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Variance of a white noise

  1. Jun 29, 2012 #1

    I have a pretty simple question which I thought I do not need to make a topic about, but Google is actually not helping, which is surprising. So here it goes:

    How can white noise have infinite power if its variance is finite?

    As far as I am aware, the following is always valid for a stationary zero-mean random process X which is classified as white noise (i.e. flat power spectrum)

    [itex]R_{x}(t)|_{t=0}= power = E[X^2(t)] = \sigma^2 \cdot \delta(t)|_{t=0} = infinite = Var[X(t)] = \sigma^2 = finite[/itex]

    assuming that the statisics of the random process are anything with the finite variance, for example, Gaussian distribution. So, yeah, I'm looking at the AWGN.

    So, what gives?

    And I am aware of the physique of the realistic white processes, however I'm purely interested in the theoretical point of view here, so let's assume that this white process does have an infinite power. How is that possible when we also assumed that its variance is finite?

    Many thanks in advance.
  2. jcsd
  3. Jun 29, 2012 #2


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    I don't understand what you wrote, so I will address only your question about the power of AWGN being infinite. The definition of white noise is that it has a constant power spectral density (PSD). Integrating over all frequencies gives infinite power. The catch is that the noise is never white over frequencies from 0 to infinity. At very high frequencies, for instance, quantum effects take over, and in practice circuits and systems roll off long before that point. When we say that the power is sigma^2, that is the PSD integrated over the bandwidth of our system, measurement device, etc. which is always finite.
  4. Jun 30, 2012 #3
    Hey, thanks for the answer. Let me try to make it clearer

    [itex]R_{x}(\tau) = E[X(t)X(t+\tau)][/itex]
    [itex]R_{x}(0) = E[X^{2}(t)] = \sigma^2[/itex]
    [itex]R_{x}(\tau) = \int S(f) e^{j2\pi ft}df[/itex]
    [itex]R_{x}(0) = \int S(f) df[/itex]

    The white noise is defined by having a flat power spectral density over the whole range of frequencies. So, that means, from the last formula, that it has an infinite power.
    To the best of my knowledge, unrelated to this, we can get a power of the process as given in the second formula, which is variance. Variance is a statistical parameter, independent of the stochastic parameters of the process. In other words, we can have various white processes (Poisson, Gaussian, etc.) as long as the power spectrum is flat.

    So, if the variance is finite, it means that the power is finite. On the other hand, by the very definition, the power of the white noise is infinite. Thus, the confusion arises :D
    I hope I made it clearer now? From my POV, that is.

    Also, I understand that this is an unrealistic scenario, but like I said in the end of the first post, I am mainly interested to see where my reasoning went wrong. AWGN may not exist, but I use it all the time in theory, so I'd like to understand it better.
  5. Jun 30, 2012 #4

    Stephen Tashi

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    If you are treating white noise as a statistical process, then each particular realization of the process over a given time interval may have a different power. On the other hand if you had an large collection of oscillators with a continuum of frequencies and a constant amplitude and you turned them all on, you would get a (theoretically) deterministic signal. Perhaps there are two different definitions of "power" involved in your question.
  6. Jul 1, 2012 #5


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    I think the issue is where you write
    [tex]R_{x}(t)|_{t=0}= power = E[X^2(t)] = \sigma^2 \cdot \delta(t)|_{t=0} = infinite = Var[X(t)] = \sigma^2 = finite.[/tex] While it is true that
    [tex]R_{x}(t)|_{t=0}=E[X^2(t)], [/tex] it is not true that [tex]E[X^2(t)]= \sigma^2 \delta(t)|_{t=0} . [/tex] Instead, the variance of a Gaussian distributed variable is finite [tex]E[X^2] = \sigma^2[/tex] without the delta function, as you wrote in your second post. This means that Gaussian noise is not white over an infinite frequency span.

    White noise, on the other hand, is a mathematical construct characterized by [tex]R(0)=c\delta(0).[/tex] It is unphysical, since only an infinitely long waveform can possess a true delta function autocorrelation. Realizable noise can therefore never be white in the mathematical sense. Thus, "white" and "Gaussian" are two different things. Putting them together as AWGN is a useful, and universally used, approximation that is valid for finite bandwidths and finite duration signals.

    As a final note, it apparently can be demonstrated with mathematical rigor that a white frequency distribution cannot be represented by a valid probabilistic distribution.

    EDIT: The RHS of the last equation should read [tex]c\delta(\tau).[/tex]
    Last edited: Jul 1, 2012
  7. Jul 1, 2012 #6
    Ah, I see. That makes sense.

    Could you perhaps elaborate just a little bit more on
    as to why it has to be infinitely long? I'm not sure I see it.

    But other than that, I am all good! Thanks!
  8. Jul 1, 2012 #7


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    Certainly, I'm happy to expand. A signal x(t) of finite duration T has a finite autocorrelation function at zero lag [tex]R(0)=\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt[/tex](and it's finite elsewhere, too, of course). To see this, recognize that the integral is the signal energy, which is finite for a finite-duration signal.

    R(0) already differs from the delta function form of the autocorrelation of a signal with a white spectrum, which (since it is a delta function) must be infinite at the origin. Does this make sense?
  9. Jul 6, 2012 #8
    Sorry, I was away these few days. Yeah, that makes sense completely, thanks again :)
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