fft and random sequence

generating 2 random sequences with randn with same variance then applying fft to them would result in 2 sequences of non = variances why is that

Answer 1

[the cyclist]

Because the sample variance is not necessarily equal to the population variance.

[yasser]

please explain more

[the cyclist]

Here is the simplest possible explanation I can think of. Imagine you use the command

randn(2,1)

to draw a sample with two values from a distribution that is N(0,1). (In other words, a normal distribution with population mean of 0 and standard deviation of 1.)

Suppose that draw gives you values of [0.7824 -0.5000]. The sample mean of that sample is 0.1412. The sample standard deviation is 0.9068.

Do you get upset that the sample mean is not exactly zero, and that the sample standard deviation is not exactly 1? Hopefully not. It's a random sample, with random variation.

Similarly in your example, there is random variation.

Answer 2

[Star Strider]

The Fourier transform integrates over the original x variable, so the Fourier transform is a function of the (complex) frequency variable only. You are taking the mean and standard deviation (variance) of F(w) rather than f(x).

So if you have different lengths of the vectors in the original ( x or t ) domain, the means and variances should be close to the same in that domain. Without appropriate scaling for different lengths or sampling frequencies, the means and variances (taken over the frequency variable) of the Fourier transforms of those data would not be the same in the complex frequency domain.

I’m not certain what you did, so this is only theoretical. The analytical maths are not trivial, and neither the Symbolic Math Toolbox or my back-of-the-envelope analysis yielded a definitive answer.

Answer 3

[yasser]

thanks u all