Reply by Rune Allnor February 23, 20062006-02-23
Dilip V. Sarwate wrote:
> "Ikaro" <ikarosilva@hotmail.com> wrote in message > news:1140536417.286233.272270@g44g2000cwa.googlegroups.com... > > > > It seems like the standard definitions of WN *should* require the > > process with zero-mean,otherwise it's not white... > > > > This thread seems to be converging to agreement with Randy's > assertion that the common definitions of white noise seem not > to necessarily imply that the process has zero mean. Thus, it > is necessary to explicitly state that the mean of a white noise > process is zero. Indeed, according to Randy, the definitions > of white noise that have given by Brown and Papoulis are not > complete, and should also include a requirement that the mean > be zero. > > So, let us consider a process X(t) with nonzero mean m and > autocorrelation function Rxx(t). Let us also suppose that > Rxx(t) does not depend on the value of m in any way; in fact, > let's take Rxx(t) = N0 delta(t) so that X(t) is a white noise > process with nonzero mean as per the definition that > everybody seems to agree on. > > Let Z(t) = X(t) - m. Then, I think we are all in agreement that > Z(t) is a zero-mean process. Is it a white noise process? Well, > Rzz(t) = Rxx(t) - m^2 = N0 delta(t) - m^2. So, it doesn't seem > to fit anybody's definition of a white noise process since ....Bzzzt... > it doesn't satisfy the requirement that Z(t1) and Z(t2) are > uncorrelated if t1 and t2 are distinct time instants (Papoulis), and > since the power spectral density is phizz(f) = N0 - m^2 delta(f) > ...Bzzzt... the process Z(t) doesn't satisfy the requirement of > constant spectral density (Brown). So, Z(t) and X(t) *both* > cannot be called white noise processes. > > In fact, getting back to X(t), its autocovariance function Cxx(t) > is the same as Rzz(t) and thus the *autocovariance* function of > the process X(t) does seem to depend on the mean m. Somehow > this does not seem quite right; covariances should not depend on > the mean(s) since we are subtracting them off.... Perhaps someone > (Randy?) would care to shed further light on this curious result.... > > Hope this confuses (it certainly is not meant to help; it is > intended to throw fuel on the flames! :-) )
Eh... again, I have not read very many posts in this thread so I may be stating opinions that have already been discused. Randy asked about the connection between the mean and and correlation properties of a process. I tried to argue that the mean has an insignificant (albeit not vanishing) effect on the autocorrelation and thus can be disregarded, while you point out a paradox. Jerry and Ikaro discuss various technical details.
>From the above discussion I tend to adopt the view that "white
noise" baiscally is a white elephant. It is not well-defined as f -> infinite, since this violates the requirement that the signal has finite energy, and thus can not be analyzed in a meaningful way by the Fourier transform. It follows that we can basically do what we like at DC, since we are trying to fit a square peg to a round hole anyway. I think the view have been aired previously on comp.dsp that "white noise" means "noise that is white within the relevant bandwidth." I think that seems reasonable. If one agrees with that, the decision what to do basically depends on whether DC is "relevant" or not. If it is, use the covariance instead of the autocorrelation, and all is well. For a textbook author, either consitently use the autocovariance, or discuss "autocorrelation of zero-mean signals". I prefer the former. Rune
Reply by February 22, 20062006-02-22
Ikaro wrote:
> Hi Maurice, > > I still don't understand... for example using your terminology above > lets assume we have a > WN with nonzero mean that is the result of a dc component added to a > zero-mean normal WN: > > y= x + c > > Where x *is* normal zero-mean WN. Then using the properties you > mentioned above we have (with E[x(t)]=0): > > Ry(t1,t2) =E[y(t1)*y(t2)] > > =E[ (x(t1)+c)*(x(t2)+c) ] > > =E[x(t1)*x(t2) + c*x(t1) + cx(t2) + c^2] > > =E[x(t1)*x(t2)] + c^2 > > =E[x(t1)]E[x(t2)] + c^2 = c^2 for t1!=t2 and using 0 = E[XY] > - E[X]E[Y] > > = Rx(0) + c^2 for t1=t2 > > note that because x is WN zero mean then R(0)=var != 0 then (using > delta as the impulse function): > > Ry(t1,t2)= var*delta(t1-t2) + c^2 %note that the process is still > WS stationary > > But it's Ppwer spectrum is now > > Sy(w)= var + c2*delta(w) != constant as a function of w > > which is not flat anymore, thus non-white. Unless I am making a mistake > ;) > it seems like we are missing something from the definition.
Well Ikaro, you actually proved my case. The power spectrum density describes the variance of a process with respect to frequency. delta(t - t0) = 0 for t != t0. This is part of the definition of delta. Therefore, delta(w) = 0 for w != 0. This means Sy(w)= var + c2*delta(w) = var, for w != 0, a constant for all w != 0
Reply by Ikaro February 21, 20062006-02-21
Hi,

>This thread seems to be converging to agreement with Randy's >assertion that the common definitions of white noise seem not >to necessarily imply that the process has zero mean.
Actually I (as well as dvsarw) was saying just the opposite. If the process is WN than it *must* be zero mean. There can't be a non-zero mean WN process.
>So, let us consider a process X(t) with nonzero mean m and >autocorrelation function Rxx(t). Let us also suppose that >Rxx(t) does not depend on the value of m in any way; in fact, >let's take Rxx(t) = N0 delta(t)
That's not possible, a constant mean will result in a constant term (m^2) in the correlation matrix (see several posts above). -Ikaro
Reply by February 21, 20062006-02-21
"Ikaro" <ikarosilva@hotmail.com> wrote in message 
news:1140536417.286233.272270@g44g2000cwa.googlegroups.com...
> > It seems like the standard definitions of WN *should* require the > process with zero-mean,otherwise it's not white... >
This thread seems to be converging to agreement with Randy's assertion that the common definitions of white noise seem not to necessarily imply that the process has zero mean. Thus, it is necessary to explicitly state that the mean of a white noise process is zero. Indeed, according to Randy, the definitions of white noise that have given by Brown and Papoulis are not complete, and should also include a requirement that the mean be zero. So, let us consider a process X(t) with nonzero mean m and autocorrelation function Rxx(t). Let us also suppose that Rxx(t) does not depend on the value of m in any way; in fact, let's take Rxx(t) = N0 delta(t) so that X(t) is a white noise process with nonzero mean as per the definition that everybody seems to agree on. Let Z(t) = X(t) - m. Then, I think we are all in agreement that Z(t) is a zero-mean process. Is it a white noise process? Well, Rzz(t) = Rxx(t) - m^2 = N0 delta(t) - m^2. So, it doesn't seem to fit anybody's definition of a white noise process since ....Bzzzt... it doesn't satisfy the requirement that Z(t1) and Z(t2) are uncorrelated if t1 and t2 are distinct time instants (Papoulis), and since the power spectral density is phizz(f) = N0 - m^2 delta(f) ...Bzzzt... the process Z(t) doesn't satisfy the requirement of constant spectral density (Brown). So, Z(t) and X(t) *both* cannot be called white noise processes. In fact, getting back to X(t), its autocovariance function Cxx(t) is the same as Rzz(t) and thus the *autocovariance* function of the process X(t) does seem to depend on the mean m. Somehow this does not seem quite right; covariances should not depend on the mean(s) since we are subtracting them off.... Perhaps someone (Randy?) would care to shed further light on this curious result.... Hope this confuses (it certainly is not meant to help; it is intended to throw fuel on the flames! :-) )
Reply by Jerry Avins February 21, 20062006-02-21
Ikaro wrote:
> Hi Jerry, > > Even if the DC bandwidth is zero we don't have a paradox because > according to the post above: > > Ry(t1,t2)= var*delta(t1-t2) + c^2 > Sy(w)= var + c2*delta(w) != constant as a function of w > > Now, from a theoretical point of view, the delta function has zero > bandwidth, but a finite area of one. So the DC component is *not* zero, > and in the case of non-zero mean, higher than the other avegare > spectrum. > > It seems like the standard definitions of WN *should* require the > process with zero-mean,otherwise it's not white...
There's the paradox. -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by Ikaro February 21, 20062006-02-21
Hi Jerry,

Even if the DC bandwidth is zero we don't have a paradox because
according to the post above:

Ry(t1,t2)= var*delta(t1-t2) + c^2
Sy(w)= var + c2*delta(w) !=  constant as a function of w

Now, from a theoretical point of view, the delta function has zero
bandwidth, but a finite area of one. So the DC component is *not* zero,
and in the case of non-zero mean, higher than the other avegare
spectrum.

It seems like the standard definitions of WN *should* require the
process with zero-mean,otherwise it's not white...

Reply by Jerry Avins February 21, 20062006-02-21
Rune Allnor wrote:
> Randy Yates wrote: > >>Let Z(t) be a white-noise (stationary) random process. >>Can we conclude that Z(t) has zero mean? > > > I haven't read comp.dsp for a couple of weeks now, and this thread has > gone > on for a while. I am sure somebody else already have come up with this, > > but here are my 2c anyway: > > There is no connection between the noise being Gaussian and it being > zero mean. Knowing that the noise is zero mean does not imply it > being Gaussian. Knowing that the noise is Gaussian does not > imply that it is 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. > > > We can separate the noise as > > Z(t) = z(t) + m_z > > where z(t) is the zero mean stationary process and m_z is the > man of Z(t). The autocorrelation becomes > > E[Z(t)^2] = E[(z(t)+m_z)^2] = E[&#2013266088;z(t)^2] + 2E[z(t)]m_z + m_z^2 > = E[z(t)^2] + m_z^2 > > since by assumption, E[z(t)]==0. > > So the only effect of adding a DC term to the zero mean > Gaussian process is that the autocorrelation is offset by a > constant value. It is no drama by that, but it does provide a > motivation why the autocovariance may be a preferred quantity > in the analysis of Gaussian processes.
Presumably, an essential property of AWGN is equal power per unit bandwidth at all frequencies. In other words, every band (f + delta_f), delta_f being constant and f being arbitrary, has the same power. This, and the assumption that the power density is not infinite, implies that power ->zero as delta_f -> zero. I ask again, what is the bandwidth of DC? If zero, as I believe, we have a paradox. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by Rune Allnor February 21, 20062006-02-21
Randy Yates wrote:
> Let Z(t) be a white-noise (stationary) random process. > Can we conclude that Z(t) has zero mean?
I haven't read comp.dsp for a couple of weeks now, and this thread has gone on for a while. I am sure somebody else already have come up with this, but here are my 2c anyway: There is no connection between the noise being Gaussian and it being zero mean. Knowing that the noise is zero mean does not imply it being Gaussian. Knowing that the noise is Gaussian does not imply that it is 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.
We can separate the noise as Z(t) =3D z(t) + m_z where z(t) is the zero mean stationary process and m_z is the man of Z(t). The autocorrelation becomes E[Z(t)^2] =3D E[(z(t)+m_z)^2] =3D E[=A8z(t)^2] + 2E[z(t)]m_z + m_z^2 =3D E[z(t)^2] + m_z^2 since by assumption, E[z(t)]=3D=3D0. So the only effect of adding a DC term to the zero mean Gaussian process is that the autocorrelation is offset by a constant value. It is no drama by that, but it does provide a motivation why the autocovariance may be a preferred quantity in the analysis of Gaussian processes. Rune
Reply by Ikaro February 21, 20062006-02-21
Hi Maurice,

I still don't understand... for example using your terminology above
lets assume we have a
WN with nonzero mean that is the result of a dc component added to a
zero-mean normal WN:

y= x + c

Where x *is* normal zero-mean WN. Then using the properties you
mentioned above we have (with E[x(t)]=0):

Ry(t1,t2)         =E[y(t1)*y(t2)]

                     =E[ (x(t1)+c)*(x(t2)+c) ]

	     =E[x(t1)*x(t2) + c*x(t1) + cx(t2) + c^2]

	     =E[x(t1)*x(t2)] + c^2

	     =E[x(t1)]E[x(t2)] + c^2 = c^2     for t1!=t2  and using 0 = E[XY]
- E[X]E[Y]

	     = Rx(0) + c^2                         for t1=t2

note that because x is WN zero mean then R(0)=var != 0 then (using
delta as the impulse function):

Ry(t1,t2)= var*delta(t1-t2) + c^2      %note that the process is still
WS stationary

But it's Ppwer spectrum is now

Sy(w)= var + c2*delta(w) !=  constant as a function of w

which is not flat anymore, thus non-white. Unless I am making a mistake
;)
it seems like we are missing something from the definition.

Reply by Randy Yates 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!= t2
I'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 definition
Again, 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