MATLAB Answers


Double precision changes to complex double after calculation

Asked by mashtine on 30 Jan 2015
Latest activity Commented on by Guillaume
on 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))];


You should not use i or j as the names of loop variables, as these are both names for the inbuilt imaginary unit .
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).

Sign in to comment.

2 Answers

Answer by Andreas Goser on 30 Jan 2015
 Accepted Answer

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.

  1 Comment

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.

Sign in to comment.

Answer by Guillaume
on 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.


Show 1 older comment
Thought I could accept two answers! Sorry there.
You can also vote for answers with the triangle on the left. This gives a little bit of credit.

Sign in to comment.