# Double precision changes to complex double after calculation

18 vues (au cours des 30 derniers jours)
mashtine le 30 Jan 2015
Commenté : Guillaume le 30 Jan 2015
Hi everyone,
I have a matrix with very simple data in double precision but after doing a calculation (listed below) some of the numbers, not all, change to complex numbers. Even the timestamp which is not used in the calculation changes to complex. The data that changes appears to be the same as the data that doesn't change.
Any ideas?
for i = 1:length(ws);
if L(i,1) >= -500 && L(i,1) <= -12
wsstd_uns(i,1) = timestamp(i,1);
wsstd_uns(i,2) = fric(i,1).*(0.35*((-(BLH(i,1)./(vK.*L(i,1)))).^(2/3)) + (2 -(10./BLH(i,1)))).^(1/2);
wsstd_uns(i,3:5) = [ws(i,1),wgst(i,1),(wgst(i,1)-ws(i,1))];
elseif L(i,1) >= 0 && L(i,1) <= 500
wsstd_s(i,1) = timestamp(i,1);
wsstd_s(i,2) = 2.*fric(i,1).*((1 -(10./BLH(i,1))).^(1/2));
wsstd_s(i,3:5) = [ws(i,1),wgst(i,1),(wgst(i,1)-ws(i,1))];
end
end
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Stephen23 le 30 Jan 2015
You should not use i or j as the names of loop variables, as these are both names for the inbuilt imaginary unit .
Guillaume le 30 Jan 2015
Actually, as per the tip section of the doc you've linked:
• Since i is a function, it can be overridden and used as a variable. However, it is best to avoid using i and j for variable names if you intend to use them in complex arithmetic.
• For speed and improved robustness in complex arithmetic, use 1i and 1j instead of i and j.
In other words, unless you use i or j as the imaginary unit, it doesn't matter. And if you do use i or j as the imaginary unit, you shouldn't and should use 1i or 1j instead (which can't be used as a variable).

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### Réponse acceptée

Andreas Goser le 30 Jan 2015
Theory one: You use i as a variable. As this is a "reserved word" for complex calculations, there may be an unexpected effect.
Theory two: One of your functions or variables shadows the real MATLAB command and does something unexpected. Like you would assign plot=1 the then try to use the plot command.
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mashtine le 30 Jan 2015
Two good theories. Finding the root issue is the problem but thanks Andreas. I think I will just use real() to eliminate the imaginary values as they are 0 anyways.

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### Plus de réponses (1)

Guillaume le 30 Jan 2015
Well, you take the square root and cubic roots of some numbers so, if these numbers are negative, you'll get some complex numbers.
As for your timestamp, are you sure it's complex, that is imag(x) ~= 0. The real numbers in a matrix containing complex numbers are displayed as complex but with an imaginary part equal to 0.
##### 4 commentairesAfficher 2 commentaires plus anciensMasquer 2 commentaires plus anciens
Guillaume le 30 Jan 2015
You can also vote for answers with the triangle on the left. This gives a little bit of credit.
Andreas Goser le 30 Jan 2015
I voted :-D

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