A Low Peak-to-average Ratio Secure Transmission Method Based on U Matrix Transformation in Orthogonal Time and Frequency Space System
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摘要: 为了降低正交时频空(OTFS)系统峰均比(PAPR)并且提升系统安全性,该文设计了一种基于酉矩阵变换的低峰均功率比OTFS安全传输方法。在该方法中,通过无线信道的时延多普勒(DD)域产生初始密钥,并将其作为混沌系统初始值进一步产生混沌序列。利用混沌序列进行酉矩阵设计,使得经过酉矩阵变换后的符号完全被混淆,具有类噪声的随机特性。此外通过索引控制酉矩阵选择,发射端将不同酉矩阵变换得到的OTFS时域信号进行排序并选择PAPR最低的信号进行发送。合法接收方获得索引值后可以正确解密和解调,而窃听者即使获得索引值信息,由于其没有相应的加密酉矩阵,为此无法正确解密。理论分析和仿真结果表明,所提方法在保证系统可靠性的前提下有效降低OTFS系统的PAPR。此外经过酉矩阵变换后的星座图呈现球状混乱,这使得调制方式和信息得以隐蔽,增大了窃听者的解密难度,系统的安全性得到保证。Abstract: In order to reduce the Peak-to-Average Power Ratio (PAPR) and improve the security of the Orthogonal Time and Frequency Space (OTFS) system, a low PAPR secure transmission method based on the U matrix transformation in OTFS system is proposed in this paper. In this method, the initial key is generated through the Delay-Doppler (DD) domain of wireless channel, which is used to generate further chaotic sequences. The U matrix is designed by the chaotic sequence, which makes the symbols after the U matrix transformation are completely confused and noise-like. Besides, the U matrix selections can be controlled by the index. The transmitter sortes the OTFS time domain signals obtained from different U matrix transformations and selectes the signal with the lowest PAPR for transmission. The encrypted signal can be correctly obtained by the legitimate receiver after obtaining the index value. However, the eavesdropper cannot decrypt the information even if he obtained the transmitted index value. The simulation results show that the proposed scheme can reduce the PAPR of OTFS system effectively while ensuring the system reliability. In addition, the constellation diagram the U matrix transformation becomes spherical chaos, which makes the modulation method and information hidden. The decryption difficulty of the eavesdropper is greatly increased, and the security of the system is effectively enhanced.
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图 15 SNR=30 dB所提方案与文献[31]方案的窃听者星座图对比
算法1 安全矩阵生成算法 输入:混沌序列$ {S_0} $; 输出:安全矩阵${{\boldsymbol{U}}}$; (1) 将$ {S_0} $分成$W$块序列${S_1},{S_2}, \cdots {S_i} \cdots {S_W}$; (2) 从$i = 1$到${W^2}$循环执行(3)~(5) (3) $S'_i = {\text{hash} }({S_i})$; (4) ${\theta _i} = 2{\pi }(S'_i\bmod \lambda )/\lambda$; (5) 结束循环 (6) 利用旋转矢量$ {\theta } $构建$ W \times W $的矩阵${ {{\boldsymbol{U}}}' }$; (7) 对${ {{\boldsymbol{U}}}'}$进行Gram正交化得到${{\boldsymbol{U}}}$ (8) 返回${{\boldsymbol{U}}}$。 -
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