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Where ever we see in the text books authors explicitly mentions zero mean white Gaussian noise and in practice also noise processes are zero mean.But if some process is Wide sense stationary but non zero mean can it still be white sense stationary.

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  • $\begingroup$ The definition here says it has zero mean. $\endgroup$
    – Glen_b
    Commented Aug 12, 2017 at 9:41
  • $\begingroup$ "white sense stationary" - did you mean wide-sense stationary? $\endgroup$
    – Ami Tavory
    Commented Aug 12, 2017 at 16:00

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White noise is called so because the defining property is that the power spectral density of the process is a constant: $S_X(f) = K, -\infty < f < \infty$. The autocorrelation function then is $R_X(t) = K\delta(t)$ where $\delta(t)$ is the Dirac delta. All this is, of course, a stretch in that the notion of autocorrelation function and power spectral density are supposed to apply to finite-power processes -- processes for which $E[X^2(t)]$ is finite for all $t$ -- and here we have a process for which the power is infinite. But, we pretend that the theory is nonetheless applicable, and use it to calculate that the power spectral density of the output process from a linear time-invariant filter with transfer function $H(f)$ is $$S_Y(f) = S_X(f)|H(f)|^2, -\infty < f < \infty,$$ which result matches up reasonably well with experimental observations. So, even though we are using a fictional object (white noise) to start with, the end result is reasonable, and so we adopt the white noise model as something works, no matter how shaky the foundation.

Turning to your question, yes, white noise does have zero mean even if this is not explicitly stated because if the mean were nonzero, then the power spectral density would have a Dirac delta at $f=0$ which disrupts the flatness of the power spectral density.

Note: white light (from which the adjective is borrowed) has equal power at all wavelengths in the visible band whereas white noise has equal power at all frequencies. In that sense, white noise is a bit of a misnomer, but the usage is too well established to change.

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