# Is White Noise Necessarily Zero-Mean?

Started by February 15, 2006
Randy Yates wrote:
> Jerry Avins <jya@ieee.org> writes:
>
>
>>2) it is infinite everywhere.
>
>
> What a mind trip! We should come up with a name for this,
> like "screamin' white noise". Not only is this
> infinite power, it is infinitely infinite power!!! (?)

It would surely solve the present energy crisis, perhaps creating a new one.

Jerry
--
Engineering is the art of making what you want from things you can get.
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"Ikaro" <ikarosilva@hotmail.com> writes:

> The DC component will *not* be zero because because its the spectrum
> of x(t)^2 and *not* x(t). So on x(t)^2 has no negative components.
>
> DC=E[x(t)^2)]= var

Nope, it's not the spectrum of x(t)^2,

It's the Fourier transform of Rxx(t), which is a different thing
altogether.

Ciao,

Peter K.

--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."


p.kootsookos@remove.ieee.org (Peter K.) writes:

> "Ikaro" <ikarosilva@hotmail.com> writes:
>
> > The DC component will *not* be zero because because its the spectrum
> > of x(t)^2 and *not* x(t). So on x(t)^2 has no negative components.
> >
> > DC=E[x(t)^2)]= var
>
> Nope, it's not the spectrum of x(t)^2,
>
> It's the Fourier transform of Rxx(t), which is a different thing
> altogether.

However, (responding to my own post... must be going mad), I see why
you're saying that (again, it's still incorrect):

The spectrum is:

FT[ E[x(t)x(t+tau)] ] = 0   (tau != 0)
= var (tau == 0)

Ciao,

Peter K.
--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."


Sorry, I meant:

DC= FFT( E[x(t)^2] )


Oops that's wrong too. You were right before.
But the idea was that x(t)^2 does have a dc component even if x(t)
doesn't.


"Ikaro" <ikarosilva@hotmail.com> writes:

> Sorry, I meant:
>
> DC= FFT( E[x(t)^2] )

Ya, that makes more sense.

Thanks,

Peter K.

--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."


I don't think this is right:

FT[ E[x(t)x(t+tau)] ] = 0   (tau != 0)
= var (tau == 0)

E[x(t)x(t+tau)]  is  delta function in time not in frequency:

E[x(t)x(t+tau)]  = 0   (tau != 0)
= var (tau == 0)

The FT of a delta function is constant in amplitude over all
frequencies.


"Ikaro" <ikarosilva@hotmail.com> writes:

> I don't think this is right:
>
> FT[ E[x(t)x(t+tau)] ] = 0   (tau != 0)
>                       = var (tau == 0)

> E[x(t)x(t+tau)]  is  delta function in time not in frequency:
>
> E[x(t)x(t+tau)]  = 0   (tau != 0)
>                       = var (tau == 0)

Yup!

>  The FT of a delta function is constant in amplitude over all
> frequencies.

Yup^3!

:-)

Ciao,

Peter K.

--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."


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).

Perhaps I'm confused by what you mean by "component".

> 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.

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.

Clearly, with the autocorrelation being C.\delta(t), there will be a
non-zero component of the PSD across all frequencies.

Like I said, I'm still trying to formulate a response; this isn't
clear to me yet. :-)

Ciao,

Peter K.

--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."


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.
>