I need help with Symbolic Differentiation Error

Question

Shiv Karpoor 2023-2-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1915505-i-need-help-with-symbolic-differentiation-error

评论： Shiv Karpoor 2023-2-20

采纳的回答： Dyuman Joshi

在 MATLAB Online 中打开

Hello MATLAB community,

I am trying to do partial differentiation to find a 6x3 (Jr) matrix, but I am getting an error which says:

" Second argument must be a variable or a nonnegative integer specifying the number of differentiations."

I tried to play around by changing the code, but the problem is still the same.

Below is my code:

syms phi theta psi Psi
h = 110.75; 
d2r = pi/180;
r2d = 180/pi;
ROT = [(cos(phi)*cos(psi))-(cos(theta)*sin(phi)*sin(psi)) -(cos(phi)*sin(psi))-(cos(theta)*cos(psi)*sin(phi)) (sin(phi)*sin(theta));...
    (cos(psi)*sin(phi))+(cos(phi)*cos(theta)*sin(psi)) (cos(phi)*cos(theta)*cos(psi)-sin(phi)*sin(psi)) -(cos(phi)*sin(theta));...
    (sin(theta)*sin(psi)) (cos(psi)*sin(theta)) cos(theta)];
px = (h/2)*(ROT(1,1)-ROT(2,2));
py = -h*ROT(2,1);
eqn = ROT(1,2) == ROT(2,1);
Psi = eqn;
pz = input('Please enter value for pz: ');
phi = input('Please enter value for phi: ')*d2r;
theta = input('Please enter value for theta: ')*d2r;
if phi >= 0
    psi = -phi ; 
elseif phi <= 0
    psi = -phi ;  
end
jr = [diff(px, pz), diff(px, phi), diff(px, theta); diff(py, pz), diff(py, phi), diff(py, theta); 1, 0, 0; 0, 1, 0; 0, 0, 1; diff(Psi, pz), diff(Psi, phi), diff(Psi, theta)];
Jr = double(subs(jr,{phi,theta,psi},{phi,theta,psi}))

You can assume the values for pz, phi, theta to be 435, 5, and 10, respectively.

Can anyone please help me with this problem.

Kind regards,

Shiv

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Dyuman Joshi 2023-2-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1915505-i-need-help-with-symbolic-differentiation-error#answer_1175325

编辑：Dyuman Joshi 2023-2-20

在 MATLAB Online 中打开

You are over-writing the symboic variables, psi and theta by taking input from the user. Rename the variables which you will use for substitution

Also, the if-else condition block seems redundant, as regardless of the value of phi, psi = -phi

syms phi theta psi Psi
h = 110.75; 
d2r = pi/180;
r2d = 180/pi;
ROT = [(cos(phi)*cos(psi))-(cos(theta)*sin(phi)*sin(psi)) -(cos(phi)*sin(psi))-(cos(theta)*cos(psi)*sin(phi)) (sin(phi)*sin(theta));...
    (cos(psi)*sin(phi))+(cos(phi)*cos(theta)*sin(psi)) (cos(phi)*cos(theta)*cos(psi)-sin(phi)*sin(psi)) -(cos(phi)*sin(theta));...
    (sin(theta)*sin(psi)) (cos(psi)*sin(theta)) cos(theta)];
px = (h/2)*(ROT(1,1)-ROT(2,2));
py = -h*ROT(2,1);
eqn = ROT(1,2) == ROT(2,1);
Psi = eqn;
pz = 435;      %input('Please enter value for pz: ');
phi0 = 5;       %input('Please enter value for phi: ')*d2r;
theta0 = 10;    %input('Please enter value for theta: ')*d2r;
if phi0 >= 0
    psi0 = -phi0 ; 
elseif phi0 <= 0
    psi0 = -phi0 ;  
end
jr = [diff(px, pz), diff(px, phi), diff(px, theta); diff(py, pz), diff(py, phi), diff(py, theta); 1, 0, 0; 0, 1, 0; 0, 0, 1; diff(Psi, pz), diff(Psi, phi), diff(Psi, theta)];
Jr = double(subs(jr,{phi,theta,psi},{phi0,theta0,psi0}))
Jr = 6×3
  -25.2772   55.4023   25.2772
   16.3887   76.5384  -16.3887
    1.0000         0         0
         0    1.0000         0
         0         0    1.0000
    1.0000         0    1.0000

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Shiv Karpoor 2023-2-20

Hi Dyuman,

Thank you for the response.

This is actually half of the original code, the pz, phi, and theta values keep changing with for loop in the original script, and the input function for pz, phi, theta in this half-script helps verify values quickly.

But anyways, I understood your solution and it worked well.

Thank you.

请先登录，再进行评论。