Hi Carsten,
On 2022-05-14 11:38, Carsten Andrich wrote:
Hi Magnus,
On 14.05.22 08:59, Magnus Danielson via time-nuts wrote:
Do note that the model of no correlation is not correct model of
reality. There is several effects which make "white noise" slightly
correlated, even if this for most pratical uses is very small
correlation. Not that it significantly changes your conclusions, but
you should remember that the model only go so far. To avoid aliasing,
you need an anti-aliasing filter that causes correlation between
samples. Also, the noise has inherent bandwidth limitations and
futher, thermal noise is convergent because of the power-distribution
of thermal noise as established by Max Planck, and is really the
existence of photons. The physics of it cannot be fully ignored as
one goes into the math field, but rather, one should be aware that
the simplified models may fool yourself in the mathematical exercise.
Thank you for that insight. Duly noted. I'll opt to ignore the
residual correlation. As was pointed out here before, the 5 component
power law noise model is an oversimplification of oscillators, so the
remaining error due to residual correlation is hopefully negligible
compared to the general model error.
Indeed. My comment is more to point out details which becomes relevant
for those attempting to do math exercises and prevent unnecessary insanity.
Yes, I keep riminding that the 5 component power law noise model is just
that, only a model, and it does not really respect the "Leeson effect"
(actually older) of resonator folding of noise, which becomes a
systematic connection of noise of different slopes.
Here you skipped a few steps compared to your other derivation. You
should explain how X[k] comes out of Var(Re(X[k])) and Var(Im(X[k])).
Given the variance of X[k] and E{X[k]} = 0 \forall k, it follows that
X[k] = Var(Re{X[k]})^0.5 * N(0, 1) + 1j * Var(Im{X[k]})^0.5 * N(0, 1)
because the variance is the scaling of a standard Gaussian N(0, 1)
distribution is the square root of its variance.
Reasonable. I just wanted it to be complete in the thread.
This is a result of using real-only values in the complex Fourier
transform. It creates mirror images. Greenhall uses one method to
circumvent the issue.
Can't quite follow on that one. What do you mean by "mirror images"?
Do you mean that my formula for X[k] is missing the complex conjugates
for k = N/2+1 ... N-1? Used with a regular, complex IFFT the
previously posted formula for X[k] would obviously generate complex
output, which is wrong. I missed that one, because my implementation
uses a complex-to-real IFFT, which has the complex conjugate implied.
However, for a the regular, complex (I)FFT given by my derivation, the
correct formula for X[k] should be the following:
{ N^0.5 * \sigma * N(0, 1) , k = 0, N/2
X[k] = { (N/2)^0.5 * \sigma * (N(0, 1) + 1j * N(0, 1)), k = 1 ... N/2 - 1
{ conj(X[N-k]) , k = N/2 + 1
... N - 1
If you process a real-value only sample list by the complex FFT, as you
did, you will have mirror fourier frequencies of opposite sign. This
comes as e^(i*2*pi*f*t)+e^(-i*2*pi*f*t) is only real. Rather than using
the optimization to remove half unused inputs (imaginary) and half
unused outputs (negative frequencies) with N/2 size transform, you can
use the N-size transform more straightforward and accept the losses for
simplicity of clarity. This is why Greenhall only use upper half
frequencies.
Cheers,
Magnus
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