Running my code in a loop to make it more efficient

x0 = 0;
y0 = 0;
Sr = [x0 y0];            %Source coordinates
SL = 220;                %Source Level 
a = 0.25;                %Attenuation 
IL = 180;                %Impact Level
v = 2;                   %speed
t = 5;                   %seconds
s = v*t;                 %metres
n = 5;

I am modelling a flee response based on random initial 2D coordinated positions.

z = [randi([-10 10],1,n); randi([-10 10],1,n)]';

This produces a 5x2 matrix with x and y values being equal to shown below

x = z(:,1);
y = z(:,2);

I have calculated the distance between each coordinate in the matrix and the source at the origin to produce a 5x1 matrix.

d = sqrt((x-x0).^2+(y-y0).^2);
r = d;

From this a Transmission Loss (TL) can be calculated from the source to the position

TL = 20*log10(r) + a*r;

And the Recieive Level (RL) at this position can be calculated

RL = SL - TL;

This value of RL is compared to the Impact Level (IL) and if RL is higher than IL the model run agains.

To calculate the next position, assuming a flee response the gradient will remain the same along the line. This produces a 5x1 matrix

m = (y-y0)./(x-x0);

To calculate the new coordinates of the new position, a 5x2 matrix

z = [x,y] + diag(s./sqrt(1+m.^2))*[ones(size(x)),m];

Again where the x and y values

x = z(:,1);
y = z(:,2);

The new distance between the positions can be calculated using speed and time, therefore

r2 = r + s;

A side note; as he position changes, its the previous distance plus s (i.e. r3 = r2 + s, r4 = r3 +s)

Calculating TL at the new position

TL2 = 20*log10(r2) + a*r2;

Calculating RL at the new position

RL2 = SL - TL2;

Compare this RL to the IL.

The problem is, I can keep coding my model in this way and keep repeating however this is not very efficient and if conditions such as n, SL or IL change, this method will become very messy and i will not know how many times this loop needs to be run for.

Therfore i need a loop to run this code, until RL is lower than IL, but also be able to change n and caculate RL at each of these positions every 's' distance away. I would like RL to be stored to be used further along, and then also use the new position coordinates onto a plot, showcasing the flee response. I have included my current code below to show how i have done this.

x0 = 0;
y0 = 0;
Sr = [x0 y0];            %Source coordinates
SL = 220;                %Source Level
a = 0.25;                %Attenuation
IL = 180;                %Impact Level
v = 2;                   %Speed
t = 5;                   %seconds
s = v*t;                 %metres
n = 5;

z1 = [randi([-10 10],1,n); randi([-10 10],1,n)]';
x1 = z1(1:n,1);                                 %this shows the first 5 numbers in column 1
y1 = z1(1:n,2);                                 %this shows the first 5 numbers in column 2
negative_values = x1<0;
x1(negative_values) = -1*x1(negative_values);   %convert negative x values to positives

d = sqrt((x1-x0).^2+(y1-y0).^2);          %Pythagoras Theorem for distance between the 2 (Source and Reciever) points
m = (y1-y0)./(x1-x0);                     %Gradient between the points
r1 = d;

TL1 = 20*log10(r1) + a*r1;  %Transmission Loss
RL1 = SL - TL1;             %Recieve Level

if RL1 < IL
    disp('No Impact') 
else
    disp('Impact')
end

Impact

z2 = [x1,y1] + diag(s./sqrt(1+m.^2))*[ones(size(x1)),m];    %New coordinates for new positions 5s later at a speed of 1.5m/s
x2 = z2(1:n,1);
y2 = z2(1:n,2);
r2 = d + s;                              %distance from source to new receive position

TL2 = 20*log10(r2) + a*r2;  
RL2 = SL - TL2;            

if RL2 < IL
    disp('No Impact') 
else
    disp('Impact')
end

Impact

z3 = [x2,y2] + diag(s./sqrt(1+m.^2))*[ones(size(x1)),m];  
x3 = z3(1:n,1);
y3 = z3(1:n,2);
r3 = d + 2*s;   

TL3 = 20*log10(r3) + a*r3;  
RL3 = SL - TL3;            

if RL3 < IL
    disp('No Impact') 
else
    disp('Impact')
end

Impact

z4 = [x3,y3] + diag(s./sqrt(1+m.^2))*[ones(size(x1)),m];    
x4 = z4(1:n,1);
y4 = z4(1:n,2);
r4 = d + 3*s;   

TL4 = 20*log10(r4) + a*r4;  
RL4 = SL - TL4;            

if RL4 < IL
    disp('No Impact') 
else
    disp('Impact')
end

Impact

z5 = [x4,y4] + diag(s./sqrt(1+m.^2))*[ones(size(x1)),m];    
x5 = z5(1:n,1);
y5 = z5(1:n,2);
r5 = d + 4*s;   

TL5 = 20*log10(r5) + a*r5;  
RL5 = SL - TL5;            

if RL5 < IL
    disp('No Impact') 
else
    disp('Impact')
end

No Impact

X1 = [x1 x2 x3 x4 x5]';
Y1 = [y1 y2 y3 y4 y5]';

X1(:,negative_values) = -1*X1(:,negative_values);
figure
hold on;
plot(X1,Y1, '-*')

title('Flee Response')
xlabel('distance (m)')
ylabel('distance (m)')
ax = gca;                       %the change of the axes location to origin
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';

Answer 1

[Karim]

You can create a for loop, and just save the variables directly into matrices, see below for an example with 10 iterations and 7 flees. I tried no to change the algorithm as you implemented it, however, the bit with the 'negative_values' is a bit weird at first glance.

x0 = 0;
y0 = 0;
Sr = [x0 y0];            %Source coordinates
SL = 220;                %Source Level
a = 0.25;                %Attenuation
IL = 180;                %Impact Level
v = 2;                   %Speed
t = 5;                   %seconds
s = v*t;                 %metres
n = 7;

iter = 10;

X = zeros(iter,n);
Y = zeros(iter,n);

% currZ(:,1) --> this stores the x coordinates 
% currZ(:,2) --> this stores the y coordinates 

for i = 1:iter
    
    if i == 1
        % starting condition - random position start
        currZ = [randi([-10 10],n,1) randi([-10 10],n,1)];
        
        negative_values = currZ(:,1) < 0;
        currZ(negative_values,1) = -1*currZ(negative_values,1);   % convert negative x values to positives
        
        d = sqrt((currZ(:,1)-x0).^2+(currZ(:,2)-y0).^2); % Pythagoras Theorem for distance between the 2 (Source and Reciever) points
        m = (currZ(:,2)-y0)./(currZ(:,1)-x0); %Gradient between the points
    else
        % update with speed
        currZ = currZ + diag(s./sqrt(1+m.^2))*[ones(n,1),m]; % New coordinates for new positions 5s later at a speed of 1.5m/s
    end
    
    r = d + (i-1)*s;  
    
    TL = 20*log10(r) + a*r;  %Transmission Loss
    RL = SL - TL;             %Recieve Level
    
    if RL < IL
        disp('No Impact')
    else
        disp('Impact')
    end
    
    % save result
    X(i,:) = currZ(:,1);
    Y(i,:) = currZ(:,2);
    
end

Impact
Impact
Impact
Impact

No Impact
No Impact
No Impact
No Impact
No Impact
No Impact

X(:,negative_values) = -1*X(:,negative_values);

figure
plot(X,Y, '-*')
title('Flee Response')
xlabel('distance (m)')
ylabel('distance (m)')
ax = gca;                       %the change of the axes location to origin
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';

[Adil Saeed]

Just looking through this code, i found some differences in comparison. The currZ should represent an x and y value between the interval -10 and 10, however from what i can see in this code, it shows a much higher values for this which is affecting the overall transmission loss and therefore the recieved level. However on the graph plot, the points do show in it is between this interval.

This affects currZ from what i can see is using the wrong stored values in the distance (d), gradient (m) and affecting r which is in the transmission loss equation.

Hope this makes sense

[Karim]

I made an edit of the code, to move the computation of m and d inside the first iteration section.

For the currZ values, notice that currZ is initialized as

currZ = [randi([-10 10],n,1) randi([-10 10],n,1)];

Thus the initial values at the first iteration will be between -10 and +10 for both x and y. However, after the second iteration currZ is overwritten, since at this point we no longer need the initial values. Thus currZ becomes

currZ = currZ + diag(s./sqrt(1+m.^2))*[ones(n,1),m];

which is the equivalent as the following part from your original code

z2 = [x1,y1] + diag(s./sqrt(1+m.^2))*[ones(size(x1)),m];  

and this repeats itself at each iteration. Thus when the loop stops currZ will hold the values for the last iteration. Which can be out of the interval -10 +10.

[Adil Saeed]

your explanation makes much more sense as the values override, the previous ones for currZ within the iterations. The edit you made is exactly what i changed it to last night, so thank you for this confirmation.

[Adil Saeed]

how can i store all the values for the Recieve Level that have been overwritten within a matrix so in this case a 7x10 matrix. i.e. at each position i can see what the RL is rather than at the end of the iteration

[Karim]

Before the loop starts, place:

RL = zeros( n ,iter );

inside the loop change the RL lines into:

RL(:,i) = SL - TL; % Recieve Level

if RL(:,i) < IL
    

This way you create a matrix where each column stores the result of an iteration.