When calculating the standard deviation of arrays of complex numbers c, why is std(c) not equal to std(abs(c))?

However, I don't think the Peter's answer gets to the heart of my question - it addresses something slightly different (unless I am missing something?)

My confusion is this: if I look at how the std() function handles a complex number it does the 'squaring' operation by multiplying by the complex conjugate i.e. if c=a+ib then it uses c* x c = a^2+b^2 in its calculation. But this is just the same thing that abs(c) does. So why do you get different answers?

Thanks again.

请先登录，再进行评论。

Answer 2

seackone 2018-5-1

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/306316-when-calculating-the-standard-deviation-of-arrays-of-complex-numbers-c-why-is-std-c-not-equal-to-s#answer_318136

在 MATLAB Online 中打开

Hi, I also want to calculate the standard deviation of a complex number.. At my research, I found your question. But std(c) and std(abs(c)) is not the same!

For example:

A = [1+1i*1 1+1i*2]

The mean of A is:

mean(A) = 1/2 * (1+1i*1 + 1+1i*2) = 1+1i*1.5

Now you can calculate the standard deviation..

But std(abs(A)) is an other equation:

abs(A) = [1.412 2.2361]

and the mean:

mean(abs(A)) = 1.8251

you got a different matrix and a different mean, so the result of the standard deviation isn't the same. If you want, you can calculate the example to the end and the results are: