3
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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

$$\text{SNR}=\frac{\text{SNR}_1\text{SNR}_2}{\text{SNR}_1+\text{SNR}_2+1}\tag{6}$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Jun 11 '16 at 15:04
  • $\begingroup$ @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 '16 at 16:55
  • $\begingroup$ 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 $X_2^2$. $\endgroup$ – Dilip Sarwate Jun 11 '16 at 17:38
  • $\begingroup$ @DilipSarwate: Yes, looks like everything is sorted out now! $\endgroup$ – Matt L. Jun 11 '16 at 17:54
  • $\begingroup$ This assumes the signal component is correlated; would be good to also see the result for simply multiplying two independent noise sources $\endgroup$ – Dan Boschen Dec 10 '19 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.