal of the target P that receives of antenna A is written as

$S_A(\hat{t},t_m)={\mathrm{\σ}}_P{w}_r(\hat{t}-\frac{R_{P\_A}\left(t_m\right)}{c})w_a(t_m-\frac{y_n}{v})\exp \left[\mathrm{j\π\γ}{\left(\hat{t}-\frac{R_{P\_A}\left(t_m\right)}{c}\right)}^2\right]\exp [-j\frac{2\mathrm{\π}}{\mathrm{\λ}}{R}_{P\_A}\left(t_m\right)]---\left(6\right)$

Wherein,

Be the fast time, λ is signal wavelength, and c is the light velocity, and γ is signal frequency modulation rate, w _r() and w _a() is respectively the distance and bearing window function, σ _PBe the reflection coefficient of target P,

The echo fundamental frequency signal of formula (6) is transformed to apart from frequency domain-orientation time domain from two-dimensional time-domain,

$S_A(f_r,t_m)={\mathrm{\σ}}_P{w}_r\left(f_r\right)w_a(t_m-\frac{y_n}{v})\exp [-\mathrm{j\π}\frac{{f_r}^2}{\mathrm{\γ}}]\exp [-j\frac{2\mathrm{\π}}{c}{R}_{P\_A}\left(t_m\right)(f_r+f_c)]---\left(7\right)$

F wherein _cBe carrier frequency, f _rBe frequency of distance, according to formula (5), formula (7) multiply by formula (8) phase term data are carried out unified linearity walk the normal moveout correction row distance of going forward side by side and process to pulse pressure:

$H_{\mathrm{LRMC}}(f_r,t_m)=\exp [-j2\mathrm{\π}\frac{v\sin {\mathrm{\θt}}_m}{c}\left(f_r{+f}_c\right)]\exp \left[\mathrm{j\π}\frac{{f_r}^2}{\mathrm{\γ}}\right].---\left(8\right)$

On the basis of technique scheme, described step 3 comprises,

Transform to two-dimensional frequency with Range Walk Correction with apart from the signal after pulse pressure, and use following phase term to carry out range migration correction and secondary range pulse pressure

$H_{\mathrm{rg}}(f_r,f_a)=\exp [-\mathrm{j\π}\frac{2\mathrm{\λ}{R}_s{\left(\frac{\mathrm{\λ}{f}_a}{2v\cos \mathrm{\θ}}\right)}^2}{c^2{\left(\sqrt{1-{\left(\frac{\mathrm{\λ}{f}_a}{2v\cos \mathrm{\θ}}\right)}^2}\right)}^3}{f_r}^2]\exp \left[j\frac{2\mathrm{\π}{R}_s}{c}{\left(\frac{\mathrm{\λ}{f}_a}{2v\cos \mathrm{\θ}}\right)}^2{f}_r\right]---\left(9\right)$

Signal is carried out distance to inverse Fourier transform, and the signal expression that obtains the distance-Doppler territory is as follows

$S_A(\hat{t},f_a)={\mathrm{\&sigma;}}_P\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{2({\mathrm{<figure-callout\; id="0"\; label="R"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">R</figure-callout>}}_0+y_n\sin \mathrm{\&theta;})}{c})\right]a_a\left(\frac{{\mathrm{<figure-callout\; id="0"\; label="R"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">R</figure-callout>}}_0\mathrm{\&lambda;}{f}_a}{{2v}^2\sqrt{1-{(f_a/f_{\mathrm{aM}})}^2}}\right)---\left(10\right)$

$\&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}}{v\cos \mathrm{\&theta;}}{R}_0\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]\exp [-j2\mathrm{\&pi;}{f}_a\frac{y_n}{v}]$

Wherein, Δ f _rBe transmitted signal bandwidth, f _aBe the orientation frequency,

R _sBe the ray oblique distance at scene center line place, finish echoed signal distance to processing.

On the basis of technique scheme, described step 4 comprises,

Formula (10) is rewritten as the form of the signal of the target P 1 that can represent simultaneously antenna A, B and receive, P2

$S_M(\hat{t},f_a)={\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_M\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]w_a\left(\frac{R_{N\_M\_0}\mathrm{\&lambda;}{f}_a}{4v^2\sqrt{1-{(f_a/f_{\mathrm{aM}})}^2}}\right)---\left(11\right)$

$\&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}}{v\cos \mathrm{\&theta;}}{R}_{N\_M\_0}\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]\exp [-j2\mathrm{\&pi;}{f}_a\frac{y_n}{v}]$

N=1 wherein, 2 expression target P ₁With P ₂, and M=A, B, expression antenna A and B.R _{N_M}Be four oblique distances in formula (1) and (2), and R _{N_M_0}Represent t _m=0 o'clock oblique distance, namely

$R_{N\_M\_0}\left\{\begin{array}{c}{R}_{1\_A\_0}=2\sqrt{x_0^2+y_0^2+H^2}\\ R_{2\_A\_0}=2\sqrt{x_0^2+y_0^2+H^2}\\ R_{1\_B\_0}=\sqrt{x_0^2+y_0^2+H^2}+\sqrt{{(x_0-d)}^2+y_0^2+H^2}\\ R_{2\_B\_0}=\sqrt{x_0^2+y_0^2+H^2}+\sqrt{{(x_0+d)}^2+y_0^2+H^2}\end{array}\right.---\left(12\right)$

First phase term in the formula (11) is the orientation modulation item, and it is expanded into f _aTaylor series get

$\exp [-j\frac{\mathrm{\&pi;}}{V\cos \mathrm{\&theta;}}{R}_{N\_M\_0}\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\&ap;\exp [-j\frac{2\mathrm{\&pi;}{R}_{N\_M\_0}}{\mathrm{\&lambda;}}(1-\frac{1}{2}\frac{f_a^2}{f_{\mathrm{aM}}^2}-\frac{1}{8}\frac{f_a^4}{f_{\mathrm{aM}}^4})]---\left(13\right)$

Construct an orientation matched filter that does not have constant term

$H_{\mathrm{az}}(\hat{t},f_a)=\exp \left[j\frac{2\mathrm{\&pi;}{R}_{N\_M\_0}}{\mathrm{\&lambda;}}(-\frac{1}{2}\frac{f_a^2}{f_{\mathrm{aM}}^2}-\frac{1}{8}\frac{f_a^4}{f_{\mathrm{aM}}^4})\right]---\left(14\right)$

The focusedimage that obtains after the orientation matched filtering is

$S_M(\hat{t},t_m)={\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_M\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[\mathrm{\&Delta;}{f}_a(t_m-\frac{y_n}{v})\right]---\left(15\right)$

$\exp [-j\frac{2\mathrm{\&pi;}{R}_{N\_M\_0}}{\mathrm{\&lambda;}}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]$

Δ f wherein _aFor the orientation to doppler bandwidth.

On the basis of technique scheme, described step 5 comprises,

Four oblique distance formula have one common in the formula (12)

Include a common phase in four focused image of formula (15) expression

With formula (16) phase term disappear common phase and excess phase $\exp (-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;})$

$H_{\mathrm{phase}}=\exp \left[j\frac{2\mathrm{\&pi;}\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}}{\mathrm{\&lambda;}}\right]\exp \left(j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}\right)---\left(16\right)$

Obtain signal

$S_M(\hat{t},t_m)={\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_M\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[\mathrm{\&Delta;}{f}_a(t_m-\frac{Y_n}{V})\right]$

$\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_M\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}].---\left(17\right)$

On the basis of technique scheme, described step 6 comprises,

Write out respectively the expression formula of aliasing signal in two antennas by (17)

$\left\{\begin{array}{c}{S}_A(\hat{t},t_m)=\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_A\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[\mathrm{\&Delta;}{f}_a(t_m-\frac{y_n}{V})\right]\\ \&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_A\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]\\ S_B(\hat{t},t_m)=\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_B\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[\mathrm{\&Delta;}{f}_a(t_m-\frac{y_n}{V})\right]\\ \&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_B\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]\end{array}\right.---\left(18\right)$

Take out the peak value Peak of impulse response in (18) _AAnd Peak _B, because the peak value of sinc function is 1, the observation signal vector S that can obtain the line arrays reception is

$S=\left[\begin{array}{c}{\mathrm{Peak}}_A\\ {\mathrm{Peak}}_B\end{array}\right]$

$=\left[\begin{array}{c}\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_A\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]\\ \underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_B\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]\end{array}\right]$

$=\left[\begin{array}{c}{\mathrm{\&sigma;}}_1{e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}}+{\mathrm{\&sigma;}}_2{e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}}\\ {\mathrm{\&sigma;}}_1{e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\left(\sqrt{{(x_0-d)}^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}\right)}+{\mathrm{\&sigma;}}_2{e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\left(\sqrt{{(x_0+d)}^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}\right)}\end{array}\right]---\left(19\right)$

$=\left[\begin{array}{c}{\mathrm{\&sigma;}}_1{a}_{1\_A}+{\mathrm{\&sigma;}}_2{a}_{2\_A}\\ {\mathrm{\&sigma;}}_1{a}_{1\_B}+{\mathrm{\&sigma;}}_2{a}_{2\_B}\end{array}\right]$

$=\left[\begin{array}{c}{a}_{1\_A},a_{2\_A}\\ a_{1\_B},a_{2\_B}\end{array}\right]\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]$

[a wherein _{1_A}, a _{2_A}] ^T[a _{1_B}, a _{2_B}] ^TBe respectively P ₁With P ₂Steering vector,

The method that forms by wave beam suppresses respectively from Y-axis one side signal, and obtains the signal of Y-axis opposite side, if make array beams point to P ₁Weighing vector be W ₁, point to P ₂Weighing vector be W ₂, then weighting matrix is W=[W ₁, W ₂],

Find the solution the expression formula of W, from formula (19), obtain system of equations

$\left\{\begin{array}{c}{\mathrm{Peak}}_A={\mathrm{\&sigma;}}_1{a}_{1\_A}+{\mathrm{\&sigma;}}_2{a}_{2\_A}\\ {\mathrm{Peak}}_B={\mathrm{\&sigma;}}_1{a}_{1\_B}+{\mathrm{\&sigma;}}_2{a}_{2\_B}\end{array}\right.---\left(20\right)$

(20) group of solving an equation obtains

$\left\{\begin{array}{c}{\mathrm{\&sigma;}}_1=\frac{1}{b}[{\mathrm{Peak}}_A{a}_{2\_B}-{\mathrm{Peak}}_B{a}_{2\_A}]\\ {\mathrm{\&sigma;}}_2=\frac{1}{b}[-{\mathrm{Peak}}_A{a}_{1\_B}+{\mathrm{Peak}}_B{a}_{1\_A}]\end{array}\right.---\left(21\right)$

B=a wherein _{1_A}a _{2_B}-a _{2_A}a _{1_B},

Write out W by (21) ₁For

${\mathrm{<figure-callout\; id="1"\; label="W"\; filenames="HDA00003502797200022.png"\; state="\{\{state\}\}">W</figure-callout>}}_1=\frac{1}{b}\left[\begin{array}{c}{a}_{2\_B}\\ -a_{2\_A}\end{array}\right]---\left(22\right)$

And W ₂For

$W_2=\frac{1}{b}\left[\begin{array}{c}-a_{1\_B}\\ a_{1\_A}\end{array}\right]---\left(23\right)$

Then weighting matrix W is

$W=[{\mathrm{<figure-callout\; id="1"\; label="W"\; filenames="HDA00003502797200022.png"\; state="\{\{state\}\}">W</figure-callout>}}_1,W_2]=\frac{1}{b}\left[\begin{array}{c}{a}_{2\_B},-a_{1\_B}\\ -a_{2\_A},a_{1\_A}\end{array}\right]---\left(24\right)$

To process fuzzy about the aliasing signal solution be

$W^{H}S=W^H\left[\begin{array}{c}{a}_{1\_A},a_{2\_A}\\ a_{1\_B},a_{2\_B}\end{array}\right]\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]=\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]---\left(25\right).$

Compared with prior art, the present invention at first looks side ways respectively imaging to the radar echo signal that double antenna receives, the SAR image that blurs about obtaining, then the respective pixel in two sub-pictures is taken out, the phase differential that utilizes the wave path-difference between double antenna to cause is set up steering vector, fuzzy about the mode that forms by wave beam is untied, obtain respectively the image of the left and right sides.Simulation results show, the method is simple and easy to do, is applicable to the processing requirements of high-speed flight applicator platform.

[description of drawings]

Fig. 1 is that aircraft of the present invention carries double antenna Forward-looking SAR ambiguity solution algorithm SAR geometric model figure;

Fig. 2 is Y-axis of the present invention right side scene Squint SAR geometric relationship figure;

Fig. 3 is that aircraft of the present invention carries double antenna Forward-looking SAR ambiguity solution algorithm process process flow diagram;

Fig. 4 is point target set-up mode figure of the present invention;

Fig. 5 is antenna A imaging results figure;

Fig. 6 is antenna B imaging results figure;

Fig. 7 is that antenna A is to central point imaging results figure;

Fig. 8 is that antenna B is to central point imaging results figure;

Fig. 9 is the Y-axis left-side images;

Figure 10 is the Y-axis image right;

Figure 11 is Y-axis left side scene center point diagram;

Figure 12 is Y-axis right side scene center point diagram.

[embodiment]

Please refer to Fig. 1 to Fig. 3, the simple aircraft of a kind of algorithm carries double antenna Forward-looking SAR ambiguity solution algorithm, and it comprises,

Step 1): set up double antenna Forward-looking SAR imaging model;

Step 2): echo is carried out processing apart from FFT, proofread and correct then that linearity is walked about and apart from pulse pressure;

Step 3): the orientation is processed to FFT, then carries out secondary range compression and range curvature and proofreaies and correct;

Step 4): process and carry out Azimuth Compression and process apart from IFFT;

Step 5): phase compensation;

Step 6): the phase differential that utilizes the wave path-difference between double antenna to cause is set up steering vector, by fuzzy about the mode solution of wave beam formation, obtains respectively the image of the left and right sides.

Please refer to Fig. 1, among the figure Texas tower with speed v along O ' Y ' unaccelerated flight.Two antennas are housed on the platform, are respectively antenna A and antenna B.When radar was worked, antenna A transmitted, and A and B receive echoed signal simultaneously.Antenna distance is d.If λ is the wavelength that transmits, for avoiding graing lobe, d≤λ/2 are arranged then.Make t _mFor the slow time, work as t _m=0 o'clock, Texas tower was positioned at O ', highly was H, this moment antenna A be positioned at (0,0, H), antenna B be positioned at (d, 0, H).Step 1 comprises, establishing described double antenna is antenna A and antenna B, and establishing in the wave beam irradiation area in aircraft dead ahead has P ₁And P ₂Two point targets, P ₁And P ₂With the Y-axis left-right symmetric, coordinate is respectively (x0, y0,0) and (x0, y0,0);

Target P ₁, P ₂And the round trip oblique distance equation between the antenna A is

$\left.\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA00003502797200022.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_A}\left(t_m\right)=2\sqrt{x_0^2+{(v{t}_m-y_0)}^2+H^2}\\ R_{2\_A}\left(t_m\right)=2\sqrt{{(-x_0)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\end{array}\right\}---\left(1\right),$

P ₁, P ₂And the round trip oblique distance equation between the antenna B is

$\left.\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA00003502797200022.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_B}\left(t_m\right)=\sqrt{x_0^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}+\sqrt{{(x_0-d)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\\ R_{2\_B}\left(t_m\right)=\sqrt{x_0^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}+\sqrt{{(x_0+d)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\end{array}\right\}---\left(2\right).$

Please refer to Fig. 2, step 2 comprises, aircraft along Y-axis with the speed v linear uniform motion.The P (x, y, 0) that sets up an office is point target in the scene, through t _mAfter time, aircraft moves to the A' point, and its Y-axis coordinate is vt _m, the distance between the target P of setting up an office and the C point is y _n, Ze Cong ⊿ A'PC' can get the round trip oblique distance RP_A (t between P and antenna A _m) be

$R_{P\_A}\left(t_m\right)=2\sqrt{x^2+{({\mathrm{vt}}_m-y)}^2+H^2}$

$=2\sqrt{{({\mathrm{vt}}_m-y_n-y_c)}^2+{R_B}^2}---\left(3\right)$

$=2\sqrt{{({\mathrm{vt}}_m-y_n)}^2+{\mathrm{<figure-callout\; id="0"\; label="R"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">R</figure-callout>}}_0^2-2R_0({\mathrm{vt}}_m-y_n)\sin \mathrm{\&theta;}}$

Wherein $R_B=\sqrt{x^2+H^2},$ ${\mathrm{<figure-callout\; id="0"\; label="R"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">R</figure-callout>}}_0=\sqrt{x^2+y_c^2+H^2},$ θ is the angle of squint.

With formula (3) at y _nThe place makes Taylor series expansion, keeps (vt _m-y _n) following of secondary,

$R_{P\_A}\left(t_m\right)\&ap;2[R_0-({\mathrm{vt}}_m-y_n)\sin \mathrm{\&theta;}+\frac{v^2{\cos }^2\mathrm{\&theta;}}{2R_0}{({\mathrm{vt}}_m-y_n)}^2]---\left(4\right)$

Formula (4) is rewritten as another identical form of precision

$R_{P\_A}\left(t_m\right)=2\sqrt{{\mathrm{<figure-callout\; id="0"\; label="R"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">R</figure-callout>}}_0^2+{\cos }^2\mathrm{\&theta;}{({\mathrm{vt}}_m-y_n)}^2}-2({\mathrm{vt}}_m-y_n)\sin \mathrm{\&theta;}---\left(5\right)$

Then the echoed signal of the target P that receives of antenna A is written as

$S_A(\hat{t},t_m)={\mathrm{\&sigma;}}_P{w}_r(\hat{t}-\frac{R_{P\_A}\left(t_m\right)}{c})w_a(t_m-\frac{y_n}{v})\exp \left[\mathrm{j\&pi;\&gamma;}{\left(\hat{t}-\frac{R_{P\_A}\left(t_m\right)}{c}\right)}^2\right]\exp [-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}{R}_{P\_A}\left(t_m\right)]---\left(6\right)$

Wherein,

Be the fast time, λ is signal wavelength, and c is the light velocity, and γ is signal frequency modulation rate, w _r() and w _a() is respectively the distance and bearing window function, σ _PBe the reflection coefficient of target P,

The echo fundamental frequency signal of formula (6) is transformed to apart from frequency domain-orientation time domain from two-dimensional time-domain,

$S_A(f_r,t_m)={\mathrm{\&sigma;}}_P{w}_r\left(f_r\right)w_a(t_m-\frac{y_n}{v})\exp [-\mathrm{j\&pi;}\frac{{f_r}^2}{\mathrm{\&gamma;}}]\exp [-j\frac{2\mathrm{\&pi;}}{c}{R}_{P\_A}\left(t_m\right)(f_r+f_c)]---\left(7\right)$

F wherein _cBe carrier frequency, f _rBe frequency of distance, according to formula (5), formula (7) multiply by following phase term can carry out unified linearity to data and walk the normal moveout correction row distance of going forward side by side and process to pulse pressure:

$H_{\mathrm{LRMC}}(f_r,t_m)=\exp [-j2\mathrm{\&pi;}\frac{v\sin {\mathrm{\&theta;t}}_m}{c}\left(f_r{+f}_c\right)]\exp \left[\mathrm{j\&pi;}\frac{{f_r}^2}{\mathrm{\&gamma;}}\right]---\left(8\right)$

Transform to two-dimensional frequency with Range Walk Correction with apart from the signal after pulse pressure, then use following phase term to carry out range migration correction and secondary range pulse pressure

$H_{\mathrm{rg}}(f_r,f_a)=\exp [-\mathrm{j\&pi;}\frac{2\mathrm{\&lambda;}{R}_s{\left(\frac{\mathrm{\&lambda;}{f}_a}{2v\cos \mathrm{\&theta;}}\right)}^2}{c^2{\left(\sqrt{1-{\left(\frac{\mathrm{\&lambda;}{f}_a}{2v\cos \mathrm{\&theta;}}\right)}^2}\right)}^3}{f_r}^2]\exp \left[j\frac{2\mathrm{\&pi;}{R}_s}{c}{\left(\frac{\mathrm{\&lambda;}{f}_a}{2v\cos \mathrm{\&theta;}}\right)}^2{f}_r\right]---\left(9\right)$

Signal is carried out distance to inverse Fourier transform, and the signal expression that obtains the distance-Doppler territory is as follows

$S_A(\hat{t},f_a)={\mathrm{\&sigma;}}_P\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{2({\mathrm{<figure-callout\; id="0"\; label="R"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">R</figure-callout>}}_0+y_n\sin \mathrm{\&theta;})}{c})\right]a_a\left(\frac{{\mathrm{<figure-callout\; id="0"\; label="R"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">R</figure-callout>}}_0\mathrm{\&lambda;}{f}_a}{{2v}^2\sqrt{1-{(f_a/f_{\mathrm{aM}})}^2}}\right)---\left(10\right)$

$\&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}}{v\cos \mathrm{\&theta;}}{R}_0\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]\exp [-j2\mathrm{\&pi;}{f}_a\frac{y_n}{v}]$

Wherein, Δ f _rBe transmitted signal bandwidth, f _aBe the orientation frequency,

R _sBe the ray oblique distance at scene center line place, finish echoed signal distance to processing.

On the basis of technique scheme, described step 3 comprises,

Formula (10) is rewritten as the form of the signal of the target P 1 that can represent simultaneously antenna A, B and receive, P2

$S_M(\hat{t},f_a)={\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_M\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]w_a\left(\frac{R_{N\_M\_0}\mathrm{\&lambda;}{f}_a}{4v^2\sqrt{1-{(f_a/f_{\mathrm{aM}})}^2}}\right)---\left(11\right)$

$\&CenterDot;\exp [-j\frac{\mathrm{\&pi;}}{v\cos \mathrm{\&theta;}}{R}_{N\_M\_0}\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]\exp [-j2\mathrm{\&pi;}{f}_a\frac{y_n}{v}]$

N=1 wherein, 2 expression target P ₁With P ₂, and M=A, B, expression antenna A and B.R _{N_M}Be four oblique distances in formula (1) and (2), and R _{N_M_0}Represent t _m=0 o'clock oblique distance, namely

$R_{N\_M\_0}\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA00003502797200022.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_A\_0}=2\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}\\ R_{2\_A\_0}=2\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}\\ {\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA00003502797200022.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_B\_0}=\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}+\sqrt{{(x_0-d)}^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}\\ R_{2\_B\_0}=\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}+\sqrt{{(x_0+d)}^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}\end{array}\right.---\left(12\right)$

First phase term in the formula (11) is the orientation modulation item, and it is expanded into f _aTaylor series get

$\exp [-j\frac{\mathrm{\&pi;}}{V\cos \mathrm{\&theta;}}{R}_{N\_M\_0}\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\&ap;\exp [-j\frac{2\mathrm{\&pi;}{R}_{N\_M\_0}}{\mathrm{\&lambda;}}(1-\frac{1}{2}\frac{f_a^2}{f_{\mathrm{aM}}^2}-\frac{1}{8}\frac{f_a^4}{f_{\mathrm{aM}}^4})]---\left(13\right)$

Can notice, first in the formula (13) after the Taylor expansion is constant term, and corresponding target is to the distance of antenna.For different antennas and different target, this constant term is different, namely has different wave path-differences.Therefore we next consider to utilize this difference to P ₁And P ₂Ambiguity solution.

Because therefore constant term can construct an orientation matched filter that does not have constant term to the not impact of orientation focusing effect

$H_{\mathrm{az}}(\hat{t},f_a)=\exp \left[j\frac{2\mathrm{\&pi;}{R}_{N\_M\_0}}{\mathrm{\&lambda;}}(-\frac{1}{2}\frac{f_a^2}{f_{\mathrm{aM}}^2}-\frac{1}{8}\frac{f_a^4}{f_{\mathrm{aM}}^4})\right]---\left(14\right)$

The focusedimage that can obtain after the orientation matched filtering is

$S_M(\hat{t},t_m)={\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_M\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[\mathrm{\&Delta;}{f}_a(t_m-\frac{y_n}{v})\right]---\left(15\right)$

$\exp [-j\frac{2\mathrm{\&pi;}{R}_{N\_M\_0}}{\mathrm{\&lambda;}}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]$

Δ f wherein _aFor the orientation to doppler bandwidth.

Can find, four oblique distance formula have one common in formula (12) Therefore include a common phase in four focused image of formula (15) expression

With following phase term with this common phase and excess phase $\exp (-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;})$ Disappear together

$H_{\mathrm{phase}}=\exp \left[j\frac{2\mathrm{\&pi;}\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}}{\mathrm{\&lambda;}}\right]\exp \left(j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}\right)---\left(16\right)$

Obtain signal

$S_M(\hat{t},t_m)={\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_M\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[\mathrm{\&Delta;}{f}_a(t_m-\frac{Y_n}{V})\right]$

$\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_M\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]---\left(17\right)$

Wherein the peak value of sinc function is corresponding with the reflection coefficient of point target, and the physical location of the position of sinc function in image and point target is corresponding, because antenna A, signal is aliasing among the B, can be write out respectively by (17) expression formula of aliasing signal in two antennas

$\left\{\begin{array}{c}{S}_A(\hat{t},t_m)=\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_A\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[\mathrm{\&Delta;}{f}_a(t_m-\frac{y_n}{V})\right]\\ \&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_A\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]\\ S_B(\hat{t},t_m)=\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\sin c\left[\mathrm{\&Delta;}{f}_r(\hat{t}-\frac{R_{N\_B\_0}+2y_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[\mathrm{\&Delta;}{f}_a(t_m-\frac{y_n}{V})\right]\\ \&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_B\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]\end{array}\right.---\left(18\right)$

Take out the peak value Peak of impulse response in (18) _AAnd Peak _B, because the peak value of sinc function is 1, the observation signal vector S that can obtain the line arrays reception is

$S=\left[\begin{array}{c}{\mathrm{Peak}}_A\\ {\mathrm{Peak}}_B\end{array}\right]$

$=\left[\begin{array}{c}\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_A\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]\\ \underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_B\_0}+\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2})}{\mathrm{\&lambda;}}]\end{array}\right]$

$=\left[\begin{array}{c}{\mathrm{\&sigma;}}_1{e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}}+{\mathrm{\&sigma;}}_2{e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\sqrt{x_0^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}}\\ {\mathrm{\&sigma;}}_1{e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\left(\sqrt{{(x_0-d)}^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}\right)}+{\mathrm{\&sigma;}}_2{e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\left(\sqrt{{(x_0+d)}^2+{\mathrm{<figure-callout\; id="0"\; label="y"\; filenames="HDA00003502797200012.png"\; state="\{\{state\}\}">y</figure-callout>}}_0^2+H^2}\right)}\end{array}\right]---\left(19\right)$

$=\left[\begin{array}{c}{\mathrm{\&sigma;}}_1{a}_{1\_A}+{\mathrm{\&sigma;}}_2{a}_{2\_A}\\ {\mathrm{\&sigma;}}_1{a}_{1\_B}+{\mathrm{\&sigma;}}_2{a}_{2\_B}\end{array}\right]$

$=\left[\begin{array}{c}{a}_{1\_A},a_{2\_A}\\ a_{1\_B},a_{2\_B}\end{array}\right]\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]$

[a wherein _{1_A}, a _{2_A}] ^T[a _{1_B}, a _{2_B}] ^TBe respectively P ₁With P ₂Steering vector,

Obviously, for formula (19), can suppress respectively from Y-axis one side signal by the method that wave beam forms, and obtain the signal of opposite side, if make array beams point to P ₁Weighing vector be W ₁, point to P ₂Weighing vector be W ₂, then weighting matrix is W=[W ₁, W ₂],

Find the solution the expression formula of W, from formula (19), obtain a system of equations

$\left\{\begin{array}{c}{\mathrm{Peak}}_A={\mathrm{\&sigma;}}_1{a}_{1\_A}+{\mathrm{\&sigma;}}_2{a}_{2\_A}\\ {\mathrm{Peak}}_B={\mathrm{\&sigma;}}_1{a}_{1\_B}+{\mathrm{\&sigma;}}_2{a}_{2\_B}\end{array}\right.---\left(20\right)$

Separating this system of equations obtains

$\left\{\begin{array}{c}{\mathrm{\&sigma;}}_1=\frac{1}{b}[{\mathrm{Peak}}_A{a}_{2\_B}-{\mathrm{Peak}}_B{a}_{2\_A}]\\ {\mathrm{\&sigma;}}_2=\frac{1}{b}[-{\mathrm{Peak}}_A{a}_{1\_B}+{\mathrm{Peak}}_B{a}_{1\_A}]\end{array}\right.---\left(21\right)$

B=a wherein _{1_A}a _{2_B}-a _{2_A}a _{1_B},

Write out W by (21) ₁For

${\mathrm{<figure-callout\; id="1"\; label="W"\; filenames="HDA00003502797200022.png"\; state="\{\{state\}\}">W</figure-callout>}}_1=\frac{1}{b}\left[\begin{array}{c}{a}_{2\_B}\\ -a_{2\_A}\end{array}\right]---\left(22\right)$

And W ₂For

$W_2=\frac{1}{b}\left[\begin{array}{c}-a_{1\_B}\\ a_{1\_A}\end{array}\right]---\left(23\right)$

Then weighting matrix W is

$W=[{\mathrm{<figure-callout\; id="1"\; label="W"\; filenames="HDA00003502797200022.png"\; state="\{\{state\}\}">W</figure-callout>}}_1,W_2]=\frac{1}{b}\left[\begin{array}{c}{a}_{2\_B},-a_{1\_B}\\ -a_{2\_A},a_{1\_A}\end{array}\right]---\left(24\right)$

So, process fuzzy about the aliasing signal solution is

$W^{H}S=W^H\left[\begin{array}{c}{a}_{1\_A},a_{2\_A}\\ a_{1\_B,}{a}_{2\_B}\end{array}\right]\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]=\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]---\left(25\right).$

Equation (25) illustrates that we can be by solving respectively P in the impulse response peak value of wave beam formation by aliasing ₁And P ₂Two symmetric points target reflection coefficient σ separately ₁With σ ₂In like manner, each pixel in the aliased image is being carried out after wave beam forms, we will obtain respectively the image of Y-axis both sides.

Emulation experiment

The below verifies that with Computer Simulation the high-speed aircraft of carrying carries double antenna Forward-looking SAR ambiguity solution algorithm complexity.Simulation parameter is as shown in table 1.

Table 1 simulation parameter

Radar is operated in X-band, wavelength 0.03m.Antenna A, the spacing d of B are half of wavelength X.Body flies before Y-axis with the speed of 500m/s at the height of 1km.The scene center on Y-axis right side is positioned at (176,2500,0), and the scene center in Y-axis left side is positioned at (176,2500,0), and left and right sides scene center oblique distance is 2698.33m.

The used point target set-up mode of emulation as shown in Figure 4.Take Y-axis as axis of symmetry, in the wave beam range of exposures, place 10 symmetrical in twos point targets.Wherein, be placed to five cross point targets on the Y-axis right side, five point targets with their symmetries are placed in the Y-axis left side.Ten point targets are labeled as respectively P ₁, P ₂, P ₃... P ₁₀, their reflection coefficient is made as respectively σ _n=n, n=1,2,3 ... 10.

It should be noted that, for the ease of being responded, observes the point target Two-dimensional Pulsed after the imaging, we with the beam angle of antenna directions X arrange larger, it is larger that distance between the corresponding point target also arranges, to keep certain distance between each Two-dimensional Pulsed response after the assurance imaging, do not affect us to the analysis of experimental result.

The below utilizes institute's algorithm of carrying that point target is carried out imaging reconciliation Fuzzy Processing.

At first with carrying algorithm the echoed signal that A, B two antennas receive is carried out respectively imaging, result such as Fig. 5 and shown in Figure 6.Because in the signal that antenna A, B receive, about the mutual aliasing of signal of the point target of Y-axis symmetry, so the impulse response that in Fig. 5 and Fig. 6, only produced five aliasings of our ten point targets arranging.The P of the other mark of impulse response among Fig. 5 and Fig. 6 _n+ P _m(n=1,2,, 5, m=6,7,, 10) and represent that this impulse response is that impulse response aliasing by which two point target forms.

Here it should be noted that the imaging results of gained is the oblique distance plane picture after the SAR imaging, so the middle point target of the position relationship between the impulse response and Fig. 4 is different at the position relationship of XOY plane among Fig. 5 and Fig. 6.

For the peak value of ranging pulse response, we take out the impulse response of scene center point and carry out careful observation from Fig. 5 and Fig. 6, such as Fig. 7 and shown in Figure 8.Can find out from Fig. 7 and Fig. 8, the envelope shape of the impulse response behind the aliasing still is two-dimentional sinc function, and its peak value is respectively 373.12 and 358.89.Consider that the SAR imaging process carried out linear transformation to signal amplitude, the concrete numerical value of these two peak values is the relation between the reflecting point target reflection factor directly, so we get its ratio to two peak amplitudes, gets 373.12:358.89=1.04:1.By before point target setting as can be known because two scene center point target aliasings, so their reflection coefficient mutually the ratio after the stack should be 7:7=1:1.Obviously, the ratio between the impulse response peak amplitude that obtains of emulation experiment is consistent with the ratio between default reflection coefficient.

Then, to the carrying out peak value measurement and obtain ratio between them of each impulse response among Fig. 5 and Fig. 6, its result is as shown in table 2.Ratio between ratio and default point target reflection coefficient is compared, can find out that the aliasing of impulse response meets theoretic analysis.

The impulse response measurement result of table 2 aliasing

Then, use the ambiguity solution method of carrying that Fig. 5 and Fig. 6 are carried out ambiguity solution processing, the image of gained such as Fig. 9 and shown in Figure 10 behind the ambiguity solution.Wherein, Fig. 9 is the Y-axis left-side images, and Figure 10 is the Y-axis image right.The P of the other mark of impulse response among the figure _n(n=1,2,10) represent this impulse response is by which point target to be produced.

Similarly, from Fig. 9 and Figure 10, take out the impulse response of scene center point and carry out careful observation, obtain Figure 11 and Figure 12.

Can be found out by Figure 11 and Figure 12, the envelope shape of the scene center point behind the ambiguity solution still remains two-dimentional sinc function, but its peak change is 320.16 and 53.56, and its ratio 320.16:53.56=5.96:1 compares with the reflectivity ratio 6:1 that presets, and its ratio is almost completely consistent.

The peak value of each impulse response among survey sheet 9 and Figure 10 again, its result is as shown in table 3.Can find out after measured value asked its ratio, its ratio is almost completely consistent with preset value after rounding up.As seen, process through ambiguity solution, the point target of Y-axis left and right sides aliasing has successfully been untied.

Impulse response peak value measurement result behind table 3 ambiguity solution

It should be noted that above imaging results is the oblique distance face image with certain geometric deformation, the coordinate of each point and relative position and each point are different at relative position and the coordinate on ground.Can carry out geometric distortion correction if need to obtain in the imaging results on distance plane.

We can find out by above simulation result, by the ambiguity solution method based on wave beam formation of this paper invention, can blur about untiing easily under the double antenna forward-looking mode.

Claims (7)

1. a high-speed aircraft carries double antenna Forward-looking SAR ambiguity solution algorithm, it is characterized in that: it comprises,

Step 1): set up double antenna Forward-looking SAR imaging model;

Step 2): echo is carried out processing apart from FFT, proofread and correct then that linearity is walked about and apart from pulse pressure;

Step 3): the orientation is processed to FFT, then carries out secondary range compression and range curvature and proofreaies and correct;

Step 4): process and carry out Azimuth Compression and process apart from IFFT;

Step 5): phase compensation;

Step 6): the phase differential that utilizes the wave path-difference between double antenna to cause is set up steering vector, by fuzzy about the mode solution of wave beam formation, obtains respectively the image of the left and right sides.

2. a kind of high-speed aircraft as claimed in claim 1 carries double antenna Forward-looking SAR ambiguity solution algorithm, it is characterized in that: described step 1 comprises,

If described double antenna is antenna A and antenna B, establishing in the wave beam irradiation area in aircraft dead ahead has P ₁And P ₂Two point targets, P ₁And P ₂With the Y-axis left-right symmetric, coordinate is respectively (x0, y0,0) and (x0, y0,0);

Target P ₁, P ₂And the round trip oblique distance equation between the antenna A is

$\left.\begin{array}{c}{R}_{1\_A}\left(t_m\right)=2\sqrt{x_0^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\\ R_{2\_A}\left(t_m\right)=2\sqrt{{(-x_0)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\end{array}\right\}---\left(1\right),$

P ₁, P ₂And the round trip oblique distance equation between the antenna B is

$\left.\begin{array}{c}{R}_{1\_B}\left(t_m\right)=\sqrt{x_0^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}+\sqrt{{(x_0-d)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\\ R_{2\_B}\left(t_m\right)=\sqrt{x_0^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}+\sqrt{{(x_0+d)}^2+{({\mathrm{vt}}_m-y_0)}^2+H^2}\end{array}\right\}---\left(2\right)$

Wherein, H is the carrier aircraft height, and v is carrier aircraft speed, t _mBe the slow time.

3. a kind of high-speed aircraft as claimed in claim 1 carries double antenna Forward-looking SAR ambiguity solution algorithm, it is characterized in that: described step 2 comprises,

The P (x, y, 0) that sets up an office is point target in the scene, through t _mAfter time, aircraft moves to the A' point, and its Y-axis coordinate is vt _m, the distance between the target P of setting up an office and the C point is y _n, Ze Cong ⊿ A'PC' can get the round trip oblique distance RP_A (t between P and antenna A _m) be

$R_{P\_A}\left(t_m\right)=2\sqrt{x^2+{({\mathrm{vt}}_m-y)}^2+H^2}$

$=2\sqrt{{({\mathrm{vt}}_m-y_n-y_c)}^2+{R_B}^2}---\left(3\right)$

$=2\sqrt{{({\mathrm{vt}}_m-y_n)}^2+R_0^2-{2R}_0({\mathrm{vt}}_m-y_n)\sin \mathrm{\&theta;}}$

Wherein $R_B=\sqrt{x^2+H^2},$ $R_0=\sqrt{x^2+y_c^2+H^2},$ θ is the angle of squint.

With formula (3) at y _nThe place makes Taylor series expansion, keeps (vt _m-y _n) following of secondary,

$R_{P\_A}\left(t_m\right)\&ap;2[R_0-({\mathrm{vt}}_m-y_n)\sin \mathrm{\&theta;}+\frac{v^2{\cos }^2\mathrm{\&theta;}}{{2R}_0}{({\mathrm{vt}}_m-y_n)}^2]---\left(4\right)$

Formula (4) is rewritten as another identical form of precision

$R_{P\_A}\left(t_m\right)=2\sqrt{R_0^2+{\cos }^2\mathrm{\&theta;}{({\mathrm{vt}}_m-y_n)}^2}-2({\mathrm{vt}}_m-y_n)\sin \mathrm{\&theta;}---\left(5\right)$

Then the echoed signal of the target P that receives of antenna A is written as

$S_A(\hat{t},t_m)={\mathrm{\&sigma;}}_P{w}_r(\hat{t}-\frac{R_{P\_A}\left(t_m\right)}{c})w_a(t_m-\frac{y_n}{v})\exp \left[\mathrm{j\&pi;\&gamma;}{(\hat{t}-\frac{R_{P\_A}\left(t_m\right)}{c})}^2\right]\exp [-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}{R}_{P\_A}\left(t_m\right)]$

(6)

Wherein,

Be the fast time, λ is signal wavelength, and c is the light velocity, and γ is signal frequency modulation rate, w _r(.) and w _a(.) is respectively the distance and bearing window function, σ _PBe the reflection coefficient of target P,

The echo fundamental frequency signal of formula (6) is transformed to apart from frequency domain-orientation time domain from two-dimensional time-domain,

$S_A(f_r,t_m)={\mathrm{\&sigma;}}_P{w}_r\left(f_r\right)w_a(t_m-\frac{y_n}{v})\exp \left[\mathrm{j\&pi;}\frac{f_r^2}{\mathrm{\&gamma;}}\right]\exp [-j\frac{2\mathrm{\&pi;}}{c}{R}_{P\_A}\left(t_m\right)(f_r+f_c)---\left(7\right)$

F wherein _cBe carrier frequency, f _rBe frequency of distance, according to formula (5), formula (7) multiply by formula (8) phase term data are carried out unified linearity walk the normal moveout correction row distance of going forward side by side and process to pulse pressure:

$H_{\mathrm{LRMC}}(f_r,t_m)=\exp [-j2\mathrm{\&pi;}\frac{v\sin {\mathrm{\&theta;t}}_m}{c}(f_r+f_c)]\exp \left[\mathrm{j\&pi;}\frac{f_r^2}{\mathrm{\&gamma;}}\right].---\left(8\right)$

4. a kind of high-speed aircraft as claimed in claim 3 carries double antenna Forward-looking SAR ambiguity solution algorithm, it is characterized in that: described step 3 comprises,

Transform to two-dimensional frequency with Range Walk Correction with apart from the signal after pulse pressure, and use following phase term to carry out range migration correction and secondary range pulse pressure

$H_{\mathrm{rg}}(f_r,f_a)=\exp [-\mathrm{j\&pi;}\frac{{2\mathrm{\&lambda;R}}_s{\left(\frac{{\mathrm{\&lambda;f}}_a}{2v\cos \mathrm{\&theta;}}\right)}^2}{c^2{\left(\sqrt{1-{\left(\frac{{\mathrm{\&lambda;f}}_a}{2v\cos \mathrm{\&theta;}}\right)}^2}\right)}^3}{f_1}^2]\exp \left[j\frac{{2\mathrm{\&pi;R}}_s}{c}{\left(\frac{{\mathrm{\&lambda;f}}_a}{2v\cos \mathrm{\&theta;}}\right)}^2{f}_r\right]---\left(9\right)$

Signal is carried out distance to inverse Fourier transform, and the signal expression that obtains the distance-Doppler territory is as follows

$S_A(\hat{t},f_a)={\mathrm{\&sigma;}}_P\sin \mathrm{cp}\left[{\mathrm{\&Delta;f}}_r(\hat{t}-\frac{2(R_0+y_n\sin \mathrm{\&theta;})}{c})\right]a_a\left(\frac{R_0\mathrm{\&lambda;}{f}_a}{{2v}^2\sqrt{1-{(f_a/f_{\mathrm{aM}})}^2}}\right)$ $\&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}}{v\cos \mathrm{\&theta;}}{R}_0\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]\exp [-j2{\mathrm{\&pi;f}}_a\frac{y_n}{v}]$

(10)

Wherein, Δ f _rBe transmitted signal bandwidth, f _aBe the orientation frequency, R _sBe the ray oblique distance at scene center line place, finish echoed signal distance to processing.

5. a kind of high-speed aircraft as claimed in claim 4 carries double antenna Forward-looking SAR ambiguity solution algorithm, it is characterized in that: described step 4 comprises,

Formula (10) is rewritten as the form of the signal of the target P 1 that can represent simultaneously antenna A, B and receive, P2

$S_M(\hat{t},f_a)={\mathrm{\&sigma;}}_N\sin c\left[{\mathrm{\&Delta;f}}_r(\hat{t}-\frac{R_{N\_M\_0}+{2y}_n\sin \mathrm{\&theta;}}{c})\right]w_a\left(\frac{R_{N\_M\_0}{\mathrm{\&lambda;f}}_a}{{4v}^2\sqrt{1-{(f_a/f_{\mathrm{aM}})}^2}}\right)$

(11)

$.\exp [-j\frac{\mathrm{\&pi;}}{v\cos \mathrm{\&theta;}}{R}_{N\_M\_0}\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]\exp [-j{2\mathrm{\&pi;f}}_a\frac{y_n}{v}]$

N=1 wherein, 2 expression target P ₁With P ₂, and M=A, B, expression antenna A and B.R _{N_M}Be four oblique distances in formula (1) and (2), and R _{N_M_0}Represent t _m=0 o'clock oblique distance, namely

$R_{N\_M\_0}\left\{\begin{array}{c}{R}_{1\_A\_0}=2\sqrt{x_0^2+y_0^2+H^2}\\ R_{2\_A\_0}=2\sqrt{x_0^2+y_0^2+H^2}\\ R_{1\_B\_0}=\sqrt{x_0^2+y_0^2+H^2}+\sqrt{{(x_0-d)}^2+y_0^2+H^2}\\ R_{2\_B\_0}=\sqrt{x_0^2+y_0^2+H^2}+\sqrt{{(x_0+d)}^2+y_0^2+H^2}\end{array}\right.---\left(12\right)$

First phase term in the formula (11) is the orientation modulation item, and it is expanded into f _aTaylor series get

$\exp [-j\frac{\mathrm{\&pi;}}{V\cos \mathrm{\&theta;}}{R}_{N\_M\_0}\sqrt{f_{\mathrm{aM}}^2-f_a^2}]\&ap;\exp [-j\frac{{2\mathrm{\&pi;R}}_{N\_M\_0}}{\mathrm{\&lambda;}}(1-\frac{1}{2}\frac{f_a^2}{f_{\mathrm{aM}}^2}-\frac{1}{8}\frac{f_a^4}{f_{\mathrm{aM}}^4})]---\left(13\right)$

Construct an orientation matched filter that does not have constant term

$H_{\mathrm{az}}(\hat{t},f_a)=\exp \left[j\frac{{2\mathrm{\&pi;R}}_{N\_M\_0}}{\mathrm{\&lambda;}}(-\frac{1}{2}\frac{f_a^2}{f_{\mathrm{aM}}^2}-\frac{1}{8}\frac{f_a^4}{f_{\mathrm{aM}}^4})\right]---\left(14\right)$

The focusedimage that obtains after the orientation matched filtering is

$S_M(\hat{t},t_m)={\mathrm{\&sigma;}}_N\sin c\left[{\mathrm{\&Delta;f}}_r(\hat{t}-\frac{R_{N\_M\_0}+{2y}_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[{\mathrm{\&Delta;f}}_a(t_m-\frac{y_n}{v})\right]$

(15)

$\exp [-j\frac{{2\mathrm{\&pi;R}}_{N\_M\_0}}{\mathrm{\&lambda;}}]\exp [-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}]$

Δ f wherein _aFor the orientation to doppler bandwidth.

6. a kind of high-speed aircraft as claimed in claim 5 carries double antenna Forward-looking SAR ambiguity solution algorithm, it is characterized in that: described step 5 comprises,

Four oblique distance formula have one common in the formula (12)

Include a common phase in four focused image of formula (15) expression With formula (16) phase term disappear common phase and excess phase $\exp (-j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;})$

$H_{\mathrm{phase}}=\exp \left[j\frac{2\mathrm{\&pi;}\sqrt{x_0^2+y_0^2+H^2}}{\mathrm{\&lambda;}}\right]\exp \left(j\frac{4\mathrm{\&pi;}}{\mathrm{\&lambda;}}{y}_n\sin \mathrm{\&theta;}\right)---\left(16\right)$

Obtain signal

$S_M(\hat{t},t_m)={\mathrm{\&sigma;}}_N\sin c\left[{\mathrm{\&Delta;f}}_r(\hat{t}-\frac{R_{N\_M\_0}+{2y}_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[{\mathrm{\&Delta;f}}_a(t_m-\frac{Y_n}{V})\right]$

。(17)

$\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_M\_0}+\sqrt{x_0^2+y_0^2+H^2})}{\mathrm{\&lambda;}}]$

7. a kind of high-speed aircraft as claimed in claim 6 carries double antenna Forward-looking SAR ambiguity solution algorithm, it is characterized in that: described step 6 comprises,

Write out respectively the expression formula of aliasing signal in two antennas by (17)

$\left\{\begin{array}{c}{S}_A(\hat{t},t_m)=\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\sin c\left[{\mathrm{\&Delta;f}}_r(\hat{t}-\frac{R_{N\_A\_0}+{2y}_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[{\mathrm{\&Delta;f}}_a(t_m-\frac{y_n}{V})\right]\\ \&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_A\_0}+\sqrt{x_0^2+y_0^2+H^2})}{\mathrm{\&lambda;}}]\\ S_B(\hat{t},t_m)=\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\sin c\left[{\mathrm{\&Delta;f}}_r(\hat{t}-\frac{R_{N\_B\_0}+{2y}_n\sin \mathrm{\&theta;}}{c})\right]\sin c\left[{\mathrm{\&Delta;f}}_a(t_m-\frac{y_n}{V})\right]\\ \&CenterDot;\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_B\_0}+\sqrt{x_0^2+y_0^2{H}^2})}{\mathrm{\&lambda;}}\end{array}\right.---\left(18\right)$

Take out the peak value Peak of impulse response in (18) _AAnd Peak _B, because the peak value of sinc function is 1, the observation signal vector S that can obtain the line arrays reception is

$S=[\left.\begin{array}{c}{\mathrm{Peak}}_A\\ {\mathrm{Peak}}_B\end{array}\right]$

$=\left[\begin{array}{c}\underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\exp [-j\frac{2\mathrm{\&pi;}(R_{N\_A\_0}+\sqrt{x_0^2+y_0^2+H^2})}{\mathrm{\&lambda;}}]\\ \underset{N}{\overset{N=\mathrm{1,2}}{\mathrm{\&Sigma;}}}{\mathrm{\&sigma;}}_N\exp [-j\frac{2\mathrm{\&pi;}\left(R_{N\_B\_0}\sqrt{x_0^2+y_0^2+H^2}\right)}{\mathrm{\&lambda;}}]\end{array}\right]$

$=\left[\begin{array}{c}{{\mathrm{\&sigma;}}_{1}e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\sqrt{x_0^2+y_0^2+H^2}}+{{\mathrm{\&sigma;}}_{2}e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\sqrt{x_0^2+y_0^2+H^2}}\\ {{\mathrm{\&sigma;}}_{1}e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\left(\sqrt{{(x_0-d)}^2+y_0^2+H^2}\right)}+{{\mathrm{\&sigma;}}_{2}e}^{-j\frac{2\mathrm{\&pi;}}{\mathrm{\&lambda;}}\left(\sqrt{{(x_0+d)}^2+y_0^2+H^2}\right)}\end{array}\right]$ (19)

$=\left[\begin{array}{c}{\mathrm{\&sigma;}}_1{a}_{1\_A}+{\mathrm{\&sigma;}}_2{a}_{2\_A}\\ {\mathrm{\&sigma;}}_1{a}_{1\_B}+{\mathrm{\&sigma;}}_2{a}_{2-B}\end{array}\right]$

$=\left[\begin{array}{c}{a}_{1\_A},a_{2-A}\\ a_{1\_B},a_{2-B}\end{array}\right]\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]$

[a wherein _{1_A}, a _{2_A}] ^T[a _{1_B}, a _{2_B}] ^TBe respectively P ₁With P ₂Steering vector,

The method that forms by wave beam suppresses respectively from Y-axis one side signal, and obtains the signal of Y-axis opposite side, if make array beams point to P ₁Weighing vector be W ₁, point to P ₂Weighing vector be W ₂, then weighting matrix is W=[W ₁, W ₂],

Find the solution the expression formula of W, from formula (19), obtain system of equations

$\left\{\begin{array}{c}{\mathrm{Peak}}_A={\mathrm{\&sigma;}}_1{a}_{1\_A}+{\mathrm{\&sigma;}}_2{a}_{2\_A}\\ {\mathrm{Peak}}_B={\mathrm{\&sigma;}}_1{a}_{1\_B}+{\mathrm{\&sigma;}}_2{a}_{2\_B}\end{array}\right.---\left(20\right)$

Solving equations (20) obtains

$\left\{\begin{array}{c}{\mathrm{\&sigma;}}_1=\frac{1}{b}[{\mathrm{Peak}}_A{a}_{2\_B}-{\mathrm{Peak}}_B{a}_{2\_A}]\\ {\mathrm{\&sigma;}}_2=\frac{1}{b}[-{\mathrm{Peak}}_A{a}_{1\_B}+{\mathrm{Peak}}_B{a}_{1\_A}]\end{array}\right.---\left(21\right)$

B=a wherein _{1_A}a _{2_B}-a _{2_A}a _{1_B},

Write out W by (21) ₁For

$W_1=\frac{1}{b}\left[\begin{array}{c}{a}_{2\_B}\\ -a_{2\_A}\end{array}\right]---\left(22\right)$

And W ₂For

$W_2=\frac{1}{b}\left[\begin{array}{c}-a_{1\_B}\\ a_{1\_A}\end{array}\right]---\left(23\right)$

Then weighting matrix W is

$W=[W_1,W_2]=\frac{1}{b}\left[\begin{array}{c}{a}_{2\_B},-a_{1\_B}\\ -a_{2\_A,}{a}_{1\_A}\end{array}\right]---\left(24\right)$

To process fuzzy about the aliasing signal solution be

$W^{H}S=W^H\left[\begin{array}{c}{a}_{1\_A},a_{2\_A}\\ a_{1\_B},a_{2\_B}\end{array}\right]\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]=\left[\begin{array}{c}{\mathrm{\&sigma;}}_1\\ {\mathrm{\&sigma;}}_2\end{array}\right]---\left(25\right).$

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