does WGN(white Gaussian noise) imple zero mean?

Tim Wescott <tim@wescottnospamdesign.com> writes:

> Randy Yates wrote:
> 
> > Tim Wescott <tim@wescottnospamdesign.com> writes:
> >
> 
> >>Randy Yates wrote:
> >>
> >>
> >>>Tim Wescott <tim@wescottnospamdesign.com> writes:
> >>>
> >>>
> >>>>kiki wrote:
> >>>>
> >>>>
> >>>>
> >>>>>Hi all,
> >>>>>I read through several books but did not get clarification on
> >>>>>whether WGN(white Gaussian noise process) imply zero mean or not...
> >>>>
> >>>>OK.  Think.
> >>>>
> >>>>Hmm.  The definition of a white noise process is that the PSD is 1
> >>>>everywhere.  OK, I understand that.
> >>>>
> >>>>If I know the PSD of a function I can find the expected power between
> >>>>any two frequencies by just integrating the PSD over that interval.
> >>>>OK, I've read my books, I understand that.
> >>>>
> >>>>Now, DC means the frequency interval between 0 and 0 (technically
> >>>>between 0- and 0+).  Integrating 1 between 0 and 0 I get -- ZERO!  WOW!
> >>>
> >>>Hey Tim,
> >>>You get the same result when integrating between 1- and 1+, or
> >>>253,392- and 253,392+, etc., and we know have power at there. What
> >>>have you answered, then?
> >>
> >>Well, _I've_ answered that there's no DC content (which is another way
> >>of saying zero mean), when you take DC to it's mathematical limit
> >>(note that white noise will appear to have DC content if you only
> >>observe it for a finite amount of time, such as the time from the big
> >>bang to right now).  _You've_ extended this to show that you can pick
> >>any one, zero-bandwidth, filter and find no energy there.
> > No, you've shown that there is no power there. There is indeed energy
> 
> > there since, for white noise, Sxx(w) at w = w0 is strictly greater
> > than zero for any value of w0 (including 0), and the units of power
> > spectral density are [joules] ([watts/Hz] == [joules]). One obtains
> > power upon integration of the Sxx(w) (no matter how small of an
> > integration interval is chosen) since \int_{w_0-}^{w_0+} Sxx(w) dw has
> > units of [joules] * [1/seconds], i.e., power.
> >
> 
> Oy -- good point.  Geeze these limits-to-infinity things get tricky.

If I had a dime for every time I posted an erroroneous statement
in this newsgroup...

> There must be energy there because if you integrate a white noise
> process the variance of the result goes up with the integration
> time. But if you take the average of the white noise process (average
> = integral / integration time) then the variance goes _down_ with the
> integration time, eventually going to zero as the integration time
> goes to infinity.

Yes, agreed.

> So; zero mean, infinite energy.

No, I don't agree! See the post I'm about to make in response to 
my first response to kiki. 

> I was confused about this stuff, too, and asking questions didn't
> clear it up.  What _did_ clear it up (for the most part; see your
> comment above) was thinking about it.  I had the advantage that I take
> long bike rides, and for some reason it really worked for me to ponder
> these questions while riding.  

Yup, I'm a fello ponderer too. So much so that many people think I'm
an idiot ("what difference does the wherefore and the why make? just 
DO IT!"). 

> This is why I'm trying to get the guy pulled away from Matlab
> simulations.

YES! An excellent goal. Personally I have found Matlab to be useful
in learning only after considerable effort has already been spent
with pencil and paper (and bicycle, if that's your thing), and then
only sometimes.

It's off-topic, but it is this exact sort of thing, this sort of 
"the deeper brain cells aren't exercised with a computer" thing,
that makes me boil every time I hear about what a good thing
computers are in school and how smart kids are these days because
they know how to use a computer. Bah!
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

Randy Yates <yates@ieee.org> writes:

> "kiki" <lunaliu3@yahoo.com> writes:
> 
> > Hi all,
> >
> > I read through several books but did not get clarification on whether 
> > WGN(white Gaussian noise process) imply zero mean or not...
> 
> Hi Kiki,
> 
> Now you've got me wondering. On one hand, I've heard the term "zero-mean
> additive white Gaussian noise" many times, but on the other hand, "white"
> implies a flat PSD, which in term implies that there is some power at DC.
> So I can't answer your question.

Kiki I hope you're reading this new post,

I can now partially answer your question. In order to do this, first
realize that when someone speaks of a zero-mean, white Gaussian noise
process, they're talking about a random process with an underlying
distribution, i.e., for each point in time t, the random process x(t)
has a specific probability density function f(t, s). IT IS THIS
UNDERLYING PDF THAT IS ZERO-MEAN. 

However, that still doesn't completely resolve the issue (at least in
my mind it doesn't). A random process is "ergodic" if its time-wise
statistics are the same as its ensemble-wise statistics. So we have a
dilemma when postulating a zero-mean, white, ergodic random noise
process because ensemble-wise the mean is zero while time-wise the
mean is non-zero (since there's non-zero energy at DC). I still don't
know how to resolve THIS problem!

--Randy


> 
> > Another confusion I have is that the definition of WGN is it has flat power 
> > spectrum density, let's say S(f)=1, then Rx(t)=delta(t) is its 
> > autocorrelation function, I don't see how people say the power of this noise 
> > process is E((x(t))^2)=sigma_x, 
> 
> Rxx(t) is defined to be
> 
>   Rxx(tau)= E[x(t)*x(t-tau)] 
> 
> for a real random process x(t). Then, by definition, 
> 
>   E[x^2(t)] = E[x(t) * x(t-0)]
>             = Rxx(0)
>             = undefined (infinity)
> 
> when Rxx(t) = delta(t). Thus you're contradicting yourself somewhat. 
> 
> A truly white-noise process does have infinite power (hence the Dirac
> delta function in the autocorrelation), but most transistors I know
> of burn out after a few gigawatts, so we usually speak of a band-limited
> white noise process, i.e., a process which has a PSD Sxx(w) = c, |w| < a,
> and in which case the power is finite and Rxx(0) = a*c/pi.
> -- 
> %  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

-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

"kiki" <lunaliu3@yahoo.com> wrote in message
news:cqlbr7$a4m$1@news.Stanford.EDU...
> Hi all,
>
> I read through several books but did not get clarification on whether
> WGN(white Gaussian noise process) imply zero mean or not...
>
> Another confusion I have is that the definition of WGN is it has flat
power
> spectrum density, let's say S(f)=1, then Rx(t)=delta(t) is its
> autocorrelation function, I don't see how people say the power of this
noise
> process is E((x(t))^2)=sigma_x, something like that... the power should be
> infinite, right?
>
> Any clarifications? Thanks a lot!
>
>
>
That's because I believe most of the time the PSD for white noise is quoted
incorrectly. It should be

sigma squared/(fs) and it is flat from fs/2 to -fs/2 where fs is the
sampling freq.The area under the PSD is the total average power or sigma
squared. (including negative frequencies). If you time-average the data
(noise) you should get a variance of sigma squared (assuming the mean to be
zero).Hence the time average (sum of squares/N) is the same as the
statistical average E[x^2].
As for autocorrelation, the impulse is there to denote that it is just a
spike.If you take the Fourier Transform of the autocorrleation function you
should get PSD (Wiener-Kinchen Theorem). So you can work it out for
yourself - however - please note the difference between PSD and the AREA
UNDER IT which is the total average power.

Country Chiel

Randy Yates wrote:
> 
> Randy Yates <yates@ieee.org> writes:
> 
> > "kiki" <lunaliu3@yahoo.com> writes:
> >
> > > Hi all,
> > >
> > > I read through several books but did not get clarification on whether
> > > WGN(white Gaussian noise process) imply zero mean or not...
> >
> > Hi Kiki,
> >
> > Now you've got me wondering. On one hand, I've heard the term "zero-mean
> > additive white Gaussian noise" many times, but on the other hand, "white"
> > implies a flat PSD, which in term implies that there is some power at DC.
> > So I can't answer your question.
> 
> Kiki I hope you're reading this new post,
> 
> I can now partially answer your question. In order to do this, first
> realize that when someone speaks of a zero-mean, white Gaussian noise
> process, they're talking about a random process with an underlying
> distribution, i.e., for each point in time t, the random process x(t)
> has a specific probability density function f(t, s). IT IS THIS
> UNDERLYING PDF THAT IS ZERO-MEAN.

Yup. Note that "Gaussian" white noise doesn't exist, since its
variance would be infinite. But if you allow distrubutions with
infinite variance (thus allowing delta functions), then ok ...
 
> However, that still doesn't completely resolve the issue (at least in
> my mind it doesn't). A random process is "ergodic" if its time-wise
> statistics are the same as its ensemble-wise statistics. So we have a

First note that white noise is ergodic since it's stationary and the 
covariance vanishes at infinity.

> dilemma when postulating a zero-mean, white, ergodic random noise
> process because ensemble-wise the mean is zero while time-wise the
> mean is non-zero (since there's non-zero energy at DC). I still don't
> know how to resolve THIS problem!

I don't get how it follows that the time-average (mean) would be
infinite?
R(0) being infinite doesn't imply that the mean should be infinite.

Wilbert

>Hi all,
>
>I read through several books but did not get clarification on whether 
>WGN(white Gaussian noise process) imply zero mean or not...
>

We can all agree that white implies Rxx(tau)=dirac(tau)*sigma^2,
where Rxx(tau)=E[x(t)*x(t+tau)].
(We assume x real for notational simplicity)

Assume now that x has zero mean, and look at a new process
y=x+c, where c is a non-zero constant. We now want to prove that y cannot
be white and hence that a white process has to have zero mean.

We find: Ryy(tau)=E[y(t)*y(t+tau)]
=E[(x(t)+c)*((x(t+tau)+c))]=E[x(t)*x(t+tau)]+E[x(t)*c]+E[x(t+tau)*c]
+E(c*c) = Rxx(tau) + 0 + 0 + c^2.

This implies that Ryy(tau)=dirac(tau)*sigma^2 +c^2,
i.e not white.

Hope this helps.

Best regards,
UniTel

		
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unitel wrote:

>>Hi all,
>>
>>I read through several books but did not get clarification on whether 
>>WGN(white Gaussian noise process) imply zero mean or not...
>>
> 
> 
> We can all agree that white implies Rxx(tau)=dirac(tau)*sigma^2,
> where Rxx(tau)=E[x(t)*x(t+tau)].
> (We assume x real for notational simplicity)
> 
> Assume now that x has zero mean, and look at a new process
> y=x+c, where c is a non-zero constant. We now want to prove that y cannot
> be white and hence that a white process has to have zero mean.
> 
> We find: Ryy(tau)=E[y(t)*y(t+tau)]
> =E[(x(t)+c)*((x(t+tau)+c))]=E[x(t)*x(t+tau)]+E[x(t)*c]+E[x(t+tau)*c]
> +E(c*c) = Rxx(tau) + 0 + 0 + c^2.
> 
> This implies that Ryy(tau)=dirac(tau)*sigma^2 +c^2,
> i.e not white.
> 
> Hope this helps.
> 
> Best regards,
> UniTel

I like that proof. It's elegant in its simplicity.

> This message was sent using the Comp.DSP web interface on
> www.DSPRelated.com

Nice.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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>Another confusion I have is that the definition of WGN is it has flat
power 
>spectrum density, let's say S(f)=1, then Rx(t)=delta(t) is its 
>autocorrelation function, I don't see how people say the power of this
noise 
>process is E((x(t))^2)=sigma_x, something like that... the power should
be 
>infinite, right?
>
>Any clarifications? Thanks a lot!
>
For a continuous white noise process the power is infinite.

When people talk about the power of white noise processes they are
referring to bandlimited white noise processes, i.e. S(f) is non-zero only
for
-W<f<W.

Then you can integrate your power spectrum density function and get a
finite result.

Often white noise processes are bandlimited because of receiver filters.

Hope this helps!

Best regards,
UniTel

		
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>Hi Kiki,
>
>Now you've got me wondering. On one hand, I've heard the term "zero-mean
>additive white Gaussian noise" many times, but on the other hand,
"white"
>implies a flat PSD, which in term implies that there is some power at
DC.
>So I can't answer your question.

No it does not imply power at DC as I will try to explain.

In its general form the power spectrum density should be written:
S(w) = S'(w) + sum(2*P_n*dirac(e^jw-e^jw_n)).

Above S'(w) represents the continous part of the spectrum and the
sum represents the discrete part of the spectrum. The discrete part,
i.e. the diracs are present if and only if there are periodic components
in the process.

Let's now see what happens to the spectrum with a dc component.

With a dc component:
A dc component represents a periodic component (all periods are ok).
Thus we will have a dirac in the spectrum and we will have power at dc,
because diracs can be integrated.

Also note that diracs are of unbounded height and thus our spectrum is no
longer flat. This is also a proof of why we cannot not have dc components
if we want our process to be white.
		
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>In its general form the power spectrum density should be written:
>S(w) = S'(w) + sum(2*P_n*dirac(e^jw-e^jw_n)).

Should be a pi in the last term =>
S(w) = S'(w) + sum(2*pi*P_n*dirac(e^jw-e^jw_n)).

The sum ranges from n=1->N where N is the number of periodic components in
the process.
		
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