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how can i fix this error "Subscript indices must either be real positive integers or logicals".in the following program?
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e1=input('enter the value of ei1') e2=input('enter the value of ei2') w=1.5:0.01:1.6; k1=0.25; k2=0.25; a=0.0005; ne=3.19; r=1.5*10^-6; l=2*pi*r; kn=(2*pi*ne*l)./w; b=sqrt(1-k1); c=sqrt(1-k2); f=sqrt(k1-k2); g=sqrt(-1)*kn*l; ed=e2((c-b*exp(-a.*l-g))./(1-b*c*exp(-a.*l-g)))+e1((-f*exp(a*l/2-g/2))./(1-b*c*exp(-a.*l-g))); y=abs(ed); z=y^2; plot(w,z)
1 件のコメント
Jan
2013 年 9 月 13 日
If you format your code properly, the reading is much easier.
When you post the complete error message, we do not have to guess, where the problem occurs.
回答 (2 件)
Roger Stafford
2013 年 9 月 13 日
Your problem presumably occurs at the line
ed=e2(...) + e1(...);
If you intended e2 and e1 to be vectors, which is how matlab is interpreting them, the quantities inside the parentheses need to be real, positive integers since they are interpreted as indices, whereas you have complex fractional values there.
If you intended them to be some kind of function, your method of using 'input' for that purpose didn't achieve that. What kind of entities do you actually have in mind for e2 and e1 as a user input?
0 件のコメント
Image Analyst
2013 年 9 月 13 日
Perhaps you meant:
ed=e2.*((c-b*exp(-a.*l-g))./(1-b*c*exp(-a.*l-g)))+e1.*((-f*exp(a*l/2-g/2))./(1-b*c*exp(-a.*l-g)));
Note, I don't know what values e2 took and what e1 took but for the values I plugged in, ed was complex and then it bombed at the z=y^2 line, so you might need to fix that too.
2 件のコメント
Image Analyst
2013 年 9 月 13 日
Do what people do in situations like this. Break your equation up into smaller terms and see what each one is like:
e1=0
e2=1
w=1.5:0.01:1.6;
k1=0.25;
k2=0.25;
a=0.0005;
ne=3.19;
r=1.5*10^-6;
l=2*pi*r;
kn=(2*pi*ne*l)./w;
b=sqrt(1-k1);
c=sqrt(1-k2);
f=sqrt(k1-k2);
g=sqrt(-1)*kn*l;
term1 = c-b*exp(-a.*l-g)
term2 = 1-b*c*exp(-a.*l-g)
ed=e2.*((c-b*exp(-a.*l-g))./(1-b*c*exp(-a.*l-g)))+e1.*((-f*exp(a*l/2-g/2))./(1-b*c*exp(-a.*l-g)));
y=abs(ed);
z=y.^2;
plot(w,z)
I don't think the line is straight. I think it's just that you are looking at such a small chunk of the curve that it looks straight over that really really small portion of it.
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