Is it possible to estimate what the white noise floor should be, and after cross-correlation?

Question

I am simply sending a signal from a function generator, split it into two signals, and then send it into two inputs of an oscilloscope which can interface with my PC. I take the FFT of these two signal inputs and see a noise floor around 1e-7 V, but how can I know if this is the right noise floor value?

If I was to send this same signal, duplicate it again, and input it into this oscilloscope. Cross-correlate these two signals in the frequency domain (multiply with a conjugate), then I should observe a lower noise floor as the noise of the systems (function generator, oscilloscope, wires) should cancel out. But I observe a noise floor of around 1e-14 V which is really low... so I am not sure how to validate this, if it works or not.

edit: reworded my question better with right terminologies

Answer 1

I assume you notation "^-7V" refers to a voltage noise level of $1E-7$ volts, and then when you say it is in the frequency domain, then I assume this is the average of the FFT magnitude for all the bins that are not part of your signal (the noise floor). This average noise level is completely dependent on the number of bins in your FFT.

I do not understand clearly what your processing would be to "cross-correlate in the frequency domain", but based on your results I assume you are simply multiplying and then taking an FFT (which is not "cross-correlate in the frequency domain). If this was the case, and it was a complex conjugate multiplication then you would get the cancellation you describe, and to the degree the input signals are correlated, and the system is matched then you would likely be seeing the noise floor of your instrument.

Generally, when you perform an actual cross correlation between two noise signals, if it is the same noise at the same rms level, then the correlation result will go up in magnitude by a factor of 2. If the two noise sources are independent and identically distributed, then the correlation result will go up by $\sqrt{2}$. (Independent noise sources sum directly by their variance, correlated noise sources sum directly by their standard deviations).