Adding a varying value to a for-loop

I have a function that has some constant and some varying inputs.

t=0.2; %constant
NT =300; %constant
ti = 1:NT;
B = exp(-5*t); %constant
A = 1-exp(-5.*t); %constant
pinit = 0.242; %initial p value
r = ones(1,NT); %input step constant througout, gives vector 

%_________________________________________

for k = 1:NT

      if k<=11
          c1(k) = 0;

      else 
          c1(k) = (B*c1(k-1))   -   (pinit*c1(k-11))*(A)   +   (pinit*r(k-11))*(A);

      end

end

The varying values are as follows: The r input is a 1x300 vector, all 1's. The ti input is a 1x300 vector, over time.

and gives an output of c(t), a 1x300 vector from the eqn.

I now want to change pinit from a contant, set at 0.242 to a range of values [0:0.01:2.99] however, when I change it, I get the error that the c(t) output must have the same dimensions as the eqn, which I understand, but I thought by setting P to have a 1x300 matrix, this would work.

Can someone please help me modify my code so that pinit is a varying value?

Answer 1

[KSSV]

t=0.2; %constant
NT =300; %constant
ti = 1:NT;
B = exp(-5*t); %constant
A = 1-exp(-5.*t); %constant
% pinit = 0.242; %initial p value
pinit =  [0:0.01:2.99]; %initial p value
r = ones(1,NT); %input step constant througout, gives vector
%_________________________________________
c1 = zeros(length(pinit),length(NT)) ;
for i = 1:length(pinit)
    for k = 1:NT
        if k<=11
            c1(i,k) = 0;
        else
            c1(i,k) = (B*c1(i,k-1))   -   (pinit(i)*c1(i,k-11))*(A)   +   (pinit(i)*r(k-11))*(A);
        end
    end
end

Note that the above can be vectorized.....as you are beginner...I suggest you to run from loops and then go to vectorizing this.

[Chris Matthews]

Sorry, I'm not sure what you mean. Does that mean I can calculate c(t) using only vector arithmetic rather than a for-loop?

But thanks for the code addition, works like I wanted it to. Have a good arvo.

[KSSV]

Yes..you can avoid for loops...nby vector arithmetic. And Thanks is accepting the answer. :)