一种相位差估计的相频匹配方法

摘要

本发明涉及信号处理领域，特别是一种相位差估计的相频匹配方法。本发明的适用对象为两路正弦信号的相位差估计，包括以下步骤：首先，根据正弦信号的和差角公式，利用两路正弦信号和信号频率实现两路正弦信号的90°相移，获得两路正弦信号的正交分量；其次，对两路正弦信号及其正交分量进行滑动互相关，获得互相关信号；然后，对两路正弦信号及其正交分量进行滑动自相关，获得自相关信号，实现自相关信号和互相关信号的频率匹配；最后，对互相关信号和自相关信号进行互相关，得到互相关函数，利用互相关函数相位的加权平均获得相位差估计值。本发明涉及的两路正弦信号相位差估计方法能够充分运用两路正弦信号的信息，提高正弦信号相位差估计精度，改善相位差估计的抗噪性能。

主权项

一种相位差估计的相频匹配方法，其特征在于：适用对象为两路正弦信号的相位差估计。该方法包括以下步骤：第一步：根据正弦信号的和差角公式，利用两路正弦信号x_N(n)、y_N(n)和信号频率ω计算得到两路正弦信号的正交分量和为：$\left\{\begin{array}{c}{\hat{x}}_N\left(n\right)=\frac{x_N(n-b)-x_N\left(n\right)\cos \left(\mathrm{b\omega }\right)}{\sin \left(\mathrm{b\omega }\right)}\\ {\hat{y}}_N\left(n\right)=\frac{y_N(n-b)-y_N\left(n\right)\cos \left(\mathrm{b\omega }\right)}{\sin \left(\mathrm{b\omega }\right)}\end{array}\right.---\left(1\right)$第二步：对两路正弦信号x_N(n)、y_N(n)及其正交分量进行互相关，获得互相关信号r_1，N(k)和r_2，N(k)为：$\left\{\begin{array}{c}{r}_{1,N}\left(k\right)=\frac{1}{N_0-b-k}\underset{n=1+b}{\overset{N_0-k}{\Sigma }}[x_N\left(n\right)y_N(n+k)+{\hat{x}}_N\left(n\right){\hat{y}}_N(n+k)]\\ r_{2,N}\left(k\right)=\frac{1}{N_0-b-k}\underset{n=1+b}{\overset{N_0-k}{\Sigma }}[x_N\left(n\right){\hat{y}}_N(n+k)-{\hat{x}}_N\left(n\right)y_N(n+k)]\end{array}\right.k=\mathrm{0,1}...N_0-b-1---\left(2\right)$式中N₀为信号长度。第三步：对两路正弦信号及其正交分量进行自相关，得到自相关信号为：$\left\{\begin{array}{c}{h}_{1,N}\left(k\right)=\frac{1}{N_0-b-k}\underset{n=1+b}{\overset{N_0-k}{\Sigma }}[x_N\left(n\right)x_N(n+k)+{\hat{x}}_N\left(n\right){\hat{x}}_N(n+k)+y_N\left(n\right)y_N(n+k)+{\hat{y}}_N\left(n\right){\hat{y}}_N(n+k)]\\ h_{2,N}\left(k\right)=\frac{1}{N_0-b-k}\underset{n=1+b}{\overset{N_0-k}{\Sigma }}[x_N\left(n\right){\hat{x}}_N(n+k)-{\hat{x}}_N\left(n\right)x_N(n+k)+x_N\left(n\right){\hat{x}}_N(n+k)-{\hat{x}}_N\left(n\right)x_N(n+k)]\end{array}\right.---\left(3\right)$其中h_1，N(0)＝1，h_2，N(0)＝0，k＝0，1…N₀‑b‑1。第四步：对自相关信号h_1，N(k)、h_2，N(k)和互相关信号r_1，N(k)、r_2，N(k)进行互相关，得到互相关函数，利用互相关函数相位的加权平均获得相位差估计值Δθ为：$\mathrm{\Delta \theta }={\theta }_2-{\theta }_1=\underset{k=0}{\overset{N_0-b-1}{\Sigma }}w\left(k\right)\frac{r_{2,N}\left(k\right)h_{1,N}\left(k\right)-r_{1,N}\left(k\right)h_{2,N}\left(k\right)}{r_{1,N}\left(k\right)h_{1,N}\left(k\right)+r_{2,N}\left(k\right)h_{2,N}\left(k\right)}---\left(4\right)$式中w(k)表示权重值，$w\left(k\right)=\frac{N_0-b-k}{\underset{k=0}{\overset{N_0-b-1}{\Sigma }}(N_0-b-k)},k=\mathrm{0,1,2}...N_0-b-1.$第五步：利用式(5)和(6)计算得到两路正弦信号x_N+1和y_N+1的正交分量和为$\left\{\begin{array}{c}{\hat{x}}_{N+1}\left(N_0\right)=\frac{x_{N+1}(N_0-b)-x_{N+1}\left(N_0\right)\cos \left(\mathrm{b\omega }\right)}{\sin \left(\mathrm{b\omega }\right)}\\ {\hat{y}}_{N+1}\left(N_0\right)=\frac{y_{N+1}(N_0-b)-y_{N+1}\left(N_0\right)\cos \left(\mathrm{b\omega }\right)}{\sin \left(\mathrm{b\omega }\right)}\end{array}\right.---\left(5\right)$$\left\{\begin{array}{c}{\hat{x}}_{N+1}(1:N_0-1)={\hat{x}}_N(2:N_0)\\ {\hat{y}}_{N+1}(1:N_0-1)={\hat{y}}_N(2:N_0)\end{array}\right.---\left(6\right)$利用式(7)计算得到第N+1时刻两路正弦信号及其正交分量的第k点互相关信号r_1，N+1(k)和r_2，N+1(k)为：$\left\{\begin{array}{c}{r}_{1,N+1}\left(k\right)=r_{1,N}\left(k\right)-\frac{1}{N_0-b-k}[a_{11}-a_{12}]\\ r_{2,N+1}\left(k\right)=r_{2,N}\left(k\right)-\frac{1}{N_0-b-k}[a_{21}-a_{22}]\end{array}\right.---\left(7\right)$式中$a_{11}=x_N(1+b)y_N(1+b+k)+{\hat{x}}_N(1+b){\hat{y}}_N(1+b+k),$$a_{12}=x_{N+1}(N_0-k)y_{N+1}\left(N_0\right)+{\hat{x}}_{N+1}(N_0-k){\hat{y}}_{N+1}\left(N_0\right),$$a_{21}=x_N(1+b){\hat{y}}_N(1+b+k)-{\hat{x}}_N(1+b)y_N(1+b+k),$$a_{22}=x_{N+1}(N_0-k){\hat{y}}_{N+1}\left(N_0\right)-{\hat{x}}_{N+1}(N_0-k)y_{N+1}\left(N_0\right).$计算得到第N+1时刻第k点自相关信号h_1，N+1(k)和h_2，N+1(k)为：$\left\{\begin{array}{c}{h}_{1,N+1}\left(k\right)=h_{1,N}\left(k\right)-\frac{1}{N_0-b-k}[b_{11}-b_{12}]\\ h_{2,N+1}\left(k\right)=h_{2,N}\left(k\right)-\frac{1}{N_0-b-k}[b_{21}-b_{22}]\end{array}\right.---\left(8\right)$式中$b_{11}=x_N(1+b)x_N(1+b+k)+{\hat{x}}_N(1+b){\hat{x}}_N(1+b+k)+y_N(1+b)y_N(1+b+k)+{\hat{y}}_N(1+b){\hat{y}}_N(1+b+k),$$b_{12}=x_{N+1}(N_0-k)x_{N+1}\left(N_0\right)+{\hat{x}}_{N+1}(N_0-k){\hat{x}}_{N+1}\left(N_0\right)+y_{N+1}(N_0-k)y_{N+1}\left(N_0\right)+{\hat{y}}_{N+1}(N_0-k){\hat{y}}_{N+1}\left(N_0\right),$$b_{21}=x_N(1+b){\hat{x}}_N(1+b+k)-{\hat{x}}_N(1+b)x_N(1+b+k)+y_N(1+b){\hat{y}}_N(1+b+k)-{\hat{y}}_N(1+b)y_N(1+b+k),$$b_{22}=x_{N+1}(N_0-k){\hat{x}}_{N+1}\left(N_0\right)-{\hat{x}}_{N+1}(N_0-k)x_{N+1}\left(N_0\right)+y_{N+1}(N_0-k){\hat{y}}_N\left(N_0\right)-{\hat{y}}_{N+1}(N_0-k)y_{N+1}\left(N_0\right).$取N＝N+1，并重复步骤第四步到第五步，直到计算到最后一点采样信号。