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. �����������������������������������������������������������������������
Is White Noise Necessarily Zero-Mean?
Started by ●February 15, 2006
Reply by ●February 17, 20062006-02-17
Reply by ●February 18, 20062006-02-18
"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)]= varNope, 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."
Reply by ●February 18, 20062006-02-18
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."
Reply by ●February 18, 20062006-02-18
Reply by ●February 18, 20062006-02-18
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.
Reply by ●February 18, 20062006-02-18
"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."
Reply by ●February 18, 20062006-02-18
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.
Reply by ●February 18, 20062006-02-18
"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)Yup, my bad.> 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."
Reply by ●February 19, 20062006-02-19
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. Comments? 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."
Reply by ●February 19, 20062006-02-19
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. -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr
