Matrices Dimensions Not Consistent when t=0:10

Hi,

I don't understand why it says it's not consistent. It's a 4*4 matrix. Only the element depents on t grows, but not the whole matrix?

So if I want to grow it, I have to use for loop? Thanks!

t= 1:10;
psm_x_dsr = [1 0 0 0.25*(1-cos(pi*t));
             0 1 0 0.25*(1-sin(pi*t)); 
             0 0 1 0; 
             0 0 0 1];

Error using vertcat

Dimensions of matrices being concatenated are not consistent.

Error in main (line 15)

psm_x_dsr = [1 0 0 0.25*(1-cos(pi*t)); 0 1 0 0.25*(1-sin(pi*t)); 0 0 1 0; 0 0 0 1]';

Answer 1

[KSSV]

YOu need to store the result into a 3D matrix.

t= 1:10;
nt = length(t) ;
psm_x_dsr = zeros(4,4,nt) ;
for i = 1:nt
    psm_x_dsr(:,:,i) = [1 0 0 0.25*(1-cos(pi*t(i)));
        0 1 0 0.25*(1-sin(pi*t(i)));
        0 0 1 0;
        0 0 0 1];
end

Answer 2

[Walter Roberson]

t=1:10 does not tell MATLAB that t is a scalar variable whose values will be decided later but will be in the range 1 through 10.

t=1:10 tells MATLAB that you want to right now make t a vector with the values 1 2 3 4 5 6 7 8 9 10 stored in it.

with t being a vector pi*t is a vector and cos() of that is a vector and 1 minus cos is a vector and 1/4 times that is a vector .

Therefore [1 0 0 0.25etc] is a vector of length 13 not a vector of length 4.

If you want placeholder variables use the symbolic toolbox or create a function