函数模型
{ X k = Φ k / k − 1 X k − 1 + Γ k − 1 W k − 1 Z k = H k X k + V k \left\{\begin{array}{l} \boldsymbol{X}_{k}=\boldsymbol{\Phi}_{k / k-1} \boldsymbol{X}_{k-1}+\boldsymbol{\Gamma}_{k-1} \boldsymbol{W}_{k-1} \\ \boldsymbol{Z}_{k}=\boldsymbol{H}_{k} \boldsymbol{X}_{k}+\boldsymbol{V}_{k} \end{array}\right. {Xk=Φk/k−1Xk−1+Γk−1Wk−1Zk=HkXk+Vk
其中
[ Z k ( 1 ) Z k ( 2 ) ⋮ Z k ( N ) ] = [ H k ( 1 ) H k ( 2 ) ⋮ H k ( N ) ] X k + [ V k ( 1 ) V k ( 2 ) ⋮ V k ( N ) ] R k = [ R k ( 1 ) R k ( 2 ) ⋱ R k ( N ) ] \begin{array}{l} {\left[\begin{array}{c} \boldsymbol{Z}_{k}^{(1)} \\ \boldsymbol{Z}_{k}^{(2)} \\ \vdots \\ \boldsymbol{Z}_{k}^{(N)} \end{array}\right]=\left[\begin{array}{c} \boldsymbol{H}_{k}^{(1)} \\ \boldsymbol{H}_{k}^{(2)} \\ \vdots \\ \boldsymbol{H}_{k}^{(N)} \end{array}\right] \boldsymbol{X}_{k}+\left[\begin{array}{c} \boldsymbol{V}_{k}^{(1)} \\ \boldsymbol{V}_{k}^{(2)} \\ \vdots \\ \boldsymbol{V}_{k}^{(N)} \end{array}\right]} \\ \boldsymbol{R}_{k}=\left[\begin{array}{cccc} \boldsymbol{R}_{k}^{(1)} & & & \\ & \boldsymbol{R}_{k}^{(2)} & & \\ & & \ddots & \\ & & & \boldsymbol{R}_{k}^{(N)} \end{array}\right] \end{array} Zk(1)Zk(2)⋮Zk(N) = Hk(1)Hk(2)⋮Hk(N) Xk+ Vk(1)Vk(2)⋮Vk(N) Rk= Rk(1)Rk(2)⋱Rk(N)
序贯滤波执行框图:
如果历元间量测噪声相关,量测噪声不是对角阵,它总可以作平方根分解(Cholesky分解)
R k = L k L k T \boldsymbol{R}_{k}=\boldsymbol{L}_{k} \boldsymbol{L}_{k}^{\mathrm{T}} Rk=LkLkT
使用 L k − 1 \boldsymbol{L}_{k}^{-1} Lk−1 同时左乘量测方程两边, 得:
L k − 1 Z k = L k − 1 H k X k + L k − 1 V k \boldsymbol{L}_{k}^{-1} \boldsymbol{Z}_{k}=\boldsymbol{L}_{k}^{-1} \boldsymbol{H}_{k} \boldsymbol{X}_{k}+\boldsymbol{L}_{k}^{-1} \boldsymbol{V}_{k} Lk−1Zk=Lk−1HkXk+Lk−1Vk
得到新的量测方程:
Z k ∗ = H k ∗ X k + V k ∗ \boldsymbol{Z}_{k}^{*}=\boldsymbol{H}_{k}^{*} \boldsymbol{X}_{k}+\boldsymbol{V}_{k}^{*} Zk∗=Hk∗Xk+Vk∗
其中:
Z k ∗ = L k − 1 Z k , H k ∗ = L k − 1 H k , V k ∗ = L k − 1 V k \quad \boldsymbol{Z}_{k}^{*}=\boldsymbol{L}_{k}^{-1} \boldsymbol{Z}_{k}, \quad \boldsymbol{H}_{k}^{*}=\boldsymbol{L}_{k}^{-1} \boldsymbol{H}_{k}, \quad \boldsymbol{V}_{k}^{*}=\boldsymbol{L}_{k}^{-1} \boldsymbol{V}_{k} Zk∗=Lk−1Zk,Hk∗=Lk−1Hk,Vk∗=Lk−1Vk
从而可以将量测方差阵对角化(单位阵!):
R k ∗ = E [ V k ∗ ( V k ∗ ) T ] = E [ ( L k − 1 V k ) ( L k − 1 V k ) T ] = L k − 1 E [ V k V k T ] ( L k − 1 ) T = L k − 1 R k ( L k − 1 ) T = I \begin{aligned} \boldsymbol{R}_{k}^{*} & =\mathrm{E}\left[\boldsymbol{V}_{k}^{*}\left(\boldsymbol{V}_{k}^{*}\right)^{\mathrm{T}}\right]=\mathrm{E}\left[\left(\boldsymbol{L}_{k}^{-1} \boldsymbol{V}_{k}\right)\left(\boldsymbol{L}_{k}^{-1} \boldsymbol{V}_{k}\right)^{\mathrm{T}}\right] \\ & =\boldsymbol{L}_{k}^{-1} \mathrm{E}\left[\boldsymbol{V}_{k} \boldsymbol{V}_{k}^{\mathrm{T}}\right]\left(\boldsymbol{L}_{k}^{-1}\right)^{\mathrm{T}}=\boldsymbol{L}_{k}^{-1} \boldsymbol{R}_{k}\left(\boldsymbol{L}_{k}^{-1}\right)^{\mathrm{T}}=\boldsymbol{I} \end{aligned} Rk∗=E[Vk∗(Vk∗)T]=E[(Lk−1Vk)(Lk−1Vk)T]=Lk−1E[VkVkT](Lk−1)T=Lk−1Rk(Lk−1)T=I
满足序贯滤波模型 { X k = Φ k l k − 1 X k − 1 + Γ k − 1 W k − 1 Z k ∗ = H k ∗ X k + V k ∗ \left\{\begin{array}{l}\boldsymbol{X}_{k}=\boldsymbol{\Phi}_{k l k-1} \boldsymbol{X}_{k-1}+\boldsymbol{\Gamma}_{k-1} \boldsymbol{W}_{k-1} \\ \boldsymbol{Z}_{k}^{*}=\boldsymbol{H}_{k}^{*} \boldsymbol{X}_{k}+\boldsymbol{V}_{k}^{*}\end{array}\right. {Xk=Φklk−1Xk−1+Γk−1Wk−1Zk∗=Hk∗Xk+Vk∗