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现有均匀场往往基于阵列天线的特殊排布, 通过平顶波束赋形在角远场区域或者通过点聚焦在近场区域生成的, 生成的均匀场直接受制于阵列排布形态且无法灵活调控. 提出了一种基于角谱域和改进时间反演方法相结合的均匀场生成方法, 该方法不受阵列排布的限制, 能够以同一阵列排布形态, 在包括近场区域在内的任意位置, 生成指定大小、形状以及偏转角度的多种均匀场. 首先理论解析了本方法不受阵列排布限制的原因; 其次数值验证了固定阵列排布形态灵活生成多种均匀场的能力; 最后引入时间反演方法, 并做出反演信号幅度倒数加权的改进, 解决了上述均匀场在生成过程中由幅度衰减和相位延迟带来均匀场平坦度恶化等问题. 研究结果表明, 合成场质量与其对应角谱域包络的主瓣和副瓣信息有关, 且生成任意均匀场必须包含至少1/2的角谱域主瓣信息和1/2的副瓣信息. 本方法能够灵活调控一维和二维均匀场的位置、大小、形状以及偏转角度, 为灵活生成均匀场提供了一条新思路.Existing uniform fields are usually based on the special arrangement of the array antenna. The uniform fields generated by flat-top beam shaping in angular far-field area or by point focusing in near-field area are directly subject to the array configuration and cannot be flexibly controlled. This paper presents a method of generating uniform field based on the combination of angular spectral domain and improved time reversal technique. This method is not limited by the array arrangement. It can generate a uniform field of specified size, shape and deflection angle in the same array arrangement at any position, including the near-field region. In this work, the reason why this method is not limited by array arrangement is explained theoretically. Secondly, the ability of the fixed array configuration to generate multiple uniform fields is validated numerically. Finally, the time-reversal technique of reversal signal amplitude reciprocal weighting is introduced. The problem of deterioration of uniform field flatness, caused by amplitude decay and phase delay during the generation of uniform field, is solved through this technology. The results show that the quality of the synthesized field is related to the main lobe and sidelobe information of its corresponding angular spectrum domain envelope, and the generated any uniform field must contain at least half of the angular spectrum domain main lobe information and half of the sidelobe information. This method can flexibly control the position, size, shape and deflection angle of one-dimensional and two-dimensional uniform field, which provides a new way to flexibly generating uniform fields.
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Keywords:
- angular spectrum domain/
- time reversal/
- uniform field
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目标场 空域表达式 角谱域表达式 θn 1 $ E(x, 15\lambda ) = \left\{ {\begin{aligned} &{1, \;\;|x| \leqslant 1.5\lambda } \\ &{0, \;\;{\text{others}}}\end{aligned}} \right. $ $\widetilde{E} ({k_x}) = 2\dfrac{{\sin (3{k_x}\lambda /2)}}{{{k_x}}}$ ${\theta _n} = {\rm{arccos}}\left( {\dfrac{{(n - 16)\lambda /2}}{{\sqrt {{{\left[ {(n - 16)\lambda /2} \right]}^2} + {{\left( {15\lambda } \right)}^2}} }}} \right)$ 2 $E(x, 10\lambda ) = \left\{ {\begin{aligned} &{1, \;\;{{ - 3}}\lambda < x < \lambda } \\ &{0, \;\;{\text{others}}}\end{aligned}} \right.$ $\widetilde{E} ({k_x}) = 2\dfrac{{\sin (2{k_x}\lambda /2)}}{{{k_x}}}$ $ {\theta _n} = {\rm{arccos}}\left( {\dfrac{{(n - 16)\lambda /2 + 2\lambda }}{{\sqrt {{{\left[ {(n - 16)\lambda /2 + 2\lambda } \right]}^2} + {{\left( {10\lambda } \right)}^2}} }}} \right) $ 3 $ E(x, 10\lambda ) = \left\{ {\begin{aligned} &1 , \;\;{ - 3\lambda \leqslant x \leqslant - \lambda }\\ &{1, }\;\;{\lambda \leqslant x \leqslant 3\lambda {\text{ }}} \\ &{0, }\;\;{{\text{others }}}\end{aligned}} \right. $ $\widetilde{E} ({k_x}) = 2\dfrac{{\sin (3{k_x}\lambda /2) - \sin ({k_x}\lambda /2)}}{{{k_x}}}$ ${\theta _n} = {\rm{arccos}}\left( {\dfrac{{(n - 16)\lambda /2}}{{\sqrt {{{\left[ {(n - 16)\lambda /2} \right]}^2} + {{\left( {10\lambda } \right)}^2}} }}} \right)$ 4 $ E(x, y) = \left\{ {\begin{aligned} &{1,\;\; y = x + 13\lambda , {\text{ }}} \\ & ~~~- 3 - \dfrac{{\sqrt 2 }}{2} < \dfrac{x}{\lambda} < - 3 + \dfrac{{\sqrt 2 }}{2};\\ &0, \;\;{\text{others}} \end{aligned}} \right. $ $\widetilde{E} ({k_x}) = 2\dfrac{{\sin (2{k_x}\lambda /2)}}{{{k_x}}}$ ${\theta _n} = {\rm{arccos}}\left( {\dfrac{{(n - 16)\lambda /2 + 3\lambda }}{{\sqrt {{{\left[ {(n - 16)\lambda /2 + 3\lambda } \right]}^2} + {{\left( {10\lambda } \right)}^2}} }}} \right) + \dfrac{{\text{π }}}{4}$ 5 $ E(x, y) = \left\{ {\begin{aligned}& 1, \;\; y = x + 17.5\lambda ,\\ & ~~~- 8.5\lambda < x < - 6.5\lambda; \\ &0,\;\; {\text{others}} \end{aligned}} \right. $ $\widetilde{E} ({k_x}) = 2\dfrac{{\sin (2\sqrt 2 {k_x}\lambda /2)}}{{{k_x}}}$ ${\theta _n} = {\rm{arccos}}\left( {\dfrac{{(n - 16)\lambda /2 + 7.5\lambda }}{{\sqrt {{{\left[ {(n - 16)\lambda /2 + 7.5\lambda } \right]}^2} + {{\left( {10\lambda } \right)}^2}} }}} \right) + \dfrac{{\text{π }}}{4}$ 
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目标场 空域表达式 角谱域表达式 1 $ E(x, y, 10\lambda ) = \left\{ \begin{aligned} &{1, }\;\;{\left| x \right| \leqslant \lambda , \left| y \right| \leqslant \lambda } \\ & {0, }\;\;{{\text{others }}} \end{aligned} \right. $ $\widetilde E ({k_x}, {k_y}) = 4\dfrac{{\sin (2{k_x}\lambda /2)}}{{{k_x}}} \cdot \dfrac{{\sin (2{k_y}\lambda /2)}}{{{k_y}}}$ 2 $ E(x, y, 10\lambda ) = \left\{ {\begin{aligned} &{1, }\;\;{\begin{aligned} &{2\lambda \leqslant x \leqslant 4\lambda {\text{ }}} \\ &{ - 0.5\lambda \leqslant y \leqslant 2.5\lambda } \end{aligned}} \\ & {0, }\;\;{{\text{others }}} \end{aligned}} \right. $ $\widetilde E ({k_x}, {k_y}) = 4\dfrac{{\sin (2{k_x}\lambda /2)}}{{{k_x}}} \cdot \dfrac{{\sin (3{k_y}\lambda /2)}}{{{k_y}}}$ 3 $ E(x, y, z) = \left\{ \begin{aligned} &{1, }\;\;{\begin{aligned} &{\sqrt 3 z = - x + 28\lambda } \\ &{6\lambda \leqslant x \leqslant 8\lambda } \\ & { - \lambda \leqslant y \leqslant \lambda {\text{ }}} \end{aligned}} \\ & {0, }\;\;{{\text{others }}} \end{aligned} \right.$ $\widetilde E ({k_x}, {k_y}) = 4\dfrac{{\sin (2{k_x}\lambda /2)}}{{{k_x}}} \cdot \dfrac{{\sin (2{k_y}\lambda /2)}}{{{k_y}}}$ 下载:
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目标场 θn φn 1 $ {\rm{ arcsin}}\left( {\dfrac{{\sqrt {{{[({n_x} - 11)\lambda /2]}^2} + {{[({n_y} - 11)\lambda /2]}^2}} }}{{\sqrt {{{[({n_x} - 11)\lambda /2]}^2} + {{[({n_y} - 11)\lambda /2]}^2} + {{\left( {10\lambda } \right)}^2}} }}} \right) $ ${\rm{ arctan}}\left( {\dfrac{{({n_x} - 6)\lambda /2}}{{({n_y} - 6)\lambda /2}}} \right)$ 2 $ {\rm{ arcsin}}\left( {\dfrac{{\sqrt {{{\left[ {({n_x} - 6)\lambda /2 - 3\lambda } \right]}^2} + {{\left[ {({n_y} - 6)\lambda /2 - \lambda } \right]}^2}} }}{{\sqrt {{{\left[ {({n_x} - 6)\lambda /2 - 3\lambda } \right]}^2} + {{\left[ {({n_y} - 6)\lambda /2 - \lambda } \right]}^2} + {{\left( {10\lambda } \right)}^2}} }}} \right) $ $ {\rm{ arctan}}\left( {\dfrac{{({n_x} - 11)\lambda /2 - 3\lambda }}{{({n_y} - 11)\lambda /2 - \lambda }}} \right) $ 3 ${\rm{ arcsin}}\left( {\dfrac{{\sqrt {{\text{d}}{x_n^2} + {{\left[ {({n_y} - 11)\lambda /2} \right]}^2}} }}{{\sqrt {{\text{d}}{x_n^2} + {{\left[ {({n_y} - 11)\lambda /2} \right]}^2} + {\text{d}}{z_n^2}} }}} \right) ^*$ ${\rm{ arctan}}\left( {\dfrac{{{\text{d}}{x_n}}}{{({n_y} - 11)\lambda /2}}} \right) ^*$ 注: *其中 $ {\text{d}}{x_n} = \dfrac{{{{\left[ {\dfrac{{\left( {{n_x} - 11} \right)\lambda /2}}{{\cos ({\text{π }}/6)}}} \right]}^2} + \bigg\{ {{\left( {7\sqrt 3 \lambda } \right)}^2} + {{\left[ {\left( {{n_x} - 11} \right)\lambda /2} \right]}^2} \bigg\} - \left[ {7\sqrt 3 \lambda + \left( {{n_x} - 11} \right)\lambda /2\tan ({\text{π }}/6)} \right]}}{{\dfrac{{\left( {{n_x} - 11} \right)\lambda }}{{\cos ({\text{π }}/6)}}}} $,
${\text{d}}{z_n} = \sqrt {{{\left( {7\sqrt 3 \lambda } \right)}^2} + {{\left[ {\left( {{n_x} - 11} \right)\lambda /2} \right]}^2} - {\text{d}}{x_n^2}} $.下载:
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