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DC Component of White Noise?

Started by porterboy August 11, 2004
White noise has a PSD = No/2 for all frequencies w. Does this hold
true at w = 0? In this case does that mean that white noise must have
a DC component. What about zero mean white noise? Does this have:
PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta?

I am simulating a baseband OFDM system and I want to add system noise
at the appropriate level for a given SNR. Do I add noise on all tones
equally? Given that the DC tone is not used, should I also turn off
the noise on this tone? I am assuming a zero-mean white-noise channel.
porterboy76@yahoo.com (porterboy) writes:

> White noise has a PSD = No/2 for all frequencies w. Does this hold > true at w = 0? In this case does that mean that white noise must have > a DC component. What about zero mean white noise? Does this have: > PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta? > > I am simulating a baseband OFDM system and I want to add system noise > at the appropriate level for a given SNR. Do I add noise on all tones > equally? Given that the DC tone is not used, should I also turn off > the noise on this tone? I am assuming a zero-mean white-noise channel.
What happens if you integrate a finite PSD in a 0-Hz bandwidth (say, at DC)??? What does that tell you about the presence of a DC "tone"? -- % Randy Yates % "She's sweet on Wagner-I think she'd die for Beethoven. %% Fuquay-Varina, NC % She love the way Puccini lays down a tune, and %%% 919-577-9882 % Verdi's always creepin' from her room." %%%% <yates@ieee.org> % "Rockaria", *A New World Record*, ELO http://home.earthlink.net/~yatescr
porterboy wrote:

> White noise has a PSD = No/2 for all frequencies w. Does this hold > true at w = 0? In this case does that mean that white noise must have > a DC component. What about zero mean white noise? Does this have: > PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta? > > I am simulating a baseband OFDM system and I want to add system noise > at the appropriate level for a given SNR. Do I add noise on all tones > equally? Given that the DC tone is not used, should I also turn off > the noise on this tone? I am assuming a zero-mean white-noise channel.
Yes, this holds true at w = 0. Because the power spectral density is finite everywhere if you narrow your filter down to zero bandwidth then you will pass no noise with a white noise input. The integration from -infinity to infinity that you perform on white noise to get the mean has zero bandwidth, so you get zero mean. The only way to see a non-zero mean is if your PSD has an impulse at DC. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
a zero mean process could have no DC component in it. PSD is a
statitistical curve and not the spectrum itself. PSD non zero at DC
doesnt meant there is a net DC component in noise. If that be the
case, I should be able to generate power from noise. To understand PSD
better, you should see that if u integrate PSD over the entire
spectrum, u get the variance of the process. that is PSD gives you a
picture of how much each part of the spectrum contrubutes to the
variance. that means you should add noise of same variance to all the
tones as white noise has uniform PSD

amar

le.com>...
> White noise has a PSD = No/2 for all frequencies w. Does this hold > true at w = 0? In this case does that mean that white noise must have > a DC component. What about zero mean white noise? Does this have: > PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta? > > I am simulating a baseband OFDM system and I want to add system noise > at the appropriate level for a given SNR. Do I add noise on all tones > equally? Given that the DC tone is not used, should I also turn off > the noise on this tone? I am assuming a zero-mean white-noise channel.
after some thought, I feel I could be wrong. I am confused. cud
somebody enliten pls.

amar

porterboy76@yahoo.com (porterboy) wrote in message news:<c4b57fd0.0408110121.55376b70@posting.google.com>...
> White noise has a PSD = No/2 for all frequencies w. Does this hold > true at w = 0? In this case does that mean that white noise must have > a DC component. What about zero mean white noise? Does this have: > PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta? > > I am simulating a baseband OFDM system and I want to add system noise > at the appropriate level for a given SNR. Do I add noise on all tones > equally? Given that the DC tone is not used, should I also turn off > the noise on this tone? I am assuming a zero-mean white-noise channel.
amara vati wrote:
> after some thought, I feel I could be wrong. I am confused. cud > somebody enliten pls.
Enlighten or more likely, further confuse. From a probability perspective, think of constant voltage drawn from some sort of Gaussian constant voltage generator where the mean is zero and the variance is 1 volt. The actual voltage you measure is not likely to be zero, but if you repeat the experiment enough times the average of all the measurements will tend toward zero as you continue to repeat the experiment.
> > amar > > porterboy76@yahoo.com (porterboy) wrote in message news:<c4b57fd0.0408110121.55376b70@posting.google.com>... > >>White noise has a PSD = No/2 for all frequencies w. Does this hold >>true at w = 0? In this case does that mean that white noise must have >>a DC component. What about zero mean white noise? Does this have: >>PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta? >> >>I am simulating a baseband OFDM system and I want to add system noise >>at the appropriate level for a given SNR. Do I add noise on all tones >>equally? Given that the DC tone is not used, should I also turn off >>the noise on this tone? I am assuming a zero-mean white-noise channel.
porterboy wrote:

> White noise has a PSD = No/2 for all frequencies w. Does this hold > true at w = 0? In this case does that mean that white noise must have > a DC component. What about zero mean white noise? Does this have: > PSD(w) = No/2(1 - d(0)) where d(w) is the Kronecker Delta? > > I am simulating a baseband OFDM system and I want to add system noise > at the appropriate level for a given SNR. Do I add noise on all tones > equally? Given that the DC tone is not used, should I also turn off > the noise on this tone? I am assuming a zero-mean white-noise channel.
It will help if you clarify your understanding of what PDF, PSD, and related quantitise mean. Ignore the acronyms; consider a marksman aiming at a piece of graph paper. His aim is perturbed by so many independent disturbances that the deviation of his shots is Gaussian. He is a very good marksman, so the centroid of his cluster is right at his point of aim: the origin. Consider now the probability of a shot falling at some X location (ignore Y position) and draw this curve. Smooth it to indicate what it would look like if he had fired a very large number of shots. I think you can see that it would be symmetric about the Y axis, have its greatest value at X=0, and continue out very far at low amplitude. In short, a classical bell curve. Reading the curve, what is the probability that a shot will fall precisely on the Y axis? Zero, of course. How can that be? Jerry -- ... the worst possible design that just meets the specification - almost a definition of practical engineering. .. Chris Bore &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Jerry Avins wrote:
...
> It will help if you clarify your understanding of what PDF, PSD, and > related quantitise mean. Ignore the acronyms; consider a marksman aiming > at a piece of graph paper. His aim is perturbed by so many independent > disturbances that the deviation of his shots is Gaussian. He is a very > good marksman, so the centroid of his cluster is right at his point of > aim: the origin. Consider now the probability of a shot falling at some > X location (ignore Y position) and draw this curve. Smooth it to > indicate what it would look like if he had fired a very large number of > shots. I think you can see that it would be symmetric about the Y axis, > have its greatest value at X=0, and continue out very far at low > amplitude. In short, a classical bell curve. Reading the curve, what is > the probability that a shot will fall precisely on the Y axis? Zero, of > course. How can that be?
Does that mean no shot will ever fall on the Y axis? Or conversly: A shot entered the X axis at, say, point 0.5 - what is the probability of that? And what does this tell us about DC content of Gaussian white noise? Consider the signal s(n) = 1 for all n. Could this be an instance of zero mean Gaussian white noise process? If so, how does it differ (in probability) from other particular instance where the mean converges to zero as n -> infinity?
> > Jerry
Andor wrote:

> Jerry Avins wrote: > ... > >>It will help if you clarify your understanding of what PDF, PSD, and >>related quantitise mean. Ignore the acronyms; consider a marksman aiming >>at a piece of graph paper. His aim is perturbed by so many independent >>disturbances that the deviation of his shots is Gaussian. He is a very >>good marksman, so the centroid of his cluster is right at his point of >>aim: the origin. Consider now the probability of a shot falling at some >>X location (ignore Y position) and draw this curve. Smooth it to >>indicate what it would look like if he had fired a very large number of >>shots. I think you can see that it would be symmetric about the Y axis, >>have its greatest value at X=0, and continue out very far at low >>amplitude. In short, a classical bell curve. Reading the curve, what is >>the probability that a shot will fall precisely on the Y axis? Zero, of >>course. How can that be? > > > Does that mean no shot will ever fall on the Y axis? Or conversly: A > shot entered the X axis at, say, point 0.5 - what is the probability > of that? And what does this tell us about DC content of Gaussian white > noise? > > Consider the signal s(n) = 1 for all n. Could this be an instance of > zero mean Gaussian white noise process? If so, how does it differ (in > probability) from other particular instance where the mean converges > to zero as n -> infinity? > > >>Jerry
It means that in theory there will be no shot where the exact centroid of the bullet lies over the exact centroid of the y axis -- the same for any other finite set of points on the target that have zero area. What this tells us about the DC content of the Gaussian white noise is that because DC is a single point on the frequency axis when you integrate the PDF from 0 to 0 there will be no area under the curve (finite number * zero length = zero). If you mean n as being sample time then s(n) = 1 for all n has a mean value of 1, not zero, so it cannot be an instance of a zero mean Gaussian white noise process. If you were on the grounds of a University math department you could claim that it's a Gaussian process with mean = 1 and sigma = 0, but that's degenerate. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Andor wrote:

> Jerry Avins wrote: > ... > >>It will help if you clarify your understanding of what PDF, PSD, and >>related quantitise mean. Ignore the acronyms; consider a marksman aiming >>at a piece of graph paper. His aim is perturbed by so many independent >>disturbances that the deviation of his shots is Gaussian. He is a very >>good marksman, so the centroid of his cluster is right at his point of >>aim: the origin. Consider now the probability of a shot falling at some >>X location (ignore Y position) and draw this curve. Smooth it to >>indicate what it would look like if he had fired a very large number of >>shots. I think you can see that it would be symmetric about the Y axis, >>have its greatest value at X=0, and continue out very far at low >>amplitude. In short, a classical bell curve. Reading the curve, what is >>the probability that a shot will fall precisely on the Y axis? Zero, of >>course. How can that be? > > > Does that mean no shot will ever fall on the Y axis?
I don't address that. My intention is to consider (plot) only the X value of position, ignoring Y.
> Or conversly: A > shot entered the X axis at, say, point 0.5 - what is the probability > of that? And what does this tell us about DC content of Gaussian white > noise?
I think the OP's error arose from failing to distinguish probability from probability density. I tried to stimulate an example that requires the distinction.
> Consider the signal s(n) = 1 for all n. Could this be an instance of > zero mean Gaussian white noise process? If so, how does it differ (in > probability) from other particular instance where the mean converges > to zero as n -> infinity?
Jerry -- ... the worst possible design that just meets the specification - almost a definition of practical engineering. .. Chris Bore &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;