index must be a positive integer or logical.
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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??
3 个评论
Marta Salas
2014-3-11
编辑:Marta Salas
2014-3-11
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?
采纳的回答
Chris C
2014-3-11
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
2 个评论
更多回答(2 个)
dpb
2014-3-11
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.
0 个评论
Marta Salas
2014-3-11
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?
7 个评论
dpb
2014-3-11
Well, read and do a little thinking on your own...fix the upper limit on the for loop to be 1:length(n)
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