Not your typical vertcat error. Weird behaviour with it.

Question

Mario Malic 2023-5-11

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1961544-not-your-typical-vertcat-error-weird-behaviour-with-it

编辑： VBBV 2023-5-11

采纳的回答： VBBV

nlworkspace.mat

在 MATLAB Online 中打开

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)
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.

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Answer 1

VBBV 2023-5-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1961544-not-your-typical-vertcat-error-weird-behaviour-with-it#answer_1233099

在 MATLAB Online 中打开

nlworkspace.mat

load("nlworkspace.mat")
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

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Mario Malic 2023-5-11

编辑：Mario Malic 2023-5-11

Oh my... I thought I was going crazy. I should take some time off. 😂

Thank you.

VBBV 2023-5-11

编辑：VBBV 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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