Turning imaginary numbers into real numbers

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Colin Lynch
Colin Lynch 2018-2-12
评论： Walter Roberson 2018-2-12
Hello there!
I am attempting to find the points where n = S with the following equations. Whenever S goes into the imaginary plane though, I want it to convert to S = 1. Even though I have a check for that in this code, for some reason MATLAB seems to be ignoring that bit and converting S into 0 instead. How can I fix this?
for k = 1:1:100
n = (100 / k);
s = sqrt((((.25*(n-(.5*(100-n))))/(n))-.25)/(-((n - .5*(100-n)))/(n)));
if (0<=s) && (s<=1)
s = s;
elseif s > 1
s = 1;
elseif (isnan(s)==1)
s = 1;
elseif (isreal(s)==0)
s = 1;
else
s = 1;
end
RTDlineark{k} = [k];
RTDlinearkn{k} = [n];
RTDlinearks{k} = [s];
end

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采纳的回答

Birdman
Birdman 2018-2-12
To get rid of that, your code can be improved as follows:
for k = 1:1:100
n = (100 / k);
s = sqrt((((.25*(n-(.5*(100-n))))/(n))-.25)/(-((n - .5*(100-n)))/(n)));
if (0<s) && (s<=1)
s = s;
elseif s > 1
s = 1;
elseif (real(s)>=0) && imag(s)~=0
s = 1;
end
RTDlineark{k} = [k];
RTDlinearkn{k} = [n];
RTDlinearks{k} = [s];
end
  1 个评论
Walter Roberson
Walter Roberson 2018-2-12
This happens to work with the data, but still suffers from the problem of using < and <= operators on complex quantities, as those operators are defined to ignore the complex component.

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更多回答（2 个）

Andrei Bobrov
Andrei Bobrov 2018-2-12
k = 1:100;
n = 100 ./ k;
s = sqrt( (.25*(n-.5*(100-n))./n-.25)./-(n - .5*(100-n))./n );
t = abs(imag(s)) > 0;
s(t) = t(t) + 0;
Walter Roberson
Walter Roberson 2018-2-12
The relative comparison operations ignore the imaginary component. You need to do the test for imaginary first.

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