If I have two noisy signals where the noise is not correlated and I know the SNR of each signal, how would I find the SNR after multiplying the two signals together?
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The multiplication of the two noisy signal gives
$$(x_1+n_1)(x_2+n_2)=x_1x_2+x_1n_2+x_2n_1+n_1n_2=x+n\tag{1}$$
with the desired signal
$$x=x_1x_2\tag{1}$$
and the noise part
$$n=x_1n_2+x_2n_1+n_1n_2\tag{2}$$
Assuming all signals are independent of each other and have zero mean, we get for the signal power
$$\sigma^2_{x}=\sigma_{x_1}^2\sigma_{x_2}^2\tag{3}$$
and for the noise power
$$\sigma_n^2=\sigma_{x_1}^2\sigma^2_{n_2}+\sigma^2_{x_2}\sigma^2_{n_1}+\sigma^2_{n_1}\sigma^2_{n_2}\tag{4}$$
For the total SNR you get
$$\text{SNR}=\frac{\sigma_x^2}{\sigma_n^2}=\frac{\sigma_{x_1}^2\sigma_{x_2}^2}{\sigma_{x_1}^2\sigma^2_{n_2}+\sigma^2_{x_2}\sigma^2_{n_1}+\sigma^2_{n_1}\sigma^2_{n_2}}\tag{5}$$
With $\text{SNR}_1=\sigma_{x_1}^2/\sigma_{n_1}^2$ and $\text{SNR}_2=\sigma_{x_2}^2/\sigma_{n_2}^2$ this can be rewritten as
$$\text{SNR}=\frac{\text{SNR}_1\text{SNR}_2}{\text{SNR}_1+\text{SNR}_2+1}\tag{6}$$
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$\begingroup$ You might need to say that the signals are independent, not just uncorrelated, in order to claim that $\sigma^2_{x}=\sigma_{x_1}^2\sigma_{x_2}^2$ because what you are asserting is that $E[(X_1X_2)^2] = E[X_1^2]E[X_2^2]$. This factorization does not work even if $X_1^2$ and $X_2^2$ are uncorrelated signals because the squared signals don't have zero mean. $\endgroup$ Jun 11, 2016 at 15:04
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$\begingroup$ @DilipSarwate: You're right about the independence, thanks for pointing that out! But I don't agree with your last remark: if $X_1^2$ and $X_2^2$ are uncorrelated then $E[X_1^2X_2^2]=E[X_1^2]E[X_2^2]$ must hold (simply by the definition of uncorrelatedness as vanishing covariance). $\endgroup$– Matt L.Jun 11, 2016 at 16:55
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$\begingroup$ You are right about my second point; what I was thinking of was than uncorrelated $X_1$ and $X_2$ do not imply uncorrelated $X_1^2$ and $X_2^2$, whereas independent $X_1$ and $X_2$ imply independent $X_1^2$ and $X_2^2$. $\endgroup$ Jun 11, 2016 at 17:38
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$\begingroup$ @DilipSarwate: Yes, looks like everything is sorted out now! $\endgroup$– Matt L.Jun 11, 2016 at 17:54
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$\begingroup$ This assumes the signal component is correlated; would be good to also see the result for simply multiplying two independent noise sources $\endgroup$ Dec 10, 2019 at 16:44