Why is the derivative of a matrix not of the same order?
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Hi
I need to take the derivative of the "Q" matrix of joints in Matlab for a robot. but using the derivative method here
I see that the derivative of a 3x3 matrix is a 2x3 matrix! why? should it be so?
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Walter Roberson
2022년 11월 17일
For the case where rotations are no rotations in z, but there are two joints moving around and with angular rotational velocities and :
syms x y z t alpha_1 alpha_2 cx_1 cy_1 cz_1 cx_2 cy_2 cz_2 theta theta_1 theta_2
R = @(theta) [cos(theta) sin(theta) 0 0; -sin(theta) cos(theta) 0 0; 0 0 1 0; 0 0 0 1]
T = @(tx, ty, tz) [1 0 0 tx; 0 1 0 ty; 0 0 1 tz; 0 0 0 1]
RotAround = @(theta, cx, cy, cz) T(-cx, -cy, -cz) * R(theta) * T(cx, cy, cz)
Q1 = RotAround(theta_1 + alpha_1 * t, cx_1, cy_1, cz_1)
Q2 = RotAround(theta_2 + alpha_2 * t, cx_2, cy_2, cz_2)
Q = simplify(Q1 * Q2)
eqns = Q*[x;y;z;0]
dx = simplify(diff(eqns(1), t))
dy = simplify(diff(eqns(2), t))
dz = simplify(diff(eqns(3), t))
You could extend this to include rotations in Z or to include additional joints.
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Bruno Luong
2022년 11월 15일
편집: Bruno Luong
2022년 11월 15일
I guess you mistaken between
- numerical finite difference https://fr.mathworks.com/help/matlab/ref/diff.html
- and symbolic derivative https://fr.mathworks.com/help/symbolic/diff.html
Use the fist is just the difference of 3 rows of Q 3x3 matrix, there fore returns 2x3 matrix. It is NOT the derivative/
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Bruno Luong
2022년 11월 15일
Nah, instead of taking the derivative wrt the robot joint (where the matrix depends on), he takes the difference of rows. Completely wrong methodology.
The discrepency is in the wrong computation methodology, nothing to do with accuracy of numerical method or step size.
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