基于Relax算法的星載高分寬幅成像方法研究
doi: 10.11999/JEIT180596
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南京航空航天大學(xué)電子信息工程學(xué)院 ??南京 ??210000
Research on Spaceborne High Resolution Wide Swath Imaging Method Based on Relax Algorithm
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College of Electronic Information Engineering, Nanjing University of Aeronautics and Astronautics, Nanjing 210000, China
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摘要:
現(xiàn)代星載合成孔徑雷達(dá)(SAR)系統(tǒng)要求同時(shí)具備高分辨率和寬測(cè)繪帶的能力,而傳統(tǒng)單通道星載SAR系統(tǒng)在分辨率和測(cè)繪帶兩個(gè)重要指標(biāo)之間存在固有矛盾,因此方位向多通道的方法被提出并用于解決上述問(wèn)題。該文在分析方位向多通道回波模型的基礎(chǔ)上,結(jié)合Relax算法的特點(diǎn),提出了一種基于Relax算法的星載SAR高分寬幅(HRWS)成像方法,并給出了新方法的詳細(xì)迭代流程。通過(guò)點(diǎn)目標(biāo)回波仿真,并與傳統(tǒng)的方位向多通道HRWS重建方法進(jìn)行對(duì)比,驗(yàn)證了新方法的可靠性和有效性。
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關(guān)鍵詞:
- 星載合成孔徑雷達(dá) /
- 高分寬幅 /
- Relax算法
Abstract:The modern spaceborne SAR system requires both high resolution and wide swath, and the conventional single channel spaceborne SAR system has a contradiction between the two important indexes, the azimuth multichannel method is proposed and used to solve the above problem. Based on the analysis of the azimuth multichannel echo model and the characteristics of the Relax algorithm, a spaceborne SAR High Resolution Wide Swath (HRWS) imaging method is proposed, and the iterative process of the new method is described in detail. By the simulation of point target echo, and comparing with the traditional azimuth multichannel HRWS reconstruction methods, the reliability and effectiveness of the proposed method are verified.
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表 1 公式中用到的符號(hào)說(shuō)明
符號(hào) 說(shuō)明 $M$ 通道個(gè)數(shù) $r$ 距離變量,$r = {\rm c}t/2$ c 光速 $t$ 距離向時(shí)間變量 ${f_{\rm d}}$ 多普勒頻率 $\vartheta $ 目標(biāo)參數(shù)向量,$\vartheta = {[{T_0}, {R_0}, {\sigma ^2}, \varTheta ]^{\rm{T}}}$ ${T_0}$ 最小斜距對(duì)應(yīng)的時(shí)間 ${R_0}$ 最小斜距$\min \{ R(T\,)\} = R({T_0}) = {R_0}$ ${\sigma ^2}$ 目標(biāo)功率 $\varTheta $ 目標(biāo)相位 ${f\!_{p}}$ 各通道的脈沖重復(fù)頻率 $\varOmega $ 2次相位項(xiàng)的變化率 $R(T\,)$ 雷達(dá)與目標(biāo)之間的距離 $T\,$ 慢時(shí)間變量 ${{u}}[{f_{\rm d}}, \vartheta ]$ 天線(xiàn)與目標(biāo)間的瞄準(zhǔn)線(xiàn)向量 ${A_m}$ 第$m$個(gè)通道的雙向天線(xiàn)方向圖 ${d_m}$ 第$m$個(gè)通道的等效相位中心 下載: 導(dǎo)出CSV
表 2 基于Relax算法的模糊抑制
For $m = 1: \rm Na$ For $n = 1:{\rm Nr}$
%遍歷所有距離-多普勒單元(其中,$\rm Na$表示多普勒單元的個(gè)數(shù),
$\rm Nr$表示距離單元的個(gè)數(shù))
${Z_{{\rm{rec_{-}1}}}}(r, {f_{\rm d}} + p{f\!_{p}}) = \frac{{{{a}}_p^{\rm{H}}\left( {{f_{\rm d}}} \right){{Z}}\left( {r, {f_{\rm d}}} \right)}}{M}$ %初始化操作
%第1次迭代
$\begin{align}{{{Z}}_{p_{-}1}}\left( {r, {f_{\rm d}}} \right) =& {{Z}}\left( {r, {f_{\rm d}}} \right) \\&- \sum\limits_{i = - \left( {M - 1} \right)/2, i \ne p}^{\left( {M - 1} \right)/2} {{{{a}}_i}\left( {{f_{\rm d}}} \right)} \cdot {Z_{{\rm{rec_{-}1}}}}\left( {r, {f_{\rm d}} + p{f\!_{p}}} \right)\end{align}$
${\hat Z_{{\rm{rec_{-}1}}}}\left( {r, {f_{\rm d}} + p{f\!_{p}}} \right) = \frac{{{{a}}_p^{\rm{H}}\left( {{f_{\rm d}}} \right){{{Z}}_{p_{-}1}}\left( {r, {f_{\rm d}}} \right)}}{M}$
如果未滿(mǎn)足收斂條件或者未達(dá)到迭代次數(shù)
%第$k$次迭代($k \ge 2$)
$\begin{align}{{{Z}}_{p_{-}k}}\left( {r, {f_{\rm d}}} \right) =& {{Z}}\left( {r, {f_{\rm d}}} \right) \\& - \sum\limits_{i = - \left( {M - 1} \right)/2, i \ne p}^{\left( {M - 1} \right)/2} {{{{a}}_i}\left( {{f_{\rm d}}} \right)} \cdot {\hat Z_{{{{\rm rec}_{-}k - 1}}}}\left( {r, {f_{\rm d}} + p{f\!_{p}}} \right)\end{align}$
${\hat Z_{{{{\rm rec}_{-}k}}}}\left( {r, {f_{\rm d}} + p{f\!_{p}}} \right) = \frac{{{{a}}_p^{\rm{H}}\left( {{f_{\rm d}}} \right){{{Z}}_{p_{-}k}}\left( {r, {f_{\rm d}}} \right)}}{M}$
End
End
End下載: 導(dǎo)出CSV
表 3 不同重建方法的性能對(duì)比
算法類(lèi)別 PSLR(dB) 方位向分辨率(m) SNR(dB) SANR(dB) 矩陣求逆法 13.99 1.66 39.46 27.08 最大信號(hào)法 13.99 1.66 40.33 12.97 Relax法 13.82 1.66 50.56 21.22 下載: 導(dǎo)出CSV
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