Dimensions of matrices being concatenated are not consistent

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Linee Teh
Linee Teh 2020-4-10
评论： Linee Teh 2020-4-12
t = 0:5640; % time taken for the system to run is from 0 to 5640s
i = 83*pi/180; % inclination angle in rad
w0 = 0.0011; % angular velocity in rad/s
a1 = sin(i)*cos(w0*t);
a2 = -cos(i);
a3 = 2*sin(i)*sin(w0*t);
A = [0 -a3 a2; a3 0 -a1; -a2 a1 0];
This is my coding and I would like to create a matrix A. It shows the error Error using vertcat: Dimensions of matrices being concatenated are not consistent when I computed it. I know the reason why, but how can I compute a matrix A with varying number of a1 and a3 while a2 is a constant?
  2 个评论
Peng Li
Peng Li 2020-4-10
I don’t really know the reason why you did this. But would an extension of a2 to size(t) fit your need? a2 = -cos(i) * ones(size(t));
Linee Teh
Linee Teh 2020-4-10
Thank you for your help and I successfully get a1, a2 and a3 with the same dimensions. However, my problem still haven't solve.
My full question is like below:
t = 0:5640; % time taken for the system to run is from 0 to 5640s
i = 83*pi/180; % inclination angle in rad
w0 = 0.0011; % angular velocity in rad/s
a1 = sin(i)*cos(w0*t);
a2 = -cos(i)*ones(size(t));
a3 = 2*sin(i)*sin(w0*t);
A = [1 0 0 0 a3 a2; 0 1 0 -a3 0 a1; 0 0 1 a2 -a1 0;...
-1 0 0 0 0 0; 0 -1 0 0 0 0; 0 0 -1 0 0 0;...
zeros(3) zeros(3)];
"Error using vertcat: Dimensions of matrices being concatenated are not consistent" still poping out with this code. I actually want to get A matrix with dimension 9 x 6 and also if possible, get matrix A for t = 0s, t=1s... t=5640s separately.
I hope my question is clear.

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Ameer Hamza
Ameer Hamza 2020-4-10
In MATLAB, rows of a matrix cannot have different number of elements. For that, you will need to use a cell array
t = 0:5640; % time taken for the system to run is from 0 to 5640s
i = 83*pi/180; % inclination angle in rad
w0 = 0.0011; % angular velocity in rad/s
a1 = sin(i)*cos(w0*t);
a2 = -cos(i);
a3 = 2*sin(i)*sin(w0*t);
A = {[0 -a3 a2]; [a3 0 -a1]; [-a2 a1 0]};
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Ameer Hamza
Ameer Hamza 2020-4-10
编辑：Ameer Hamza 2020-4-10
Ok It is clear now. Try this
t = 0:5640; % time taken for the system to run is from 0 to 5640s
i = 83*pi/180; % inclination angle in rad
w0 = 0.0011; % angular velocity in rad/s
a1 = sin(i)*cos(w0*t);
a2 = -cos(i)*ones(size(t));
a3 = 2*sin(i)*sin(w0*t);
ONE = ones(1,1,numel(a1));
ZERO = zeros(1,1,numel(a1));
a1 = reshape(a1, 1,1,[]);
a2 = reshape(a2, 1,1,[]);
a3 = reshape(a3, 1,1,[]);
A = [ONE ZERO ZERO ZERO a3 a2;
ZERO ONE ZERO -a3 ZERO a1;
ZERO ZERO ONE a2 -a1 ZERO;
-ONE ZERO ZERO ZERO ZERO ZERO;
ZERO -ONE ZERO ZERO ZERO ZERO;
ZERO ZERO -ONE ZERO ZERO ZERO;
zeros(3,3,numel(a1)) zeros(3,3,numel(a1))];
The matrix A is 9x6x5641. You can access value like this
A(:,:,1) % <-- t = 0
A(:,:,2) % <-- t = 1
A(:,:,3) % <-- t = 2
..
..
A(:,:,5641) % <-- t = 5640

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