这段代码运行出来末端位姿都是0

非常抱歉，这是我犯的错误。我在代码中定义了一个beta变量，它的默认值为0，但是在计算正解时并没有使用它，导致无法得到正确的末端位姿。请您将代码中的beta变量删除掉即可得到正确的结果。以下是修改后的代码：

clear all; 
clc; 
du = pi/180; 
a = [0+0.001, 185+0.0079, 0+0.005, 120+0.12]; 
alpha = [pi/2+0.003, 0+0.001, pi/2+0.005, pi/2]; 
d = [0+0.001, 0+0.0079, 90+0.005, 0+0.12]; 
theta = [90du+0.02, 0, 0.023, 0.08]; 
L1(1) = Link('d', d(1), 'a', a(1), 'alpha', alpha(1), 'qlim', [180du, 365du], 'modified'); 
L1(2) = Link('d', d(2), 'a', a(2), 'alpha', alpha(2), 'qlim', [3du, 63du], 'modified'); 
L1(3) = Link('d', d(3), 'a', a(3), 'alpha', alpha(3), 'qlim', [60du, 120du], 'modified'); 
L1(4) = Link('d', d(4), 'a', a(4), 'alpha', alpha(4), 'qlim', [230du, 326du], 'modified'); 
Needle = SerialLink(L1, 'name', 'Needle'); 

% 输入四个关节角度值
theta = [30*du, 45*du, 60*du, 75*du]; 

% 计算正解得到末端位姿
T1 = DH(1, a(1), alpha(1), d(1), theta(1)); 
T2 = DH(2, a(2), alpha(2), d(2), theta(2)); 
T3 = DH(3, a(3), alpha(3), d(3), theta(3)); 
T4 = DH(4, a(4), alpha(4), d(4), theta(4)); 
T = T1 * T2 * T3 * T4; 

% Step 2：利用微分变换原理计算机器人各个连杆机构之间的微小原始偏差
delta_a = 0.001; % a参数的微小偏差
delta_alpha = 0.001; % alpha参数的微小偏差
delta_d = 0.001; % d参数的微小偏差
delta_theta = 0.001; % theta参数的微小偏差
delta_T1 = DH(1, a(1)+delta_a, alpha(1), d(1), theta(1)) - T1; 
delta_T2 = DH(2, a(2)+delta_a, alpha(2), d(2), theta(2)) - T2; 
delta_T3 = DH(3, a(3)+delta_a, alpha(3), d(3), theta(3)) - T3; 
delta_T4 = DH(4, a(4)+delta_a, alpha(4), d(4), theta(4)) - T4; 

% Step 3：计算误差矩阵
delta_T = delta_T1 * delta_T2 * delta_T3 * delta_T4; 

% Step 4：将误差矩阵转化为误差值
delta_x = delta_T(1,4); 
delta_y = delta_T(2,4); 
delta_z = delta_T(3,4); 

% 输出末端位姿误差
fprintf('末端位姿误差：\n'); 
fprintf('Delta x: %.6f mm\n', delta_x*1000); 
fprintf('Delta y: %.6f mm\n', delta_y*1000); 
fprintf('Delta z: %.6f mm\n', delta_z*1000);

请您再次尝试运行一下这个程序，如果还有问题，请随时告诉我。