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摘要: 射频掩护是最早的雷达主动抗干扰措施之一,其通过在雷达脉冲信号之前发射不同频率的掩护脉冲来诱导敌方干扰机,实现抗干扰。近年来,随着抗干扰需求更加迫切,射频掩护技术进一步发展,最具代表性的是采用非连续谱信号作为掩护信号,但掩护信号的能量利用率仍存在提升空间。针对此问题,该文在非连续谱掩护信号基础上提出了一种离散谱掩护信号,建立了恒模和频谱幅度联合约束下的波形设计优化问题,通过交替向量乘子法以及频谱塑形算法求解,生成频谱离散、能量聚集的掩护信号。仿真结果表明,在能量和带宽相同的情况下,离散谱掩护信号相比于非连续谱掩护信号具有更高的频谱幅度,提升5~12 dB;在能量相同,频谱幅度接近的情况下,离散谱掩护信号能覆盖更大的频谱范围,实现了更好的抗干扰掩护效果。
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关键词:
- 射频掩护 /
- 非连续谱信号 /
- 抗干扰 /
- 波形设计 /
- 交替向量乘子法(ADMM)
Abstract: Radio Frequency-screen is one of the earliest radar active antijamming measures. It achieves antijamming by transmitting cover pulses of different frequencies before the radar pulse signal to induce enemy jammers. As the demand for antijamming measures has become increasingly urgent in recent years, Radio Frequency-screen technology has been further developed. The most representative is the use of discontinuous spectrum signals as a cover signal. However, energy utilization for sending the cover signal can be improved further. To address this problem, this paper proposes a discrete spectrum cover signal based on the discontinuous spectrum cover signal and establishes the waveform design function under the joint constraint of constant modulus and spectral amplitude. The cover signal with discrete spectrum and energy aggregation is generated using the Alternating Direction Method of Multipliers (ADMM) and spectrum shaping algorithm solution. The simulation results show that the discrete spectrum cover signal has a higher spectral amplitude of approximately 5~12 dB than the discontinuous-spectrum cover signal for the same energy and bandwidth. Moreover, the discrete spectrum cover signal can cover a larger spectral range with the same energy and close spectral amplitude, realizing a better antijamming cover effect. -
1 恒模约束下离散谱波形设计算法
1. Discrete spectrum waveform design algorithm under constant modulus constraint
输入:R, y 初始化: $ {{\bar {\boldsymbol{a}}}}(0),{\boldsymbol{\bar x}}(0),{{\boldsymbol{\bar f}}_p}(0),{{\boldsymbol{\lambda}} _1}(0),{{\boldsymbol{\lambda}} _{2,p}}(0) $ 1:for $t = 0,1, \cdots ,{N_{{\text{iter}}}}$,其中 ${N_{{\text{iter}}}}$为预先设定的最大迭代次数。 2: 更新 ${\boldsymbol{\bar a}}(t + 1)$: $ {\boldsymbol{\bar a}}(t + 1) = - \dfrac{1}{2}{\boldsymbol{R}}_1^{ - 1}{{\boldsymbol{d}}_1}(t + 1) $ 3: 更新 ${\boldsymbol{\bar x}}(t + 1)$: $ \left[ {\begin{array}{*{20}{c}} {{{\bar x}_k}} \\ {{{\bar x}_{k + K}}} \end{array}} \right] = - \dfrac{{\left[ {\begin{array}{*{20}{c}} {{d_{2,k}}(t + 1)} \\ {{d_{2,k + K}}(t + 1)} \end{array}} \right]}}{{\left\| {\left[ {\begin{array}{*{20}{c}} {{d_{2,k}}(t + 1)} \\ {{d_{2,k + K}}(t + 1)} \end{array}} \right]} \right\|}},\;k = 0,1, \cdots, K - 1 $ 4: 更新 ${{\boldsymbol{\bar f}}_p}(t + 1)$: $ {\bar {\boldsymbol{f}}_p}(t + 1) = \left\{ \begin{aligned} & {\dfrac{{\sqrt {(U + L)/2 + r} }}{{\left\| {{{{{\stackrel \frown{\boldsymbol f} }}}_p}(t + 1)} \right\|}}{{{{\stackrel \frown{\boldsymbol f} }}}_p}(t + 1),{\text{ }}\left\| {{{{{\stackrel \frown{\boldsymbol f} }}}_p}(t + 1)} \right\| \ge \sqrt {(U + L)/2 + r} } \\ & {\dfrac{{\sqrt {(U + L)/2 - r} }}{{\left\| {{{{{\stackrel \frown{\boldsymbol f} }}}_p}(t + 1)} \right\|}}{{{{\stackrel \frown{\boldsymbol f} }}}_p}(t + 1),{\text{ }}\left\| {{{{{\stackrel \frown{\boldsymbol f} }}}_p}(t + 1)} \right\| \le \sqrt {(U + L)/2 - r} } \\ & {{{{{\stackrel \frown{\boldsymbol f} }}}_p}(t + 1),{\text{ }}\sqrt {(U + L)/2 - r} < \left\| {{{{{\stackrel \frown{\boldsymbol f} }}}_p}(t + 1)} \right\| < \sqrt {(U + L)/2 + r} } \end{aligned} \right. $ 5: 更新 $ {{\boldsymbol{\lambda}} _1}(t + 1) $, $ {{\boldsymbol{\lambda}} _{2,p}}(t + 1) $: $ {{\boldsymbol{\lambda}} _1}(t + 1) = {{\boldsymbol{\lambda}} _1}(t) + \rho ({\boldsymbol{\bar x}} - {\boldsymbol{\bar V\bar a}}) $
$ {{\boldsymbol{\lambda}} _{2,p}}(t + 1) = {{\boldsymbol{\lambda}} _{2,p}}(t) + \rho ({{\boldsymbol{\bar f}}_p} - {{\boldsymbol{\bar v}}_p}{\boldsymbol{\bar a}}),p = 1,2, \cdots ,P $6:end 当 $t = {N_{{\text{iter}}}}$或同时满足 $\left\| {{\boldsymbol{\bar x}}(t + 1) - {\boldsymbol{\bar V\bar a}}(t + 1)} \right\| < {\delta _1}$, $ \left\| {{{{\boldsymbol{\bar f}}}_p}(t + 1) - {{{\boldsymbol{\bar v}}}_p}{\boldsymbol{\bar a}}(t + 1)} \right\| < {\delta _2} $ 7:对得到的时域序列 ${\boldsymbol{\bar x}}$进行频谱塑形 8:for $t = 0,1, \cdots ,{N_{{\text{iter}}}}$,其中 ${N_{{\text{iter}}}}$为预先设定的最大迭代次数 9: 更新 $ {\boldsymbol{\theta}} (t + 1) $: $ {\boldsymbol{\theta}} (t + 1) = \arg \{ {{\boldsymbol{F}}^{\text{H}}}{\boldsymbol{x}}\} $ 10: 更新 ${\boldsymbol{x}}(t + 1)$: $ {\boldsymbol{x}}(t + 1) = {{\text{e}}^{{\text{j}}\arg ({\boldsymbol{F}}({\boldsymbol{y}}{{\text{e}}^{{\text{j}}{\boldsymbol{\theta}} }}))}} $ 11:end 当 $t = {N_{{\text{iter}}}}$或 $ \left\| {{{\boldsymbol{F}}^{\text{H}}}{\boldsymbol{x}} - {\boldsymbol{y}}{{\text{e}}^{{\text{j}}{\boldsymbol{\theta}} }}} \right\|_2^2 < {\delta _3} $时,获得波形序列x 表 1 评价指标
Table 1. Evaluation Indicators
指标 符号 计算公式 通带平均幅度 $ {A_{{\rm{fw}}}} $ $ {A_{{\rm{fw}}}} = \displaystyle\sum\nolimits_{p = 1}^P {\left\| {{{\boldsymbol{f}}_p}} \right\|/P} $ 阻带最大幅度 $ {\epsilon}_{1} $ $ {\epsilon}_{1}=\underset{s}{\mathrm{max}}\Vert {f}_{s}\Vert $ 通带起伏度 ${\sigma ^2}$ ${\sigma ^2} = {\displaystyle\sum\nolimits_{p = 1}^P {(\left\| {{{\boldsymbol{f}}_p}} \right\| - {A_{{\rm{fw}}}})} ^2}/P$ 消耗干扰资源比例 $\eta $ $\eta = {W_{{\text{waste}}}}/{W_{{\text{all}}}}$ 表 2 掩护信号参数
Table 2. Cover signal parameters
参数 数值 离散谱掩护信号带宽( ${\text{MHz}}$) $ [50{\text{ 200}}] $ 离散谱掩护信号时宽( ${\text{μs}}$) $ 25 $ 离散谱掩护信号阻带区域( ${\text{MHz}}$) 25~30 离散谱掩护信号谱线间隔(MHz) 0.5 非连续谱掩护信号带宽(MHz) 50 非连续谱掩护信号时宽(μs) 25 非连续谱掩护信号阻带区域(MHz) 25~30 采样率(MHz) 600 表 3 发射信号参数
Table 3. Transmit signal parameters
参数 数值 载频( ${\text{GHz}}$) 1 时宽( $ {\text{μs}} $) $ 25 $ 脉冲重复间隔( $ {\text{μs}} $) $ 100 $ 采样率( ${\text{MHz}}$) ${\text{600}}$ 脉冲个数 $50$ 离散谱掩护信号带宽( ${\text{MHz}}$) ${\text{200}}$ 非连续谱掩护信号带宽( ${\text{MHz}}$) ${\text{50}}$ 窄带信号带宽( ${\text{MHz}}$) ${\text{5}}$ 探测信号与掩护信号有效辐射功率之比 1:4 表 4 抗干扰场景参数
Table 4. Anti-interference scene parameters
参数 数值 目标距离( ${\text{km}}$) $10$ 目标速度( ${\text{m/s}}$) $60$ 干信比( ${\text{dB}}$) $45$ 表 5 抗干扰场景参数
Table 5. Anti-interference scene parameters
参数 数值 目标距离( ${\text{km}}$) $10$ 目标速度( ${\text{m/s}}$) $60$ 干扰假目标距离( ${\text{km}}$) ${\text{9}}{\text{.8}}$ 干扰假目标速度( ${\text{m/s}}$) $62$ 切片时长( $ {\text{μs}} $) 1 转发次数 5 干信比( ${\text{dB}}$) ${\text{30}}$ 灵巧干扰中噪声信号带宽( ${\text{MHz}}$) $10$ -
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