Why is the derivative of a matrix not of the same order?

Hi

I need to take the derivative of the "Q" matrix of joints in Matlab for a robot. but using the derivative method here

https://de.mathworks.com/help/matlab/ref/diff.html

I see that the derivative of a 3x3 matrix is a 2x3 matrix! why? should it be so?

[Stephen23]

"I see that the derivative of a 3x3 matrix is a 2x3 matrix! why?"

Lets consider a simpler example: the difference between 8 and 5 is 3. Note that 3 is not two values, it is only one value.

"should it be so?"

Yes.

[Les Beckham]

I'm not familiar with the Q matrix for a robot, but, if it is similar to a direction cosine matrix for rigid body motion (such as for defining the attitude of an aircraft), you can't just take the derivative of the matrix to determine the derivatives of the motion variable.

So, perhaps this is not a Matlab question but, rather, a robotic motion theory question.

[Farzad Torabi]

Thank you very much for your point

[Les Beckham]

You are quite welcome.

Answer 1

[Star Strider]

Use the gradient function instead of diff.

[Bruno Luong]

Nah, usually in robotics the 3x3 matrix represents the (local) frame that is needed to be differentiable to q vector (joints).

It can not be computesd by gradient.

[Star Strider]

@Farzad Torabi — See if gradient will do what you want.

[Farzad Torabi]

Thank you very much! sure

[Star Strider]

My pleasure!

[Farzad Torabi]

It worked!! Many Thanks!!!

[Star Strider]

As always, my pleasure!

Answer 2

[Walter Roberson]

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]

R = function_handle with value:
    @(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]

T = function_handle with value:
    @(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)

RotAround = function_handle with value:
    @(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)

Q1 = 

Q2 = RotAround(theta_2 + alpha_2 * t, cx_2, cy_2, cz_2)

Q2 = 

Q = simplify(Q1 * Q2)

Q = 

eqns = Q*[x;y;z;0]

eqns = 

dx = simplify(diff(eqns(1), t))

dx = 

dy = simplify(diff(eqns(2), t))

dy = 

dz = simplify(diff(eqns(3), t))

dz = 

0

You could extend this to include rotations in Z or to include additional joints.

Answer 3

[Bruno Luong]

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/

[farzad]

And what I have is a numerical matrix therefore I can't use syms. So why should there be such a drastic difference and what is the solution to my problem?

[Walter Roberson]

There should be such a drastic difference because the numeric diff function was never designed to be derivative, and was always defined as repeated steps of A(2:end)-A(1:end-1) along the appropriate dimension repeated the requested number of times.

Derivatives can only be calculated for continuous functions, and are calculated by theory and proofs. If what you have instead is only discrete states then you can only estimate the derivative and would need to know the step sizes to do so. You do not pass the step sizes into numeric diff() so it cannot possibly be calculating derivatives.

[Bruno Luong]

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.

[Farzad Torabi]

Bruno, since you seem to have experience in robotics, please advice me which method is correct when I have a numerical matrix of Q