Not your typical vertcat error. Weird behaviour with it.

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.

Answer 1

[VBBV]

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

[Mario Malic]

Okay,

so the issue is in the element commented below.

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];

What is this behaviour, why does it work when there is space between '+' and 'u(1)'?

[Stephen23]

"why does it work when there is space between '+' and 'u(1)'?"

You can see this for yourself:

[1 -2]

ans = 1×2

1 -2

[1 - 2]

ans = -1

[VBBV]

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]

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

Thank you.

[VBBV]

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