How can I subtract this matrix in this function?

Hello, I'm having some difficulty with implementing a matrix into a particular function I'm using in the code below:

 % function dy = Verhulst_KT(t,y)
dy = zeros(1,1);
dy(1) =  r*y(:)*(1-y(1)/(300 + 100*sin(pi/6*t-pi/2))-H );
 end

Where H is defined by:

 for i = 1:61
    s = ones(1,60);
    s(i) = rem(T(i),mod);
    if s(i) < 10 && s(i) > 3
        H(i) = 160;
    else
        H(i) = 0;
    end
end

I've tried referring to individual positions in H within the function and to the whole matrix itself but either a single element is read which is not what I'm after or I get the error "In an assignment A(:) = B, the number of elements in A and B must be the same." I was just hoping I could get some quick assistance in what is hopefully a fairly simple problem. Thank you in advance.

EDIT: I'm using ode45 to solve the equation itself, T is a row matrix from 0:60 and r is a constant. In hindsight, I probably could've done with pasting in the whole code but I wanted to avoid unnecessary content, apologies for any confusion caused.

[Stephen23]

What is T ? What values are you calling Verhulst_KT with ?

[Jan]

A simpler method to obtain H:

S = rem(T, m);  % Do not use "mod" as a name of a variable
H = and(S > 3, S < 10) * 160;

Answer 1

[John BG]

Tyler

1.- in order to have your V_KT function working, to prevent the error A(I)=B you mention and others i had to change it to the following:

    function dy = V_KT(t,y) % your Verhulst_KT()
    dy = zeros(1,1);
    r=ones(1,length(y))
    % H noise whatever not deflined
    % dy(1) =  r*y(:)*(1-y(1)./(300 + 100*sin(pi/6*t-pi/2))-H );
    dy =  r*y(:)*(1-y(1)./(300 + 100*sin(pi/6*t-pi/2)));
    end

2.- regarding H, what are you trying to do with it, embed data or noise in the phase?

H has to have same size as t, your T otherwise when attempting to combine them inside the sin(pi/6 ..) MATLAB will say things referring to size mismatch.

3.-

you need to pass H in the function dy=V_KT(t,y,H)

otherwise how is dy going to have anything to do with H?

If you find this answer of any help solving your question, please click on the thumbs-up vote link,

thanks in advance

John

[e-mail]

[Tyler Clarke]

I added H as a variable for the function as suggested in part 3 and redefined H in an if statement within the function also therefore making it a variable rather than a matrix and it turned out perfect. Thanks for your assistance.