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If noise on a signal is randomly fluctuating in time domain around its mean following a Gaussian normal distribution (Gaussian noise), would this be equivalent to an constant intensity in the frequency space (white noise)?

Can we somehow relate the frequency distribution to the amplitude distribution?

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    \$\begingroup\$ They are orthogonal terms : one is an amplitude distribution, another is a frequency distribution. \$\endgroup\$ Jun 11 '20 at 14:37
  • \$\begingroup\$ What still confused me is that in the describtion of the Matlab Simulink Block for Band-limited White Noise it is written that the block outputs Nnrmally distributed random numbers. So does that mean the same misconception is at play there? ch.mathworks.com/help/simulink/slref/bandlimitedwhitenoise.html \$\endgroup\$
    – Manumerous
    Jun 13 '20 at 20:13
  • \$\begingroup\$ You can create band-limited noise with a normal amplitude distribution. \$\endgroup\$ Jun 13 '20 at 20:35
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No, they are completely orthogonal concepts. The probability distribution says nothing about the frequency content, and the power distribution across frequency says nothing about the sample probability distribution. You have to specify both.

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As Dave (and Brian) said: two totally different concepts. One doesn't imply the other. This is homework, and you should research it well! Getting the difference between (auto)correlation/PSD and amplitude distribution straight is a critical thing. If this isn't clear to, you should probably ask your professor/teacher (if you have one) – it's easier to explain if one has a didactic "framework" to work with.

There's one thing that's special about Gaussian noise w.r.t. to correlation, and that if random variables (for example, noise measurements from different times) are jointly uncorrelated (and that's a big restriction!), then they are independent.

For all other distributions, lack of correlation does not imply independence.

This is a property about circularly symmetric gaussian (\$\sim\mathcal{CN}\$) noise that allows us to do a lot of mathematical transforms on it (e.g. correcting the phase of a received signal) and still have independent noise components, and that is what's necessary for a lot of estimators to actually work optimally. So, hurray for circularly symmetric gaussian noise!

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Gaussian noise definitely does not imply white noise, because Gaussian noise can have an arbitrary (not necessarily flat) frequency spectrum.

However, contrary to the other answers, there is a sense in which white noise implies Gaussian noise, if the noise is white to arbitrarily high frequencies (arbitrarily small time scales). Or more practically, if our measurements average the noise over time intervals much longer than its correlation time. In this case, the central limit theorem says that the measured noise amplitude, being composed of many independent contributions (with finite variance for physical reasons), converges to a Gaussian distribution.

EDIT: How much longer than the correlation time is needed for the central limit theorem to converge depends on the statistics of the noise. John Doty's comment points out that it does not happen quickly for white noise consisting of pulses that follow a Poisson process. In this case, the amplitude has a highly skewed distribution that is mostly concentrated on zero. This is a "worst case" for the central limit theorem. Averaging over a few pulse widths (correlation times) doesn't make it Gaussian; we have to average over longer than the mean interval between pulses. When we do this, we start to get a less skewed Poisson distribution that is approximately Gaussian. So it still holds that if measurements are averaged over long enough times, white noise looks Gaussian.

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  • \$\begingroup\$ Not true. Geiger counter clicks have a white spectrum up to a cutoff determined by the width of the pulses, but they don't have a Gaussian amplitude distribution. \$\endgroup\$
    – John Doty
    Jun 12 '20 at 21:13
  • \$\begingroup\$ @JohnDoty Thanks, good caveat -- I have edited to clarify. \$\endgroup\$
    – nanoman
    Jun 12 '20 at 23:46
  • \$\begingroup\$ What happens if the amplitude of the clicks follows a Cauchy distribution? Hint: Cauchy is a non-Gaussian stable distribution. \$\endgroup\$
    – John Doty
    Jun 12 '20 at 23:51
  • \$\begingroup\$ @JohnDoty My answer mentions the assumption that noise typically has finite variance for physical reasons (e.g., energy). In any case, I am not suggesting a rigorous formulation here, just saying that in practice white noise often becomes Gaussian at long time scales (low frequency components), so the concepts aren't completely unrelated. \$\endgroup\$
    – nanoman
    Jun 13 '20 at 0:18
  • \$\begingroup\$ Beware. Although the variance of a Cauchy distribution doesn't converge, a finite collection of pulses with Cauchy distributed amplitudes will always have finite energy. Long-tailed distributions of this sort are quite common in physical situations. Mandelbrot became famous by pointing this out. \$\endgroup\$
    – John Doty
    Jun 13 '20 at 12:01
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"White" implies the independence of signals in time, "Gaussian" implies the probability distribution of the momentary value at an individual point of time. Pretty much orthogonal.

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I'd like to add something which the other answers haven't mentioned yet.

It is true that a gaussian distribution is not the same thing as a uniform white-noise distribution. They can be related however. True white noise occurs in a resistor as a result of the brownian motion. It kind of implies that the imaginary part of a complex resistance is zero: in a + ib, b has to equal 0.

Non-zero imaginary are caused by capacitance / inductivities, and will cause low-pass or high-pass behaviour, and thus, the spectral distribution is no longer uniform, but still "linear"

However, non-linear components, such as diodes, can shape a uniform distribution in a way that it becomes a gaussian distributed random value. Or any other distribution, depending on what non-linearity is at play.

Because from an engineering perspective, it sometimes is not very important how the spectral composition looks like, we may want to calculate one number to express a noise level instead. This can done by integrating over the whole (or parts of the) spectrum.

Afterwards, it is impossible to tell wheter the integrated value came from what was originally gaussian, or uniform, or anything else. Assuming we integrated over a a gaussian distribution, we can always find an uniform distribution that is scaled so that its integral matches the integral of our gaussian distribution.

This has immediate merit for our calculations: We can then assume a complex non-linear component is just a resistance after all, which may simplify calculations. I recall that sometimes this is done by the manufacturer already and being given in the datasheet.

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