雅可比伪逆求六自由度机械臂逆运动学问题，matlab代码

以下是使用雅可比伪逆求解六自由度机械臂逆运动学问题的matlab代码：

% 机械臂DH参数
L1 = 0.5;
L2 = 0.5;
L3 = 0.5;
L4 = 0.5;
L5 = 0.5;
L6 = 0.5;

% 初始机械臂关节角度
theta1 = 0;
theta2 = 0;
theta3 = 0;
theta4 = 0;
theta5 = 0;
theta6 = 0;

% 目标末端执行器位置
x = 0.5;
y = 0.5;
z = 0.5;

% 迭代次数
num_iter = 100;

% 迭代误差
epsilon = 0.001;

for i = 1:num_iter
    % 计算机械臂末端执行器位置
    T = forward_kinematics(theta1, theta2, theta3, theta4, theta5, theta6);
    px = T(1,4);
    py = T(2,4);
    pz = T(3,4);
    
    % 计算机械臂雅可比矩阵
    J = jacobian(theta1, theta2, theta3, theta4, theta5, theta6);
    
    % 计算雅可比矩阵的伪逆
    J_pinv = pinv(J);
    
    % 计算机械臂末端执行器位置误差
    error = norm([x;y;z] - [px;py;pz]);
    
    % 如果误差小于迭代误差，则停止迭代
    if error < epsilon
        break;
    end
    
    % 计算关节角度增量
    delta_theta = J_pinv * ([x;y;z] - [px;py;pz]);
    
    % 更新关节角度
    theta1 = theta1 + delta_theta(1);
    theta2 = theta2 + delta_theta(2);
    theta3 = theta3 + delta_theta(3);
    theta4 = theta4 + delta_theta(4);
    theta5 = theta5 + delta_theta(5);
    theta6 = theta6 + delta_theta(6);
end

% 显示最终关节角度
fprintf('theta1 = %.2f\n', theta1);
fprintf('theta2 = %.2f\n', theta2);
fprintf('theta3 = %.2f\n', theta3);
fprintf('theta4 = %.2f\n', theta4);
fprintf('theta5 = %.2f\n', theta5);
fprintf('theta6 = %.2f\n', theta6);

% 正向运动学函数
function T = forward_kinematics(theta1, theta2, theta3, theta4, theta5, theta6)
    % 机械臂DH参数
    L1 = 0.5;
    L2 = 0.5;
    L3 = 0.5;
    L4 = 0.5;
    L5 = 0.5;
    L6 = 0.5;
    
    % 计算机械臂正向运动学矩阵
    T01 = DH(L1, 0, pi/2, theta1);
    T12 = DH(L2, 0, 0, theta2);
    T23 = DH(0, L3, pi/2, theta3);
    T34 = DH(L4, 0, -pi/2, theta4);
    T45 = DH(L5, 0, pi/2, theta5);
    T56 = DH(0, L6, 0, theta6);
    
    T = T01 * T12 * T23 * T34 * T45 * T56;
end

% DH函数
function T = DH(a, d, alpha, theta)
    T = [cos(theta), -sin(theta)*cos(alpha), sin(theta)*sin(alpha), a*cos(theta);
         sin(theta), cos(theta)*cos(alpha), -cos(theta)*sin(alpha), a*sin(theta);
         0, sin(alpha), cos(alpha), d;
         0, 0, 0, 1];
end

% 雅可比矩阵函数
function J = jacobian(theta1, theta2, theta3, theta4, theta5, theta6)
    % 机械臂DH参数
    L1 = 0.5;
    L2 = 0.5;
    L3 = 0.5;
    L4 = 0.5;
    L5 = 0.5;
    L6 = 0.5;
    
    % 计算机械臂正向运动学矩阵
    T01 = DH(L1, 0, pi/2, theta1);
    T12 = DH(L2, 0, 0, theta2);
    T23 = DH(0, L3, pi/2, theta3);
    T34 = DH(L4, 0, -pi/2, theta4);
    T45 = DH(L5, 0, pi/2, theta5);
    T56 = DH(0, L6, 0, theta6);
    T06 = T01 * T12 * T23 * T34 * T45 * T56;
    
    % 计算机械臂末端执行器位置
    px = T06(1,4);
    py = T06(2,4);
    pz = T06(3,4);
    
    % 计算机械臂末端执行器旋转矩阵
    R06 = T06(1:3,1:3);
    
    % 计算雅可比矩阵
    Jv = [cross([0;0;1], [px;py;pz]), cross(R06(:,1), [px;py;pz]), cross(R06(:,2), [px;py;pz]), cross(R06(:,3), [px;py;pz]), R06(:,3), [0;0;0]];
    Jw = [0, 0, 0, 0, 0, 1;
          0, 0, 0, 0, -1, 0;
          0, 0, 0, 1, 0, 0;
          0, 0, -1, 0, 0, 0;
          0, 1, 0, 0, 0, 0;
          -1, 0, 0, 0, 0, 0];
    J = [Jv; Jw];
end

在代码中，`forward_kinematics()`函数用于计算机械臂的正向运动学，`DH()`函数用于计算机械臂DH参数的变换矩阵，`jacobian()`函数用于计算机械臂的雅可比矩阵，主函数中的迭代部分使用了雅可比伪逆方法求解逆运动学问题。最终，程序输出了机械臂六个关节的最终角度。