弱相关非高斯信道下的伪码捕获方法

摘要

本发明提供的是一种弱相关非高斯信道下的伪码捕获方法。针对弱相关非高斯噪声环境，将伪码捕获等价为假设检验问题，将弱相关非高斯噪声建模为一阶滑动平均SαS噪声模型，利用局部最佳检测算法推导出弱相关非高斯噪声环境下的伪码捕获检测统计量，得到弱相关噪声模型下伪码二维捕获结构及其简化形式，并与传统的伪码捕获结构进行了性能仿真对比，仿真结果表明该捕获结构在相关非高斯噪声环境下检测性能有较大幅度的提高。

主权项

1、一种弱相关非高斯信道下的伪码捕获方法，其特征是：通过局部最佳算法算出的统计量与捕获判决门限进行比较来验证是否捕获到直扩信号；如果小于门限，则滑动本地伪码相位重新进行判断比较直至捕获到信号；高斯加性噪声下常规平方和(SS)检测器结构为$T_{\mathrm{SS}}(X^I,X^Q)={\left({\Sigma }_{i=1}^M{X}_i^I\right)}^2+{\left({\Sigma }_{i=1}^M{X}_i^Q\right)}^2;$其中统计量是把捕获结构中两支路的联合概率密度函数应用局部最佳检测算法得到，具体方法为：$f_{X^I,X^Q}(X^I,X^Q)$在θ＝0处的一阶导数为：$\frac{f_{X^I,X^Q}(X^I,X^Q)}{\mathrm{d\theta }}{|}_{\theta =0}=E_{\phi }\left\{\underset{i=1}{\overset{N_b}{\Sigma }}{f}_{N^I,N^Q}^{\prime }(N_i^I,N_i^Q)\underset{j=1,j\ne i}{\overset{N_b}{\Pi }}{f}_{N^I,N^Q}^{\prime }(N_j^I,N_j^Q){|}_{\theta =0}\right\}$其中：$E_{\phi }\left\{\frac{d{f}_{N^I,N^Q}(N_i^I,N_i^Q)}{\mathrm{d\theta }}{|}_{\theta =0}\right\}=E_{\phi }\{-\cos \phi \frac{\partial f_{N^I,N^Q}(N_i^I,N_i^Q)}{\partial N_i^I}-\sin \phi \frac{\partial f_{N^I,N^Q}(X_i^I,X_i^Q)}{\partial N_i^Q}\}=0$在θ＝0处的一阶导数为0，求在θ＝0处的二阶导$\frac{d^2{f}_{Y^I,Y^Q}(Y^I,Y^Q)}{d{\theta }^2}{|}_{\theta =0}=E_{\phi }\{\underset{i=1}{\overset{M}{\Sigma }}[f_{{\Lambda }^I,{\Lambda }^Q}^{\prime \prime }({\Lambda }_i^I,{\Lambda }_i^Q)\underset{j=1,j\ne i}{\overset{M}{\Pi }}{f}_{{\Lambda }^I,{\Lambda }^Q}^{\prime \prime }({\Lambda }_j^I,{\Lambda }_j^Q)$$+f_{{\Lambda }^I,{\Lambda }^Q}^{\prime }({\Lambda }_i^I,{\Lambda }_i^Q)\underset{i=1}{\overset{M}{\Sigma }}{f}_{{\Lambda }^I,{\Lambda }^Q}^{\prime }({\Lambda }_j^I,{\Lambda }_j^Q)\underset{k=1,k\ne i,j}{\overset{M}{\Pi }}{f}_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_k^I,{\Lambda }_k^Q)\left]\right\}$其中：$\frac{d^2{f}_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{d{\theta }^2}={\cos }^2\phi C_i^2\frac{{\partial }^2{f}_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{{(\partial {\Lambda }_i^I)}^2}+{\sin }^2\phi C_i^2\frac{{\partial }^2{f}_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{{(\partial {\Lambda }_i^Q)}^2}$$+\cos \phi {\sin \mathrm{\phi C}}_i^2\frac{{\partial }^2{f}_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{\partial {\Lambda }_i^I\partial {\Lambda }_i^Q}+\sin \phi \cos \phi C_i^2\frac{{\partial }^2{f}_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{\partial {\Lambda }_i^Q\partial {\Lambda }_i^I}$在θ＝0处的二阶导函数中，第一项表示为：$E_{\phi }\left\{\underset{i=1}{\overset{M}{\Sigma }}{f}_{{\Lambda }^I,{\Lambda }^Q}^{\prime \prime }({\Lambda }_i^I,{\Lambda }_i^Q)\underset{j=1,j\ne i}{\overset{M}{\Pi }}{f}_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_j^I,{\Lambda }_j^Q)\right\}{|}_{\theta =0}=\frac{1}{2}\underset{i=1}{\overset{M}{\Sigma }}{C}_i^2\{h\left(Y_i^I\right)+h\left(Y_i^Q\right)\}\underset{j=1}{\overset{M}{\Pi }}{f}_{{\Lambda }^I,{\Lambda }^Q}(Y_j^I,Y_j^Q)$其中：$h\left(Y_i^b\right)=\frac{1}{f_{{\Lambda }^I,{\Lambda }^Q}(Y_i^I,Y_i^Q)}\times \frac{{\partial }^2{f}_{{\Lambda }^I,{\Lambda }^Q}(Y_i^I,Y_i^Q)}{\partial {(Y_i^b)}^2}$b＝{I，Q}对应同相正交支路；同样$f_{{\Lambda }^I,{\Lambda }^Q}^{\prime }({\Lambda }_i^I,{\text{Λ}}_i^Q)f_{{\Lambda }^I,{\Lambda }^Q}^{\prime }({\Lambda }_j^I,{\text{Λ}}_j^Q)={\cos }^2\phi C_i{C}_j\frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{\partial {\Lambda }_i^I}\frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_j^I,{\Lambda }_j^Q)}{\partial {\Lambda }_j^I}$${+\sin }^2\phi C_i{C}_j\frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{\partial {\Lambda }_i^I}\frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_j^I,{\Lambda }_j^Q)}{\partial {\Lambda }_j^Q}$$+\cos \phi {\sin \mathrm{\phi C}}_i{C}_j[\frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{\partial {\Lambda }_i^I}\frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_j^I,{\Lambda }_j^Q)}{\partial {\Lambda }_j^Q}+\frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{\partial {\Lambda }_i^Q}\frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_i^I,{\Lambda }_i^Q)}{\partial {\Lambda }_i^I}]$在θ＝0处的二阶导函数中的第二项表示为：$E_{\phi }\left\{\underset{i=1}{\overset{M}{\Sigma }}\underset{j=1,j\ne i}{\overset{M}{\Sigma }}{f}_{{\Lambda }^I,{\Lambda }^Q}^{\prime }({\Lambda }_i^I,{\text{Λ}}_i^Q)f_{{\Lambda }^I,{\Lambda }^Q}^{\prime }({\Lambda }_j^I,{\text{Λ}}_j^Q)\underset{k=1,k\ne i,j}{\overset{M}{\Pi }}{f}_{{\Lambda }^I,{\Lambda }^Q}({\Lambda }_k^I,{\Lambda }_k^Q)\right\}{|}_{\theta =0}$$=\frac{1}{2}\underset{i=1}{\overset{M}{\Sigma }}\underset{j=1,j\ne i}{\overset{M}{\Sigma }}{C}_i{C}_j\{g\left(Y_i^I{\text{)g(Y}}_j^I\right)+g\left(Y_j^Q\right)g\left({\text{Y}}_j^Q\right)\}\times \underset{k=1}{\overset{M}{\Pi }}{f}_{{\Lambda }^I,{\Lambda }^Q}(Y_k^I,Y_k^Q)$其中：$g\left(Y_i^b\right)=\frac{1}{f_{{\Lambda }^I,{\Lambda }^Q}(Y_i^I,Y_i^Q)}\times \frac{\partial f_{{\Lambda }^I,{\Lambda }^Q}(Y_i^I,Y_i^Q)}{\partial Y_i^b}$b＝{I，Q}对应同相正交支路；把上述函数带入$f_{Y^I,Y^Q}(Y^I,{\text{Y}}^Q){|}_{\theta =0}=\underset{k=1}{\overset{M}{\Pi }}{f}_{{\Lambda }^I,{\Lambda }^Q}(Y_k^I,Y_k^Q)$得到局部最佳统计量T_LO。