Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say you have two white noise signals with different variation amplitudes A1 and A2 as shown in this beautiful Excel graph:

White noise signals

Ignoring the DC offset as it's been represented here, how do you relate the amplitudes A1 and A2 to the magnitude of the Fourier coefficients after a Fourier transform (as shown in the diagram below)?

Fourier transforms of white noise signals

In other words, is it possible to relate A1 to Mag1 and A2 to Mag2? Can this even be done analytically or will it require a bit of simulation?

Any help is much appreciated!

share|cite|improve this question
Most recent thought is that Parseval's theorem might have something to do with it - will follow up if I get hit with a bolt of inspiration anytime soon... – Ned Yoxall Feb 12 '13 at 15:33
Look at the Fourier Power Spectrum for your time-series. – Killercam Feb 12 '13 at 15:51
I'm looking for a general relationship between the variation (in time) and the Fourier coefficient in the case of white noise - I don't see how looking at the power spectrum in one particular case could be useful? – Ned Yoxall Feb 12 '13 at 16:01
To what end? I don't see how the coefficent can give you any information here. For a given signal, the power spectrum gives a plot of the portion of a signal's power (energy per unit time) falling within given frequency bins. – Killercam Feb 12 '13 at 16:09
The background for this question is related to some experimental work I'm doing with lasers. To pick out a very small signal, my measurements take place at a specific point in frequency space (roughly 600kHz). There will, however, be some noise caused by the laser at this frequency. I'd like to work out at what point the laser noise (which I'm assuming to be white) becomes bigger than the signal at 600kHz. The laser trace in time is easily measurable, but that needs to be Fourier transformed to get the noise value. The question above is find how much amplitude variation in time is tolerable. – Ned Yoxall Feb 12 '13 at 16:19

It can be done analytically, but numerical results depend on what conventions you use to define the Fourier transform.

You have $\eta(t)$ a random variate. I assume that you mean short range correlated noise, so that $\langle \eta(t) \eta(t') \rangle = \sigma^2 \delta(t-t')$ ($\langle . \rangle$ indicates an ensemble average).

The spectral amplitude is a complex valued random variate $$z(\omega) = \int e^{i \omega t} \eta(t) dt$$

which has a first moment $$ \langle z(\omega) \rangle = \int e^{i \omega t} \langle \eta(t) \rangle dt = \langle \eta \rangle \delta(\omega) $$

If this is non-zero, you can subtract it off from the noise data; so from here I'll assume zero mean noise data.

Now, the second moment of this data is given by $$ \langle \lvert z(\omega) \rvert^2 \rangle = \int \int e^{i \omega t } e^{-i \omega t'} \langle \eta(t) \eta(t') \rangle = \int \sigma^2 dt $$

There are a few concepts in this derivation:

  • Note the interchange in order between doing the integrals, and doing the ensemble averaging; this is almost always mathematically justified.
  • The second moment equation comes from explicitly writing out $\lvert z(\omega) \rvert^2=z(\omega) z^*(\omega)$, and then simplifying using the $\delta$ function correlation structure of the noise.
  • As a formal, indefinite integral, the final expression is not completely defined; I tend to interpret as the integral from $[-L/2,L/2]$ for some arbitrarily large, but finite $L$.

The key results are that the expected intensity of the power spectrum: 1. is linearly proportional to the variance of the noise, and 2. grows linearly with the integration time.

Secondarily, depending on where you put the factor of $2 \pi$ involved in the Fourier transform, you may need to account for it in your noise spectrum.

For discretely sampled data, essentially the same logic applies, but with the integrals replaced by discrete sums. The key thing in this case is knowing how/where your FFT library applies factors of $1/N$; when doing this kind of thing, I often explicitly check what is going on by injecting simulated zero-mean, unit-variance Gaussian noise into the spectral processing chain just to verify what comes out. Once you know the response to unit-variance noise, you scale that result by the observed noise variance.

share|cite|improve this answer
presumably the OP will use the same conventions for each signal, so the analytical relationship does not depend on the conventions used. Hopefully, the same sampling rate and time interval for sampling will be used each time, too. – joseph f. johnson Feb 12 '13 at 18:37
That point about thinking of the Fourier transform as being over a finite range, and then taking that range to infinity as a limit, is actually really important. If you don't do that you won't even get the dimensions right. You have to multiply the Fourier transform integral by $1/\sqrt{L}$ to get a spectral density with the right dimensions. – DanielSank Dec 25 '14 at 14:07

The answer depends whether you have a discrete or continuous signal and wether you have white noise or not:

1) continuous signal (i.e. f(t)) and white noise

  • No, there is no analytical relationship. This is because the variances/standard deviations A1 and A2 are infinite

2) discrete signal (sampled with sample rate $f_s$) and white noise

  • The (one sided) noise spectral density is $S_{xx} = 2 A_{1}^2 / f_s$ with $A_1$ the standard deviation of the noise

3) non white noise (with effective bandwidth $B$):

  • The (one sided) noise spectral density is $S_{xx} = A_{1}^2 / B$ with $A_1$ the standard deviation of the noise

Here the engineering convention is assumed, that the frequency is non-angular and the the noise spectral density is one-sided. This is what spectrum analyzer instruments output. Theorists (also in engineering) usually use the two-sided convention, which just omits the above factor of two.

share|cite|improve this answer
yes, that is a typo. I changed it. thanks – Andreas H. Feb 12 '13 at 17:41

The answer is very simple. If the amplitude of the noise is multiplied by a factor of D, then the Fourier transform has to be multiplied by the same factor. (the different conventions make no difference since obviously you are going to use the same conventions for each signal.)

But I assume you mean power spectrum, not Fourier transform, since that's what spectral analysers measure and it's what your graph looks like. If the amplitude of the noise is multiplied by a factor of D, its power is multiplied by $D^2$, which gives you the relationship sought. Note that in this case, the variance of the noise would go up by $D^2$, so, As @Dave says, the power goes up linearly with the variance, like I said.

Worries about infinite variance are irrelevant since white noise is an idealisation, really you probably have some kind of smoothed pink noise, and your graphs from your analyser are clearly finite. And in mathematical theory, we still distinguish between the Dirac Delta function $\delta$ and its multiples, $D\delta$, anyway, so the analytical relationship holds good for infinity, too.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.