Randy Yates wrote:> Hi Peter, > > p.kootsookos@remove.ieee.org (Peter K.) writes: > > >>Randy Yates <yates@ieee.org> writes: >> >> >>>Your error is in the logic "zero mean ==> DC component of the spectrum >>>is zero." >>> >>>The PSD is a power *density* - the power per unit Hz. When a random >>>signal has a "component" at some frequency \omega_c, we mean that there >>>is a non-zero power at that one specific frequency. That means that >>>the power *density* at that frequency goes to infinity, i.e., >>>\Phi_xx(\omega) = f(\omega) + a*\delta(\omega - \omega_c), where >>>f(\omega) is a function defining the PSD at frequencies not at >>>\omega_c. >> >>This bit doesn't make sense to me... you seem to be saying that either >>the PSD is infinite or the PSD is zero... and nothing in between >>(which is not the case, but that's where the way I read the above >>takes me). > > > I can see how I might be confused to say that. What I mean is that if > the PSD is NOT infinite (that is, if it is finite, whether that is > non-zero or zero) at a specific frequency, then there is no > "component" at that frequency. > > >>Perhaps I'm confused by what you mean by "component". > > > I mean there is a sinusoid present at that frequency (and the frequency > could be zero to represent a DC level). > > >>>However, a PSD in which a certain band of frequencies, say B, are >>>non-zero but at the same time non-infinite (i.e., have no delta >>>functions) has some non-zero power, say P, over that band B. Thus we >>>have >>> >>> P/B = watts/Hz >>> = (joules/second)/Hz >>> = (joules/second) * (second) >>> = joules >>> >>>In other words, we have some *energy* at DC, but no power. >>> >>>Howz that? ... :) >> >>Whereas you and Jerry look at this as the difference between power and >>energy signals, I can't really see it. >> >>To me, DC has infinite energy (finite power) and so does white noise, >>so I don't see that as the difference between the two. > > > Oh come now Peter. White noise has finite power? You know better! > In fact that is the difference - a "component" has infinite energy > but finite power; white noise has neither finite energy nor > finite power. > > >>I'm starting to think it's more related to the difference between >>first order (mean) and second order statistics >>(autocovariance/autocorrelation). >> >>My "zero mean" <-> "DC component is zero" MUST be true if what you're >>talking about is the Fourier transform of the noise signal, provided >>the signal is ergodic: >> >>FT[ x(t) ] = \int x(t) exp(-i*wt) dt (give or take a scale factor) >> >>which, for w=0 is just \int x(t) dt. >> >>However, we're not talking about the Fourier transform of the signal >>when it comes to the PSD. We're really talking about the Fourier >>transform of the AUTOCORRELATION of the signal. > > > Yes! > > >>Clearly, with the autocorrelation being C.\delta(t), there will be a >>non-zero component of the PSD across all frequencies. >> >>Comments? > > > Right. I think we're very close to agreeing.Sometimes, the concept of negative frequencies can be useful for more than simplicity of calculation, and I think this is one of them. Ideally, the spectrum of white noise stretches from -oo to +oo, and ought not be altered by being shifted a few Hz one way or another. Imagine that we shift the spectrum +100 Hz, and that there had been a DC offset before the shift. The result of that shift would certainly not be white noise, so the signal, whatever it was, hadn't been white noise to start with. Q.E.D.? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Is White Noise Necessarily Zero-Mean?
Started by ●February 15, 2006
Reply by ●February 19, 20062006-02-19
Reply by ●February 20, 20062006-02-20
Jerry Avins wrote:> Peter K. wrote: > >> Randy Yates wrote: >> >> >>> Let Z(t) be a white-noise (stationary) random process. >>> Can we conclude that Z(t) has zero mean? >> >> >> >> Suppose it is zero mean. >> >> Therefore the DC component of the spectrum is zero. >> >> But white noise has a constant (flat, non-zero) spectrum. >> >> Discuss. >> >> :-) > > > :-) indeed! White noise has constant power per unit of bandwidth. > How much power does DC have? > > JerryJerry, I thought you were about to deliver the deciding argument here which would run something like this: The bandwidth of DC is zero because "DC" is one single frequency point, not a frequency interval. If we have noise-power present with a given power per unit bandwidth then the product of this power density times the bandwidth is zero. So, the level of the DC component is zero, i.e. there is no DC component present in white noise. (Question for those arguing the opposite: If DC is present, what polarity would it be?) Regards, John
Reply by ●February 20, 20062006-02-20
John Monro <johnmonro@optusnet.com.au> writes:> Jerry Avins wrote: >> Peter K. wrote: >> >>> Randy Yates wrote: >>> >>> >>>> Let Z(t) be a white-noise (stationary) random process. >>>> Can we conclude that Z(t) has zero mean? >>> >>> >>> >>> Suppose it is zero mean. >>> >>> Therefore the DC component of the spectrum is zero. >>> >>> But white noise has a constant (flat, non-zero) spectrum. >>> >>> Discuss. >>> >>> :-) >> :-) indeed! White noise has constant power per unit of bandwidth. >> How much power does DC have? >> Jerry > > Jerry, > I thought you were about to deliver the deciding argument here which > would run something like this: > > The bandwidth of DC is zero because "DC" is one single frequency > point, not a frequency interval. If we have noise-power present with > a given power per unit bandwidth then the product of this power > density times the bandwidth is zero. So, the level of the DC > component is zero, i.e. there is no DC component present in white > noise. > > (Question for those arguing the opposite: If DC is present, what > polarity would it be?)Can't tell from the PSD for the same reason you can't tell the phase of a sinusoid. It's a magnitude response only. -- % Randy Yates % "Ticket to the moon, flight leaves here today %% Fuquay-Varina, NC % from Satellite 2" %%% 919-577-9882 % 'Ticket To The Moon' %%%% <yates@ieee.org> % *Time*, Electric Light Orchestra http://home.earthlink.net/~yatescr
Reply by ●February 20, 20062006-02-20
John Monro wrote:> Jerry Avins wrote: > >> Peter K. wrote: >> >>> Randy Yates wrote: >>> >>> >>>> Let Z(t) be a white-noise (stationary) random process. >>>> Can we conclude that Z(t) has zero mean? >>> >>> >>> >>> >>> Suppose it is zero mean. >>> >>> Therefore the DC component of the spectrum is zero. >>> >>> But white noise has a constant (flat, non-zero) spectrum. >>> >>> Discuss. >>> >>> :-) >> >> >> >> :-) indeed! White noise has constant power per unit of bandwidth. >> How much power does DC have? >> >> Jerry > > > Jerry, > I thought you were about to deliver the deciding argument here which > would run something like this: > > The bandwidth of DC is zero because "DC" is one single frequency point, > not a frequency interval. If we have noise-power present with a given > power per unit bandwidth then the product of this power density times > the bandwidth is zero. So, the level of the DC component is zero, i.e. > there is no DC component present in white noise.I thought I made that argument by implication. I should be more explicit.> (Question for those arguing the opposite: If DC is present, what > polarity would it be?)Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 20, 20062006-02-20
Randy Yates <yates@ieee.org> writes:> Hi Peter, > > p.kootsookos@remove.ieee.org (Peter K.) writes: > > > > > To me, DC has infinite energy (finite power) and so does white noise, > > so I don't see that as the difference between the two. > > Oh come now Peter. White noise has finite power? You know better!Yup, I was getting confused between discrete and continuous time.> > Right. I think we're very close to agreeing. >Yup! Thanks for humoring me! Ciao, Peter K. -- "And he sees the vision splendid of the sunlit plains extended And at night the wondrous glory of the everlasting stars."
Reply by ●February 20, 20062006-02-20
p.kootsookos@remove.ieee.org (Peter K.) writes:> Randy Yates <yates@ieee.org> writes: > >> Hi Peter, >> >> p.kootsookos@remove.ieee.org (Peter K.) writes: >> >> > >> > To me, DC has infinite energy (finite power) and so does white noise, >> > so I don't see that as the difference between the two. >> >> Oh come now Peter. White noise has finite power? You know better! > > Yup, I was getting confused between discrete and continuous time. > >> >> Right. I think we're very close to agreeing. >> > > Yup! Thanks for humoring me!It is I who should be thanking you, Peter. You and Dilip are the ones that have most inspired me to try to understand this question. -- % Randy Yates % "Maybe one day I'll feel her cold embrace, %% Fuquay-Varina, NC % and kiss her interface, %%% 919-577-9882 % til then, I'll leave her alone." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr
Reply by ●February 20, 20062006-02-20
Jerry Avins wrote:> John Monro wrote: > >> Jerry Avins wrote: >> >>> Peter K. wrote: >>> >>>> Randy Yates wrote: >>>> >>>> >>>>> Let Z(t) be a white-noise (stationary) random process. >>>>> Can we conclude that Z(t) has zero mean? >>>> >>>> >>>> >>>> >>>> >>>> Suppose it is zero mean. >>>> >>>> Therefore the DC component of the spectrum is zero. >>>> >>>> But white noise has a constant (flat, non-zero) spectrum. >>>> >>>> Discuss. >>>> >>>> :-) >>> >>> >>> >>> >>> :-) indeed! White noise has constant power per unit of bandwidth. >>> How much power does DC have? >>> >>> Jerry >> >> >> >> Jerry, >> I thought you were about to deliver the deciding argument here which >> would run something like this: >> >> The bandwidth of DC is zero because "DC" is one single frequency >> point, not a frequency interval. If we have noise-power present with >> a given power per unit bandwidth then the product of this power >> density times the bandwidth is zero. So, the level of the DC >> component is zero, i.e. there is no DC component present in white >> noise. > > > I thought I made that argument by implication. I should be more explicit >Yes,on reflection I see that. You probably should be more explicit because the point you made is really is the deciding argument, and yet it has gone largely un-noticed.>> (Question for those arguing the opposite: If DC is present, what >> polarity would it be?) > > > Jerry
Reply by ●February 20, 20062006-02-20
Randy Yates wrote:> Let Z(t) be a white-noise (stationary) random process. > Can we conclude that Z(t) has zero mean? > > My thought is: yes. The intuitive reason that comes to > mind (and it may be wrong!) is this: If Z(t) has a > non-zero mean, then there would be some amount of > correlation between samples due to the means. Thus > the autocorrelation would be a delta function. > -- > % Randy Yates % "Bird, on the wing, > %% Fuquay-Varina, NC % goes floating by > %%% 919-577-9882 % but there's a teardrop in his eye..." > %%%% <yates@ieee.org> % 'One Summer Dream', *Face The Music*, ELO > http://home.earthlink.net/~yatescrSoon as I go out of town for a few days, you come up with an interesting topic, and I get in on it late. To answer directly, white noise does NOT need to be zero mean. A process is white noise if the values x(t1) and x(t2) are uncorrelated for all t1 != t2. This does NOT mean the correlation is zero! A process is ORTHOGONAL if R(t1,t2) = 0 for every t1,t2 t1!= t2 R is the correlation matrix. A process is UNCORRELATED if the covariance matrix C(t1,t2) = 0 for every t1,t2 t1!= t2 C(X,Y) = E[(X - MUx) (Y - MUy)] MUx is expected value of X = mean MUy is expected value of Y = mean This is the familiar C(X,Y) = E[XY] - E[X]E[Y] The reason we normally talk of a zero-mean process when talking about white noise, is that we generally are talking about additive white noise. To make things easier, we use GAUSSIAN distributed processes. This is so for several reasons: 1. The mean, median, and mode are all equal. 2. The process is symmetrical about the mean 3. We can get a zero-mean process merely by subtracting the mean. 4. And most importantly, the process is TOTALLY defined by the mean and variance. If we can merely subtract the mean to get a zero-mean process, then the covariance matrix becomes C(X,Y) = E[XY], which is identically the correlation matrix. Unfortunately, we teach Gaussian, zero-mean processes so much, many students think the correlation matrix R(X,Y) = 0 defines a white process, when actually C(X,Y) = 0 is the definition Maurice Givens Givens Burnet, Inc. Chicago, IL
Reply by ●February 20, 20062006-02-20
Randy Yates wrote:> John Monro <johnmonro@optusnet.com.au> writes: > > >>Jerry Avins wrote: >> >>>Peter K. wrote: >>> >>> >>>>Randy Yates wrote: >>>> >>>> >>>> >>>>>Let Z(t) be a white-noise (stationary) random process. >>>>>Can we conclude that Z(t) has zero mean? >>>> >>>> >>>> >>>>Suppose it is zero mean. >>>> >>>>Therefore the DC component of the spectrum is zero. >>>> >>>>But white noise has a constant (flat, non-zero) spectrum. >>>> >>>>Discuss. >>>> >>>>:-) >>> >>>:-) indeed! White noise has constant power per unit of bandwidth. >>>How much power does DC have? >>>Jerry >> >>Jerry, >>I thought you were about to deliver the deciding argument here which >>would run something like this: >> >>The bandwidth of DC is zero because "DC" is one single frequency >>point, not a frequency interval. If we have noise-power present with >>a given power per unit bandwidth then the product of this power >>density times the bandwidth is zero. So, the level of the DC >>component is zero, i.e. there is no DC component present in white >>noise. >> >>(Question for those arguing the opposite: If DC is present, what >>polarity would it be?) > > > Can't tell from the PSD for the same reason you can't tell the phase > of a sinusoid. It's a magnitude response only.Point taken! Regards,John
Reply by ●February 20, 20062006-02-20
maury001@core.com writes:> Randy Yates wrote: >> Let Z(t) be a white-noise (stationary) random process. >> Can we conclude that Z(t) has zero mean? >> >> My thought is: yes. The intuitive reason that comes to >> mind (and it may be wrong!) is this: If Z(t) has a >> non-zero mean, then there would be some amount of >> correlation between samples due to the means. Thus >> the autocorrelation would be a delta function. >> -- >> % Randy Yates % "Bird, on the wing, >> %% Fuquay-Varina, NC % goes floating by >> %%% 919-577-9882 % but there's a teardrop in his eye..." >> %%%% <yates@ieee.org> % 'One Summer Dream', *Face The Music*, ELO >> http://home.earthlink.net/~yatescr > > > Soon as I go out of town for a few days, you come up with an > interesting topic, and I get in on it late.Hey Maury, good to hear from you! I don't think we've communicated since the conference!> To answer directly, white noise does NOT need to be zero mean. A > process is white noise if the values x(t1) and x(t2) are uncorrelated > for all t1 != t2.Yes, that is *one* definition. See the paper I wrote on this just a couple days ago: http://www.digitalsignallabs.com/white.pdf I think we are agreeing quite nicely.> This does NOT mean the correlation is zero!True.> A process is ORTHOGONAL if R(t1,t2) = 0 for every t1,t2 t1!= t2 > R is the correlation matrix.I've never seen "orthogonal process" defined. Usually it is that two random variables are orthogonal if cov(X,Y) = 0.> A process is UNCORRELATED if the covariance matrix C(t1,t2) = 0 for > every t1,t2 t1!= t2I've seen it defined as E[X*Y] = E[X] * E[Y]. Same damn thing though.> C(X,Y) = E[(X - MUx) (Y - MUy)] > MUx is expected value of X = mean > MUy is expected value of Y = mean > > This is the familiar C(X,Y) = E[XY] - E[X]E[Y]Yes.> The reason we normally talk of a zero-mean process when talking about > white noise, is that we generally are talking about additive white > noise. To make things easier, we use GAUSSIAN distributed processes. > This is so for several reasons: > > 1. The mean, median, and mode are all equal. > 2. The process is symmetrical about the mean > 3. We can get a zero-mean process merely by subtracting the mean. > 4. And most importantly, the process is TOTALLY defined by the mean > and variance. > > If we can merely subtract the mean to get a zero-mean process, then the > covariance matrix becomes C(X,Y) = E[XY], which is identically the > correlation matrix. > > Unfortunately, we teach Gaussian, zero-mean processes so much, many > students think the correlation matrix R(X,Y) = 0 defines a white > process, when actually C(X,Y) = 0 is the definitionAgain, see the paper. It is defined by at least one author (Papoulis) that way, but another author (Brown) states it another way. But basically we agree - by *either* definition you cannot conclude that the mean is zero. -- % Randy Yates % "Maybe one day I'll feel her cold embrace, %% Fuquay-Varina, NC % and kiss her interface, %%% 919-577-9882 % til then, I'll leave her alone." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr
