Randy, The same applies to Papoulis definition as well. Note that because they are uncorrelated samples: E[x(t)x(t+tau)]= delta(tau)*var So that just like before: (E[x(t)])^2 = E[x(t)^2] -var = E[R(0)] - var = var -var = 0 Because: E[R(0)] = E[delta(0)*var]= var . The two definitions are the same and they automatically imply zero mean for the process (unless I made a mistake :) ).

# Is White Noise Necessarily Zero-Mean?

Started by ●February 15, 2006

Reply by ●February 17, 20062006-02-17

Reply by ●February 17, 20062006-02-17

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

Reply by ●February 17, 20062006-02-17

"Ikaro" <ikarosilva@hotmail.com> writes:> Hi, > > I think Brown's definition of WN gives zero mean for the process > (unless I did a mistake). > >>From his definition a WN (stationary with constant psd) : > > E[x(t)]=mx %constant > R(x(t2),x(t1))= R(tau) > Sx= var (constant psd) which implies -> R(tau)=delta(tau)*var > > So that using this equation for the mean > > var = E[x(t)^2] - (E[x(t)])^2 > > we get > > (E[x(t)])^2 = E[x(t)^2] -var = E[R(0)] - var = var -var = 0Hi Ikaro, Thanks for responding. I think your error is near the end of the very last line. Rxx(0) != var unless X is zero mean. -- % Randy Yates % "I met someone who looks alot like you, %% Fuquay-Varina, NC % she does the things you do, %%% 919-577-9882 % but she is an IBM." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr

Reply by ●February 17, 20062006-02-17

"Ikaro" <ikarosilva@hotmail.com> writes:> Randy, > > The same applies to Papoulis definition as well. Note that because they > are uncorrelated samples: > > E[x(t)x(t+tau)]= delta(tau)*varI think this is your error here. Papoulis says nothing about the power spectrum (i.e., that it is constant, and therefore that the autocorrelation is the form you state above), just that the process is uncorrelated. -- % Randy Yates % "...the answer lies within your soul %% Fuquay-Varina, NC % 'cause no one knows which side %%% 919-577-9882 % the coin will fall." %%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO http://home.earthlink.net/~yatescr

Reply by ●February 17, 20062006-02-17

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 -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●February 17, 20062006-02-17

"Peter K." <p.kootsookos@iolfree.ie> writes:> 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.Ahh, but that's another issue, isn't it Peter? In the spirit, though, here goes: 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. 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? ... :) -- % Randy Yates % "I met someone who looks alot like you, %% Fuquay-Varina, NC % she does the things you do, %%% 919-577-9882 % but she is an IBM." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr

Reply by ●February 17, 20062006-02-17

Randy Yates wrote:> "Peter K." <p.kootsookos@iolfree.ie> writes: > > >>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. > > > Ahh, but that's another issue, isn't it Peter? In the spirit, though, > here goes: > > 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. > > 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? ... :)That seems much like what I wrote above more succinctly but less elegantly. (Boltzmann: Elegance is for tailors!) Putting what we both wrote differently: If a noise process has a non-zero mean, its PSD has an impulse at 0 Hz. There are then two possibilities: 1) it isn't white; and 2) it is infinite everywhere. Most of the time, (1) will be the case. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●February 17, 20062006-02-17

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!!! (?) -- % Randy Yates % "The dreamer, the unwoken fool - %% Fuquay-Varina, NC % in dreams, no pain will kiss the brow..." %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Eldorado Overture', *Eldorado*, ELO http://home.earthlink.net/~yatescr

Reply by ●February 17, 20062006-02-17

Jerry Avins <jya@ieee.org> writes:> [...] > That seems much like what I wrote above more succinctly but less > elegantly. (Boltzmann: Elegance is for tailors!) Putting what we both > wrote differently: > > If a noise process has a non-zero mean, its PSD has an impulse at 0 > Hz. There are then two possibilities: 1) it isn't white; and 2) it is > infinite everywhere. Most of the time, (1) will be the case.How about this for a succinct explanation: PSD has units of watts/Hz, and watts/Hz = (joules/second) * second = joules. i.e., PSD is the *energy* spectrum. There is *energy* at DC, but no power. -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO http://home.earthlink.net/~yatescr

Reply by ●February 17, 20062006-02-17

Hi Randy,> > > > The same applies to Papoulis definition as well. Note that because they > > are uncorrelated samples: > > > > E[x(t)x(t+tau)]= delta(tau)*var > > I think this is your error here. Papoulis says nothing about the power > spectrum (i.e., that it is constant, and therefore that the > autocorrelation is the form you state above), just that the process is > uncorrelated.>From using the fact that the samples are uncorrelated:Rx(t1,t2) = E[ x(t1)x(t2) ] = E[ x(t1)x(t2) ] = 0 for t1 not equal to t2 and E[ x(t1)x(t2) ] = E[x(t)^2] = R(0) fot t1=t2 So that: Rx(t1,t2)=delta(t1-t2)*R(t1-t2)>From FFT properties we know that the FFT(delta)= constant.So that: Sx= R(0) But you are right about the last line where I set: R(0)=var Is there are way that this can be true without refering to the equation var= R(0) - mx^2 and without refering to independence between sample? My guess is that it is not and we *must* assume the samples are indepent. Which is a big assumption they omitted from their definition.