Not your typical vertcat error. Weird behaviour with it.
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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
load("nlworkspace.mat");
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)
(-((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
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];
采纳的回答
load("nlworkspace.mat")
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
x(2)
(-((k1+k2)*x(1))/m1) + ((k2*x(3))/m1) - (((d1+d2)*x(2))/m1) + ((d2*x(4))/m1) +u(1)
x(4)
x(3)
x(4)
(d2*x(2)/m2) - (d2*x(4)/m2) + (k2*x(1)/m2) - (k2*x(3)/m2) + 0
(-((k1+k2)*x(1))/m1) + ((k2*x(3))/m1) - (((d1+d2)*x(2))/m1) + ((d2*x(4))/m1) +u(1)
(d2*x(2)/m2) - (d2*x(4)/m2) + (k2*x(1)/m2) - (k2*x(3)/m2) + 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]
5 个评论
Mario Malic
2023-5-11
"why does it work when there is space between '+' and 'u(1)'?"
You can see this for yourself:
[1 -2]
[1 - 2]
Mathematical operators have precedence when used in equations or expressions.
When a space is separating between 2 variables is not provided , even though prefixed with operator they are treated as 2 different elements in a matrix. That's how matlab evaluates expressions and interprets the operators
Mario Malic
2023-5-11
编辑：Mario Malic
2023-5-11
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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