给我一个matlab求机械臂工作空间的代码

以下是一个简单的matlab代码，用于求解机械臂的工作空间：

clc;
clear;
% 机械臂关节角度范围
theta1 = -pi:pi/20:pi;
theta2 = -pi/2:pi/20:pi/2;
theta3 = -pi/2:pi/20:pi/2;
theta4 = -pi/2:pi/20:pi/2;
theta5 = -pi/2:pi/20:pi/2;
theta6 = -pi:pi/20:pi;
% DH参数
d1 = 0.5; % mm
a2 = 0.5; % mm
d3 = 0.3; % mm
a3 = 0.8; % mm
d5 = 0.4; % mm
% 运动学正解
T = @(theta1,theta2,theta3,theta4,theta5,theta6) ...
    [cos(theta1)*cos(theta2+theta3+theta4)*cos(theta5+theta6)-sin(theta1)*sin(theta5+theta6),-cos(theta1)*cos(theta2+theta3+theta4)*sin(theta5+theta6)-sin(theta1)*cos(theta5+theta6),cos(theta1)*sin(theta2+theta3+theta4),cos(theta1)*(d3*sin(theta2)+a2*sin(theta2+theta3)+a3*sin(theta2+theta3+theta4));sin(theta1)*cos(theta2+theta3+theta4)*cos(theta5+theta6)+cos(theta1)*sin(theta5+theta6),-sin(theta1)*cos(theta2+theta3+theta4)*sin(theta5+theta6)+cos(theta1)*cos(theta5+theta6),sin(theta1)*sin(theta2+theta3+theta4),sin(theta1)*(d3*sin(theta2)+a2*sin(theta2+theta3)+a3*sin(theta2+theta3+theta4));-sin(theta2+theta3+theta4)*cos(theta5+theta6),sin(theta5+theta6)*cos(theta2+theta3+theta4),cos(theta2+theta3+theta4),d1+d3*cos(theta2)+a2*cos(theta2+theta3)+a3*cos(theta2+theta3+theta4)-d5;0,0,0,1];

x = [];
y = [];
z = [];
% 遍历所有关节角度组合
for i = 1:length(theta1)
    for j = 1:length(theta2)
        for k = 1:length(theta3)
            for l = 1:length(theta4)
                for m = 1:length(theta5)
                    for n = 1:length(theta6)
                        % 计算末端执行器的位置
                        P = T(theta1(i),theta2(j),theta3(k),theta4(l),theta5(m),theta6(n))*[0;0;0;1];
                        x(end+1) = P(1);
                        y(end+1) = P(2);
                        z(end+1) = P(3);
                    end
                end
            end
        end
    end
end

% 绘制工作空间
scatter3(x,y,z,'.');
xlabel('X');
ylabel('Y');
zlabel('Z');
title('机械臂工作空间');

需要根据机械臂的具体DH参数和关节角范围进行修改。