基于三阶RC的卡尔曼滤波SOC估算模型.zip

%% Mybatterymodel: A 1-input 1-output 4-state system with non-zero D C0=110; %Ah SOC0=0.97; Temp=[0 1.7 4.4 5.0 7.2 10 12.8 15.6 18.3 18.9 19.4 20 20.6 21.1 21.7 22.2 22.8 23.4 23.9 24.5 25 25.6 26.1 26.7 27.2 27.8 28.3 28.9 29.4 30 30.6 31.1 31.6 32.2 35 37.8 40 40.6 43.3 45]; Temp_eff=[1.398 1.35 1.3 1.289 1.25 1.19 1.15 1.11 1.08 1.072 1.064 1.056 1.048 1.04 1.034 1.029 1.023 1.017 1.011 1.006 1 0.994 0.987 0.98 0.976 0.972 0.968 0.964 0.96 0.956 0.952 0.948 0.944 0.94 0.93 0.91 0.894 0.89 0.88 0.872]; R0=[0.00113 0.00114 0.00115 0.00112 0.00106 0.00111 0.00092 0.00095 0.00099 0.00094]; R1=[0.0088 0.0086 0.0083 0.0081 0.0077 0.0078 0.0075 0.0071 0.0067 0.0064]; C1=[53800 53820 53830 53840 53840 53850 53860 53860 53870 53880]; R2=[0.0005 0.0008 0.0012 0.0013 0.0015 0.0017 0.0018 0.0023 0.0025 0.0024]; C2=[5700 5718 5723 5738 5749 5757 5782 5794 5812 5826]; R3=[0.0012 0.0023 0.0045 0.0054 0.0074 0.0083 0.0094 0.0089 0.0012 0.0028]; C3=[3737.6 3925.9 3028 3113.4 3203 3400.4 3438.8 3628 3849.8 3537.9]; for i=1:10 A = [1-1/(R1(i)*C1(i)) 0 0 0 0 1-1/(R2(i)*C2(i)) 0 0 0 0 1-1/(R3(i)*C3(i)) 0 0 0 0 1]; B = [1/C1(i) 1/C2(i) 1/C3(i) 0]; C = [-1 -1 -1 1]; D =-R0(i); % process noise variance Q=diag([0.05^2 0.02^2 0.03^2 0.05^2]); % measurement noise variance R=0.05; % random initial state x0 = [0 0 0 0.97]'; % Kalman filter set up % The same model A_ =A; B_ = B; C_ = C; D_ = D; Q_ = Q; R_ = R; % However, zeros initial state x1 = [0 0 0 0.97]'; % Initial state covariance P1 = eye(4); end