jacobian from trigonometric function

Question

ves dim 2016-3-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/274399-jacobian-from-trigonometric-function

编辑： Walter Roberson 2016-3-20

Hi have made this code to calculate the jacobian but the result presents some complex number

is it because matlab convert trigonometric like (cos = eix+e-ix/2) and if it 's that how can i ha ve a trigonometric expression or is it any bug in the code

syms t1;
syms t2;
syms t3;
syms t4;
T1=[cos(t1) -sin(t1) 0 0;sin(t1) cos(t1) 0 0;0 0 1 0;0 0 0 1];
T2=[cos(t2) -sin(t2) 0 90;0 0 1 0;-sin(t2) cos(t2) 0 0;0 0 0 1];
T3=[cos(t3) -sin(t3) 0 0;0 0 1 70;-sin(t3) -cos(t3) 0 0;0 0 0 1];
T4=[cos(t4) -sin(t4) 0 0;0 0 -1 320;sin(t4) -cos(t4) 0 0;0 0 0 1];
T5=[1 0 0 260;0 0 1 0;0 1 0 0;0 0 0 1];
%calcul
T=T1*T2;
T=T*T3;
T=T*T4;
T=T*T5;
px=T(1,4);
py=T(2,4);
pz=T(3,4);
psi=atan2(-T(2,3),T(3,3));
a=(T(2,3)*T(2,3))+(T(3,3)*T(3,3));
phi=atan2(T(1,3),sqrt(a));
teta=atan2(-T(1,3),T(1,1));
J=jacobian([px,py,pz,psi,phi,teta],[t1,t2,t3,t4]);