Unbiased Self-synchronous Scrambler Identification Based on Log Conditional Likelihood Ratio
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摘要: 为克服现有无偏自同步扰码识别算法在低信噪比(SNR)下存在适应性差的缺点,该文提出一种基于对数条件似然比的软判决识别方法。该方法首先构建了线性分组码自同步加扰和卷积码自同步加扰的对偶向量积线性约束方程;然后推导了基于软判决的对数条件似然比函数衡量方程的成立概率,并分析了其均值和方差的分布特性;最后通过2元假设和推导的相应判别门限来完成两种自同步加扰的识别。仿真结果表明,所提算法能够在低信噪比下完成生成多项式的识别,具有较好的适应能力,与基于求解代价函数的识别方法相比,在信噪比低于3 dB时的算法识别率得到较大提高,识别率为90%时,约有3 dB的性能增益。Abstract: To overcome the drawback of poor adaptability of existing unbiased self-synchronous scrambling code recognition algorithms at low Signal-to-Noise Ratios (SNR), a soft-judgement recognition method based on the log conditional likelihood ratio is proposed. Firstly, the linear constraint equations for the pairwise even-vector product of the self-synchronous scrambler of linear grouping codes and the self-synchronous scrambler of convolutional codes are constructed, and then the log likelihood ratio function is introduced, the log conditional likelihood ratio function based on the soft judgment is constructed, and the distribution characteristics of its mean and variance are analyzed. Finally the identification of self-synchronous scrambler of linear grouping codes and self-synchronous scrambler of convolutional codes is accomplished through binary assumption and the derived coresponding judgement threshold value. The simulations show that the proposed algorithm is able to complete the recognition of generating polynomials at low signal-to-noise ratios, and has a good low signal-to-noise adaptation capability. Compared with the recognition method based on solving the cost function, the recognition rate of the algorithm is greatly improved at signal-to-noise ratios below 3 dB, and when the recognition rate is 90%, the proposed algorithm achieves a performance gain of about 3 dB.
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1. 引言
在实际通信系统中,为了确保传输的数据码流中0和1的均衡性,信号发送端会使用线性反馈移位寄存器(Linear Feedback Shift Register, LFSR)对发送码流进行扰乱处理[1],经过扰乱处理的信源信号具有伪随机特性,极大增强了信号传输的保密性,确保了信息被安全地传输;同时,扰乱处理提高了信号的位定时恢复能力,频谱更适用于基带传输[2–5]。因此,扰码参数识别技术对非合作方信息的获取、密码的破译起着重要的作用[6,7]。
同步扰码的盲识别包括生成多项式和扰码初态的识别[8–11]。自同步扰码的盲识别是进行生成多项式的确定。自同步扰码由于其保密性强,且需要将扰码序列作为LFSR的输入,因此其生成多项式的识别更加复杂。针对有偏自同步扰码的研究主要基于信源的有偏性[12–14]。针对无偏自同步扰码识别,文献[15]提出了一种利用加扰序列游程属性差识别无偏自同步扰码的方法,该算法未考虑误码和信道噪声的影响,同时数据量低于4 000 bit时,识别率几乎为0。为了改善算法的识别率,文献[16]利用秩特性识别出线性分组码的码长。该方法的主要缺点是抗误码性较差。针对非2元域的线性分组码加扰识别,文献[17]提出了一种误码条件下的里所码Reed-Solomon (CodesRS)码自同步加扰识别方法,和文献[16]方法一样,当误码率(Bit Error Rate, BER)大于0.01时,秩特性不明显[18],这时码长估计会失效,无法进一步达到多项式识别的目的。为了降低算法的数据量需求,文献[19]将线性分组码自同步加扰扩展到卷积码自同步加扰领域,提出了一种软判决下的卷积码自同步加扰识别方法,利用改进的烟花搜索算法最小化代价函数完成了反馈系数的求解。该方法所需数据量较少,但是在求解代价函数的过程中用到了近似处理,较低信噪比(Signal to Noise Ratio, SNR)的识别性能有待进一步提高,同时未考虑线性分组码自同步加扰的情况。
综上所述,软判决识别方法在低SNR下的识别率更佳,但是现有软判决算法的识别性能有待进一步提高,而且缺少一种线性分组码自同步加扰软判决识别算法。基于上述不足,本文提出一种利用对数条件似然比进行线性分组码自同步加扰和卷积码自同步加扰的联合识别方法。首先构建线性分组码自同步加扰和卷积码自同步加扰对偶向量积的线性约束方程,然后引入对数似然比函数,构建了基于软判决的对数条件似然比函数,并分析了其均值和方差的分布特性,建立2元假设和推导了相应的判别门限完成了线性分组码自同步加扰和卷积码自同步加扰的识别。与以往算法相比,本算法综合考虑了线性分组码自同步加扰和卷积码自同步加扰,同时本算法在低信噪比下的识别率得到了较大的提升。
2. 无偏自同步扰码原理和模型
2.1 自同步扰码原理
扰码的加扰基础为LFSR,自同步扰码加扰过程可以表示为
yk=xk⊕L∑i=1⊕ciyk−i (1) 其中, ∑⊕表示模2累加, ci为反馈多项式的系数。ci的取值为0或1,c0=cL=1。
扰码的生成多项式可以表示为
f(x)=1+c1x+c2x2+⋯+cLxL (2) 2.2 无偏自同步扰码模型
现有方法的无偏自同步扰码模型主要考虑了单一纠错码加扰,并没有考虑线性分组码自同步加扰和卷积码自同步加扰的识别。本文充分考虑此两种无偏自同步扰码识别,模型图如图2所示。
图2中,{mk}为有偏的信源序列,{xk}为线性分组码序列或卷积码序列,信道噪声为高斯白噪声,均值为0,方差为σ2。
3. 无偏自同步扰码识别模型
3.1 基于对偶向量的线性约束方程的建立
针对(n,k)线性分组码,任意一段码长的码元序列与校验向量的乘积为0向量,构造系数矩阵为
x=[x1x2⋯xnxn+1xn+2⋯x2n⋮⋮⋱⋮x(K−1)n+1x(K−1)n+2⋯xKn] (3) 其中,N为扰码序列长度,K=⌊N/Nnn⌋,为N/Nnn的向下取整。
假设线性分组码的一个校验向量为h=[h1h2⋯hn]T,根据扰码解扰过程,可得
[L∑i=0ci⋅y1−iL∑i=0ci⋅y2−i⋯L∑i=0ci⋅yn−iL∑i=0ci⋅y(n+1)−iL∑i=0ci⋅y(n+2)−i⋯L∑i=0ci⋅y2n−i⋮⋮⋱⋮L∑i=0ci⋅y(K−1)n+1−iL∑i=0ci⋅y(K−1)n+2−i⋯L∑i=0ci⋅yKn−i]⋅[h1h2⋯hn]T=[00⋯0]T (4) 考虑式(4)所有的行。生成多项式系数的线性约束方程可变换为
[rKmin (5) 其中, {K_{\min }} = \left\lceil {{L \mathord{\left/ {\vphantom {L n}} \right. } n}} \right\rceil ,\left\lceil \cdot \right\rceil 为向上取整
当自同步加扰前的编码方式为\left( {{n_2},{k_2},m} \right)卷积码时,同理可得线性约束方程为
\left[ {\begin{array}{*{20}{c}} {r_{{M_{\min }} \cdot {n_2} - L + 1}'}&{r_{{M_{\min }} \cdot {n_2} - L + 2}'}& \cdots &{r_{\left( {{M_{\min }} + 1} \right) \cdot {n_2} + 1}'} \\ {{r_{\left( {{M_{\min }}{\text{ + }}1} \right) \cdot {n_2} - L + 1}}}&{{r_{\left( {{M_{\min }}{\text{ + }}1} \right) \cdot {n_2} - L + 2}}}& \cdots &{r_{\left( {{M_{\min }}{\text{ + 2}}} \right) \cdot {n_2} + 1}'} \\ \vdots & \vdots & \ddots & \vdots \\ {r_{\left( {M - 1} \right) \cdot {n_2} - L + 1}'}&{r_{\left( {M - 1} \right) \cdot {n_2} - L + 2}'}& \cdots &{r_{\left( {M - 1} \right) \cdot {n_2} + 1}'} \end{array}} \right] \cdot {\left[ {\begin{array}{*{20}{c}} {{c_L}}&{{c_{L - 1}}}& \cdots &{{c_0}} \end{array}} \right]^{\text{T}}} = {\left[ {\begin{array}{*{20}{c}} 0&0& \cdots &0 \end{array}} \right]^{\text{T}}} (6) 3.2 对数条件似然比函数的建立与推导
以线性分组码为例,为了方便说明,取式(5)的第j行,系数矩阵进行累加和变换后的第j行\left( {r_{j \cdot n - L + 1}}, {r_{j \cdot n - L + 2}}, \cdots , {r_{j \cdot n + 1}} \right)记为{{\boldsymbol{r}}_j},生成多项式系数\left( {c_L}, {c_{L - 1}}, \cdots ,{c_1},{c_0} \right)记为{\boldsymbol{c}},根据式(5),有式(7)成立
{{\boldsymbol{r}}_j} \cdot {{\boldsymbol{c}}^{\text{T}}} = 0 (7) 定义式(7)的条件概率形式为
P\left( {\sum\limits_{t = 1}^L { \oplus {r_{j,t}} \cdot {c_t} = 0|{{\boldsymbol{z}}_j}} } \right) (8) 其中,{{\boldsymbol{r}}_j}为接收序列码元{{\boldsymbol{z}}_j}对应的软判决码元。
假设{r_{j,i}}为抽头为1时对应的{{\boldsymbol{r}}_j}的码元位置,下面引入对数似然比的概念,对数条件似然比定义为
\begin{split} & L\left( {\sum\limits_{t = 1}^L { \oplus {r_{j,t}} \cdot {c_t} = 0|{{\boldsymbol{z}}_j}} } \right) \\ & \quad = \ln \left( {\frac{{P\left( {\displaystyle\sum\limits_{t = 1}^L { \oplus {r_{j,t}} \cdot {c_t} = 0|{{\boldsymbol{z}}_j}} } \right)}}{{P\left( {\displaystyle\sum\limits_{t = 1}^L { \oplus {r_{j,t}} \cdot {c_t} = 1|{{\boldsymbol{z}}_j}} } \right)}}} \right) \end{split} (9) L\left( \cdot \right)可以表示截获码元的可靠程度,P( \displaystyle\sum\nolimits_{t = 1}^L \oplus {r_{j,t}}\cdot {c_t} = 0|{{\boldsymbol{z}}_j} )越大,P\left( {\displaystyle\sum\nolimits_{t = 1}^L { \oplus {r_{j,t}} \cdot {c_t} = 1|{{\boldsymbol{z}}_j}} } \right)越小,对应的生成多项式与序列{{\boldsymbol{r}}_j}校验关系成立的可能性越大,L\left( \cdot \right)的值越大。
假设调制方式为二进制相移键控 (Binary Phase Shift Keying, BPSK),幅度A取1, 下面对P\left( {\displaystyle\sum\nolimits_{t = 1}^L { \oplus {r_{j,t}} \cdot {c_t} = 1|{{\boldsymbol{z}}_j}} } \right)的计算进行推导,给出定理1。
定理1 对于2元域的两个独立的随机变量\alpha 和\beta ,两者有如式(9)的概率关系成立[20]
\begin{split} P\left( {\alpha \oplus \beta = 1} \right) = & \frac{1}{2} - \frac{1}{2}\left( {1 - 2P\left( {\alpha = 1} \right)} \right) \\ & \cdot \left( {1 - 2P\left( {\beta = 1} \right)} \right) \end{split} (10) 根据定理1可知
P\left( {{r_{j - i}} = 1} \right) = \frac{1}{2} - \frac{1}{2}\prod\limits_{t = 1}^n {\left( {1 - 2{h_t} \cdot P\left( {{y_{j - i + t - 1}} = {\text{1}}} \right)} \right)} (11) 当截获的加扰软判决为 {z_j} ,对应的硬判决扰码为 {y_j} ,可以得到码元后验概率[14]为
P\left( {{y_j} = 1|{z_j}} \right) = \frac{{{{\text{e}}^{{{2{z_j}} \mathord{\left/ {\vphantom {{2{z_j}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}}}}{{{{\text{e}}^{{{2{z_j}} \mathord{\left/ {\vphantom {{2{z_j}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}} + 1}} (12) 可得 P\left( {\displaystyle\sum\nolimits_{t = 1}^L { \oplus {r_{j,t}} \cdot {c_t} = 1|{{\boldsymbol{z}}_j}} } \right) 的表达式为
P\left( {\sum\limits_{t = 1}^L { \oplus {r_{j,t}} \cdot {c_t} = 1|{{\boldsymbol{z}}_j}} } \right) = \frac{1}{2} - \frac{1}{2}\prod\limits_{i = 0}^L {\left( {1 - {c_i} \cdot \left( {1 - \prod\limits_{t = 1}^n {\left( {1 - 2{h_t} \cdot \frac{{{{\text{e}}^{{{2{z_{j - i + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j - i + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}}}}{{{{\text{e}}^{{{2{z_{j - i + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j - i + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}} + 1}}} \right)} } \right)} \right)} (13) 将式(13)代入式(8),考虑所有的行,并取平均值,可得平均对数条件似然函数 \bar \varUpsilon 为
\bar \varUpsilon = \frac{1}{{K - {K_{\min }} + 1}} \cdot \sum\limits_{j = {K_{\min }}}^{K} {\ln \left( {\frac{{1 + \displaystyle\prod\limits_{\left\{ {i|{c_i} = 1} \right\}} {\left( {\displaystyle\prod\limits_{\left\{ {t|{h_t} = 1} \right\}} {\left( {1 - 2 \cdot \dfrac{{{{\text{e}}^{{{2{z_{j - i + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j - i + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}}}}{{{{\text{e}}^{{{2{z_{j + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}} + 1}}} \right)} } \right)} }}{{1 - \displaystyle\prod\limits_{\left\{ {i|{c_i} = 1} \right\}} {\left( {\displaystyle\prod\limits_{\left\{ {t|{h_t} = 1} \right\}} {\left( {1 - 2 \cdot \dfrac{{{{\text{e}}^{{{2{z_{j - i + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j - i + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}}}}{{{{\text{e}}^{{{2{z_{j + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}} + 1}}} \right)} } \right)} }}} \right)} (14) 通过遍历生成多项式,当 \bar \varUpsilon 取最大值时所对应的多项式即为正确的生成多项式。下面对\bar \varUpsilon 的统计特性进行分析,并建立相应的判别门限。
3.3 判别门限的设定
建立两种假设:
{\mathcal{H}_0} :{\boldsymbol{c}}为错误的生成多项式系数;
{\mathcal{H}_1} :{\boldsymbol{c}}为正确的生成多项式系数。
假设 {\mathcal{H}_0} 成立时,式(23)的校验关系不成立,对应的E\left( {\bar \varUpsilon |{\mathcal{H}_0}} \right) = 0。此时,\left( {r_{j \cdot n - L + 1}},{r_{j \cdot n - L + 2}}, \cdots , {r_{j \cdot n + 1}} \right)中的码元为随机的序列,假设抽头数为m,所有可能的组合情况为
\left( {V|{\mathcal{H}_0}} \right) = \sum\limits_{j = 0}^m {C_m^j} (15) 其中,C_m^n = \dfrac{{n \cdot \left( {n - 1} \right) \cdots \left( {n - m + 1} \right)}}{{m \cdot \left( {m - 1} \right) \cdots 1}}。
不妨记
\ln \left( {\frac{{1 + \Re \left( z \right)}}{{1 - \Re \left( z \right)}}} \right) = \ln \left( {\frac{{1 + \displaystyle\prod\limits_{\left\{ {i|{c_i} = 1} \right\}} {\left( {\displaystyle\prod\limits_{\left\{ {t|{h_t} = 1} \right\}} {\left( {1 - 2 \cdot \dfrac{{{{\text{e}}^{{{2{z_{j - i + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j - i + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}}}}{{{{\text{e}}^{{{2{z_{j + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}} + 1}}} \right)} } \right)} }}{{1 - \displaystyle\prod\limits_{\left\{ {i|{c_i} = 1} \right\}} {\left( {\displaystyle\prod\limits_{\left\{ {t|{h_t} = 1} \right\}} {\left( {1 - 2 \cdot \dfrac{{{{\text{e}}^{{{2{z_{j - i + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j - i + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}}}}{{{{\text{e}}^{{{2{z_{j + t - 1}}} \mathord{\left/ {\vphantom {{2{z_{j + t - 1}}} {{\sigma ^2}}}} \right. } {{\sigma ^2}}}}} + 1}}} \right)} } \right)} }}} \right) (16) 根据方差的定义可得
\begin{split} {\text{Var}}\left( {\bar \varUpsilon |{{\mathcal{H}}_0}} \right) =\,& \sum\limits_{j = 0}^m {\frac{{C_m^j}}{{\left( {V|{\mathcal{H}_0}} \right)}}} \cdot {\left( {\int\limits_{ - \infty }^\infty {{{\left( {\ln \left( {\frac{{1 + \Re \left( z \right)}}{{1 - \Re \left( z \right)}}} \right)} \right)}^2} \cdot P\left( {z|y = 1} \right){\mathrm{d}}x} } \right)^j}\\ & \cdot {\left( {\int\limits_{ - \infty }^\infty {{{\left( {\ln \left( {\frac{{1 + \Re \left( z \right)}}{{1 - \Re \left( z \right)}}} \right)} \right)}^2} \cdot P\left( {z|y = 0} \right){\mathrm{d}}x} } \right)^{m - j}} \end{split} (17) 假设 {\mathcal{H}_1} 成立时,式(23)的校验关系成立,所有的组合情况为
\left( {V|{\mathcal{H}_1}} \right) = \sum\limits_{j = 0}^{\left\lfloor {m/2} \right\rfloor } {C_m^{2j}} (18) 其中,\left\lfloor {\cdot} \right\rfloor 表示向下取整。
同理可得假设 {\mathcal{H}_1} 成立下,均值E\left( {\bar \varUpsilon |{\mathcal{H}_1}} \right)和方差{\mathrm{Var}}\left( {\bar \varUpsilon |{\mathcal{H}_1}} \right) 为
E\left( {\bar \varUpsilon |{\mathcal{H}_1}} \right) = \sum\limits_{j = 0}^{\left\lfloor {{m \mathord{\left/ {\vphantom {m 2}} \right. } 2}} \right\rfloor } {\frac{{C_m^{2j}}}{{\left( {V|{\mathcal{H}_1}} \right)}}} \cdot {\left( {\int\limits_{ - \infty }^\infty {\ln \left( {\frac{{1 + \Re \left( z \right)}}{{1 - \Re \left( z \right)}}} \right)} \cdot P\left( {z|y = 1} \right){\mathrm{d}}x} \right)^{2j}} \cdot {\left( {\int\limits_{ - \infty }^\infty {\ln \left( {\frac{{1 + \Re \left( z \right)}}{{1 - \Re \left( z \right)}}} \right) \cdot P\left( {z|y = 0} \right){\mathrm{d}}x} } \right)^{m - 2j}} (19) \begin{split} {\text{Var}}\left( {\bar \varUpsilon |{\mathcal{H}_1}} \right) = \,& E\left( {\bar \varUpsilon |{\mathcal{H}_1}} \right) - E{\left( {\bar \varUpsilon |{\mathcal{H}_1}} \right)^2} = \sum\limits_{j = 0}^{\left\lfloor {{m / 2}} \right\rfloor } {\frac{{C_m^{2j}}}{{\left( {V|{\mathcal{H}_1}} \right)}}} \cdot {\left( {\int\limits_{ - \infty }^\infty {{{\left( {\ln \left( {\frac{{1 + \Re \left( z \right)}}{{1 - \Re \left( z \right)}}} \right)} \right)}^2} \cdot P\left( {z|y = 1} \right){\mathrm{d}}x} } \right)^{2j}} \\ & \cdot {\left( {\int\limits_{ - \infty }^\infty {{{\left( {\ln \left( {\frac{{1 + \Re \left( z \right)}}{{1 - \Re \left( z \right)}}} \right)} \right)}^2} \cdot P\left( {z|y = 0} \right){\mathrm{d}}x} } \right)^{m - 2j}} - E{\left( {\bar \varUpsilon |{\mathcal{H}_1}} \right)^2} \end{split} (20) 记{\mu _0} \;=\; E\left( {\bar \varUpsilon |{\mathcal{H}_0}} \right), \sigma _0^2 \;=\; {{\text{Var}}\left( {\bar \varUpsilon |{H_0}} \right)} \;/\; \left( K \;- {K_{\min }} + 1 \right), {\mu _1} = E\left( {\bar \varUpsilon |{\mathcal{H}_1}} \right), \sigma _1^2 = {{\text{Var}}\left( {\bar \varUpsilon |{H_1}} \right)} / \left( K - {K_{\min }} + 1 \right)。
设虚警概率{P_{{\text{fa}}}}和漏警概率{P_{{\text{nd}}}}分别为
{P_{{\text{fa}}}} = \int\limits_{\boldsymbol{\varGamma }}^\infty {\frac{1}{{\sqrt {2{\pi}} {\sigma _1}}}{{\text{e}}^{{{ - {{\left( {x - {\mu _0}} \right)}^2}} \mathord{\left/ {\vphantom {{ - {{\left( {x - {\mu _0}} \right)}^2}} {\sigma _0^2}}} \right. } {\sigma _0^2}}}}{\text{d}}x} (21) {P_{{\text{nd}}}} = \int\limits_{ - \infty }^{\boldsymbol{\varGamma }} {\frac{1}{{\sqrt {2{\pi}} {\sigma _2}}}{{\text{e}}^{{{ - {{\left( {x - {\mu _1}} \right)}^2}} \mathord{\left/ {\vphantom {{ - {{\left( {x - {\mu _1}} \right)}^2}} {\sigma _1^2}}} \right. } {\sigma _1^2}}}}{\text{d}}x} (22) 极大极小错误准则下的判别门限 {{\varGamma }} 可表示为
\begin{split} & {{\varGamma }} =\\ & \frac{{{u_1}\sigma _0^2 - {u_0}\sigma _1^2 - {\sigma _0}{\sigma _1}\sqrt {\left( {{{\left( {{u_1} - {u_0}} \right)}^2} + \left( {\sigma _1^2 - \sigma _0^2} \right)\ln \left( {\dfrac{{{\sigma _1}}}{{{\sigma _0}}}} \right)} \right)} }}{{\sigma _0^2 - \sigma _1^2}} \end{split} (23) 当\bar \varUpsilon > {{\varGamma }}时,假设 {\mathcal{H}_1} 成立,遍历的多项式为生成多项式;反之,遍历的多项式不是生成多项式。
3.4 算法计算复杂度分析
假设扰码序列长度为N,遍历的码长范围为[{n_{\min }},{n_{\max }}],遍历的生成多项式个数为S,构造的序列{\boldsymbol{r}}的计算量为\left( {2{n_{\max }} - 1} \right) \cdot N,遍历一次需要进行 {O}\left( {S \cdot \left( {\left\lfloor {{N \mathord{\left/ {\vphantom {N {{n_{\max }}}}} \right. } {{n_{\max }}}}} \right\rfloor } \right)} \right) 量级的乘法和加法。假设遍历到{n_{\max }}才识别出生成多项式,则算法复杂度近似为 {O}\left( {\left( {\left\lfloor {{N \mathord{\left/ {\vphantom {N {{n_{\max }}}}} \right. } {{n_{\max }}}}} \right\rfloor } \right) \cdot \left( {{n_{\max }} - {n_{\min }}} \right) \cdot S + \left( {2{n_{\max }} - 1} \right) \cdot N} \right) 。假设基于代价函数求解的计算复杂度主要集中在烟花求解算法上,假设烟花个体维度为d,单个种群的烟花个体数为{N_{\text{f}}},爆炸火花数为{N_{\text{d}}},迭代次数为T,序列中满足约束关系的个数为 t组,则烟花搜索算法[19]的计算复杂度为 {O}\left( 3\left( {{N_{\text{f}}} + {N_{\text{d}}}} \right) \cdot T\cdot t \cdot d + \left( {2{n_{\max }} - 1} \right) \cdot N \right) 。虽然本文算法的计算复杂度高于基于代价求解函数的计算复杂度,两种算法的计算复杂度在一个量级上,算法复杂度的提升带来的是较低信噪比下识别率的显著提高,计算复杂度在可接受的范围内。
4. 仿真分析
4.1 线性分组码自同步加扰仿真验证
仿真设定线性分组码为(15,11)线性分组码,自同步扰码的生成多项式为1 + {x^3} + {x^{10}}。不同序列长度下,本文算法的识别率随SNR的变化曲线如图3所示。
从图3可以看出,在相同SNR下,扰码序列的长度越长,算法的识别性能越好,工程实际中可以通过增加序列长度来抵抗SNR带来的影响。
仿真设定扰码序列长度为600 bit,生成多项式为1 + {x^3} + {x^{10}}。不同码长n下,本文算法的识别率随SNR的变化曲线如图4所示。
从图4可以看出,相同扰码序列长度下,码长越大,算法的识别率越低,并且识别率达到100%时的SNR越大。原因是码长越大,对应的对偶向量码重越大,码元出现误判的概率越大。
仿真设定序列长度为600 bit,纠错码为 (15,11) 线性分组码、扰码的生成多项式分别为f\left( x \right) = 1 + {x^6} + {x^7}, f\left( x \right) = 1 + {x^{14}} + {x^{15}}, f\left( x \right) = 1 + {x^{18}} + {x^{23}}, f\left( x \right) = 1 + {x^{28}} + {x^{31}}以及f\left( x \right) = 1 + {x^{33}} + {x^{47}}。不同L下,本文算法的识别率随SNR的变化曲线如图5所示。
从图5可以看出,生成多项式阶数越低,算法的识别率越高。SNR为5 dB时为600 bit, 5种阶数下(15,11)线性分组码的识别率均能达到95%以上,这说明算法对生成多项式阶数具有较好的鲁棒性。
4.2 卷积码自同步加扰仿真验证
仿真设定加扰前的纠错码为 (2,1,5)卷积码,编码参数如表1所示。
表 1 卷积码参数设定(不同码重)码重w 生成多项式 3 [1,1 + {D^5}] 5 [1 + {D^5},1 + {D^4} + {D^5}] 7 [1 + {D^3} + {D^5},1 + {D^3} + {D^4} + {D^5}] 9 [1 + {D^2} + {D^4} + {D^5},1 + {D^2} + {D^3} + {D^4} + {D^5}] 11 [1 + D + {D^3} + {D^4} + {D^5},1 + D + {D^2} + {D^3} + {D^4} + {D^5}] 仿真中,码元序列长度设定为600 bit,扰码的生成多项式为f\left( x \right) = 1 + {x^{14}} + {x^{15}}。不同对偶向量码重下,算法的识别率随SNR的变化曲线如图6所示。
从图6可以看出, w越大,算法的识别率越低。原因是w越大,对偶向量{{\boldsymbol{h}}'}位置为1的个数越多,相同信噪比下,包含错误码元的概率越大,识别时错误的概率越高。实际运用中,可以寻找一个低码重的对偶向量进行序列{\boldsymbol{r}}的构造,有利于识别率的提高。
从图7可以看出,编码约束长度的变化对算法识别率的影响不大。原因是序列{\boldsymbol{r}}只受到对偶向量位置为1的影响,编码约束长度的增加不会带来序列的错误码元的增加,因此不受编码约束长度的影响。
仿真设定加扰前的卷积码参数如表2所示。
表 2 卷积码参数设定(不同编码约束长度)编码约束长度{\text{sL}} 生成多项式 8 [1 + {D^3},1 + {D^2} + {D^3}] 10 [1 + {D^4},1 + {D^3} + {D^4}] 12 [1 + {D^5},1 + {D^4} + {D^5}] 14 [1 + D,1 + D + {D^6}] 仿真中,码元序列长度设定为600 bit,扰码的生成多项式为f\left( x \right) = 1 + {x^{14}} + {x^{15}}。不同序列长度下,算法的识别率随SNR的变化曲线如图8所示。
仿真设定加扰前的纠错码为 (2,1,4)卷积码,卷积码的生成多项式为\left[ 1 + {D^4},1 + D + {D^2} + {D^3} + {D^4} \right],扰码的生成多项式为f\left( x \right) = 1 + {x^3} + {x^{10}}。仿真分析不同序列长度下,算法识别率随SNR的变化关系,并和基于代价函数求解的算法进行对比,仿真对比结果如图8所示。
从图8可以看出,相同SNR下,扰码序列长度越长,算法的识别率越高。主要原因是构造的约束方程中方程系数矩阵的行向量的个数为\left( M - {M_{\min }} - 1 \right) = \left\lfloor {{N / {{n_2}}}} \right\rfloor - {n_2} \cdot \left( {m + 1} \right) - \left\lceil {{L / {{n_2}}}} \right\rceil ,扰码序列长度N越长,此时系数方程矩阵行向量的个数越大,参与校验关系判决的方程的数量越多,对数条件似然比就越接近理论推导结果,此时的算法识别率越接近算法正确识别率的理论极限。相比于基于代价函数求解的算法,本文算法在低信噪比下的识别效果更好,尤其是较低信噪比时本文算法的性能提升显著。
仿真设定纠错码为 (2,1,4)卷积码,序列长度分别为600 bit,卷积码的生成多项式为\left[ 1 + {D^4}, 1 + D + {D^2} + {D^3} + {D^4} \right],扰码的生成多项式分别为f\left( x \right) = 1 + {x^3} + {x^7},f\left( x \right) = 1 + {x^2} + {x^{11}}和f\left( x \right) = 1 + x + {x^{15}}。仿真分析不同生成多项式下,算法识别率随SNR的变化关系,并和基于代价函数求解的算法进行对比,结果如图9所示。
从图9可以看出,随着阶数的升高,算法的识别率逐渐下降,原因是阶数的升高会导致基于对偶向量建立的线性约束方程中待求解向量 {\left[ {{c_L}}\;\;{{c_{L - 1}}}\;\; \cdots\;\; {{c_0}} \right]^{\text{T}}} 长度的增加,从而导致正确识别率的降低。相比于基于代价函数求解的算法,本文算法在SNR低于3 dB的识别效果提升是显著的。
5. 结束语
本文提出一种基于对数条件似然比的无偏自同步扰码识别方法,首先构建了线性分组码自同步加扰和卷积码自同步加扰的线性约束方程,然后引入了对数条件似然比函数衡量约束方程校验关系成立的概率,最后求解了判别门限完成了线性分组码自同步加扰和卷积码自同步加扰的识别。相比于现有算法,所提方法降低了所需扰码数据量,在SNR低于3 dB时,识别率得到了明显提高,同时,实现了线性分组码加扰和卷积码自同步加扰的联合识别。本文方法进一步提高了软判决下算法的识别率,但是针对纠错码自同步扰码的对偶向量的求解,目前还是一个难点问题,下一步将重点研究自同步加扰下纠错码的对偶向量求解问题,这将有助于利用软判决序列进一步求解扰码参数,提高算法的实用性。
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表 1 卷积码参数设定(不同码重)
码重w 生成多项式 3 [1,1 + {D^5}] 5 [1 + {D^5},1 + {D^4} + {D^5}] 7 [1 + {D^3} + {D^5},1 + {D^3} + {D^4} + {D^5}] 9 [1 + {D^2} + {D^4} + {D^5},1 + {D^2} + {D^3} + {D^4} + {D^5}] 11 [1 + D + {D^3} + {D^4} + {D^5},1 + D + {D^2} + {D^3} + {D^4} + {D^5}] 表 2 卷积码参数设定(不同编码约束长度)
编码约束长度{\text{sL}} 生成多项式 8 [1 + {D^3},1 + {D^2} + {D^3}] 10 [1 + {D^4},1 + {D^3} + {D^4}] 12 [1 + {D^5},1 + {D^4} + {D^5}] 14 [1 + D,1 + D + {D^6}] -
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