一种短暂非完备条件下基于多普勒测速的卫星导航定位方法

摘要

本发明一种短暂非完备条件下基于多普勒测速的卫星导航定位方法，它包括：在卫星信号完备条件下进行正常定位，并且储存相关定位信息；当卫星信号受到遮挡，可见卫星为3颗时，执行非完备条件下的定位方法，依据多普勒频偏与接收机速度、卫星速度之间的关系建立辅助方程，将非线性方程组通过泰勒展开截取一阶项线性化之后使用最小二乘法进行求解得到接收机位置及速度；当可见卫星数量恢复到4颗及4颗以上时，执行完备条件下的定位方法。实际测试表明，在数十秒的短暂非完备期间内，该方法满足导航要求的定位结果，从而提高了GNSS接收机的有效性。本发明不依赖额外传感器，不需要额外仪器仪表，亦不需要电子地图支持，计算量小，系统成本无增加。

主权项

一种短暂非完备条件下基于多普勒测速的卫星导航定位方法，其特征在于：该方法具体步骤如下：步骤一：获取数据信息：接收机需要在每个解算历元获取以下数据信息；首先接收机通过跟踪环路的输出得到接收机此时相对于3颗可见卫星的多普勒频移值f_d1，f_d2，f_d3；然后接收机通过导航电文得到3颗可见卫星的速度及3颗可见卫星的位置坐标接着接收机读取储存的前一个历元时刻解算得到的接收机的位置坐标[x₀,y₀,z₀]^T；最后将每个历元的持续时间记为Δt；步骤二：推导定位解算的非线性方程组；因为接收机位置的非线性方程组中含有接收机位置和钟差4个未知数，所以定位解算至少需要4个观测方程；当可见卫星数量为3颗时，观测方程数量不足以完成定位，因此需要用其他方法获取信息，建立辅助方程来完成非完备条件下的定位，定义此时将接收机待解算的位置坐标记为[x,y,z]^T，此时接收机待解算速度记为[v_x,v_y,v_z]^T，此时接收机的解算的钟差记为δt_u；得到接收机当前位置坐标与接收机当前速度的关系为：$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right]+\left[\begin{array}{c}{v}_x·\mathrm{\Delta t}\\ v_y·\mathrm{\Delta t}\\ v_z·\mathrm{\Delta t}\end{array}\right]---\left(1\right)$将式(1)带入伪距观测方程得3颗可见卫星的伪距方程组为：$\left\{\begin{array}{c}\sqrt{{(x^{\left(1\right)}-x_0-v_x·\mathrm{\Delta t})}^2+{(y^{\left(1\right)}-y_0-v_y·\mathrm{\Delta t})}^2+{(z^{\left(1\right)}-z_0-v_z·\mathrm{\Delta t})}^2}+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(1\right)}\\ \sqrt{{(x^{\left(2\right)}-x_0-v_x·\mathrm{\Delta t})}^2+{(y^{\left(2\right)}-y_0-v_y·\mathrm{\Delta t})}^2+{(z^{\left(2\right)}-z_0-v_z·\mathrm{\Delta t})}^2}+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(2\right)}\\ \sqrt{{(x^{\left(3\right)}-x_0-v_x·\mathrm{\Delta t})}^2+{(y^{\left(3\right)}-y_0-v_y·\mathrm{\Delta t})}^2+{(z^{\left(3\right)}-z_0-v_z·\mathrm{\Delta t})}^2}+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(3\right)}\end{array}\right.---\left(2\right)$定义下述新变量以简化上述公式：$r_j=\sqrt{{(x^{\left(j\right)}-x_0-v_x·\mathrm{\Delta t})}^2+{(y^{\left(j\right)}-y_0-v_y·\mathrm{\Delta t})}^2+{(z^{\left(j\right)}-z_0-v_z·\mathrm{\Delta t})}^2}---\left(3\right)$方程组(2)简化为$\left\{\begin{array}{c}{r}_1+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(1\right)}\\ r_2+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(2\right)}\\ r_3+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(3\right)}\end{array}\right.---\left(4\right)$做出以下定义以完成算法推导：a.定义卫星在接收机处的单位观测矢量为它的计算公式如下所示：$\overrightarrow{l^{\left(s\right)}}=\frac{1}{\sqrt{{\mathrm{\Delta x}}^2+\Delta y^2+{\mathrm{\Delta z}}^2}}\left[\begin{array}{c}\mathrm{\Delta x}\\ \mathrm{\Delta y}\\ \mathrm{\Delta z}\end{array}\right]---\left(5\right)$式中$\left[\begin{array}{c}\mathrm{\Delta x}\\ \mathrm{\Delta y}\\ \mathrm{\Delta z}\end{array}\right]=\left[\begin{array}{c}{x}^{\left(s\right)}\\ y^{\left(s\right)}\\ z^{\left(s\right)}\end{array}\right]-\left[\begin{array}{c}x\\ y\\ z\end{array}\right],$为接收机到卫星的观测向量；其中，[x^(s),y^(s),z^(s)]^T为卫星的位置坐标，[x,y,z]^T为接收机的位置坐标；b.多普勒频移值与接收机运行速度和卫星运行速度之间的关系为：$f_d=\frac{(\overrightarrow{v}-\overrightarrow{v^{\left(s\right)}})\overrightarrow{l^{\left(s\right)}}}{\lambda }---\left(6\right)$式中为接收机运行速度，值为[v_x,v_y,v_z]^T；为接收机运行速度，值为λ为卫星信号的波长；将接收机位置坐标[x,y,z]^T，3颗可见卫星位置坐标[x⁽¹⁾,y⁽¹⁾,z⁽¹⁾]^T，[x⁽²⁾,y⁽²⁾,z⁽²⁾]^T，[x⁽³⁾,y⁽³⁾,z⁽³⁾]^T带入式(5)得：$\left\{\begin{array}{c}\overrightarrow{l^{\left(1\right)}}=\frac{1}{r_1}{[x^{\left(1\right)}-x,y^{\left(1\right)}-y,z^{\left(1\right)}-z]}^t\\ \overrightarrow{l^{\left(2\right)}}=\frac{1}{r_2}{[x^{\left(2\right)}-x,y^{\left(2\right)}-y,z^{\left(2\right)}-z]}^T\\ \overrightarrow{l^{\left(3\right)}}=\frac{1}{r_3}{[x^{\left(3\right)}-x,y^{\left(3\right)}-y,z^{\left(3\right)}-z]}^T\end{array}\right.---\left(7\right)$将式(1)带入式(6)得：$\left\{\begin{array}{c}\overrightarrow{l^{\left(1\right)}}=\frac{1}{r_1}{[x^{\left(1\right)}-x_0-v_x\mathrm{\Delta t},y^{\left(1\right)}-y_0-v_y\mathrm{\Delta t},z^{\left(1\right)}-z_0-v_z\mathrm{\Delta t}]}^T\\ \overrightarrow{l^{\left(2\right)}}=\frac{1}{r_2}{[x^{\left(2\right)}-x_0-v_x\mathrm{\Delta t},y^{\left(2\right)}-y_0-v_y\mathrm{\Delta t},z^{\left(2\right)}-z_0-v_z\mathrm{\Delta t}]}^T\\ \overrightarrow{l^{\left(3\right)}}=\frac{1}{r_3}{[x^{\left(3\right)}-x_0-v_x\mathrm{\Delta t},y^{\left(3\right)}-y_0-v_y\mathrm{\Delta t},z^{\left(3\right)}-z_0-v_z\mathrm{\Delta t}]}^T\end{array}\right.---\left(8\right)$将接收机运行速度[v_x,v_y,v_z]^T，3颗可见卫星的运行速度和式(8)带入式(6)得：$\left\{\begin{array}{c}{f}_{d1}·\lambda ·r_1=[v_x-v_x^{\left(1\right)},v_y-v_y^{\left(1\right)},v_z-v_z^{\left(1\right)}]{[x^{\left(1\right)}-x_0-v_x·\mathrm{\Delta t},y^{\left(1\right)}-y_0-v_y·\mathrm{\Delta t},z^{\left(1\right)}-z_0-v_z·\mathrm{\Delta t}]}^T\\ f_{d2}·\lambda ·r_2=[v_x-v_x^{\left(2\right)},v_y-v_y^{\left(2\right)},v_z-v_z^{\left(2\right)}]{[x^{\left(2\right)}-x_0-v_x·\mathrm{\Delta t},y^{\left(2\right)}-y_0-v_y·\mathrm{\Delta t},z^{\left(2\right)}-z_0-v_z·\mathrm{\Delta t}]}^T\\ f_{d3}·\lambda ·r_3=[v_x-v_x^{\left(3\right)},v_y-v_y^{\left(3\right)},v_z-v_z^{\left(3\right)}]{[x^{\left(3\right)}-x_0-v_x·\mathrm{\Delta t},y^{\left(3\right)}-y_0-v_y·\mathrm{\Delta t},z^{\left(3\right)}-z_0-v_z·\mathrm{\Delta t}]}^T\end{array}\right.---\left(9\right)$引进下述变量以简化上述公式：$\left\{\begin{array}{c}{a}_j=x^{\left(j\right)}-x_0+v_x^{\left(j\right)}\mathrm{\Delta t}\\ b_j=y^{\left(j\right)}-y_0+v_y^{\left(j\right)}\mathrm{\Delta t}\\ c_j=z^{\left(j\right)}-z_0+v_z^{\left(j\right)}\mathrm{\Delta t}\\ e_j=v_x^{\left(j\right)}{x}_{\mathrm{po}}-v_x^{\left(j\right)}{x}^{\left(j\right)}+v_y^{\left(j\right)}{y}_o-v_y^{\left(j\right)}{y}^{\left(j\right)}+v_z^{\left(j\right)}{z}_o-v_z^{\left(j\right)}{z}^{\left(j\right)}\end{array}\right.---\left(10\right)$定义以下矩阵以将式(9)写成矩阵形式：$A=\left[\begin{array}{c}{f}_{d1}·\lambda ·r_1\\ f_{d2}·\lambda ·r_2\\ f_{d3}·\lambda ·r_3\end{array}\right],B=\left[\begin{array}{ccccccc}-\mathrm{\Delta t}& -\mathrm{\Delta t}& -\mathrm{\Delta t}& a_1& b_1& c_1& e_1\\ -\mathrm{\Delta t}& -\mathrm{\Delta t}& -\mathrm{\Delta t}& a_2& b_2& c_2& e_2\\ -\mathrm{\Delta t}& -\mathrm{\Delta t}& -\mathrm{\Delta t}& a_3& b_3& c_3& e_3\end{array}\right],C=\left[\left[\begin{array}{c}{v}_x^2\\ v_y^2\\ v_z^2\\ v_x\\ v_y\\ v_z\\ 1\end{array}\right],\begin{array}{cccccc}& & & & & \end{array}\right]$得到：A＝BC   (11)通过矩阵变换将矩阵B变换为$B=\left[\begin{array}{ccccccc}-\mathrm{\Delta t}& -\mathrm{\Delta t}& -\mathrm{\Delta t}& a_1& b_1& c_1& e_1\\ -\mathrm{\Delta t}& -\mathrm{\Delta t}& -\mathrm{\Delta t}& a_2& b_2& c_2& e_2\\ -\mathrm{\Delta t}& -\mathrm{\Delta t}& -\mathrm{\Delta t}& a_3& b_3& c_3& e_3\end{array}\right]=\left[\begin{array}{ccccccc}0& 0& 0& a_1-a_2& b_1-b_2& c_1{-c}_2& e_1-e_2\\ 0& 0& 0& a_2-a_3& b_2-b_3& c_2-c_3& e_2-e_3\\ 0& 0& 0& a_1-a_3& b_1-b_3& c_1-c_3& e_1-e_3\end{array}\right]---\left(12\right)$通过以上矩阵变换将非线性方程组(18)线性化，消除定义以下矩阵：$B^{\prime }=\left[\begin{array}{cccc}{a}_1-a_2& b_1-b_2& c_1-c_2& e_1-e_2\\ a_2-a_3& b_2-b_3& c_2-c_3& e_2-e_3\\ a_1-c_3& b_1-b_3& c_1-c_3& e_1-e_3\end{array}\right],C^{\prime }=\left[\begin{array}{c}{v}_x\\ v_y\\ v_z\\ 1\end{array}\right]$经过矩阵变换后式(11)变换为：A＝B'C'   (13)将式(2)带入式(13)消除r₁，r₂，r₃得线性方程组：$\left\{\begin{array}{c}\frac{a_1-a_2}{f_{d1}·\lambda }{v}_x+\frac{b_1-b_2}{f_{d1}·\lambda }{v}_y+\frac{c_1-c_2}{f_{d1}·\lambda }{v}_z+\frac{e_1-e_2}{f_{d1}·\lambda }+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(1\right)}\\ \frac{a_2-a_3}{f_{d2}·\lambda }{v}_x+\frac{b_2-b_3}{f_{d2}·\lambda }{v}_y+\frac{c_2-c_3}{f_{d2}·\lambda }{v}_z+\frac{e_2-e_3}{f_{d2}·\lambda }+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(2\right)}\\ \frac{a_1-a_3}{f_{d3}·\lambda }{v}_x+\frac{b_1-b_3}{f_{d3}·\lambda }{v}_y+\frac{c_1-c_3}{f_{d3}·\lambda }{v}_z+\frac{e_1-e_3}{f_{d3}·\lambda }+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(3\right)}\end{array}\right.---\left(14\right)$方程组(2)和方程组(14)联立得定位解算的非线性方程组：$\left\{\begin{array}{c}\frac{a_1-a_2}{f_{d1}·\lambda }{v}_x+\frac{b_1-b_2}{f_{d1}·\lambda }{v}_y+\frac{c_1-c_2}{f_{d1}·\lambda }{v}_z+\frac{e_1-e_2}{f_{d1}·\lambda }+c·\delta t_u={\rho }_c^{\left(1\right)}z\\ \frac{a_2-a_3}{f_{d2}·\lambda }{v}_x+\frac{b_2-b_3}{f_{d2}·\lambda }{v}_y+\frac{c_2-c_3}{f_{d2}·\lambda }{v}_z+\frac{e_2-e_3}{f_{d2}·\lambda }+c·\delta t_u={\rho }_c^{\left(2\right)}\\ \frac{a_1-a_3}{f_{d3}·\lambda }{v}_x+\frac{b_1-b_3}{f_{d3}·\lambda }{v}_y+\frac{c_1-c_3}{f_{d3}·\lambda }{v}_z+\frac{e_1-e_3}{f_{d3}·\lambda }+c·\delta t_u={\rho }_c^{\left(3\right)}\\ \sqrt{{(x^{\left(1\right)}-x_0-v_x·\mathrm{\Delta t})}^2+{(y^{\left(1\right)}-y_0-v_y·\mathrm{\Delta t})}^2+{(z^{\left(1\right)}-z_0-v_z\mathrm{\Delta t})}^2}+c·{\mathrm{\delta t}}_u={\rho }_c^{\left(1\right)}\\ \sqrt{{(x^{\left(2\right)}-x_0-v_x·\mathrm{\Delta t})}^2+{(y^{\left(2\right)}-y_0-v_y·\mathrm{\Delta t})}^2+{(z^{\left(2\right)}-z_0-v_z·\mathrm{\Delta t})}^2}+c·\delta t_u={\rho }_c^{\left(2\right)}\\ \sqrt{{(x^{\left(3\right)}-x_0-v_x·\mathrm{\Delta t})}^2+{(y^{\left(3\right)}-y_0-v_y·\mathrm{\Delta t})}^2+{(z^{\left(3\right)}-z_0-v_z·\mathrm{\Delta t})}^2}+c·\delta t_u={\rho }_c^{\left(3\right)}\end{array}\right.----\left(15\right)$式(15)中未知量为：v_x,v_y,v_z和δt_u；步骤三：将非线性方程组(15)线性化并求解：将接收机的速度和接收机时钟偏差看做由近似分量和增量分量两部分组成，即：$\left\{\begin{array}{c}{v}_x={\hat{v}}_x+\Delta v_x\\ v_y={\hat{v}}_y+\Delta v_y\\ v_z={\hat{v}}_z+\Delta v_z\\ {\mathrm{\delta t}}_u=\delta {\hat{t}}_u+\Delta {\mathrm{\delta t}}_u\end{array}\right.---\left(16\right)$对联立方程组(15)进行泰勒展开截取一阶偏导数的项，得到线性方程组为：$\left\{\begin{array}{c}{\rho }_c^{\left(1\right)}={\hat{\rho }}_c^{\left(1\right)}-\frac{x^{\left(1\right)}-x_0-{\hat{v}}_x}{{\hat{r}}_1}\mathrm{\Delta t}·{\mathrm{\Delta v}}_x-\frac{y^{\left(1\right)}-y_0-{\hat{v}}_y}{{\hat{r}}_1}\mathrm{\Delta t}·{\mathrm{\Delta v}}_y-\frac{z^{\left(1\right)}-z_0-{\hat{v}}_z}{{\hat{r}}_1}\mathrm{\Delta t}·{\mathrm{\Delta v}}_z+c·\mathrm{\Delta \delta }{t}_u\\ {\rho }_c^{\left(2\right)}={\hat{\rho }}_c^{\left(2\right)}-\frac{x^{\left(2\right)}-x_0-{\hat{v}}_x}{{\hat{r}}_2}\mathrm{\Delta t}·{\mathrm{\Delta v}}_x-\frac{y^{\left(2\right)}-y_0-{\hat{v}}_y}{{\hat{r}}_2}\mathrm{\Delta t}·{\mathrm{\Delta v}}_y-\frac{z^{\left(2\right)}-z_0-{\hat{v}}_z}{{\hat{r}}_2}\mathrm{\Delta t}·{\mathrm{\Delta v}}_z+c·\mathrm{\Delta \delta }{t}_u\\ {\rho }_c^{\left(3\right)}={\hat{\rho }}_c^{\left(3\right)}-\frac{x^{\left(3\right)}-x_0-{\hat{v}}_x}{{\hat{r}}_3}\mathrm{\Delta t}·{\mathrm{\Delta v}}_x-\frac{y^{\left(3\right)}-y_0-{\hat{v}}_y}{{\hat{r}}_3}\mathrm{\Delta t}·{\mathrm{\Delta v}}_y-\frac{z^{\left(3\right)}-z_0-{\hat{v}}_z}{{\hat{r}}_3}\mathrm{\Delta t}·\Delta v_z+c·\mathrm{\Delta \delta }{t}_u\\ {\rho }_c^{\left(1\right)}={\hat{\rho }}_c^{\left(1\right)}+\frac{a_1-a_2}{f_{d1}·\lambda }\Delta v_x+\frac{b_1-b_2}{f_{d1}·\lambda }\Delta v_y+\frac{c_1-c_2}{f_{d1}·\lambda }\Delta v_z+c·\mathrm{\Delta \delta }{t}_u\\ {\rho }_c^{\left(2\right)}={\hat{\rho }}_c^{\left(2\right)}+\frac{a_2-a_3}{f_{d2}·\lambda }\Delta v_x+\frac{b_2-b_3}{f_{d2}·\lambda }{\mathrm{\Delta v}}_y+\frac{c_2-c_3}{f_{d2}·\lambda }\Delta v_z+c·\mathrm{\Delta \delta }{t}_u\\ {\rho }_c^{\left(3\right)}={\hat{\rho }}_c^{\left(3\right)}+\frac{a_1-a_3}{f_{d3}·\lambda }\Delta v_x+\frac{b_1-b_3}{f_{d3}·\lambda }\Delta v_y+\frac{c_1-c_3}{f_{d3}·\lambda }\Delta v_z+c·\mathrm{\Delta \delta }{t}_u\end{array}\right.---\left(17\right)$其中，${\hat{r}}_j=\sqrt{{(x^{\left(j\right)}-x_0-{\hat{v}}_x·\mathrm{\Delta t})}^2+{(y^{\left(j\right)}-y_0-{\hat{v}}_y·\mathrm{\Delta t})}^2+{(z^{\left(j\right)}-z_0-{\hat{v}}_z·\mathrm{\Delta t})}^2},$${\hat{\rho }}_c^{\left(j\right)}={\hat{r}}_j+c·\delta {\hat{t}}_u$定义下述变量以简化上述公式：$\left\{\begin{array}{c}\Delta {\rho }_c^{\left(j\right)}={\hat{\rho }}_c^{\left(j\right)}-{\rho }_c^{\left(j\right)}\\ {\alpha }_{j1}=\frac{x^{\left(j\right)}-x_0-{\hat{v}}_x}{{\hat{r}}_j}\mathrm{\Delta t}\\ {\alpha }_{j2}=\frac{y^{\left(j\right)}-y_0-{\hat{v}}_y}{{\hat{r}}_j}\mathrm{\Delta t}\\ {\alpha }_{j1}=\frac{z^{\left(j\right)}-z_0-{\hat{v}}_y}{{\hat{r}}_j}\mathrm{\Delta t}\\ {\beta }_j=-\frac{1}{f_{\mathrm{dj}}·\lambda }\end{array}\right.$式(17)简化为：$\left\{\begin{array}{c}\Delta {\rho }_c^{\left(1\right)}={\alpha }_{11}·{\mathrm{\Delta v}}_x+{\alpha }_{12}·\Delta v_y+{\alpha }_{13}·\Delta v_z-c·\mathrm{\Delta \delta }{t}_u\\ \Delta {\rho }_c^{\left(2\right)}={\alpha }_{21}·\Delta v_x+{\alpha }_{22}·{\mathrm{\Delta v}}_y+{\alpha }_{33}·\Delta v_z-c·\mathrm{\Delta \delta }{t}_u\\ \Delta {\rho }_c^{\left(3\right)}={\alpha }_{31}·\Delta v_x+{\alpha }_{32}·\Delta v_y+{\alpha }_{33}·\Delta v_z-c·\mathrm{\Delta \delta }{t}_u\\ \Delta {\rho }_c^{\left(1\right)}=(a_1-a_2){\beta }_1\Delta v_x+(b_1-b_2){\beta }_1\Delta v_y+(c_1-c_2){\beta }_1\Delta v_z-c·\mathrm{\Delta \delta }{t}_u\\ \Delta {\rho }_c^{\left(2\right)}=(a_2-a_3){\beta }_2\Delta v_x+(b_2-b_3){\beta }_2\Delta v_y+(c_2-c_3){\beta }_2\Delta v_z-c\surd \mathrm{\Delta \delta }{t}_u\\ \Delta {\rho }_c^{\left(3\right)}=(a_1-a_3){\beta }_3\Delta v_x+(b_1-b_3){\beta }_3\Delta v_y+(c_1-c_3){\beta }_3\Delta v_z-c·\mathrm{\Delta \delta }{t}_u\end{array}\right.---\left(18\right)$利用以下矩阵将上式写成矩阵形式：${\mathrm{\Delta \rho }}^{\prime }=\left[\begin{array}{c}\Delta {\rho }_c^{\left(1\right)}\\ \Delta {\rho }_c^{\left(2\right)}\\ \Delta {\rho }_c^{\left(3\right)}\\ \Delta {\rho }_c^{\left(1\right)}\\ \Delta {\rho }_c^{\left(2\right)}\\ \Delta {\rho }_c^{\left(3\right)}\end{array}\right],H^{\prime }=\left[\begin{array}{cccc}{\alpha }_{11}& {\alpha }_{12}& {\alpha }_{13}& 1\\ {\alpha }_{21}& {\alpha }_{22}& {\alpha }_{23}& 1\\ {\alpha }_{31}& {\alpha }_{32}& {\alpha }_{33}& 1\\ ({\alpha }_1-{\alpha }_2){\beta }_1& (b_1-b_2){\beta }_1& (c_1-c_2){\beta }_1& 1\\ ({\alpha }_2-{\alpha }_3){\beta }_2& (b_2-b_3){\beta }_2& (c_2-c_3){\beta }_2& 1\\ (a_1-a_3){\beta }_3& (b_1-b_3){\beta }_3& (c_1-c_3){\beta }_3& 1\end{array}\right],{\mathrm{\Delta u}}^{\prime }=\left[\begin{array}{c}{\mathrm{\Delta v}}_x\\ {\mathrm{\Delta v}}_y\\ {\mathrm{\Delta v}}_z\\ -c·\mathrm{\Delta \delta }{t}_u\end{array}\right]$最后得到：Δρ'＝H'Δu'   (19)它的解是Δu'＝H^'‑1Δρ'   (20)对上式采用最小二乘法进行求解就能得到接收机的速度v_x,v_y,v_z，将接收机运行速度带入式(1)即得到接收机此时位置坐标。