INS 时间更新

2023-12-17 20:27:32

4 时间更新

下式是位置、速度、姿态的误差方程式,详细的推导可参考严恭敏的《捷联惯导算法与组合导航原理讲义》
δ λ ˙ = v E s e c L t a n L R N h δ L ? v E s e c L R N h 2 δ h + s e c L R N h δ v E δ L ˙ = ? v N R M h 2 δ h + 1 R M h δ v N δ h ˙ = δ v U v ˙ E = [ 2 ( v N ω N + v U ω U ) + v E v N s e c 2 L R N h ] δ L + v E ( v U ? v N t a n L ) R N h 2 δ h + v N t a n L ? v U R N h δ v E + ( 2 ω U + v E t a n L R N h ) δ v N ? ( 2 ω N + v E R N h ) δ v U ? f U ? N + f N ? U + ? E v ˙ N = ? v E ( 2 ω N + v E s e c 2 L R N h ) δ L + ( v N v U R M h 2 + v E 2 t a n L R N h 2 ) δ h ? 2 ( ω U + v E t a n L R N h ) δ v E ? v U R M h δ v N ? v N R M h δ v U + f U ? E ? f E ? U + ? N v ˙ U = ? [ 2 ω U v E + g e s i n 2 L ( β ? 4 β 1 c o s 2 L ) ] δ L ? ( v E 2 R N h 2 + v N 2 R M h 2 ? β 2 ) δ h + ? f N ? E + f E ? N + 2 ( ω N + v E R N h ) δ v E + 2 v N R M h δ v N ? U ? ˙ E = v N R M h 2 δ h ? 1 R M h δ v N + ( ω U + v E t a n L R N h ) ? N ? ( ω N + v E R N h ) ? U ? ε E ? ˙ N = ? ω U δ L ? v E R M h 2 δ h + 1 R N h δ v E ? ( ω U + v E t a n L R N h ) ? E ? v N R M h ? U ? ε N ? ˙ U = ( ω N + v E s e c 2 L R N h ) δ L ? v E t a n L R N h 2 δ h + t a n L R N h δ v E + ( ω N + v E R N h ) ? E + v N R M h ? N ? ε U \begin{aligned} \delta \dot{\lambda} &= \frac{v_EsecLtanL}{R_{Nh}}\delta L - \frac{v_EsecL}{R^2_{Nh}}\delta h + \frac{secL}{R_{Nh}}\delta v_E\\ \delta \dot{L} &= - \frac{v_N}{R^2_{Mh}}\delta h + \frac{1}{R_{Mh}}\delta v_N\\ \delta \dot{h} &= \delta v_U\\ \dot{v}_E &=[2(v_N\omega_N + v_U\omega_U)+\frac{v_Ev_N sec^2L}{R_{Nh}}]\delta L +\frac{v_E(v_U-v_NtanL)}{R^2_{Nh}}\delta h+ \frac{v_NtanL - v_U}{R_{Nh}}\delta_{v_E} +(2\omega_U + \frac{v_E tanL}{R_{Nh}})\delta v_N -(2\omega_N + \frac{v_E}{R_{Nh}})\delta v_U\\ & -f_U\phi_N + f_N\phi_U +\nabla_E\\ \dot{v}_N &= -v_E(2\omega_N + \frac{v_E sec^2L}{R_{Nh}})\delta L +(\frac{v_Nv_U}{R^2_{Mh}}+\frac{v^2_EtanL}{R^2_{Nh}})\delta h - 2(\omega_U + \frac{v_E tanL}{R_{Nh}})\delta v_E -\frac{v_U}{R_{Mh}}\delta v_N - \frac{v_N}{R_{Mh}}\delta v_U + f_U\phi_E - f_E\phi_U\\ &+ \nabla_N\\ \dot{v}_U &= -[2\omega_Uv_E+g_esin2L(\beta-4\beta_1cos2L)]\delta L -(\frac{v^2_E}{R^2_{Nh}} + \frac{v^2_N}{R^2_{Mh}} - \beta_2)\delta_h+-f_N\phi_E + f_E\phi_N + 2(\omega_N + \frac{v_E}{R_{Nh}})\delta v_E +\frac{2v_N}{R_{Mh}}\delta v_N\\ &\nabla_U\\ \dot{\phi}_E &=\frac{v_N}{R_{Mh}^2}\delta h - \frac{1}{R_{Mh}}\delta v_N+ (\omega_U+\frac{v_E tanL}{R_{Nh}})\phi_N - (\omega_N + \frac{v_E}{R_{Nh}})\phi_U - \varepsilon_E\\ \dot{\phi}_N &= - \omega_U\delta L -\frac{v_E}{R^2_{Mh}}\delta h + \frac{1}{R_{Nh}}\delta v_E -(\omega_U+\frac{v_E tanL}{R_{Nh}})\phi_E -\frac{v_N}{R_{Mh}}\phi_U-\varepsilon_N\\ \dot{\phi}_U &= (\omega_N+\frac{v_E sec^2L}{R_{Nh}})\delta L -\frac{v_EtanL}{R^2_{Nh}}\delta h+ \frac{tanL}{R_{Nh}}\delta v_E + (\omega_N+\frac{v_E}{R_{Nh}})\phi_E +\frac{v_N}{R_{Mh}}\phi_N - \varepsilon_U\\ \end{aligned} δλ˙δL˙δh˙v˙E?v˙N?v˙U??˙?E??˙?N??˙?U??=RNh?vE?secLtanL?δL?RNh2?vE?secL?δh+RNh?secL?δvE?=?RMh2?vN??δh+RMh?1?δvN?=δvU?=[2(vN?ωN?+vU?ωU?)+RNh?vE?vN?sec2L?]δL+RNh2?vE?(vU??vN?tanL)?δh+RNh?vN?tanL?vU??δvE??+(2ωU?+RNh?vE?tanL?)δvN??(2ωN?+RNh?vE??)δvU??fU??N?+fN??U?+?E?=?vE?(2ωN?+RNh?vE?sec2L?)δL+(RMh2?vN?vU??+RNh2?vE2?tanL?)δh?2(ωU?+RNh?vE?tanL?)δvE??RMh?vU??δvN??RMh?vN??δvU?+fU??E??fE??U?+?N?=?[2ωU?vE?+ge?sin2L(β?4β1?cos2L)]δL?(RNh2?vE2??+RMh2?vN2???β2?)δh?+?fN??E?+fE??N?+2(ωN?+RNh?vE??)δvE?+RMh?2vN??δvN??U?=RMh2?vN??δh?RMh?1?δvN?+(ωU?+RNh?vE?tanL?)?N??(ωN?+RNh?vE??)?U??εE?=?ωU?δL?RMh2?vE??δh+RNh?1?δvE??(ωU?+RNh?vE?tanL?)?E??RMh?vN???U??εN?=(ωN?+RNh?vE?sec2L?)δL?RNh2?vE?tanL?δh+RNh?tanL?δvE?+(ωN?+RNh?vE??)?E?+RMh?vN???N??εU??
将上式采用矩阵的形式可以写作:
δ x ˙ ( t ) = F ( t ) δ x ( t ) \begin{aligned} \delta\dot{\mathbf{x}}(t) = \mathbf{F}(t)\delta \mathbf{x}(t) \end{aligned} δx˙(t)=F(t)δx(t)?

其中, F \mathbf{F} F的具体形式可以参见下表。

\ δ λ \delta\lambda δλ δ L \delta L δL δ h \delta h δh δ v E \delta v_E δvE? δ v N \delta v_N δvN? δ v U \delta v_U δvU? δ ? p i t c h \delta\phi_{pitch} δ?pitch? δ ? r o l l \delta\phi_{roll} δ?roll? δ ? y a w \delta\phi_{yaw} δ?yaw?
δ λ ˙ \delta\dot{\lambda} δλ˙ v E s e c L t a n L R N h \frac{v_EsecLtanL}{R_{Nh}} RNh?vE?secLtanL? ? v E s e c L R N h 2 -\frac{v_E secL}{R_{Nh}^{2}} ?RNh2?vE?secL? s e c L R N h \frac{secL}{R_{Nh}} RNh?secL?
δ L ˙ \delta\dot{L} δL˙ ? v n R M h 2 -\frac{v_n}{R_{Mh}^{2}} ?RMh2?vn?? 1 R M h \frac{1}{R_{Mh}} RMh?1?
δ h ˙ \delta\dot{h} δh˙1
δ v E ˙ \delta\dot{v_E} δvE?˙? 2 ( v N ω N + v U ω U ) + v E v N s e c 2 L R N h 2(v_N\omega_N+v_U\omega_U)+\frac{v_Ev_Nsec^2L}{R_{Nh}} 2(vN?ωN?+vU?ωU?)+RNh?vE?vN?sec2L? v E v U ? v E v n t a n λ R N h 2 \frac{v_Ev_U-v_Ev_ntan\lambda}{R_{Nh}^2} RNh2?vE?vU??vE?vn?tanλ? v N t a n L ? v U R N h \frac{v_NtanL-v_U}{R_{Nh}} RNh?vN?tanL?vU?? 2 ω U + v E t a n L R N h 2\omega_U+\frac{v_EtanL}{R_{Nh}} 2ωU?+RNh?vE?tanL? ? ( 2 ω N + v E R N h ) -(2\omega_N+\frac{v_E}{R_{Nh}}) ?(2ωN?+RNh?vE??) ? f U -f_U ?fU? f N f_N fN?
δ v N ˙ \delta\dot{v_N} δvN?˙? ? v E ( 2 ω N + v E s e c 2 L R N h ) -v_E(2\omega_N+\frac{v_Esec^2L}{R_{Nh}}) ?vE?(2ωN?+RNh?vE?sec2L?) v N v U R M h 2 + v E 2 t a n L R N h 2 \frac{v_Nv_U}{R_{Mh}^2}+\frac{v_E^2tanL}{R_{Nh}^2} RMh2?vN?vU??+RNh2?vE2?tanL? 2 ( ω U + v E t a n L R N h ) 2(\omega_U+\frac{v_EtanL}{R_{Nh}}) 2(ωU?+RNh?vE?tanL?) ? v U R M h -\frac{v_U}{R_{Mh}} ?RMh?vU?? ? v N R M h -\frac{v_N}{R_{Mh}} ?RMh?vN?? ? f U -f_U ?fU? ? f E -f_E ?fE?
δ v ˙ U \delta\dot{v}_U δv˙U? ? 2 ω U v E -2\omega_Uv_E ?2ωU?vE? ? ( v E 2 R N h 2 + v N 2 R M h 2 ) -(\frac{v_E^2}{R_{Nh}^2}+\frac{v_N^2}{R_{Mh}^2}) ?(RNh2?vE2??+RMh2?vN2??) 2 ( ω N + v E R N h ) 2(\omega_N+\frac{v_E}{R_{Nh}}) 2(ωN?+RNh?vE??) 2 v N R M h \frac{2v_N}{R_{Mh}} RMh?2vN?? ? f N -f_N ?fN? f E f_E fE?
δ ? ˙ E \delta\dot{\phi}_E δ?˙?E? v N R M h 2 \frac{v_N}{R_{Mh}^2} RMh2?vN?? ? 1 R M h -\frac{1}{R_{Mh}} ?RMh?1? ω U + v E t a n L R N h \omega_U+\frac{v_EtanL}{R_{Nh}} ωU?+RNh?vE?tanL? ? ( ω N + v E R N h ) -(\omega_N+\frac{v_E}{R_{Nh}}) ?(ωN?+RNh?vE??)
δ ? ˙ N \delta\dot{\phi}_N δ?˙?N? ω U \omega_U ωU? v E R M h 2 \frac{v_E}{R_{Mh}^2} RMh2?vE?? 1 R M h \frac{1}{R_{Mh}} RMh?1? ? ( ω U + v E t a n L R N h ) -(\omega_U+\frac{v_EtanL}{R_{Nh}}) ?(ωU?+RNh?vE?tanL?) ? v N R M h -\frac{v_N}{R_{Mh}} ?RMh?vN??
δ ? ˙ U \delta\dot{\phi}_U δ?˙?U? ω U + v E t a n L R N h \omega_U+\frac{v_EtanL}{R_{Nh}} ωU?+RNh?vE?tanL? v E t a n L R M h 2 \frac{v_EtanL}{R_{Mh}^2} RMh2?vE?tanL? t a n L R N h \frac{tanL}{R_{Nh}} RNh?tanL? ω N + v E R N h \omega_N+\frac{v_E}{R_{Nh}} ωN?+RNh?vE?? v N R M h \frac{v_N}{R_{Mh}} RMh?vN??1
续表 a b i a s x abias_x abiasx? a b i a s y abias_y abiasy? a b i a s z abias_z abiasz? g b i a s x gbias_x gbiasx? g b i a s y gbias_y gbiasy? g b i a s z gbias_z gbiasz?
δ λ ˙ \delta\dot{\lambda} δλ˙
δ L ˙ \delta\dot{L} δL˙
δ H ˙ \delta\dot{H} δH˙
δ v ˙ E \delta\dot{v}_E δv˙E? c 11 c_{11} c11? c 12 c_{12} c12? c 13 c_{13} c13?
δ v ˙ N \delta\dot{v}_N δv˙N? c 21 c_{21} c21? c 22 c_{22} c22? c 23 c_{23} c23?
δ v ˙ U \delta\dot{v}_U δv˙U? c 31 c_{31} c31? c 32 c_{32} c32? c 33 c_{33} c33?
δ ? ˙ E \delta\dot{\phi}_E δ?˙?E? ? c 11 -c_{11} ?c11? ? c 12 -c_{12} ?c12? ? c 13 -c_{13} ?c13?
δ ? ˙ N \delta\dot{\phi}_N δ?˙?N? ? c 21 -c_{21} ?c21? ? c 22 -c_{22} ?c22? ? c 23 -c_{23} ?c23?
δ ? ˙ U \delta\dot{\phi}_U δ?˙?U? ? c 31 -c_{31} ?c31? ? c 32 -c_{32} ?c32? ? c 33 -c_{33} ?c33?

进一步考虑加速度计和角速度计的测量噪声和加速度计零偏角速度计零偏噪声的随机模型,INS误差模型及传感器误差模型的联合形式可以进一步表达为:
δ x ˙ ( t ) = F ( t ) δ x ( t ) + G ( t ) w ( t ) \begin{aligned} \delta\dot{\mathbf{x}}(t) = \mathbf{F}(t)\delta \mathbf{x}(t) + \mathbf{G}(t)\mathbf{w}(t) \end{aligned} δx˙(t)=F(t)δx(t)+G(t)w(t)?
其中,噪声向量依次为加速度计测量噪声、角速度计测量噪声、加速度计零偏噪声,角速度计零偏噪声为:
w = [ w a T , w g T , w a b T , w g b T ] T \begin{aligned} \mathbf{w} = [\mathbf{w}^T_a, \mathbf{w}^T_g, \mathbf{w}^T_{ab}, \mathbf{w}^T_{gb}]^T \end{aligned} w=[waT?,wgT?,wabT?,wgbT?]T?
噪声向量的设计矩阵为:
G = [ 0 0 0 0 C b n 0 0 0 0 C b n 0 0 0 0 I 0 0 0 0 I ] \begin{aligned} \mathbf{G}=\begin{bmatrix} 0 & 0 & 0 & 0\\ \mathbf{C}_b^n & 0 & 0 & 0\\ 0 & \mathbf{C}_b^n & 0 & 0\\ 0 & 0 & \mathbf{I} & 0\\ 0 & 0 & 0 & \mathbf{I}\\ \end{bmatrix} \end{aligned} G= ?0Cbn?000?00Cbn?00?000I0?0000I? ??

w \mathbf{w} w为噪声向量。 w ( t ) \mathbf{w}(t) w(t)的元素为白噪声,其协方差阵为: E [ w ( t ) w ( τ ) T ] = Q ( t ) δ ( t ? τ ) \begin{aligned} E[\mathbf{w}(t)\mathbf{w}(\tau)^T] = \mathbf{Q}(t)\delta(t-\tau) \end{aligned} E[w(t)w(τ)T]=Q(t)δ(t?τ)?由于捷联惯导系统一般使用高频离散采样数据,因此需要将方程转化为离散的形式:
δ x ( t k + 1 ) = Φ ( t k + 1 , t k ) δ x ( t k ) + G ( τ ) w ( τ ) d τ \begin{aligned} \delta \mathbf{x}(t_{k+1}) = \Phi(t_{k+1},t_k)\delta \mathbf{x}(t_k) + \mathbf{G}(\tau)\mathbf{w}(\tau)d\tau \end{aligned} δx(tk+1?)=Φ(tk+1?,tk?)δx(tk?)+G(τ)w(τ)dτ? Φ k \Phi_k Φk?称为状态转移矩阵。

如果 △ t k + 1 = t k + 1 ? t k \triangle t_{k+1} = t_{k+1} -t_k tk+1?=tk+1??tk?很小,因此在这段时间内可以认为 F ( t ) F(t) F(t)为常量。转移矩阵的近似计算方法如下:
Φ k = e x p ( F ( t ) △ t k + 1 ) ≈ I + F ( t ) △ t k + 1 \Phi_k = exp(F(t)\triangle t_{k+1})\approx I+F(t)\triangle t_{k+1} Φk?=exp(F(t)tk+1?)I+F(t)tk+1?状态时间更新对应的协方差更新为:
Q δ x ( t k + 1 ) = Φ ( t k + 1 , t k ) Q δ x ( t k ) Φ T ( t k + 1 + G ( τ ) Q w ( τ ) d τ G T ( τ ) \begin{aligned} Q_{\delta \mathbf{x}(t_{k+1})} = \Phi(t_{k+1},t_k)Q_{\delta \mathbf{x}(t_k)}\Phi^T(t_{k+1} + \mathbf{G}(\tau)Q_{\mathbf{w}(\tau)d\tau}\mathbf{G}^T(\tau) \end{aligned} Qδx(tk+1?)?=Φ(tk+1?,tk?)Qδx(tk?)?ΦT(tk+1?+G(τ)Qw(τ)dτ?GT(τ)?使用式子(1.6)和(1.8)即可完成时间更新。

文章来源:https://blog.csdn.net/qq_40230900/article/details/135028785
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