3D Parameters Estimation of Helicopter with Constant Speed Circular Motion Based on Single Hydrophone
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摘要: 针对空中匀速圆弧运动目标激发的水下声信号,该文采用单水听器解决该动目标3维运动参数的估计问题。首先以直升机离散线谱为声源特征,在空气-水介质中建立声源线谱特征在匀速圆弧运动下3维多普勒传播模型。然后根据多普勒频移曲线、声源运动模型以及声线传播几何关系,选取3个时间观测点计算目标多普勒频移,推导了单水听器估计空中匀速圆弧运动声源的3维参数估计算法。最后,通过仿真单水听器所接收的水声信号,验证了该算法估计匀速圆弧运动声源飞行参数的有效性和精度。Abstract: The Three-Dimensional(3D) parameter estimation algorithm of the helicopter with constant speed circular flight from the underwater acoustic data with single hydrophone is proposed. Firstly, the helicopter line spectrum is used as the characteristic of the source, and its 3D Doppler propagation model in two-layer air-water medium is established. Then, the parameters estimation for helicopter in 3D space is derived according to the Doppler frequency shift curve, the sound source motion model and the geometric model of sound propagation. Finally, the effectiveness and accuracy of the algorithm for Doppler signal with the alpha stable noise on single hydrophone is verified.
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表 1 参数估计结果
序号 $t$(s) ${f_{\text{d}}}$(Hz) ${f_0}$(Hz) $v$(m/s) $h$(m) $r$(m) 1 $\left. {\begin{array}{*{20}{ll} } { {t_A} = 1.5} \\ { {t_M} = 2} \\ { {t_C} = 2.5} \end{array} } \right\}$ $\left. {\begin{array}{*{20}{lll} } {f\left( { {t_A} } \right) = 67.12} \\ {f\left( { {t_M} } \right) = 67.9} \\ {f\left( { {t_C} } \right) = 68.66} \end{array} } \right\}$ 68.7 119.8 145.6 56 2 $\left. {\begin{array}{*{20}{ll} } { {t_A} = 2.7} \\ { {t_M} = 3.2} \\ { {t_C} = 3.7} \end{array} } \right\}$ $ \left. {\begin{array}{*{20}{llll}} {f\left( {{t_A}} \right) = 68.9} \\ {f\left( {{t_M}} \right) = 69.25} \\ {f\left( {{t_C}} \right) = 69.21} \end{array}} \right\} $ 68.5 120.1 146.8 61 3 $\left. {\begin{array}{*{20}{ll} } { {t_A} = 4} \\ { {t_M} = 4.2} \\ { {t_C} = 4.4} \end{array} } \right\}$ $ \left. {\begin{array}{*{20}{llll}} {f\left( {{t_A}} \right) = 68.99} \\ {f\left( {{t_M}} \right) = 69.79} \\ {f\left( {{t_C}} \right) = 69.56} \end{array}} \right\} $ 67.5 120.8 143.8 58 4 $\left. {\begin{array}{*{20}{ll} } { {t_A} = 3.2} \\ { {t_M} = 3.4} \\ { {t_C} = 3.6} \end{array} } \right\}$ $ \left. {\begin{array}{*{20}{llll}} {f\left( {{t_A}} \right) = 68.25} \\ {f\left( {{t_M}} \right) = 69.3} \\ {f\left( {{t_C}} \right) = 69.24} \end{array}} \right\} $ 68.4 125.2 143.1 34 5 $\left. {\begin{array}{*{20}{lll} } { {t_A} = 1.5} \\ { {t_M} = 2} \\ { {t_C} = 2.5} \end{array} } \right\}$ $ \left. {\begin{array}{*{20}{llll}} {f\left( {{t_A}} \right) = 65.58} \\ {f\left( {{t_M}} \right) = 65.66} \\ {f\left( {{t_C}} \right) = 66.24} \end{array}} \right\} $ 68.2 120.6 48.6 58 -
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