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The nearest question I can find is here, but it does not have an answer. Say I want to correlate the following two noisy channels: $x_1 = s_1 + n_1$ and $x_2 = s_2 + n_2$, where $s_1$ and $s_2$ are arbitrary signals and $n_1$ and $n_2$ are independent complex Gaussian noise (identically distributed for starters). Further, let either $M \lt N$ or $M \ll N$, with $x_1 \in \mathbb{C}^M$ and $x_2 \in \mathbb{C}^N$. The cross-correlation of the two channels yields

$$ x_1 \star x_2 = s_1 \star s_2 + s_1 \star n_2 + n_1 \star s_2 + n_1 \star n_2. $$

Perhaps I am searching with the wrong keywords, but I am having trouble finding information about $n_1 \star n_2$. What kind of distribution does it have? How much power does it contribute to the output?

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  • $\begingroup$ Since n1 and n2 are independent, they are uncorrelated regardless of the time delay (lag) between the two, therefore the cross correlation function would be 0. $\endgroup$
    – Dan Boschen
    Jun 21, 2017 at 0:49
  • $\begingroup$ @DanBoschen How can that be the case when spectrums overlap (as the spectrums for two full-band Gaussian noise distributions would)? Parseval's theorem guarantees that the time-domain will have non-zero energy. $\endgroup$
    – AnonSubmitter85
    Jun 21, 2017 at 0:56
  • $\begingroup$ If you cross-correlate two orthogonal signals (for example $\sin(\omega t)$ and $\cos(\omega t)$ the result is zero, yet the spectrums overlap. Parseval's theorem relates the spectrum and it's time domain signal. The time domain signal of the cross correlation result is 0, so the spectrum will also be zero. This doesn't mean that the signals prior to correlation have zero energy, which Parseval's relates to. $\endgroup$
    – Dan Boschen
    Jun 21, 2017 at 1:17
  • $\begingroup$ If noise spectrums are given by $N_1$ and $N_2$, then (ignoring conjugation and flipping), the output of the cross-correlation is the inverse DFT of $N_1 \cdot N_2$, no? This product has non-zero energy and, thus, its time-domain representation must also have non-zero energy. What am I missing? If the signals are orthogonal, then their spectums do not overlap by definition. But two AWGN signals are guaranteed to not be orthogonal. $\endgroup$
    – AnonSubmitter85
    Jun 21, 2017 at 1:24
  • $\begingroup$ @DanBoschen As an example of what I mean, consider the following Matlab snippet that cross-correlates the first 50 samples of a noise vector with 1000 samples of another: n = 1/sqrt(2) * (randn( 1e3, 2 ) + 1j*randn( 1e3, 2 )); q = conv( n(1:50,1), conj(flipud(n(:,2)))); The mean input power is 1 and the mean power in the output appears to range between 30 and 50. $\endgroup$
    – AnonSubmitter85
    Jun 21, 2017 at 1:32

2 Answers 2

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I don't understand why your complex vectors have different dimensions but there is an answer for a scaler case,

from:

Leo A. Goodman. On the exact variance of products. Journal of the American Statistical Association, 55(292):708–713, 1960.

Some definitions,

$$\begin{array}{c} m_{x1}= E\{ x_1 \}\\ var(x_1) = E\{ |x_1|^2\} - |m_{x1}|^2 \\ m_{x2}= E\{ x_2 \} \\ var(x_2) = E\{ |x_2|^2\} - |m_{x2}|^2 \end{array} $$ for $x_1$ and $x_2$ independent,

$$ var(x_1 x_2^{*}) = var(x_1)|m_{x2}|^2 +var(x_2^*)|m_{x1}|^2 + var(x_2^*)var(x_1) $$

You don't specify the distribution of your signals. assuming deterministic signals

N. O’Donoughue and J. M. F. Moura. On the product of independent complex gaussians. IEEE Transactions on Signal Processing, 60(3):1050– 1063, March 2012.

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  • $\begingroup$ Thanks. Those two papers were enough to figure it out. For the typical AWGN case, the output power is the product of the two input powers times the number of independent samples used in the correlation. The amplitude of the noise product is not Rayleigh distributed, but it's phase is uniform over 0 to 2pi. Either way, the central limit theorem will imply that the output is again a zero-mean complex Gaussian. $\endgroup$
    – AnonSubmitter85
    Jun 24, 2017 at 20:14
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It seems to me that one important aspect is missing (time). When you have two independent random signals and do the correlation between them over a finite time you will find a (small) but random number. When you continue to accumulate the correlation over a longer time, the correlation will get smaller and eventually reach zero for an infinite time.

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