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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.$ $\widetilde E ({k_x}, {k_y}) = 4\dfrac{{\sin (2{k_x}\lambda /2)}}{{{k_x}}} \cdot \dfrac{{\sin (2{k_y}\lambda /2)}}{{{k_y}}}$ DownLoad:
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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}} $.DownLoad:
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