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Subject:
Power spectrum of a normally distributed noise
Category: Science > Math Asked by: noisy-ga List Price: $20.00 |
Posted:
18 Jun 2002 13:07 PDT
Expires: 25 Jun 2002 13:07 PDT Question ID: 28618 |
What is the power spectrum of a signal consisting of normally distributed noise? Is there a formula that relates the standard deviation of the normally distributed noise to the power spectrum of the noise? I think I have read somwhere that the power of the signal is related to the standard deviation of the normal distribution, but I have not been able to confirm this. |
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Subject:
Re: Power spectrum of a normally distributed noise
Answered By: richard-ga on 18 Jun 2002 14:54 PDT |
Hello and thanks for your question. I'll assume that when you say normal noise you mean white (Gaussian) noise, also known as Johnson noise or thermal noise. In white noise, by definition, the power density is constant over all ranges of frequency. In other words, Noise Power= kT [delta]f or Noise Power per hertz = kT where k = Boltzmann's constant in joules per Kelvin, T = temperature in Kelvin degrees. So again, the power density is equal throughout the frequency spectrum, depending only on k and T: "Thermal Noise" http://www.atis.org/tg2k/_thermal_noise.html This relationship is fully explained (and the power spectrum that you specifically asked about is illustrated) on pages 12-13 of the following Adobe Acrobat document. I really can't summarize it here because you need to see the graphics and I'd need a Greek alphabet anyway, so please look at the following: "Power Spectra" http://www.upscale.utoronto.ca/GeneralInterest/Harrison/TimeSeries/PowerSpectra.pdf And here's a longer document that provides a more complete explanation (again, please look at the .pdf file for the equations and graphs: "Op Amp Noise Theory and Applications" http://www-s.ti.com/sc/psheets/sloa082/sloa082.pdf I should close by remarking that this is a highly complex and technical area. If you want to see more of the analysis, there's an article ironically called "Basics" that goes deeper into this correlation than I can hope to explain: "Basic Definitions and Theorems about ARIMA models" http://www.xycoon.com/basics.htm Other useful links: Chem 524 Notes http://www.chem.uic.edu/chem524/notes10/notes10.html Distributions http://astronomy.swin.edu.au/~pbourke/analysis/distributions/ Search terms used: Johnson noise "power spectrum" "standard deviation" noise "standard deviation" "power spectrum" gaussian "johnson noise" Please let me know if you have any questions after checking out these sources, and thanks again! Richard | |
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Subject:
Re: Power spectrum of a normally distributed noise
From: west-ga on 18 Jun 2002 15:23 PDT |
Dear noisy, Your recollection that noise power is related to the standard deviation of the normal distribution is correct. The relationship is as follows: The standard deviation equals the root mean square (r.m.s.) value of the normally distributed noise waveform. Therefore if the noise voltage is developed across a resistor, the noise power is simply the square of the standard deviation divided by the value of the resistor. There are good lecture notes on Random Signals and Noise at the following link. See Chapter 8 'Power Density Spectrum' equation 8-10 for the Power Density Spectrum. I hope this contributes to your understanding. |
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Re: Power spectrum of a normally distributed noise
From: west-ga on 18 Jun 2002 15:25 PDT |
Sorry noisy I forgot to give you the lin which is as follows: http://www.eb.uah.edu/ece/courses/ee420-500/ |
Subject:
Re: Power spectrum of a normally distributed noise
From: hedgie-ga on 18 Jun 2002 18:07 PDT |
Dear noisy This is just a short note, not to disagree withe my esteemed fellow researchers, but to assure you that does not have to be all that complex. Einstein said : explanation be as simple as possible .. , let's try that: The phrase you use, normally distributed noise, is a bit ambiguous, but most likely meaning is that the power spectrum of that noise is normal, aka gaussian aka Bell curve shaped. This is not the same as white noise, which means (usually) uniform (flat) distribution. So, unless you make question more exact, the best answer is: the relation is identity: half-width of the power spectrum is the standard deviation of signal around its mean. Here is a simple into to Gaussian (normal) distributions http://www.umich.edu/~orgolab/Chroma/chromahow3.html hedgie |
Subject:
Re: Power spectrum of a normally distributed noise
From: west-ga on 18 Jun 2002 20:11 PDT |
Dear hedgie Totally agree with keeping explanations as simple as possible. However I must point out that for almost all noise phenomena the Central Distribution Theorem applies. Therefore the Normal (otherwise known as Gaussian) Distribution is appropriate and tells us the likelihood of the value of the noise waveform attaining any particular magnitude. |
Subject:
Re: Power spectrum of a normally distributed noise
From: hedgie-ga on 19 Jun 2002 02:34 PDT |
Quick self-correction Yes, west-ga. The Gaussian distribution in question is most likely result of the Central Distribution Theorem which inddeed is likely to apply. I am a bit woried about the noisy-ga who did not came back to clarify and probably regrets s/he ever asked a question evoking statements such as "(r.m.s.) value of the normally distributed noise waveform .." the meaning of which, even we, the experts, may ponder and argue about. I, in my eagerness to simplify have forgotten the second half of Einstein's quote ' ..but not simpler' and said the relationship is identity. Well, depending on the meaning of the asker's term "normally distributed noise' it may actually be an inverse relationship -a mathematical version of the uncertainty principle. As a pennance, I will describe a scenario: Lets imagine that the the noisy signal is repeated measurement of some physical quantity. The measurement repeated at regular intervals of will yield a gaussian distribution of measured values, as described in any elementary error analysis recepie, e.g. http://www.physics.lsa.umich.edu/IP-LABS/Errordocs/stddev.html Now, if we do the Fourier transform of that distribution, we get a power spectrum, which is also Gaussian. There is a relationship between the half-width of these two gaussians, relation between the uncertaneity in frequency and in time: http://sepwww.stanford.edu/sep/prof/waves/rnd/paper_html/node2.html But, unless noisy cames back, we may never know what memory s/he was chasing and what scenario or experiment s/he had on mind .. |
Subject:
Re: Power spectrum of a normally distributed noise
From: noisy-ga on 19 Jun 2002 18:30 PDT |
Dear Hedgie-ga and West-ga, Thank-you for your comments, and also for the links you've suggested, I'm eager to take a look at them. I'm sorry I haven't been able to clarify my question sooner, but it took me a while to get through the material suggested by Richard-ga ;). I'll try to make the clarification less ambiguous than the question: The data that I am looking at are from an experiment very similar to the one hypothesized by Hedgie. I have a time series of values (in this specific case they are supposed to represent some voltages) and the values are assumed to be normally distributed (N(0,sd)). At first I was thinking about writing a program to generate the power spectrum of the data, but I remembered from a course long ago something about the power of the power spectrum of Gaussian noise being related the the standard deviation of the time series, however I couldn't quite remember. I thought if I could could remember the relationship between the standard deviaition of the noise in the time series to the power in the power spectra (or get someone to find it for me ;) I could save myself some programming time. So far from the materials I've read, I haven't found a formula for the relationship between power and standard deviation, however it has been ages since I took a course on this type of math, so it's quite possible I could be mistaken about that. Thanks again for your comments, they really helped! |
Subject:
Re: Power spectrum of a normally distributed noise
From: west-ga on 19 Jun 2002 20:44 PDT |
Dear noisy-ga Thank you for your comment. Having grasped the gist of your experiment and your need to find the power spectrum, I have concluded as follows: 1. Your slightly hazy memory of use of the standard deviation (sd) actually relates to the calculation of power and not to the calculation of the power spectrum. The sd does not appear to be useful in arriving at the power spectrum. 2. You should proceed to use a program to arrive at the power spectrum. A suitable program is described at the following link: http://sprott.physics.wisc.edu/cdafaq.htm |
Subject:
Re: Power spectrum of a normally distributed noise
From: west-ga on 21 Jun 2002 15:18 PDT |
Hello noisy-ga, richard-ga and hedgie-ga, Just for the record, my earlier reference to the 'Central Distribution Theorem' should have read 'Central Limit Theorem'. It's pretty obvious my meaning has been understood. Sorry for the slip. west-ga |
Subject:
Re: Power spectrum of a normally distributed noise
From: abalter-ga on 21 Jun 2002 20:52 PDT |
I just want to say that I think there has been a great deal of over-complication and misunderstanding here. Noisy just wants to know what relation there is to the standard deviation of some Gaussian noise and the power spectrum. Simply, the power spectrum of noise provides an estimate (usually good) of the CONTRIBUTION OF EACH FREQUENCY TO THE TOTAL VARIANCE. That means that the TOTAL POWER is equivalent to the VARIANCE. The variance is the square of the standard deviation. It sounds like that is all you really needed to know. Now by looking at the power spectrum, you can determine whether most of your variation is coming from high frequency events (more power at high frequencies) or from slowly changing low frequency events (more power at low frequencies). You may find that it is fairly evenly distributed over all frequencies. This would be the case of white noise mentioned before--white because it is an equal mixture of all frequencies. With that settled I would like to clear up a big mess. There is a big difference between the probability densty function, PDF, (e.g. the familiar Gaussian curve which shows how many times each even ocurrs) and the power spectrum. The PDF has absolutely no information of how the events have evolved in time. As far as the PDF is concerned, the events could have happened in ascending order, decending order or completely random. The PDF, a simple counting of events, will not change. When the PDF is normalized to the total number of events it is called a histogram. Now, the power spectrum has a lot of information about the time evolution of the process, but (almost) none about the number of individual events. If both the power spectrum and the phases of the frequency componants are known, then the PDF can be found by reconstruction the original time series. However, if only the power spectrum is known, then NOTHING AT ALL CAN BE SAID ABOUT THE PDF! For those not familiar with this fact, I suggest trying it out with a numerical simulation. Take a given power spectrum and give it different phases. Then reconstruct a timeseries through inverse fft. You will find completely different PDF's! It is worth saying one more thing about power spectra. I would like to remind everyone of the Werner-Khinchine theorem. THE POWER SPECTRUM IS THE FOURIER TRANSFORM OF THE AUTOCORRELATION FUNCTION. So, another way to view the time evolution is in terms of time lag correlations. By the properties of the Dirac delta function, or simply the definition of autocorrelation, the zero-lag autocorrelation is equal to the VARIANCE or TOTAL POWER (same thing). As a final note, I will clear up one more thing. It is possible to Fourier transform a PDF. This has ABSOLUTELY NOTHING TO DO WITH THE POWER SPECTRUM--it is not in the time domain. Suppose you have a probability distribution p(x). Then the Fourier transform P(k) is a function of wavelength, and is called the CHARACTERISTIC FUNCTION of the PDF and has some nice properties in general. In specific, the Gaussian is the unique distribution for which the shape of the PDF p(x) is the same as that of the characteristic function P(k). If p(x) = (2*pi*sigma)^-1/2 * exp[ -1/2 * ( (x - mu)/sigma )^2 ] then P(k) = integral p(x) * exp(-i*k*x) = exp( -1/2*sigma^2*k^2 + i*k*mu ) I think someone brought that up, but it has nothing to do with power spectrum (time domain) analysis. All the best to everyone (especially noisy). I hope this helps! |
Subject:
Re: Power spectrum of a normally distributed noise
From: west-ga on 22 Jun 2002 01:43 PDT |
Dear abalter-ga Firstly there is an error in the following statement in your first paragraph: " That means that the TOTAL POWER is equivalent to the VARIANCE". This statement should read " That means that the TOTAL POWER is proportional to the VARIANCE". This corrected statement was already demonstrated by the equation in my first comment on 18 June. Secondly nobody has suggested that noisy-ga tries to derive the Power Spectrum from the PDF or vice versa. Basically I suggested that noisy-ga use a program to process the data from the time series to arrive at the Power Spectrum. By the way I'm very pleased to see you confirm that the standard deviation is of no use in calculating the power spectrum! |
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