集中式KF
综合考虑主导航系统及各子导航系统,取多传感器组合导航系统的状态方程为:
Xk+1=Φk+1,kXk+ΓkWk(1)
\boldsymbol{X}_{k+1}=\boldsymbol{\Phi}_{k+1, k}\boldsymbol{X}_{k}+\boldsymbol{\Gamma}_{k}\boldsymbol{W}_{k}\tag{1}
Xk+1=Φk+1,kXk+ΓkWk(1)
各子导航系统的量测方程可表示为:
Zk+1i=Hk+1iXk+1+Vk+1i(2)
\boldsymbol{Z}_{k+1}^{i}=\boldsymbol{H}_{k+1}^{i} \boldsymbol{X}_{k+1}+\boldsymbol{V}_{k+1}^{i}\tag{2}
Zk+1i=Hk+1iXk+1+Vk+1i(2)
其中,Zk+1i∈Rmi\boldsymbol{Z}_{k+1}^{i} \in \boldsymbol{R}^{m_{i}}Zk+1i∈Rmi为量测向量,Hk+1i\boldsymbol{H}_{k+1}^{i}Hk+1i为量测矩阵,Vk+1i∈Rmi\boldsymbol{V}_{k+1}^{i} \in \boldsymbol{R}^{m_{i}}Vk+1i∈Rmi为均值为零、相互独立的高斯序列, 且:
E{[WkVki][WlT(Vli)T]}=[Qkl00Rkli]δkl \begin{aligned} \mathrm{E}\left\{\left[\begin{array}{l} \boldsymbol{W}_{k} \\ \boldsymbol{V}_{k}^{i} \end{array}\right]\left[\begin{array}{ll} \boldsymbol{W}_{l}^{\mathrm{T}} & \left(\boldsymbol{V}_{l}^{i}\right)^{\mathrm{T}} \end{array}\right]\right\}=\left[\begin{array}{cc} \boldsymbol{Q}_{kl} & 0 \\ 0 & \boldsymbol{R}_{kl}^{i} \end{array}\right] \delta_{kl} \end{aligned} E{[WkVki][WlT(Vli)T]}=[Qkl00Rkli]δkl
于是集中式卡尔曼滤波器量测方程为:
Zk+1=Hk+1Xk+1+Vk+1(3) \boldsymbol{Z}_{k+1}=\boldsymbol{H}_{k+1} \boldsymbol{X}_{k+1}+\boldsymbol{V}_{k+1}\tag{3} Zk+1=Hk+1Xk+1+Vk+1(3)
其中,
Zk+1=[(Zk+11)T,(Zk+12)T,⋯ ,(Zk+1N)T]THk+1=[(Hk+11)T,(Hk+12)T,⋯ ,(Hk+1N)T]TVk+1=[(Vk+11)T,(Vk+12)T,⋯ ,(Vk+1N)T]T \begin{aligned} \boldsymbol{Z}_{k+1} & =\left[\left(\boldsymbol{Z}_{k+1}^{1}\right)^{\mathrm{T}},\left(\boldsymbol{Z}_{k+1}^{2}\right)^{\mathrm{T}}, \cdots,\left(\boldsymbol{Z}_{k+1}^{N}\right)^{\mathrm{T}}\right]^{\mathrm{T}} \\ \boldsymbol{H}_{k+1} & =\left[\left(\boldsymbol{H}_{k+1}^{1}\right)^{\mathrm{T}},\left(\boldsymbol{H}_{k+1}^{2}\right)^{\mathrm{T}}, \cdots,\left(\boldsymbol{H}_{k+1}^{N}\right)^{\mathrm{T}}\right]^{\mathrm{T}} \\ \boldsymbol{V}_{k+1} & =\left[\left(\boldsymbol{V}_{k+1}^{1}\right)^{\mathrm{T}},\left(\boldsymbol{V}_{k+1}^{2}\right)^{\mathrm{T}}, \cdots,\left(\boldsymbol{V}_{k+1}^{N}\right)^{\mathrm{T}}\right]^{\mathrm{T}} \end{aligned} Zk+1Hk+1Vk+1=[(Zk+11)T,(Zk+12)T,⋯,(Zk+1N)T]T=[(Hk+11)T,(Hk+12)T,⋯,(Hk+1N)T]T=[(Vk+11)T,(Vk+12)T,⋯,(Vk+1N)T]T
且
E{[WkVkX~0∣0][WlTVlTX~0∣0T]}=[Qklδkl000Rklδkl00P0∣0] \begin{aligned} \mathrm{E}\left\{\left[\begin{array}{c} \boldsymbol{W}_{k} \\ \boldsymbol{V}_{k} \\ \tilde{\boldsymbol{X}}_{{0 \mid 0}} \end{array}\right]\left[\begin{array}{lll} \boldsymbol{W}_{l}^{\mathrm{T}} & \boldsymbol{V}_{l}^{\mathrm{T}} & \tilde{\boldsymbol{X}}_{{0\mid 0 }}^{\mathrm{T}} \end{array}\right]\right\}=\left[\begin{array}{ccc} \boldsymbol{Q}_{kl} \delta_{k l} & 0 & 0 \\ 0 & \boldsymbol{R}_{kl} \delta_{k l} & \\ 0 & 0 & \boldsymbol{P}_{0\mid 0} \end{array}\right] \end{aligned} E⎩⎨⎧WkVkX~0∣0[WlTVlTX~0∣0T]⎭⎬⎫=Qklδkl000Rklδkl00P0∣0
其中,
Rk+1−1=diag{(Rk+11)−1,(Rk+12)−1,⋯ ,(Rk+1N)−1}(4)
\boldsymbol{R}_{k+1}^{-1}=\operatorname{diag}\left\{\left(\boldsymbol{R}_{k+1}^{1}\right)^{-1},\left(\boldsymbol{R}_{k+1}^{2}\right)^{-1}, \cdots,\left(\boldsymbol{R}_{k+1}^{N}\right)^{-1}\right\}\tag{4}
Rk+1−1=diag{(Rk+11)−1,(Rk+12)−1,⋯,(Rk+1N)−1}(4)
集中卡尔曼滤波器的状态估计方程为:
X^k+1∣k+1=X^k+1∣k+Kk+1[Zk+1−Hk+1X^k+1∣k](5)
\hat{\boldsymbol{X}}_{k+1 \mid k+1}=\hat{\boldsymbol{X}}_{k+1 \mid k}+\boldsymbol{K}_{k+1}\left[\boldsymbol{Z}_{k+1}-\boldsymbol{H}_{k+1} \hat{\boldsymbol{X}}_{k+1 \mid k}\right]\tag{5}
X^k+1∣k+1=X^k+1∣k+Kk+1[Zk+1−Hk+1X^k+1∣k](5)
X^k+1∣k=Φk+1,kX^k∣k(6) \hat{\boldsymbol{X}}_{k+1 \mid k}=\boldsymbol{\Phi}_{k+1, k} \hat{\boldsymbol{X}}_{k \mid k}\tag{6} X^k+1∣k=Φk+1,kX^k∣k(6)
Kk+1=Pk+1∣k+1Hk+1TRk+1−1=Pk+1∣k+1[(Hk+11)T(Rk+11)−1,⋯ ,(Hk+1N)T(Rk+1N)−1]=[Kk+11,⋯ ,Kk+1N](7) \begin{aligned} \boldsymbol{K}_{k+1}&=\boldsymbol{P}_{k+1 \mid k+1} \boldsymbol{H}_{k+1}^{\mathrm{T}} \boldsymbol{R}_{k+1}^{-1}\\ &=\boldsymbol{P}_{k+1 \mid k+1}\left[\left(\boldsymbol{H}_{k+1}^{1}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{1}\right)^{-1}, \cdots,\left(\boldsymbol{H}_{k+1}^{N}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{N}\right)^{-1}\right]\\ &=\left[\boldsymbol{K}_{k+1}^{1}, \cdots, \boldsymbol{K}_{k+1}^{N}\right] \end{aligned} \tag{7} Kk+1=Pk+1∣k+1Hk+1TRk+1−1=Pk+1∣k+1[(Hk+11)T(Rk+11)−1,⋯,(Hk+1N)T(Rk+1N)−1]=[Kk+11,⋯,Kk+1N](7)
Pk+1∣k+1−1=Pk+1∣k−1+Hk+1TRk+1−1Hk+1=Pk+1∣k−1+∑i=1N(Hk+1i)T(Rk+1i)−1Hk+1i=Pk+1∣k−1+∑i=1N[(Pk+1∣k+1i)−1−(Pk+1∣ki)−1](8) \begin{aligned} \boldsymbol{P}_{k+1 \mid k+1}^{-1}&=\boldsymbol{P}_{k+1 \mid k}^{-1}+\boldsymbol{H}_{k+1}^{\mathrm{T}} \boldsymbol{R}_{k+1}^{-1} \boldsymbol{H}_{k+1}\\ &=\boldsymbol{P}_{k+1 \mid k}^{-1}+\sum_{i=1}^{N}\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{H}_{k+1}^{i}\\ &=\boldsymbol{P}_{k+1 \mid k}^{-1}+\sum_{i=1}^{N}\left[\left(\boldsymbol{P}_{k+1 \mid k+1}^{i}\right)^{-1}-\left(\boldsymbol{P}_{k+1 \mid k}^{i}\right)^{-1}\right] \end{aligned} \tag{8} Pk+1∣k+1−1=Pk+1∣k−1+Hk+1TRk+1−1Hk+1=Pk+1∣k−1+i=1∑N(Hk+1i)T(Rk+1i)−1Hk+1i=Pk+1∣k−1+i=1∑N[(Pk+1∣k+1i)−1−(Pk+1∣ki)−1](8)
Pk+1∣k=Φk+1,kPk∣kΦk+1,kT+ΓkQkΓkT(9) \boldsymbol{P}_{k+1 \mid k}=\boldsymbol{\Phi}_{k+1, k} \boldsymbol{P}_{k \mid k} \boldsymbol{\Phi}_{k+1, k}^{\mathrm{T}}+\boldsymbol{\Gamma}_{k} \boldsymbol{Q}_{k} \boldsymbol{\Gamma}_{k}^\mathrm{T}\tag{9} Pk+1∣k=Φk+1,kPk∣kΦk+1,kT+ΓkQkΓkT(9)
其中, 式(8)中 Pk+1∣k+1\boldsymbol{P}_{k+1 \mid k+1}Pk+1∣k+1 也可表示为:
Pk+1∣k+1=[I−Kk+1Hk+1]Pk+1∣k(10) \boldsymbol{P}_{k+1 \mid k+1}=\left[\boldsymbol{I}-\boldsymbol{K}_{k+1} \boldsymbol{H}_{k+1}\right] \boldsymbol{P}_{k+1 \mid k}\tag{10} Pk+1∣k+1=[I−Kk+1Hk+1]Pk+1∣k(10)
综合以上, 得到多传感器系统的集中状态估计表达式为:
X^k+1∣k+1=X^k+1∣k+∑i=1NKk+1i[Zk+1i−Hk+1iX^k+1∣ki](11) \hat{\boldsymbol{X}}_{k+1 \mid k+1}=\hat{\boldsymbol{X}}_{k+1 \mid k}+\sum_{i=1}^{N} \boldsymbol{K}_{k+1}^{i}\left[\boldsymbol{Z}_{k+1}^{i}-\boldsymbol{H}_{k+1}^{i} \hat{\boldsymbol{X}}_{k+1 \mid k}^{i}\right]\tag{11} X^k+1∣k+1=X^k+1∣k+i=1∑NKk+1i[Zk+1i−Hk+1iX^k+1∣ki](11)
分散式KF的一种表达形式
根据式(1)和式(2), 与第iii个子导航传感器对应的卡尔曼滤波方程为:
X^k+1∣k+1i=X^k+1∣ki+Kk+1i[Zk+1i−Hk+1iX^k+1∣ki](12)
\hat{\boldsymbol{X}}_{k+1 \mid k+1}^{i}=\hat{\boldsymbol{X}}_{k+1 \mid k}^{i}+\boldsymbol{K}_{k+1}^{i}\left[\boldsymbol{Z}_{k+1}^{i}-\boldsymbol{H}_{k+1}^{i} \hat{\boldsymbol{X}}_{k+1 \mid k}^{i}\right]\tag{12}
X^k+1∣k+1i=X^k+1∣ki+Kk+1i[Zk+1i−Hk+1iX^k+1∣ki](12)
Kk+1i=Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1(13) \boldsymbol{K}_{k+1}^{i}=\boldsymbol{P}_{k+1 \mid k+1}^{i}\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1}\tag{13} Kk+1i=Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1(13)
(Pk+1∣k+1i)−1=(Pk+1∣ki)−1+(Hk+1i)T(Rk+1i)−1Hk+1i(14) \left(\boldsymbol{P}_{k+1 \mid k+1}^{i}\right)^{-1}=\left(\boldsymbol{P}_{k+1 \mid k}^{i}\right)^{-1}+\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{H}_{k+1}^{i}\tag{14} (Pk+1∣k+1i)−1=(Pk+1∣ki)−1+(Hk+1i)T(Rk+1i)−1Hk+1i(14)
X^k+1∣ki=Φk+1,kX^k∣ki(15) \hat{\boldsymbol{X}}_{k+1 \mid k}^{i}=\boldsymbol{\Phi}_{k+1, k} \hat{\boldsymbol{X}}_{k \mid k}^{i}\tag{15} X^k+1∣ki=Φk+1,kX^k∣ki(15)
Pk+1∣ki=Φk+1,kPk∣kiΦk+1,kT+ΓkQkΓkT(16) \boldsymbol{P}_{k+1 \mid k}^{i}=\boldsymbol{\Phi}_{k+1, k} \boldsymbol{P}_{k \mid k}^{i} \boldsymbol{\Phi}_{k+1, k}^{\mathrm{T}}+\boldsymbol{\Gamma}_{k} \boldsymbol{Q}_{k} \boldsymbol{\Gamma}_{k}^{T}\tag{16} Pk+1∣ki=Φk+1,kPk∣kiΦk+1,kT+ΓkQkΓkT(16)
展开式(5)右边, 合并后有:
X^k+1∣k+1=[I−Kk+1Hk+1]X^k+1∣k+Kk+1Zk+1(17)
\hat{\boldsymbol{X}}_{k+1 \mid k+1}=\left[\boldsymbol{I}-\boldsymbol{K}_{k+1} \boldsymbol{H}_{k+1}\right] \hat{\boldsymbol{X}}_{k+1 \mid k}+\boldsymbol{K}_{k+1} \boldsymbol{Z}_{k+1}\tag{17}
X^k+1∣k+1=[I−Kk+1Hk+1]X^k+1∣k+Kk+1Zk+1(17)
由式10可得:
I−Kk+1Hk+1=Pk+1∣k+1Pk+1∣k−1(18)
\boldsymbol{I}-\boldsymbol{K}_{k+1} \boldsymbol{H}_{k+1}= \boldsymbol{P}_{k+1 \mid k+1} \boldsymbol{P}_{k+1 \mid k}^{-1}\tag{18}
I−Kk+1Hk+1=Pk+1∣k+1Pk+1∣k−1(18)
从式(12)可推出:
X^k+1∣k+1i=[I−Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1Hk+1i]X^k+1∣ki+Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1Zk+1i(19) \begin{aligned} \hat{\boldsymbol{X}}_{k+1 \mid k+1}^{i}=&\left[\boldsymbol{I}-\boldsymbol{P}_{k+1 \mid k+1}^{i}\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{H}_{k+1}^{i}\right] \hat{\boldsymbol{X}}_{k+1 \mid k}^{i}\\ &+\boldsymbol{P}_{k+1 \mid k+1}^{i}\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{Z}_{k+1}^{i} \end{aligned} \tag{19} X^k+1∣k+1i=[I−Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1Hk+1i]X^k+1∣ki+Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1Zk+1i(19)
这里隐含假定所有出现的矩阵求逆都是存在的, 并且 P0\boldsymbol{P}_{0}P0为非奇异的。由式(7)可得:
Kk+1Zk+1=Pk+1∣k+1Hk+1TRk+1−1Zk+1=Pk+1∣k+1[(Hk+11)T,(Hk+12)T,⋯ ,(Hk+1N)T]⋅ diag{(Rk+11)−1,(Rk+12)−1,⋯ ,(Rk+1N)−1}⋅ [(Zk+11)T,(Zk+12)T,⋯ ,(Zk+1N)T]T=Pk+1∣k+1∑i=1N(Hk+1i)T(Rk+1i)−1Zk+1i(20) \begin{aligned} \boldsymbol{K}_{k+1} \boldsymbol{Z}_{k+1} & =\boldsymbol{P}_{k+1 \mid k+1} \boldsymbol{H}_{k+1}^{\mathrm{T}} \boldsymbol{R}_{k+1}^{-1} \boldsymbol{Z}_{k+1} \\ & =\boldsymbol{P}_{k+1 \mid k+1}\left[\left(\boldsymbol{H}_{k+1}^{1}\right)^{\mathrm{T}},\left(\boldsymbol{H}_{k+1}^{2}\right)^{\mathrm{T}}, \cdots,\left(\boldsymbol{H}_{k+1}^{N}\right)^{\mathrm{T}}\right] \cdot \\ &\ \ \ \ \operatorname{diag}\left\{\left(\boldsymbol{R}_{k+1}^{1}\right)^{-1},\left(\boldsymbol{R}_{k+1}^{2}\right)^{-1}, \cdots,\left(\boldsymbol{R}_{k+1}^{N}\right)^{-1}\right\} \cdot \\&\ \ \ \ \left[\left(\boldsymbol{Z}_{k+1}^{1}\right)^{\mathrm{T}},\left(\boldsymbol{Z}_{k+1}^{2}\right)^{\mathrm{T}}, \cdots,\left(\boldsymbol{Z}_{k+1}^{N}\right)^{\mathrm{T}}\right]^{\mathrm{T}} \\ & =\boldsymbol{P}_{k+1 \mid k+1} \sum_{i=1}^{N}\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{Z}_{k+1}^{i} \end{aligned} \tag{20} Kk+1Zk+1=Pk+1∣k+1Hk+1TRk+1−1Zk+1=Pk+1∣k+1[(Hk+11)T,(Hk+12)T,⋯,(Hk+1N)T]⋅ diag{(Rk+11)−1,(Rk+12)−1,⋯,(Rk+1N)−1}⋅ [(Zk+11)T,(Zk+12)T,⋯,(Zk+1N)T]T=Pk+1∣k+1i=1∑N(Hk+1i)T(Rk+1i)−1Zk+1i(20)
对于式(20)中 (Hk+1i)T(Rk+1i)−1Zk+1i\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{Z}_{k+1}^{i}(Hk+1i)T(Rk+1i)−1Zk+1i 部分, 利用式(19)可得:
(Hk+1i)T(Rk+1i)−1Zk+1i=(Pk+1∣k+1i)−1X^k+1∣k+1i−(Pk+1∣k+1i)−1⋅[I−Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1Hk+1i]X^k+1∣ki(21) \begin{aligned} \left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{Z}_{k+1}^{i}=&\left(\boldsymbol{P}_{k+1 \mid k+1}^{i}\right)^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k+1}^{i}-\left(\boldsymbol{P}_{k+1 \mid k+1}^{i}\right)^{-1} \cdot \\ &\left[\boldsymbol{I}-\boldsymbol{P}_{k+1 \mid k+1}^{i}\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{H}_{k+1}^{i}\right] \hat{\boldsymbol{X}}_{k+1\mid k}^{i} \end{aligned} \tag{21} (Hk+1i)T(Rk+1i)−1Zk+1i=(Pk+1∣k+1i)−1X^k+1∣k+1i−(Pk+1∣k+1i)−1⋅[I−Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1Hk+1i]X^k+1∣ki(21)
类似于式(18), 有:
I−Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1Hk+1i=Pk+1∣k+1i(Pk+1∣ki)−1(22) \boldsymbol{I}-\boldsymbol{P}_{k+1 \mid k+1}^{i}\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{H}_{k+1}^{i}=\boldsymbol{P}_{k+1 \mid k+1}^{i}\left(\boldsymbol{P}_{k+1 \mid k}^{i}\right)^{-1}\tag{22} I−Pk+1∣k+1i(Hk+1i)T(Rk+1i)−1Hk+1i=Pk+1∣k+1i(Pk+1∣ki)−1(22)
将式(22)代入式(21)有:
(Hk+1i)T(Rk+1i)−1Zk+1i=(Pk+1∣k+1i)−1X^k+1∣k+1i−(Pk+1∣ki)−1X^k+1∣ki(23) \left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{Z}_{k+1}^{i}=\left(\boldsymbol{P}_{k+1 \mid k+1}^{i}\right)^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k+1}^{i}-\left(\boldsymbol{P}_{k+1 \mid k}^{i}\right)^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k}^{i}\tag{23} (Hk+1i)T(Rk+1i)−1Zk+1i=(Pk+1∣k+1i)−1X^k+1∣k+1i−(Pk+1∣ki)−1X^k+1∣ki(23)
将式(18)、式(20)、式(23)代入式(17)有:
X^k+1∣k+1=Pk+1∣k+1{Pk+1∣k−1X^k+1∣k+∑i=1N[(Pk+1∣k+1i)−1X^k+1∣k+1i−(Pk+1∣ki)−1X^k+1∣ki]}(24) \hat{\boldsymbol{X}}_{k+1 \mid k+1}=\boldsymbol{P}_{k+1 \mid k+1}\left\{\boldsymbol{P}_{k+1 \mid k}^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k}+\sum_{i=1}^{N}\left[\left(\boldsymbol{P}_{k+1 \mid k+1}^{i}\right)^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k+1}^{i}-\left(\boldsymbol{P}_{k+1 \mid k}^{i}\right)^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k}^{i}\right]\right\}\tag{24} X^k+1∣k+1=Pk+1∣k+1{Pk+1∣k−1X^k+1∣k+i=1∑N[(Pk+1∣k+1i)−1X^k+1∣k+1i−(Pk+1∣ki)−1X^k+1∣ki]}(24)
为此, 当式(12)-(16)给出了子滤波器的状态估计时, 式(24)代表了NNN个子导航系统在融合中心的最优估计的表达形式。式(24)中 Pk+1∣k+1、Pk+1∣k、X^k+1∣k\boldsymbol{P}_{k+1 \mid k+1} 、 \boldsymbol{P}_{k+1 \mid k} 、 \hat{\boldsymbol{X}}_{k+1 \mid k}Pk+1∣k+1、Pk+1∣k、X^k+1∣k 分别由式(8)、式(9)和式(6)给出。
对于与第iii个子导航传感器对应的子滤波器iii , 如果在k+1k+1k+1时刻检测到故障发生, 则该子滤波器只进行时间更新, 即 X^k+1∣k+1i=X^k+1∣ki\hat{\boldsymbol{X}}_{k+1 \mid k+1}^{i}=\hat{\boldsymbol{X}}_{k+1 \mid k}^{i}X^k+1∣k+1i=X^k+1∣ki , Pk+1∣k+1i=Pk+1∣ki\boldsymbol{P}_{k+1 \mid k+1}^{i}=\boldsymbol{P}_{k+1 \mid k}^{i}Pk+1∣k+1i=Pk+1∣ki 。此时, 根据式(24), 第iii个子滤波器不参与系统级的信息融合, 进而对第iii个子滤波器进行隔离, 并自动对系统进行重组 。
以上分析说明, 式(24)的分散模型不仅具有集中KF的最优特性, 同时具有联邦KF较好的容错特性。
总结(分散式KF两种形式)
Pk+1∣k+1−1=Pk+1∣k−1+∑i=1N(Hk+1i)T(Rk+1i)−1Hk+1iX^k+1∣k+1=Pk+1∣k+1{Pk+1∣k−1X^k+1∣k+∑i=1N[(Hk+1i)T(Rk+1i)−1Zk+1i]}(25) \begin{aligned} \boldsymbol{P}_{k+1 \mid k+1}^{-1}&=\boldsymbol{P}_{k+1 \mid k}^{-1}+\sum_{i=1}^{N}\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{H}_{k+1}^{i}\\ \hat{\boldsymbol{X}}_{k+1 \mid k+1}&=\boldsymbol{P}_{k+1 \mid k+1}\left\{\boldsymbol{P}_{k+1 \mid k}^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k}+\sum_{i=1}^{N}\left[\left(\boldsymbol{H}_{k+1}^{i}\right)^{\mathrm{T}}\left(\boldsymbol{R}_{k+1}^{i}\right)^{-1} \boldsymbol{Z}_{k+1}^{i}\right]\right\} \end{aligned} \tag{25} Pk+1∣k+1−1X^k+1∣k+1=Pk+1∣k−1+i=1∑N(Hk+1i)T(Rk+1i)−1Hk+1i=Pk+1∣k+1{Pk+1∣k−1X^k+1∣k+i=1∑N[(Hk+1i)T(Rk+1i)−1Zk+1i]}(25)
Pk+1∣k+1−1=Pk+1∣k−1+∑i=1N[(Pk+1∣k+1i)−1−(Pk+1∣ki)−1]X^k+1∣k+1=Pk+1∣k+1{Pk+1∣k−1X^k+1∣k+∑i=1N[(Pk+1∣k+1i)−1X^k+1∣k+1i−(Pk+1∣ki)−1X^k+1∣ki]}(26) \begin{aligned} \boldsymbol{P}_{k+1 \mid k+1}^{-1}&=\boldsymbol{P}_{k+1 \mid k}^{-1}+\sum_{i=1}^{N}\left[\left(\boldsymbol{P}_{k+1 \mid k+1}^{i}\right)^{-1}-\left(\boldsymbol{P}_{k+1 \mid k}^{i}\right)^{-1}\right]\\ \hat{\boldsymbol{X}}_{k+1 \mid k+1}&=\boldsymbol{P}_{k+1 \mid k+1}\left\{\boldsymbol{P}_{k+1 \mid k}^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k}+\sum_{i=1}^{N}\left[\left(\boldsymbol{P}_{k+1 \mid k+1}^{i}\right)^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k+1}^{i}-\left(\boldsymbol{P}_{k+1 \mid k}^{i}\right)^{-1} \hat{\boldsymbol{X}}_{k+1 \mid k}^{i}\right]\right\} \end{aligned} \tag{26} Pk+1∣k+1−1X^k+1∣k+1=Pk+1∣k−1+i=1∑N[(Pk+1∣k+1i)−1−(Pk+1∣ki)−1]=Pk+1∣k+1{Pk+1∣k−1X^k+1∣k+i=1∑N[(Pk+1∣k+1i)−1X^k+1∣k+1i−(Pk+1∣ki)−1X^k+1∣ki]}(26)