# Not your typical vertcat error. Weird behaviour with it.

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Mario Malic on 11 May 2023
Edited: VBBV on 11 May 2023
Hey,
I am not sure what's going on with editor, but I am having issues in understanding why does dxdt does not work properly. The error is about vertcat.
% Error using vertcat
% Dimensions of arrays being concatenated are not consistent.
Here is the code
m1 = parameters(1);
m2 = parameters(2);
k1 = parameters(3);
k2 = parameters(4);
d1 = parameters(5);
d2 = parameters(6);
% Output equation.
y = [x(1)]; % Displacement of the smaller mass
Now we execute each row of dxdt (further below) and we see the result
x(2)
ans = 50
(-((k1+k2)*x(1))/m1) + ((k2*x(3))/m1) - (((d1+d2)*x(2))/m1) + ((d2*x(4))/m1) +u(1)
ans = -416.6667
x(4)
ans = 50
(d2*x(2)/m2) - (d2*x(4)/m2) + (k2*x(1)/m2) - (k2*x(3)/m2) + 0
ans = 0
But if I want to do it this way, it doesn't work. Issue is that I have to add extra parentheses on the second element, but there should be none!
% State equations.
dxdt = [x(2); ...
(-((k1+k2)*x(1))/m1) + ((k2*x(3))/m1) - (((d1+d2)*x(2))/m1) + ((d2*x(4))/m1) +u(1); ...
x(4); ...
(d2*x(2)/m2) - (d2*x(4)/m2) + (k2*x(1)/m2) - (k2*x(3)/m2) + 0];
Error using vertcat
Dimensions of arrays being concatenated are not consistent.

VBBV on 11 May 2023
m1 = parameters(1)
m1 = 12
m2 = parameters(2)
m2 = 13
k1 = parameters(3)
k1 = 100
k2 = parameters(4)
k2 = 100
d1 = parameters(5)
d1 = 0
d2 = parameters(6)
d2 = 0
% Output equation.
y = [x(1)] % Displacement of the smaller mass
y = 50
x(2)
ans = 50
(-((k1+k2)*x(1))/m1) + ((k2*x(3))/m1) - (((d1+d2)*x(2))/m1) + ((d2*x(4))/m1) +u(1)
ans = -416.6667
x(4)
ans = 50
x(3)
ans = 50
x(4)
ans = 50
(d2*x(2)/m2) - (d2*x(4)/m2) + (k2*x(1)/m2) - (k2*x(3)/m2) + 0
ans = 0
(-((k1+k2)*x(1))/m1) + ((k2*x(3))/m1) - (((d1+d2)*x(2))/m1) + ((d2*x(4))/m1) +u(1)
ans = -416.6667
(d2*x(2)/m2) - (d2*x(4)/m2) + (k2*x(1)/m2) - (k2*x(3)/m2) + 0
ans = 0
% State equations.
dxdt = [x(2);(-((k1+k2)*x(1))/m1) + ((k2*x(3))/m1) - (((d1+d2)*x(2))/m1) + ((d2*x(4))/m1) + u(1);
x(4); (d2*x(2)/m2) - (d2*x(4)/m2) + (k2*x(1)/m2) - (k2*x(3)/m2) + 0]
dxdt = 4×1
50.0000 -416.6667 50.0000 0
Mario Malic on 11 May 2023
Edited: Mario Malic on 11 May 2023
Oh my... I thought I was going crazy. I should take some time off. 😂
Thank you.
VBBV on 11 May 2023
Edited: VBBV on 11 May 2023
As you said, it works when parenthesis is added, it's again because of operator precedence. Parenthesis () operator has the higher precedence in equation than others, so when you add a () it then delineates everything within the outermost () as ONE expression or element in matrix and evaluates it, otherwise it's treated as 2 different elements

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