It is possible to update a matrix each step?

Question

Maikel 2013-5-7

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/74955-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

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Babak 2013-5-7

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

Maikel 2013-5-8

Thank you very much, I really appreciate your comments.

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Answer 1

James Kristoff 2013-5-7

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在 MATLAB Online 中打开

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