Kalman filter execution too slow
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I have written a script for doing SOC estimation using Kalman Filter. The Recursive Parameter Estimation done before it is really faster but not the later Kalman filter. I don't know why its lot slower and 1000 iterations takes more than half an hour ! Why So ?
%Setup
clc;
clear;
D = load("data_math.mat");
%load("mar18_1.mat");
sympref('FloatingPointOutput',true);
syms R1 RC1;
C = 10.0056;
%C = 3.28; SOC becomes negative overtime ! (From IIT_data, max discharge)
%Sampled Data
%U_L = S.synthData.V; % U_L data
%I_L = -S.synthData.I; % I_L data
%Tsamp = S.synthData.t; % Sampled Time
%U_L = T.Voltage(1749:3656);
%I_L = T.Current(1749:3656);
%Tsamp = T.Time(1749:3656);
U_L = D.Datamath.Voltage;
I_L = D.Datamath.Current;
Tsamp = D.Datamath.Time;
%RLS Parameters
lamda=0.98; % forget factor
Theta = zeros(4,1); % Initial value of theta vector
P=eye(4); % Initial value of Covariance Matrix
N = length(Tsamp); % Number of samples
theta_arr = zeros(4,N-1);% For storing parameters
% Other Variables
O_CV = zeros(1,N-1);
R_O = zeros(1,N-1);
SOC_arr = zeros(1,N);
y_k_arr = zeros(1,N-1);
SOC_arr(1) = 1; % As We are considering discharging, Intially SOC will be high and let's assume it to be 1.
%psi_fn = [1 0 I_L(1) 0]; % Initial value of psi_fn (1 I_L(0) I_L(1) U_L(0))
%NOTE n=1 corresponds to t = 0, n = 2 is same as of (t=1)
% RLS for Circuit Model Parameter Estimation
for n = 2 : N
%if n != 1
psi_fn = [1 U_L(n-1) I_L(n) I_L(n-1)];
T = Tsamp(n) - Tsamp(n-1);
%end
error_rls = U_L(n) - psi_fn*Theta ; % Instantaneous error of RLS
K = P*(transpose(psi_fn))/((lamda + (psi_fn)*P*transpose(psi_fn))); % Gain vector
P= (lamda^-1)*(P - K*(psi_fn)*P); % Correlation Matrix Update
Theta = Theta + error_rls*K; % Weights Update
theta_arr(:,n-1) = Theta;
O_CV(n-1) = Theta(1)/(1-Theta(2));
R_O(n-1) = T*(Theta(4)-Theta(3))/(T+Theta(2));
SOC_arr(n) = SOC_arr(n-1) - (I_L(n) * T) /(3600*C);
y_k_arr(n-1) = (psi_fn*Theta) + error_rls;
end
% Circuit Parameters
R_O_val = R_O(N-1);
eq1 = 0 == -((T-(2*RC1))+(theta_arr(2,N-1)*(T+(2*RC1))));
%eq2 = 0 == -(((R_O_val*T)+(R1*T)+(2*R_O_val*RC1))+(theta_arr(3,n-1)*(T+(2*RC1))));
%eq3 = 0 == -((R_O_val*T)+(R1*T)-(2*R_O_val*RC1))+(theta_arr(4,n-1)*(T+(2*RC1))));
eq2 = 0 == -(((R_O_val*T)+(R1*T))+((theta_arr(3,N-1)*theta_arr(4,N-1))*(T+(2*RC1))));
S = solve(eq1,eq2);
R1_val = S.R1;
C1_val = S.RC1/S.R1 ;
%Observed Values at last: R_O = 0.0230, R1 = 2.8450, C1 = 950.5212
% Gain Scheduling/ Linearization of OCV vs SOC plot
OCV_SOC_fit = polyfit(SOC_arr(2:N),O_CV,9);
dOCV_SOC_fit = polyder(OCV_SOC_fit);
k_arr = zeros(1,10);
b_arr = zeros(1,10);
for i = 1:10
k_arr(i) = polyval(dOCV_SOC_fit,0.05*((2*i)-1));
b_arr(i) = polyval(OCV_SOC_fit,0.05*((2*i)-1)) - (k_arr(i)*((2*i)-1)*0.05);
end
% Kalman filter
% Initial values
eta = 1; % Current Efficiency
Ad = [[1-(1/(R1_val*C1_val)) 0];[0 1]];
Bd = [(1/C1_val); 1/(3600*C)];
D = R_O_val;
X_arr = zeros(2,N);
X_val = [0; 1]; %Assume initial value of V2 is unknown
P_kal = eye(2);
Q_kal = diag([0.1 0.1]);
R = 1;
%Y_k_arr = zeros(1,N);
%Y_k_arr(1) = U_L(1) - b_arr(10);
% Iterations
for n = 2:N
Cd = [1 k_arr(min(ceil(X_val(2)/0.1),10))];
Uin = [I_L(n)];
% Prediction Update
X_val = Ad*X_val + Bd*Uin;
P_kal = (Ad*P_kal*transpose(Ad));
% Kalman Gain Update
Kal_Gain = (P_kal*transpose(Cd))/((Cd*P_kal*transpose(Cd))+R);
% State Estimation Update
X_val = X_val + Kal_Gain*((U_L(n) - b_arr(min(ceil(X_val(2)/0.1),10))) - (Cd*X_val) - (D*Uin));
X_arr(:,n) = X_val;
% Error Covariance update
P_kal = (eye(2)-(Kal_Gain*Cd))*P_kal;
if mod(n,1000) == 0
disp(n);
end
end
%Plots
figure;
fsize=14; % plot text font size
%SOC_arr_rev = flip(SOC_arr,2);
plot(Tsamp(2:N),X_arr(2,2:N),'','linewidth',4);
%plot(Tsamp(2:N),y_k_arr,"r",'LineWidth',4);
%hold on;
%plot(Tsamp(2:N),U_L(2:N),"b",'LineWidth',4);
%hold off;
ylim([3 4.5]);
%ax=xticklabels;
%xticklabels(flip(ax))
xlabel('time','FontName','Times New Roman','FontSize',fsize);
ylabel('SOC','FontName','Times New Roman','FontSize',fsize);
4 commentaires
Mike Croucher
le 21 Mar 2024
Can you also share the file data_math.mat so we can run this on our own machines?
Annamalai N
le 24 Mar 2024
KALYAN ACHARJYA
le 25 Mar 2024
Modifié(e) : KALYAN ACHARJYA
le 25 Mar 2024
It performing approximately 100,000 for loop iterations in two phases, its operation duration may vary depending on array processing. Consider experimenting with smaller values of N to check performance
Annamalai N
le 25 Mar 2024
Réponse acceptée
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