It is possible to update a matrix each step?

Hi everyone, I'm trying to program a matrix dependant of the values of certain variables that depend on time (matrix K in the code below). I've programmed but I really don't know if the program is taking the first value that matches the condition (it compares x values) or it evaluates each step the values of K1, K2, K3.

I've tried before with an K(:,i+1) but it seems that it doesn't work with matrices (or I don't know how)

Please, would you give me some advice?

Thx in advance. PS: I'm Spanish please excuse me for my poor english

    function [K] = Newm3
clear;clc;
    h=1;
    x=[0;0;0];
    v=[sqrt(2*9.81*h);0;0];
    F=[0;0;0];
    M=[600 0 0;0 174.875 0;0 0 321.186];
    C=[0 0 0;0 0 0;0 0 0];
    K=[1e7 -1e7 0;-1e7 1e7+8.022e7 -8.022e7;0 -8.022e7 8.024e7];
    a=(M^-1)*(F-C*v-K*x);
    al=1/6;
    be=1/2;
    % Time step
    ti = 0. ;
    tf = 1. ;
    dt = 0.0005 ;
    t = ti:dt:tf ;
    nt = fix((tf-ti)/dt) ;
    n = length(M) ;
        for i = 1:nt
            part1=M*((1/(al*dt^2))*x(:,i)+(1/(al*dt))*v(:,i)+((1/(2*al))-1)*a(:,i));
            part2=C*((be/(al*dt))*x(:,i)+((be/al)-1)*v(:,i)+((be/al)-2)*(dt/2)*a(:,i));
            x(:,i+1)=((1/(al*(dt)^2))*M+(be/(al*dt))*C+K)^-1*...
                (F+part1+part2);
            if x(1,i+1)-x(2,i+1) > 0,
               K1=1e8;
            else
                K1=0;
            end
            dx=abs(x(2,i+1)-x(3,i+1));
            if dx<=3.848e-5,
                K2=8.018e7;
            elseif 3.848e-5<dx<1.103e-4
                K2=-1.295e8;
            elseif 1.103e-4<dx<1.821e-4
                K2=4.618e4;
            else
                K2=342.036;
            end
            if x(3,i+1)-x(3,i)<=0
                K3=1.766e4;
            elseif x(3,i+1)-x(3,i)>0
                K3=3.208e3;
            else
                K3=56.104
            end
            K=[K1 -K1 0;-K1 K1+K2 -K2;0 -K2 K2+K3];
            a(:,i+1)=(1/(al*dt^2))*(x(:,i+1)-x(:,i))-(1/(al*dt))*v(:,i)-((1/(2*al))-1)*a(:,i);
            v(:,i+1)=v(:,i)+(1-be)*dt*a(:,i)+be*dt*a(:,i+1);
        end
    figure;
    hold on
    d1=plot(t,x(1,:)) ;
    set(d1,'Color','red')
    d2=plot(t,x(2,:)) ;
    set(d2,'Color','green')
    d3=plot(t,x(3,:)) ;
    set(d3,'Color','blue')
    hold off
end

[Babak]

K changes only once at index i=8 and then stays constant.

I'm not sure if this is what you want. You may want to explain more on what you mean by: " I've programmed but I really don't know if the program is taking the first value that matches the condition (it compares x values) or it evaluates each step the values of K1, K2, K3."

In the for loop on i, when you calculate x(:,i+1), none of the parameters change as i increases except part1 and part2. Is this also what you want?

[Maikel]

Actually no, what I really want is that K1, K2 and K3 vary depending on the "if" condition stated for each one, and then build for each step with these values the matrix K.

Thx.

[Babak]

That's up to the algorithm how it's changing the values of variables every loop index increment.

Do you really mean to have F and C and part2 be always zero?

[Maikel]

yes, I don't want to update F and C but I want them in the code just in case.

Your answer then is yes? I mean for each step Matlab checks the "if" condition of K1, K2, K3 and fills K with these values?

Or I have to define something like K(i+1)

[Babak]

Yes, the variable K is updated according to K1, K2, K3 every step in the for loop with index i.

[Maikel]

Thank you very much, I really appreciate your comments.

Answer 1

[James Kristoff]

It is possible to update a matrix in each iteration of a loop. Your code is currently updating the matrix K every iteration and then using the updated values in the next loop. If you want to capture every version of this matrix, you could do something similar to the following:

myMatrix = [0,0; 0,0];
for i = 1:10
  myMatrix(:,:,i) = [i, i+1; i^2, i/2];
end

The : operator means, all values in this dimension, and since you want to create a copy of the matrix, which already has 2 dimensions, so you need to make the iterator the third dimension.

I made some comments about the code you posted below:

function [K] = Newm3
% CLEAR is not needed inside a function like this.  Functions have their
% own workspace, which initially contains only the inputs to the function,
% unless the function contains persistent variables.
% 
% clear;clc;
% 
clc;
% try to name your variables something that makes sense
h=1; %height?
x=[0;0;0]; % position?
v=[sqrt(2*9.81*h);0;0]; % velocity?  if so, I believe this equation is incorrect.
F=[0;0;0]; % force?
M=[600   0       0;...
     0 174.875   0;...
     0   0     321.186]; % Mass?
% this can also be done with C = zeros(3);
C=[0 0 0;...
   0 0 0;...
   0 0 0]; % dampener coeffiecent matrix?
K=[ 1e7 -1e7          0;...
   -1e7  1e7+8.022e7 -8.022e7;...
    0   -8.022e7      8.024e7]; % stiffness coeffiecient matrix?
a=(M^-1)*(F-C*v-K*x); % accelleration
al=1/6; %?
be=1/2; %?
% Time step
ti = 0.;
tf = 1.;
dt = 0.0005;
t = ti:dt:tf;
nt = fix((tf-ti)/dt);
% unused variable
% n = length(M);
for i = 1:nt
    part1 = M*((1/(al*dt^2))*x(:,i) + (1/(al*dt))*v(:,i) + ((1/(2*al))-1)*a(:,i));
    part2 = C*((be/(al*dt)) *x(:,i) + ((be/al)-1)*v(:,i) + ((be/al)-2)*(dt/2)*a(:,i));

      % F starts as [0;0;0] and never changes.
      x(:,i+1) = ((1/(al*(dt)^2))*M + (be/(al*dt))*C+K)^-1*(F+part1+part2);

      if x(1,i+1)-x(2,i+1) > 0,
          K1=1e8;
      else
          K1=0;
      end
      dx=abs(x(2,i+1)-x(3,i+1));
      if dx<=3.848e-5,
          K2=8.018e7;
      elseif (3.848e-5<dx) && (dx<1.103e-4)  % the previous syntax is invalid
          K2=-1.295e8;
      elseif (1.103e-4<dx) && (dx<1.821e-4)
          K2=4.618e4;
      else
          K2=342.036;
      end
      if x(3,i+1)-x(3,i)<=0
          K3=1.766e4;
      elseif x(3,i+1)-x(3,i)>0
          K3=3.208e3;
      else
          K3=56.104;
      end
      K=[ K1    -K1         0;...
         -K1  K1+K2       -K2;...
           0    -K2     K2+K3];
      a(:,i+1) = (1/(al*dt^2))*(x(:,i+1) - x(:,i))-(1/(al*dt))*v(:,i) - ((1/(2*al))-1)*a(:,i);
      v(:,i+1) = v(:,i) + (1-be)*dt*a(:,i) + be*dt*a(:,i+1);
  end
  figure;
  hold on
  d1=plot(t,x(1,:)) ;
  set(d1,'Color','red')
  d2=plot(t,x(2,:)) ;
  set(d2,'Color','green')
  d3=plot(t,x(3,:)) ;
  set(d3,'Color','blue')
  hold off
  end