基于参数外推的超宽带雷达有源极化校准技术

doi:10.12305/j.issn.1001-506X.2025.08.08

摘要/Abstract

摘要：

有源极化雷达校准器（polarimetric active radar calibrator，PARC）以其高雷达散射截面以及高信杂噪比的优点，在雷达极化校准领域中受到了广泛的应用。针对PARC系统只能工作在固定频段和有限带宽内而在超宽带雷达极化校准中使用受限的问题，提出一种基于参数外推的超宽带雷达有源极化校准技术。利用谱估计方法将校准器对应频段的极化校准参数外推至其他频段，然后对相应频段进行处理，从而拓宽有源极化校准器的工作频带范围。利用工作频率为8~12 GHz的PARC对4~16 GHz全极化雷达进行校准，原始校准参数和外推校准参数在各自频段均取得良好的极化校准效果，证明了所提方法的有效性和工程实用价值。

关键词: 极化校准, 超宽带雷达, 有源极化校准器, 参数外推

Abstract:

Polarimetric active radar calibrator (PARC) are widely used in the field of radar polarization calibration due to its high radar cross section and high signal-to-noise ratio. A parameter extrapolation based active polarization calibration technique for ultrawide-band radar is proposed to address the problem that PARC system can only operate in fixed frequency bands and limited bandwidth, which limits their use in polarization calibration. Using spectral estimation methods to extrapolate the polarization calibration parameters of the calibrator corresponding frequency band to other frequency bands, and then processing the corresponding frequency bands to broaden the operating frequency range of the active polarization calibrator. The PARC with a working frequency of 8~12 GHz is used to calibrate the 4~16 GHz fully polarized radar. The original calibration parameters and extrapolated calibration parameters achieved good polarization calibration results in their respective frequency bands, proving the effectiveness and engineering practical value of the proposed method.

Key words: polarimetric calibration, ultrawide-band radar, polarimetric active radar calibrator (PARC), parameter extrapolation

中图分类号:

TN957

杨宽, 许小剑. 基于参数外推的超宽带雷达有源极化校准技术[J]. 系统工程与电子技术, 2025, 47(8): 2498-2510.

Kuan YANG, Xiaojian XU. Active polarimetric calibration technique based on parameter extrapolation for ultrawide-band radar[J]. Systems Engineering and Electronics, 2025, 47(8): 2498-2510.

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序号	发射天线极化	接收天线极化	PSM
1	h	h	${{\boldsymbol{S}}_1} = \sqrt {\dfrac{\sigma }{{4{\text{π }}}}} \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}} \right]$
2	h	v	${{\boldsymbol{S}}_2} = \sqrt {\dfrac{\sigma }{{4{\text{π }}}}} \left[ {\begin{array}{*{20}{c}} 0&{ - 1} \\ 0&0 \end{array}} \right]$
3	v	h	$ {{\boldsymbol{S}}_3} = \sqrt {\dfrac{\sigma }{{4{\text{π }}}}} \left[ {\begin{array}{*{20}{c}} 0&0 \\ { - 1}&0 \end{array}} \right] $
4	v	v	${{\boldsymbol{S}}_4} = \sqrt {\dfrac{\sigma }{{4{\text{π }}}}} \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&1 \end{array}} \right]$

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