Understanding "voltage" and "power" in SNR of a sampled signal I'm facing a gap in my understanding of when a signal is a "voltage" signal and when it is a "power" signal that I've always managed to avoid resolving until now...  First what I think I understand, then my question after the horizontal rule:

I understand SNR to be the ratio of average signal power to RMS noise power: 
$$
SNR=\frac{S_P}{N_P}
$$

or in dB:
$$
SNR_{dB}=10\cdot\log(S_P)-10\cdot\log(N_P)
$$

When calculating SNR based on voltages, power is proportional to voltage squared ($P=V^2/R$) and the Rs drop out in the ratio so we get:
$$
SNR=\frac{{S_V}^2/R}{{N_V}^2/R}=\left(\frac{{S_V}}{{N_V}}\right)^2
$$$$
SNR_{dB}=20\cdot\log(S_V)-20\cdot\log(N_V)
$$

With the familiar 10log for power, 20log for voltage relation doing the squaring for voltage.

I have a radio signal that I receive at an antenna.  For simplicity, I assume I have a perfect receiver and noiseless amplifier so that my received signal power is $P_{Rx}$ and noise is only Johnson thermal AWGN: $N_T=k_B\cdot T\cdot B$, where $k_B$ is Boltzman's constant, T is temperature and B is bandwidth-- none of which really figure in after this.

I believe my SNR at this point to be:
$$
SNR_{RF}=\frac{P_{Rx}}{N_T}
$$



Now I sample, and this is where I start to have questions.

Ignoring quantization noise and such, I believe the analog-digital conversion process results in voltage signals-- that is, the sample values are measurements of the voltage across a resistor ladder or some such meaning the SNR of the sampled signal follows the voltage law:
$$
SNR_{ADC}=\left(\frac{S_{ADC}}{N_{ADC}}\right)^2
$$

Setting aside practical implementation losses, I believe $SNR_{ADC}=SNR_{RF}$.

Where I start to be less sure is when I start operating on the sampled signals.  Let's say I multiply the sampled signal by a delayed version of itself.  Ok, the noise terms get more complicated because I'm taking the product of two independent random variables with non-zero mean but, more fundamentally, is the result a "voltage" or a "power"?  Is there a physical explanation that will help me understand this?

That is: in order to maintain consistency among my SNR estimates, is this a 10log or 20log calculation?

By a pure units analysis, I should have voltage-squared which implies power-- but these are still ADC levels.  It would also seem odd to say that my signal is voltage-cubed if I multiply by two delayed copies...
Handling units in signal processing is tricky (what are the units of $e^{s(t)}$?). I find it useful to think about operations on signals physically, instead of trying to make sense of them mathematically.

For instance, a mixer multiplies two signals. Let's assume the signals are measured in volts. The mixer output is, obviously, also a voltage. This is true even when one of the mixer inputs is its own delayed output.

When calculating an SNR, you square the signal sample (which is a voltage value). In this case, you get a result in watts, because it turns out that the signal power is equal to its square. Physically, you could connect a watt-meter to get a reading of the signal power; you would find that, at every instant, the power turns out to be equal (or proportional) to the square of each sample.

So why, when multiplying two voltages, you get a voltage in one case (mixer) and a power in another (SNR)? I think the reason is that you interpret the result differently in each case. In the case of SNR calculation, the square of a sample physically matches its instantaneous power, so you can assign that meaning to the mathematical operation.