卡尔曼滤波器原理By_DR_CAN 学习笔记

发布时间:2024年01月23日

DR_CAN卡尔曼滤波器

Kalman Filter
Optimal Recursive Data Processing Algorithm

不确定性:
1、不存在完美的数学模型
2、系统的扰动不可控,也很难建模
3、测量传感器存在误差

Recursive Algorithm

第 K 次测量结果 z k z_k zk?
通过样本均值来估计真实值
估计值 x ^ k = 1 k ( z 1 + z 2 + . . . + z k ) = 1 k ( z 1 + z 2 + . . . + z k ? 1 ) + 1 k z k = k ? 1 k x ^ k ? 1 + 1 k z k = x ^ k ? 1 + 1 k ( z k ? x ^ k ? 1 ) \hat x_k = \frac{1}{k}(z_1+z_2+...+z_k) = \frac{1}{k}(z_1+z_2+...+z_{k-1})+\frac{1}{k}z_k = \frac{k-1}{k}\hat x_{k-1}+\frac{1}{k}z_k=\hat x_{k-1}+\frac{1}{k}(z_k-\hat x_{k-1}) x^k?=k1?(z1?+z2?+...+zk?)=k1?(z1?+z2?+...+zk?1?)+k1?zk?=kk?1?x^k?1?+k1?zk?=x^k?1?+k1?(zk??x^k?1?)

k → ∞ , 1 k → 0 , x ^ k = x ^ k ? 1 k \to \infty,\frac{1}{k} \to 0,\hat x_k=\hat x_{k-1} k,k1?0,x^k?=x^k?1? 测量结果不重要
k → 0 , 1 k → ∞ k \to 0,\frac{1}{k} \to \infty k0,k1? 测量值 z k z_k zk?作用较大

k k = 1 k k_k=\frac{1}{k} kk?=k1?,其中 k k k_k kk?为Kalman Gain
x ^ k = x ^ k ? 1 + k k ( z k ? x ^ k ? 1 ) \hat x_k = \hat x_{k-1}+k_k(z_k-\hat x_{k-1}) x^k?=x^k?1?+kk?(zk??x^k?1?)

当前估计值 = 上次估计值 + 系数 *(当前测量值-上次估计值)
估计误差 e E S T e_{EST} eEST?
测量误差 e M E A e_{MEA} eMEA?
Klaman Gain k k = e E S T k ? 1 e E S T k ? 1 + e M E A k k_k = \frac{e_{EST_{k-1}}}{e_{EST_{k-1}}+e_{MEA_k}} kk?=eESTk?1??+eMEAk??eESTk?1???

e E S T k ? 1 ? e M E A k , k k → 1 , x ^ k = z k e_{EST_{k-1}} \gg e_{MEA_k}, k_k\to1, \hat x_k=z_k eESTk?1???eMEAk??,kk?1,x^k?=zk? 估计误差大,估计值取测量值
e E S T k ? 1 ? e M E A k , k k → 0 , x ^ k = x ^ k ? 1 e_{EST_{k-1}} \ll e_{MEA_k}, k_k\to0, \hat x_k=\hat x_{k-1} eESTk?1???eMEAk??,kk?0,x^k?=x^k?1? 测量误差大,估计值取上次估计值

迭代过程

STEP1:计算 Kalman Gain k k = e E S T k ? 1 e E S T k ? 1 + e M E A k k_k = \frac{e_{EST_{k-1}}}{e_{EST_{k-1}}+e_{MEA_k}} kk?=eESTk?1??+eMEAk??eESTk?1???
STEP2:计算 x ^ k = x ^ k ? 1 + k k ( z k ? x ^ k ? 1 ) \hat x_k = \hat x_{k-1} + k_k(z_k-\hat x_{k-1}) x^k?=x^k?1?+kk?(zk??x^k?1?)
STEP3:更新 e E S T k = ( 1 ? k k ) e E S T k ? 1 e_{EST_{k}}=(1-k_k)e_{EST_{k-1}} eESTk??=(1?kk?)eESTk?1??

数学基础

正态分布和6-Sigma

通过测量获取多组数据,用样本均值代替期望,样本方差代替方差

样本方差 S 2 = 1 n ∑ i = 1 n ( x i ? x ^ ) 2 S^2 = \frac{1}{n}\sum_{i=1}^{n}{(x_i-{\hat x})^2} S2=n1?i=1n?(xi??x^)2
样本标准差 S = 1 n ∑ i = 1 n ( x i ? x ^ ) 2 S= \sqrt{ \frac{1}{n}\sum_{i=1}^{n}{(x_i-{\hat x})^2} } S=n1?i=1n?(xi??x^)2 ?

理论正态分布
在这里插入图片描述

实际修正后的对应表格

σ \sigma σ产出百分比瑕疵百分比
393.3 %6.7%
599.977%0.023 %
699.99966%0.00034%

生产设备

每日生产量3 σ \sigma σ5 σ \sigma σ6 σ \sigma σ
2001天13次21天1次4年1次

false dismissal 漏检
将某些不合格视为合格

false alarm 假警报
将某些合格视为不合格

在这里插入图片描述

Data Fusion

Measurement: z 1 = 30 g z_1=30g z1?=30g Standard Deviation: σ 1 = 2 g \sigma_1=2g σ1?=2g
Measurement: z 2 = 32 g z_2=32g z2?=32g Standard Deviation: σ 1 = 4 g \sigma_1=4g σ1?=4g
两者均遵从 Natural Distribution

估计真实值 z ^ \hat z z^
z ^ = z 1 + k ( z 2 ? z 1 ) , k ∈ [ 0 , 1 ] \hat z=z_1+k(z_2-z_1),k \in[0, 1] z^=z1?+k(z2??z1?)k[0,1]
求 k 使得 σ z ^ \sigma_{\hat z} σz^?最小
σ z ^ 2 = V a r ( z 1 + k ( z 2 ? z 1 ) ) = V a r [ ( 1 ? k ) z 1 + k z 2 ] = V a r [ ( 1 ? k ) z 1 ] + V a r ( k z 2 ) = ( 1 ? k ) 2 σ 1 2 + k 2 σ 2 2 \sigma_{\hat z}^2=Var(z_1+k(z_2-z_1)) = Var[(1-k)z_1+kz_2]=Var[(1-k)z_1]+Var(kz_2)=(1-k)^2\sigma_1^2+k^2\sigma_2^2 σz^2?=Var(z1?+k(z2??z1?))=Var[(1?k)z1?+kz2?]=Var[(1?k)z1?]+Var(kz2?)=(1?k)2σ12?+k2σ22?

d σ z ^ 2 d k = ? 2 ( 1 ? k ) σ 1 2 + 2 k σ 2 2 = 0 \frac{d\sigma_{\hat z}^2}{dk}=-2(1-k)\sigma_1^2+2k\sigma_2^2=0 dkdσz^2??=?2(1?k)σ12?+2kσ22?=0
k = σ 1 2 σ 1 2 + σ 2 2 k=\frac{\sigma_1^2}{\sigma_1^2+\sigma_2^2} k=σ12?+σ22?σ12??

将数据带入
k = σ 1 2 σ 1 2 + σ 2 2 = 4 4 + 16 = 0.2 k=\frac{\sigma_1^2}{\sigma_1^2+\sigma_2^2}=\frac{4}{4+16}=0.2 k=σ12?+σ22?σ12??=4+164?=0.2
z ^ = 30 + 0.2 ? ( 32 ? 30 ) = 30.4 \hat z=30+0.2*(32-30)=30.4 z^=30+0.2?(32?30)=30.4 最优解
σ z ^ 2 = 0. 8 2 ? 4 + 0. 2 2 ? 16 = 3.2 \sigma_{\hat z}^2=0.8^2*4+0.2^2*16=3.2 σz^2?=0.82?4+0.22?16=3.2

Covariance Matrix

将方差、协方差在一个矩阵中表现出来,体现变量间的联动关系

球员身高(x)体重(y)年龄(z)
瓦尔迪1797433
奥巴梅扬1878031
萨拉赫1757128
平均180.37530.7

方差 σ x 2 = 1 3 [ ( 179 ? 180.3 ) 2 + ( 187 ? 180.3 ) 2 + ( 175 ? 180.3 ) 2 ] = 24.89 \sigma_x^2=\frac{1}{3}[(179-180.3)^2+(187-180.3)^2+(175-180.3)^2]=24.89 σx2?=31?[(179?180.3)2+(187?180.3)2+(175?180.3)2]=24.89
σ y 2 = 14 \sigma_y^2 = 14 σy2?=14
σ z 2 = 4.22 \sigma_z^2=4.22 σz2?=4.22

协方差 σ x σ y = 1 3 [ ( 179 ? 180.3 ) ( 74 ? 75 ) + ( 187 ? 180.3 ) ( 80 ? 75 ) + ( 175 ? 180.3 ) ( 71 ? 75 ) ] = 18.7 = σ y σ x \sigma_x \sigma_y=\frac{1}{3}[(179-180.3)(74-75)+(187-180.3)(80-75)+(175-180.3)(71-75)]=18.7=\sigma_y\sigma_x σx?σy?=31?[(179?180.3)(74?75)+(187?180.3)(80?75)+(175?180.3)(71?75)]=18.7=σy?σx?
σ x σ z = 4.4 = σ z σ x \sigma_x\sigma_z=4.4=\sigma_z\sigma_x σx?σz?=4.4=σz?σx?
σ y σ z = 3.3 = σ z σ y \sigma_y\sigma_z=3.3=\sigma_z\sigma_y σy?σz?=3.3=σz?σy?

Covariance matrix P = [ σ x 2 σ x σ y σ x σ z σ y σ x σ y 2 σ y σ z σ z σ x σ z σ y σ z 2 ] P= \left[ \begin{matrix} \sigma_x^2 & \sigma_x \sigma_y & \sigma_x \sigma_z \\ \sigma_y \sigma_x & \sigma_y^2 & \sigma_y \sigma_z \\ \sigma_z\sigma_x & \sigma_z\sigma_y & \sigma_z^2 \end{matrix} \right] P= ?σx2?σy?σx?σz?σx??σx?σy?σy2?σz?σy??σx?σz?σy?σz?σz2?? ?

过度矩阵 a = [ x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3 ] ? 1 3 [ 1 1 1 1 1 1 1 1 1 ] [ x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3 ] 过度矩阵 a=\left[\begin{matrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{matrix}\right] - \frac{1}{3} \left[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}\right] \left[\begin{matrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{matrix}\right] 过度矩阵a= ?x1?x2?x3??y1?y2?y3??z1?z2?z3?? ??31? ?111?111?111? ? ?x1?x2?x3??y1?y2?y3??z1?z2?z3?? ?

P = 1 3 a T a P=\frac{1}{3}a^Ta P=31?aTa

σ x σ y \sigma_x \sigma_y σx?σy?表示协方差,不是两个标准差相乘

State Space Representation

在这里插入图片描述
m x ¨ = F ? k x ? b x ˙ m \ddot x=F-kx-b\dot x mx¨=F?kx?bx˙
F 为 Input,记为u
m x ¨ = u ? k x ? b x ˙ m \ddot x=u-kx-b\dot x mx¨=u?kx?bx˙

取 State Variable
x 1 = x x_1=x x1?=x
x 2 = x ˙ x_2=\dot x x2?=x˙

x ˙ 1 = x ˙ = x 2 \dot x_1 = \dot x = x_2 x˙1?=x˙=x2?
x ˙ 2 = x ¨ = 1 m u ? k m x ? b m x ˙ = 1 m u ? k m x 1 ? b m x 2 \dot x_2 = \ddot x=\frac{1}{m}u-\frac{k}{m}x-\frac{b}{m}\dot x=\frac{1}{m}u-\frac{k}{m}x_1-\frac{b}{m}x_2 x˙2?=x¨=m1?u?mk?x?mb?x˙=m1?u?mk?x1??mb?x2?

( x ˙ 1 x ˙ 2 ) = ( 0 1 ? k m ? b m ) ( x 1 x 2 ) + ( 0 1 m ) u \left(\begin{matrix} \dot x_1 \\ \dot x_2 \end{matrix}\right) = \left(\begin{matrix} 0 & 1 \\ -\frac{k}{m} & -\frac{b}{m} \end{matrix}\right) \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right)+ \left(\begin{matrix} 0 \\ \frac{1}{m} \end{matrix}\right)u (x˙1?x˙2??)=(0?mk??1?mb??)(x1?x2??)+(0m1??)u

记作 随时间变化 X ˙ ( t ) = A X ( t ) + B U ( t ) \dot X(t) = AX(t) + BU(t) X˙(t)=AX(t)+BU(t)

测量量Measurement
z 1 = x = x 1 位置 z_1=x=x_1 位置 z1?=x=x1?位置
z 2 = x ˙ = x 2 速度 z_2=\dot x=x_2 速度 z2?=x˙=x2?速度

( z 1 z 2 ) = ( 1 0 0 1 ) ( x 1 x 2 ) \left(\begin{matrix} z_1 \\ z_2 \end{matrix}\right) = \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) (z1?z2??)=(10?01?)(x1?x2??)

记作 Z(t) = HX(t)

整个过程充满不确定性,故状态空间方程离散化表达式为
计算值/估计值 X k = A X k ? 1 + B U k + W k ? 1 X_k = AX_{k-1}+BU_k+W_{k-1} Xk?=AXk?1?+BUk?+Wk?1?
测量值 Z k = H X k + V k Z_k=HX_k + V_k Zk?=HXk?+Vk?

W k ? 1 W_{k-1} Wk?1? 记作过程噪音 Process Noise
V k V_k Vk? 记作测量噪音 Measurement Noise

可通过前面介绍的数据融合,将估计值与测量值融合,得到更加可信赖的目标估计值

离散化推导

通过欧拉法(前向差分)将状态空间方程离散化
X ˙ = X k + 1 ? X k T = A X k + B U k \dot X = \frac{X_{k+1}-X_{k}}{T}=AX_k+BU_k X˙=TXk+1??Xk??=AXk?+BUk?
X k + 1 = ( T A + I ) X k + T B U k X_{k+1} = (TA+I)X_{k}+TBU_{k} Xk+1?=(TA+I)Xk?+TBUk?
记作 X k + 1 = Φ X k + G U k X_{k+1} = \Phi X_{k} + GU_{k} Xk+1?=ΦXk?+GUk?

X k = Φ X k ? 1 + G U k ? 1 + W k ? 1 X_k = \Phi X_{k-1} + GU_{k-1}+W_{k-1} Xk?=ΦXk?1?+GUk?1?+Wk?1?

linearization

Taylor Series

线性系统 linear system ? \Leftrightarrow ? 叠加原理 superposition

x ˙ = f ( x ) \dot x =f(x) x˙=f(x)
x 1 , x 2 x_1, x_2 x1?,x2?是解
x 3 = k 1 x 1 + k 2 x 2 ( k 1 , k 2 为 c o n s t a n t ) x_3=k_1x_1+k_2x_2(k_1,k_2为constant) x3?=k1?x1?+k2?x2?(k1?,k2?constant)
x 3 x_3 x3?是解

线性化:Taylor Series
是在某一点附近的线性化,而不是全局线性化

f ( x ) = f ( x 0 ) + f ′ ( x 0 ) 1 ! ( x ? x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x ? x 0 ) 2 + . . . + f ( n ) ( x 0 ) n ! ( x ? x 0 ) n f(x)=f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+...+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n f(x)=f(x0?)+1!f(x0?)?(x?x0?)+2!f′′(x0?)?(x?x0?)2+...+n!f(n)(x0?)?(x?x0?)n
x ? x 0 → 0 x-x_0 \rightarrow 0 x?x0?0,则 ( x ? x 0 ) 2 → 0 (x-x_0)^2 \rightarrow 0 (x?x0?)20 ( x ? x 0 ) n → 0 (x-x_0)^n \rightarrow 0 (x?x0?)n0
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x ? x 0 ) f(x) = f(x_0) + f'(x_0)(x-x_0) f(x)=f(x0?)+f(x0?)(x?x0?) f ( x ) = k 2 x + b f(x) = k_2x+b f(x)=k2?x+b

平衡点
x ¨ + x ˙ + 1 x = 1 \ddot x + \dot x + \frac{1}{x} =1 x¨+x˙+x1?=1 在平衡点(Fixed Point)附近线性化
x ¨ = x ˙ = 0 \ddot x=\dot x=0 x¨=x˙=0 则平衡点 x 0 = 1 则平衡点x_0=1 则平衡点x0?=1
around x 0 : x δ = x 0 + x d x_0:x_\delta = x_0+x_d x0?xδ?=x0?+xd?
x ¨ δ + x ˙ δ + 1 x δ = 1 \ddot x_\delta+\dot x_\delta+\frac{1}{x_\delta}=1 x¨δ?+x˙δ?+xδ?1?=1
1 x δ 的线性化 1 x δ = 1 x 0 ? 1 x 0 2 ( x δ ? x 0 ) = 1 ? x d \frac{1}{x_\delta}的线性化\frac{1}{x_\delta}=\frac{1}{x_0}-\frac{1}{x_0^2}(x_\delta-x_0)=1-x_d xδ?1?的线性化xδ?1?=x0?1??x02?1?(xδ??x0?)=1?xd?

x ¨ d + x ˙ d + 1 ? x d = 1 \ddot x_d+\dot x_d+1-x_d=1 x¨d?+x˙d?+1?xd?=1
x ¨ d + x ˙ d ? x d = 0 \ddot x_d+\dot x_d-x_d=0 x¨d?+x˙d??xd?=0

2-D

x ˙ 1 = f 1 ( x 1 , x 2 ) \dot x_1=f_1(x_1,x_2) x˙1?=f1?(x1?,x2?)
x ˙ 2 = f 2 ( x 1 , x 2 ) \dot x_2=f_2(x_1,x_2) x˙2?=f2?(x1?,x2?)

[ x ˙ 1 d x ˙ 2 d ] = [ ? f 1 ? x 1 ? f 1 ? x 2 ? f 2 ? x 1 ? f 2 ? x 2 ] ∣ x = x 0 [ x 1 d x 2 d ] \left[\begin{matrix}\dot x_{1d} \\ \dot x_{2d}\end{matrix}\right]= \left[\begin{matrix}\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2}\\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{matrix}\right]_{|x=x_0} \left[\begin{matrix} x_{1d} \\ x_{2d}\end{matrix}\right] [x˙1d?x˙2d??]=[?x1??f1???x1??f2????x2??f1???x2??f2???]x=x0??[x1d?x2d??]

x ¨ + x ˙ + 1 x = 1 \ddot x + \dot x + \frac{1}{x} =1 x¨+x˙+x1?=1
Let x 1 = x , x 2 = x ˙ x_1=x,x_2=\dot x x1?=xx2?=x˙
x ˙ 1 = x ˙ = x 2 x ˙ 2 = x ¨ = ? x ˙ ? 1 x + 1 \dot x_1=\dot x=x_2\\ \dot x_2=\ddot x=-\dot x-\frac{1}{x}+1 x˙1?=x˙=x2?x˙2?=x¨=?x˙?x1?+1

x ˙ 1 = x ˙ 2 = 0 \dot x_1=\dot x_2=0 x˙1?=x˙2?=0,平衡点 x 10 = 1 , x 20 = 0 x_{10}=1, x_{20}=0 x10?=1,x20?=0
[ x ˙ 1 d x ˙ 2 d ] = [ 0 1 1 ? 1 ] [ x 1 d x 2 d ] \left[\begin{matrix}\dot x_{1d} \\ \dot x_{2d}\end{matrix}\right]= \left[\begin{matrix}0 & 1 \\ 1 & -1 \end{matrix}\right] \left[\begin{matrix} x_{1d} \\ x_{2d}\end{matrix}\right] [x˙1d?x˙2d??]=[01?1?1?][x1d?x2d??]


x ˙ 1 d = x 2 d x ˙ 2 d = x 1 d ? x 2 d x ¨ d = x d ? x ˙ d ? x ¨ d + x ˙ d ? x d = 0 \dot x_{1d}=x_{2d}\\ \dot x_{2d}=x_{1d}-x_{2d}\\ \ddot x_d=x_d-\dot x_d \Rightarrow \ddot x_d + \dot x_d - x_d=0 x˙1d?=x2d?x˙2d?=x1d??x2d?x¨d?=xd??x˙d??x¨d?+x˙d??xd?=0

Summary

1-D
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x ? x 0 ) , x ? x 0 → 0 f(x)=f(x_0)+f'(x_0)(x-x_0), x-x_0 \rightarrow 0 f(x)=f(x0?)+f(x0?)(x?x0?),x?x0?0

2-D
[ x ˙ 1 d x ˙ 2 d ] = [ ? f 1 ? x 1 ? f 1 ? x 2 ? f 2 ? x 1 ? f 2 ? x 2 ] ∣ x = x 0 → F i x e d ? P o i n t [ x 1 d x 2 d ] \left[\begin{matrix}\dot x_{1d} \\ \dot x_{2d}\end{matrix}\right]= \left[\begin{matrix}\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2}\\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{matrix}\right]_{|x=x_0 \rightarrow Fixed \ Point} \left[\begin{matrix} x_{1d} \\ x_{2d}\end{matrix}\right] [x˙1d?x˙2d??]=[?x1??f1???x1??f2????x2??f1???x2??f2???]x=x0?Fixed?Point?[x1d?x2d??]

Step by Step Derivation of Kalman Gain

X k = A X k ? 1 + B U k ? 1 + W k ? 1 Z k = H X k + V k X_k = AX_{k-1} + BU_{k-1} + W_{k-1}\\ Z_k=HX_k+V_k Xk?=AXk?1?+BUk?1?+Wk?1?Zk?=HXk?+Vk?

A为状态矩阵 B为控制矩阵

P(W) ~ N(0, Q) ,W为过程噪声
Q 为协方差矩阵, Q = E [ W W T ] Q=E[WW^T] Q=E[WWT]

P(v) ~ N(0, R),V为测量噪声
R 为协方差矩阵, R = E [ V V T ] R=E[VV^T] R=E[VVT]

先验估计值 X ^ k ? = A X k ? 1 + B U k ? 1 \hat X_k^- = AX_{k-1}+BU_{k-1} X^k??=AXk?1?+BUk?1?
测量值预测值 Z k = H X k ? X ^ k M E A = H ? 1 Z k Z_k=HX_k \Longrightarrow \hat X_{kMEA}=H^{-1}Z_k Zk?=HXk??X^kMEA?=H?1Zk?

X ^ k = X ^ k ? + G ( H ? 1 Z k ? X ^ k ? ) , G ∈ [ 0 , 1 ] \hat X_k = \hat X_k^- + G(H^{-1}Z_k-\hat X_k^{-}), G\in[0, 1] X^k?=X^k??+G(H?1Zk??X^k??),G[0,1]
G = K k H G=K_kH G=Kk?H,带入上式可得:
X ^ k = X ^ k ? + K k ( Z k ? H X ^ k ? ) , K k ∈ [ 0 , H ? 1 ] \hat X_k = \hat X_k^{-} + K_k(Z_k-H\hat X_k^-), K_k \in [0, H^{-1}] X^k?=X^k??+Kk?(Zk??HX^k??),Kk?[0,H?1]

目标: 寻找 K k ,使得 X ^ k → X k ,即使得 V a r ( X ^ k ) 最小 K_k,使得\hat X_k \to X_k,即使得Var(\hat X_k)最小 Kk?,使得X^k?Xk?,即使得Var(X^k?)最小

e k = X k ? X ^ k e_k=X_k - \hat X_k ek?=Xk??X^k?
P ( e k ) P(e_k) P(ek?) ~ N(0, P)
P 为误差协方差矩阵, P = E ( e e T ) P=E(ee^T) P=E(eeT)
t r ( P ) 最小时,方差最小 tr(P)最小时,方差最小 tr(P)最小时,方差最小

P = E ( e e T ) = E [ ( X k ? X ^ k ) ( X k ? X ^ k ) T ] P=E(ee^T)=E[(X_k-\hat X_k)(X_k-\hat X_k)^T] P=E(eeT)=E[(Xk??X^k?)(Xk??X^k?)T]

X k ? X ^ k = X k ? [ X ^ k ? + K k ( Z k ? H X ^ k ? ) ] = X k ? X ^ k ? ? K k Z k + K k H X ^ k ? = X k ? X ^ k ? ? K k ( H X k + V k ) + K k H X ^ k ? = ( X k ? X ^ k ? ) ? K k H ( X k ? X ^ k ? ) ? K k V k = ( I ? K k H ) ( X k ? X ^ k ? ) ? K k V k = ( I ? K k H ) e k ? ? K k V k X_k-\hat X_k=X_k - [\hat X_k^- + K_k(Z_k-H\hat X_k^-)]\\ =X_k - \hat X_k^- - K_kZ_k + K_kH\hat X_k^-\\ =X_k - \hat X_k^- - K_k(HX_k+V_k) + K_kH\hat X_k^-\\ =(X_k-\hat X_k^-) - K_kH(X_k-\hat X_k^-)-K_kV_k\\ =(I-K_kH)(X_k-\hat X_k^-)-K_kV_k\\ =(I-K^kH)e_k^- -K_kV_k Xk??X^k?=Xk??[X^k??+Kk?(Zk??HX^k??)]=Xk??X^k???Kk?Zk?+Kk?HX^k??=Xk??X^k???Kk?(HXk?+Vk?)+Kk?HX^k??=(Xk??X^k??)?Kk?H(Xk??X^k??)?Kk?Vk?=I?Kk?H)(Xk??X^k??)?Kk?Vk?=(I?KkH)ek???Kk?Vk?
其中 e k ? 称为先验误差 e_k^-称为先验误差 ek??称为先验误差

E [ ( I ? K k H ) e k ? ? K k V k ] [ ( I ? K k H ) e k ? ? K k V k ] T = E [ ( I ? K k H ) e k ? ? K k V k ] [ e k ? T ( I ? K k H ) T ? V k T K k T ] = E [ ( I ? K k H ) e k ? e k ? T ( I ? K k H ) T ? ( I ? K k H ) e k ? V k T K k T ? K k V k e k ? T ( I ? K k H ) T + K k V k V k T K k T ] E[(I-K_kH)e_k^--K_kV_k][(I-K_kH)e_k^--K_kV_k]^T\\ =E[(I-K_kH)e_k^--K_kV_k][e_k^{-T}(I-K_kH)^T-V_k^TK_k^T]\\ =E[(I-K_kH)e_k^-e_k^{-T}(I-K_kH)^T-(I-K_kH)e_k^-V_k^TK_k^T-K_kV_ke_k^{-T}(I-K_kH)^T+K_kV_kV_k^TK_k^T] E[(I?Kk?H)ek???Kk?Vk?][(I?Kk?H)ek???Kk?Vk?]T=E[(I?Kk?H)ek???Kk?Vk?][ek?T?(I?Kk?H)T?VkT?KkT?]=E[(I?Kk?H)ek??ek?T?(I?Kk?H)T?(I?Kk?H)ek??VkT?KkT??Kk?Vk?ek?T?(I?Kk?H)T+Kk?Vk?VkT?KkT?]

E ( I ? K k H ) e k ? V k T K k T = ( I ? K k H ) E ( e k ? V k T ) K k T = ( I ? K k H ) E ( e k ? ) E ( V k T ) K k T = 0 E(I-K_kH)e_k^-V_k^TK_k^T\\ =(I-K_kH)E(e_k^-V_k^T)K_k^T\\ =(I-K_kH)E(e_k^-)E(V_k^T)K_k^T=0 E(I?Kk?H)ek??VkT?KkT?=(I?Kk?H)E(ek??VkT?)KkT?=(I?Kk?H)E(ek??)E(VkT?)KkT?=0

P k = ( I ? K k H ) E ( e k ? e k ? T ) ( I ? K k H ) T + K k E ( V k V k T ) K k T = ( I ? K k H ) P k ? ( I ? K k H ) T + K k R K k T = ( P k ? ? K k H P k ? ) ( I ? H T K k T ) + K k R K k T = P k ? ? P k ? H T K k T ? K k H P k ? + K k H P k ? H T K k T + K k R K k T P_k=(I-K_kH)E(e_k^-e_k^{-T})(I-K_kH)^T+K_kE(V_kV_k^T)K_k^T\\ =(I-K_kH)P_k^-(I-K_kH)^T+K_kRK_k^T\\ =(P_k^--K_kHP_k^-)(I-H^TK_k^T)+K_kRK_k^T\\ = P_k^--P_k^-H^TK_k^T-K_kHP_k^-+K_kHP_k^-H^TK_k^T+K_kRK_k^T Pk?=(I?Kk?H)E(ek??ek?T?)(I?Kk?H)T+Kk?E(Vk?VkT?)KkT?=(I?Kk?H)Pk??(I?Kk?H)T+Kk?RKkT?=(Pk???Kk?HPk??)(I?HTKkT?)+Kk?RKkT?=Pk???Pk??HTKkT??Kk?HPk??+Kk?HPk??HTKkT?+Kk?RKkT?

t r ( P k ) = t r ( P k ? ) ? 2 t r ( K k H P k ? ) + t r ( K k H P k ? H T K k T ) + t r ( K k R K k T ) tr(P_k)=tr(P_k^-)-2tr(K_kHP_k^-)+tr(K_kHP_k^-H^TK_k^T)+tr(K_kRK_k^T) tr(Pk?)=tr(Pk??)?2tr(Kk?HPk??)+tr(Kk?HPk??HTKkT?)+tr(Kk?RKkT?)

矩阵求导公式

d t r ( A B ) d A = B T \frac{dtr(AB)}{dA}=B^T dAdtr(AB)?=BT
d t r ( A B A T ) d A = 2 A B \frac{dtr(ABA^T)}{dA}=2AB dAdtr(ABAT)?=2AB

d t r ( P k ) d K k = 0 ? 2 ( H P k ? ) T + 2 K k H P k ? H T + 2 K k R = ? P k ? H T + K k ( H P k ? H T + R ) = 0 \frac{dtr(P_k)}{dK_k}=0-2(HP_k^-)^T+2K_kHP_k^-H^T+2K_kR\\ =-P_k^-H^T+K_k(HP_k^-H^T+R)=0 dKk?dtr(Pk?)?=0?2(HPk??)T+2Kk?HPk??HT+2Kk?R=?Pk??HT+Kk?(HPk??HT+R)=0

K k = P k ? H T H P k ? H T + R K_k=\frac{P_k^-H^T}{HP_k^-H^T+R} Kk?=HPk??HT+RPk??HT?
R R R 为测量噪声协方差矩阵
P k ? P_k^- Pk?? 为先验误差协方差矩阵

Prior / Posterior Error Covariance Matrix

X k = A X k ? 1 + B U k ? 1 + W k ? 1 , W X_k = AX_{k-1}+BU_{k-1}+W_{k-1},W Xk?=AXk?1?+BUk?1?+Wk?1?W~ P ( 0 , Q ) P(0, Q) P(0,Q)
Z k = H W k + V k , V Z_k = HW_k+V_k,V Zk?=HWk?+Vk?V~ P ( 0 , R ) P(0, R) P(0,R)

先验估计
X ^ k ? = A X ^ k ? 1 + B U k ? 1 \hat X_k^-=A\hat X_{k-1}+BU_{k-1} X^k??=AX^k?1?+BUk?1?

后验估计
X ^ k = X ^ k ? + K k ( Z k ? H X ^ k ? ) \hat X_k=\hat X_k^-+K_k(Z_k-H\hat X_k^-) X^k?=X^k??+Kk?(Zk??HX^k??)

卡尔曼增益
K k = P k ? H T H P k ? H T + R K_k=\frac{P_k^-H^T}{HP_k^-H^T+R} Kk?=HPk??HT+RPk??HT?

P k ? P_k^- Pk??

误差 e k = X k ? X ^ k e_k=X_k-\hat X_k ek?=Xk??X^k?
先验误差 e k ? = X k ? X ^ k ? = A X k ? 1 + B U k ? 1 + W k ? 1 ? A X ^ k ? 1 ? B U k ? 1 = A ( X k ? 1 ? X ^ k ? 1 ) + W k ? 1 = A e k ? 1 + W k ? 1 e_k^-=X_k-\hat X_k^-=AX_{k-1}+BU_{k-1}+W_{k-1}-A\hat X_{k-1}-BU_{k-1}\\ =A(X_{k-1}-\hat X_{k-1})+W_{k-1}=Ae_{k-1}+W_{k-1} ek??=Xk??X^k??=AXk?1?+BUk?1?+Wk?1??AX^k?1??BUk?1?=A(Xk?1??X^k?1?)+Wk?1?=Aek?1?+Wk?1?

P k ? = E [ e k ? e k ? T ] = E [ ( A e k ? 1 + W k ? 1 ) ( e k ? 1 T A T + W k ? 1 T ) ] = E [ A e k ? 1 e k ? 1 T A T + A e k ? 1 W k ? 1 T + W k ? 1 e k ? 1 T A T + W k ? 1 W k ? 1 T ] = A E ( e k ? 1 e k ? 1 T ) A T + E ( W k ? 1 W k ? 1 T ) = A P k ? 1 A T + Q P_k^-=E[e_k^-e_k^{-T}]\\ =E[(Ae_{k-1}+W_{k-1})(e^T_{k-1}A^T+W_{k-1}^T)]\\ =E[Ae_{k-1}e_{k-1}^TA^T+Ae_{k-1}W_{k-1}^T+W_{k-1}e_{k-1}^TA^T+W_{k-1}W_{k-1}^T]\\ =AE(e_{k-1}e_{k-1}^T)A^T+E(W_{k-1}W_{k-1}^T)\\ =AP_{k-1}A^T+Q Pk??=E[ek??ek?T?]=E[(Aek?1?+Wk?1?)(ek?1T?AT+Wk?1T?)]=E[Aek?1?ek?1T?AT+Aek?1?Wk?1T?+Wk?1?ek?1T?AT+Wk?1?Wk?1T?]=AE(ek?1?ek?1T?)AT+E(Wk?1?Wk?1T?)=APk?1?AT+Q

使用卡尔曼滤波器 估计 状态变量的值(最终五个式子)
  • 预测
    – 先验: X ^ k ? = A X ^ k ? 1 + B U k ? 1 \hat X_k^-=A\hat X_{k-1}+BU_{k-1} X^k??=AX^k?1?+BUk?1?
    – 先验误差协方差矩阵: P k ? = A P k ? 1 A T + Q P_k^-=AP_{k-1}A^T+Q Pk??=APk?1?AT+Q

  • 校正
    – 卡尔曼增益: K k = P k ? H T H P k ? H T + R K_k=\frac{P_k^-H^T}{HP_k^-H^T+R} Kk?=HPk??HT+RPk??HT?
    – 后验估计: X ^ k = X ^ k ? + K k ( Z k ? H X ^ k ? ) \hat X_k=\hat X_k^-+K_k(Z_k-H\hat X_k^-) X^k?=X^k??+Kk?(Zk??HX^k??)
    – 更新误差协方差
    P k = P k ? ? P k ? H T K k T ? K k H P k ? + K k H P k ? H T K k T + K k R K k T = P k ? ? P k ? H T K k T ? K k H P k ? + K k ( H P k ? H T + R ) K k T = P k ? ? P k ? H T K k T ? K k H P k ? + P k ? H T K k T = P k ? ? K k H P k ? = ( I ? K k H ) P k ? P_k=P_k^--P_k^-H^TK_k^T-K_kHP_k^-+K_kHP_k^-H^TK_k^T+K_kRK_k^T\\ =P_k^--P_k^-H^TK_k^T-K_kHP_k^-+K_k(HP_k^-H^T+R)K_k^T\\ =P_k^--P_k^-H^TK_k^T-K_kHP_k^-+P_k^-H^TK_k^T\\ =P_k^--K_kHP_k^-\\ =(I-K_kH)P_k^- Pk?=Pk???Pk??HTKkT??Kk?HPk??+Kk?HPk??HTKkT?+Kk?RKkT?=Pk???Pk??HTKkT??Kk?HPk??+Kk?(HPk??HT+R)KkT?=Pk???Pk??HTKkT??Kk?HPk??+Pk??HTKkT?=Pk???Kk?HPk??=(I?Kk?H)Pk??

An 2D example

excel 矩阵处理函数
矩阵相乘 mmul()
转换 transpose()
取逆 minverse()

全选F2
Ctrl+Shift+Enter

该案例需要赋初始值变量
状态变量初始值 x 1 , 0 , x 2 , 0 后验估计值初始值 x ^ 1 , 0 , x ^ 2 , 0 误差协方差矩阵 P 0 状态变量初始值 x_{1,0}, x_{2,0}\\ 后验估计值初始值 \hat x_{1,0}, \hat x_{2,0}\\ 误差协方差矩阵 P_0 状态变量初始值x1,0?,x2,0?后验估计值初始值x^1,0?,x^2,0?误差协方差矩阵P0?

Extended Kalman Filter(EKF)

理解

卡尔曼增益K:负责融合估计值与测量值,谁方差小,就更相信谁一些。

状态估计协方差矩阵P:初始状态与过程噪声有关,并且由于每次K的迭代,状态估计协方差矩阵也在迭代(因为使用了上一次的结果作为下一次的初始状态,上一次的结果来自于卡尔曼增益K对观测与估计的融合,状态估计协方差会减小)

过程噪声协方差矩阵Q:来自于世界中的不确定性,这个值怎么选,我也不太清楚,不知道如何度量世界中的不确定性。

测量误差协方差矩阵R:来自于传感器误差(我猜是可以通过试验获得,比如测量100次,然后记录数据)

其余的就是要给一个目标的初始状态,然后根据观测,不断地更新估计,得到一个稳定的K。

文章来源:https://blog.csdn.net/qq_51491920/article/details/135588176
本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。