MATLAB，四阶龙格库塔法

Question

状态方程x'=[0,1,0;0,0,1;-209.827,-161.235,-23.602]x+[0,0,1]n0. 输出方程n=[208.5,133.439,0]. 用MATLAB四阶龙格库塔法求解不同步长时图像

Answer 1

% Define the system matrices
A = [0, 1, 0;     0, 0, 1;     -209.827, -161.235, -23.602];
B = [0; 0; 1];
C = [208.5, 133.439, 0];
D = 0;

% Define the initial conditions
x0 = [0; 0; 0];

% Define the simulation time and step size
t_start = 0;
t_end = 10;
dt_vec = [0.001, 0.01, 0.1];

% Perform simulations with different step sizes
for i = 1:length(dt_vec)
    dt = dt_vec(i);
    t = t_start:dt:t_end;
    num_steps = length(t);

    % Pre-allocate memory for the output
    y = zeros(num_steps, size(C, 1));

    % Initialize the state vector
    x = x0;

    % Perform the simulation
    for j = 1:num_steps
        % Record the output
        y(j, :) = C * x;

        % Compute the k values
        k1 = A * x + B * randn;
        k2 = A * (x + 0.5 * dt * k1) + B * randn;
        k3 = A * (x + 0.5 * dt * k2) + B * randn;
        k4 = A * (x + dt * k3) + B * randn;

        % Update the state vector
        x = x + (dt / 6) * (k1 + 2 * k2 + 2 * k3 + k4);
    end

    % Plot the output
    figure;
    plot(t, y);
    xlabel('Time (s)');
    ylabel('Output');
    title(sprintf('Output with dt = %g', dt));
    legend('y_1', 'y_2', 'y_3');
end


```这个代码会求解状态方程的数值解，并且根据不同的步长绘制输出的图像。这里使用的是四阶龙格库塔法，它可以提供比较准确的结果。在这个例子中，我们采用了三个不同的步长（0.001，0.01和0.1），以便比较它们之间的差异