Help with loop and tracking values.

Hello,

I have an array full of acceleration values called B. B is calculated using the function at the bottom of the page. The rows of B correspond to the time and the columns correspond to a variable called zeta. Acceleration is calculated using these values of zeta and t that increment each iteration.

I need to figure out how to find the values of zeta which make the acceleration values in B less than or equal to 0.05 when t is greater than or equal to 0.05

How would I do this? Note: each row the time increases by 0.001 seconds, and each column of zeta increases by 0.1. Thank you.

Code:

T02=1;
tf2 = .2;
h2= 0.001;
y02 = [0, 0];
n2 = int32(ceil(tf2/h2))+1; % determine the number of steps
t2 = linspace(0, tf2, n2); % Generate a time step vector
y2 = zeros(n2,2); % Allocate the array for numerical solutions
y2(1,:) = y0'; % The first row of y is the initial condition
B = zeros(n2,21); % Allocate the array of acceleration values  
zeta2 = 0:0.1:2;
  
for ii = 1:(n2)
   for jj = 1:length(zeta2)
      k1 = func2(t2,zeta2(jj),y(ii,:))';
      k2 = func2(t2+h2/2,zeta2(jj),(y(ii,:)+ h2/2*k1))';
      y(ii+1,:) = y(ii,:) + h2*k2;
      B(ii,jj) = -10000*y(ii,1)-200*zeta2(jj)*y(ii,2)+1;
   end
end

Here is func 2 if needed:

function dy = func2(t,zeta,y)
        dy = [y(:,2);-10000*y(:,1)-200*zeta*y(:,2)+1];
end

[Image Analyst]

We can't run your code:

>> test3

Unrecognized function or variable 'y0'.

Error in test3 (line 8)

y2(1,:) = y0'; % The first row of y is the initial condition

So, what is y0?

[John Biscanti]

My apolgies for the confusion. Here is the full code. I am only concerned with the second for loop for my problem. The one with ii:n2 and jj as the length of zeta. Thank you

clc 
clear
    
    T0=1;
    tf = .01;
    h= 0.001;
    y0 = [0, 0];
    n = int32(ceil(tf/h))+1;  % determine the number of steps
    t = linspace(0, tf, n);   % Generate a time step vector
    y = zeros(n,2);           % Allocate the array for numerical solutions
    y(1,:) = y0';             % The first row of y is the initial condition               
    A = zeros(n,21);           % Allocate the array of acceleration values
    
    %----------------------------------------------------------------------
    % TODO: Implement the loop
    %----------------------------------------------------------------------
 
    
zeta = 0:0.01:0.05;
    
  for ii = 1:(n)
       
     for jj = 1:length(zeta)
        
        k1 = func2(t,zeta(jj),y(ii,:))';
        k2 = func2(t+h/2,zeta(jj),(y(ii,:)+ h/2*k1))';
        y(ii+1,:) = y(ii,:) + h*k2;
        
        A(ii,jj) = -10000*y(ii,1)-200*zeta(jj)*y(ii,2)+1;
        
    end
  end
  
 %Amax = max(A,[],'all')  %Amax
 
%A(11,6) %sanity check 
 
 
    T02=1;
    tf2 = .2;
    h2= 0.001;
    y02 = [0, 0];
    n2 = int32(ceil(tf2/h2))+1;  % determine the number of steps
    t2 = linspace(0, tf2, n2);   % Generate a time step vector
    y2 = zeros(n2,2);           % Allocate the array for numerical solutions
    y2(1,:) = y0';             % The first row of y is the initial condition               
    B = zeros(n2,21);           % Allocate the array of acceleration values
 
    
    
zeta2 = 0:0.1:2;
    
  for ii = 1:(n2)
       
     for jj = 1:length(zeta2)
        
        k1 = func2(t2,zeta2(jj),y(ii,:))';
        k2 = func2(t2+h2/2,zeta2(jj),(y(ii,:)+ h2/2*k1))';
        y(ii+1,:) = y(ii,:) + h2*k2;
        
        B(ii,jj) = -10000*y(ii,1)-200*zeta2(jj)*y(ii,2)+1;
        
        if (B <= 0.05 & B(ii) >= .05)
        
        
        
        
        end
        
 
        
        
        
        
      end
    
       
       
  end

[Image Analyst]

B is a 2-D matrix, so exactly what are your thoughts here when you reference it with no, and one, index:

        B(ii,jj) = -10000*y(ii,1)-200*zeta2(jj)*y(ii,2)+1;        
        if (B <= 0.05 & B(ii) >= .05)

[John Biscanti]

I am not sure, that was me attempting to do what I needed, but I am not super skilled with MATLAB and is why I was asking.

[John Biscanti]

I think I just thought it would be comparing acceleration values and the t values that B is using

Answer 1

[Mohith Kulkarni]

Hi, you are trying to find the values of "zeta2" for which the values in "B" are <= 0.05 for the last 151 rows(concluded this from "t">=0.05). Refer to the below code:

t05 = find(t2==0.05);  %t05 = 51, the index corresponding to time step 0.05
temp = B(t05:end,:)<=0.05; %logical matrix for B values <= 0.05 and t >= 0.05

you can retrieve zeta values for any timestep by indexing into the "zeta2". refer to below code for example:

tsq = 0.06 %time step 0.06

zeta_vals = zeta2(temp(find(t2 == tsq)-50,:))

this code should be after the loop as we are accessing the "zeta" values after computing "B" matrix values