index must be a positive integer or logical.

c1=3.29;

c2=9.90;

c3=0.77;

c4=0.20;

Ln=zeros(1,length(t));

for n=1:1:100;

      for t=0:-0.1:-1;

Ln(t)=-(c1+c2*log(n)*log(-t))-(c3+c4*log(n));

semilogx(n,Ln(t));

      end 

end

I really want t to count negative, what am I supposed to do??

[Chandrasekhar]

t cannot take negative values as it is acting as index

[Marta Salas]

You're going to define an index that it's not related with t but with the size of t. However I don't understand what you are trying to plot there. Which is the result you are expecting from that code? what do you want to plot on semilogx?

[sharif]

It is an approximate model of flat edge in communication systems where I must have 10 different lines running from n=1 to n=100

Answer 1

[Chris C]

I'm still not reall clear on what you're looking for, but try this version of Marta's code...

c1=3.29;
c2=9.90;
c3=0.77;
c4=0.20;
t = linspace(0,-1,10); %definition of t vector
n = linspace(1,100);
Ln=zeros(length(t),length(n));
for i=1:length(t)
  for j = 1:length(n);
    Ln(i,j)= -(c1+c2*log(n(j))*log(-t(i)))-(c3+c4*log(n(j)));
  end
  semilogx(n,Ln(i,:))
  hold on
end

[sharif]

yes I think this method is correct, although the graph is wrong, I have to check why it's wrong, thanks

[Chris C]

What's wrong about the graph? If it's because the lines are all blue all you have to do is populate the data first within the for loops and then graph it all at once after the loops instead of plotting a line after each iteration around i.

Answer 2

[dpb]

Keep the time as associated independent vector of same length as the solution vector but numbered from 1:N instead of trying to use it as an index.

Answer 3

[Marta Salas]

A solution is this:

c1=3.29;
c2=9.90;
c3=0.77;
c4=0.20;
Ln=zeros(1,length(t));
t= 0:-0.1:-1; %definition of t vector

for n=1:1:100;
    for i=1:length(t);
      Ln(i)= -(c1+c2*log(n)*log(-t(i)))-(c3+c4*log(n));
      semilogx(n,Ln(i)); 
    end 
end

Take into account semilogx(n,Ln(i)) is plotting a point. is this the right solution?

[sharif]

I wanted to plot for each n all values of t, i.e. when n=1, t=-0.1,-0.2,-0.3 till -1.0, couldn't achieve it with this solution :(

[dpb]

plot for each n all values of t...

t= 0:-0.1:-1; %definition of t vector

for n=1:100
  LN=log(n);
  Ln= -(c1+c2*LN.*log(-t))-(c3+c4*LN);
  semilogx(t,Ln);
  if n==1,hold on,end
end 

You can also get rid of the loop on n entirely as well...

doc meshgrid % for example

[sharif]

I will check it now thanks

[dpb]

BTW, log(0) --> -inf so you'll probably want to restrict t>0

I meant to point it out above but forgot to go back and do it after finished the other modifications...

[sharif]

This is resulting 100 lines I want only 10 lines 10 lines containing the 100 values, anyway thanks for trying to help.

[Marta Salas]

Sorry, I don't really understand what you mean about 10 lines.

[dpb]

Well, read and do a little thinking on your own...fix the upper limit on the for loop to be 1:length(n)