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卡尔曼滤波一KF

作者:互联网

卡尔曼经典公式:

\[状态一步预测: \qquad \hat{X}_{k/k-1} = \Phi_{k/k-1}\hat{X}_{k-1} \]

\[状态一步预测方差阵: \qquad P_{k/k-1}= \Phi_{k/k-1}P_{k-1}\Phi^T+\Gamma_{k-1}Q_{k-1}\Gamma^T_{k-1} \]

\[滤波增益: \qquad K_k = P_{{XZ},k/k-1}P^{-1}_{ZZ,k/k-1}\\ 或者写为:\qquad K_k = P_{k/k-1}H^T_k(H_kP_{k/k-1}H^T_k+R_k)^{-1} \]

\[状态估计:\qquad \hat{X} = \hat{X}_{k/k-1} + K_k(Z_k-H_k\hat{X}_{k/k-1}) \]

\[状态估计方差阵:\qquad P_k = (I- K_kH_k)P_{k/k-1} \]

举例如下:

一维空间的卡尔曼滤波,主要参考严老师教材

\[X_k = 0.95 \cdot X_k-1+W_{k-1}\\ y_k = X_k + V_k \]

其中均值0,单位的高斯白噪声

clear;
% 系统参数
Phik = 0.95; Bk = 1.0;Hk = 1.0;
% 噪声参数,分别是标准差和方差
q =1;r = 3;Qk = q*q;Rk = r*r;

len =100;
%含有噪声的标准正太分布
w = q*randn(len,1);v = r*randn(len,1);
xk = zeros(len,1);yk = zeros(len,1);
xk(1) =r*randn(1,1);
for k = 2:len
    xk(k) = Phik*xk(k-1) + Bk*w(k);
    yk(k)= Hk*xk(k)+v(k);    
end

 % Kalman 滤波估计,初始状态为0
 Xk = 0; 
%  第一步的误差,这个数比较大,为很大的正实数???意义
 Pxk = 100*Rk/(Hk^2*Phik^2);

for k=1:len         
    [Xk, Pxk, Kk] = kalman(Phik, Bk, Qk, Xk, Pxk, Hk, Rk, yk(k));         
    res(k,:) = [Xk,Pxk,Kk];     
end
% 稳态滤波     
ss = [Hk^2*Phik^2  Hk^2*Bk^2*Qk+Rk-Phik^2*Rk  -Bk^2*Qk*Rk];     
Px = ( - ss(2) + sqrt(ss(2)^2-4*ss(1)*ss(3)) ) / (2*ss(1));     
K = Hk*(Phik^2*Px+Bk^2*Qk)/(Hk^2*(Phik^2*Px+Bk^2*Qk)+Rk);  
G = (1-K*Hk)*Phik;
% 这里获取真值
Xk_IIR = filter(K, [1 -G], yk);     
% 作图     
subplot(1,2,1), hold off, plot(sqrt(res(:,2)),'-'), hold on, plot(res(:,3),'r:'); 
grid
xlabel('\itk'); ylabel('\it\surd P_x_k , K_k');

 subplot(122), hold off, plot(yk,'x'),
 hold on, 
 plot(xk,'m:'); 
 plot(res(:,1),'k'); 
 plot(Xk_IIR,'r-.'); 
 grid
 xlabel('\itk'); ylabel('\ity_k, x_k, x^\^_k, x^\^_k_,_I_I_R'); 

function [Xk, Pxk, Kk] = kalman(Phikk_1, Bk, Qk, Xk_1, Pxk_1, Hk, Rk, Yk)     
Xkk_1 = Phikk_1*Xk_1;
Pxkk_1 = Phikk_1*Pxk_1*Phikk_1' + Bk*Qk*Bk';

Pxykk_1 = Pxkk_1*Hk';
Pykk_1 = Hk*Pxykk_1 + Rk;

Kk = Pxykk_1*Pykk_1^-1;

Xk = Xkk_1 + Kk*(Yk-Hk*Xkk_1);
Pxk = Pxkk_1 - Kk*Pykk_1*Kk';
end

结果图如下:

在这里插入图片描述

结果如下:

误差和滤波增益很快收敛,滤波曲线存在滞后的情况,数字滤波器的相位延迟特点。

标签:Xk,KF,卡尔曼滤波,Phik,Bk,Hk,Qk,Rk
来源: https://www.cnblogs.com/wpf2022/p/16106379.html