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...

  • $\begingroup$ By multiplying you, still maintain the same unit - in the autocorrelation, one serves as function not as signal.. $\endgroup$ – Moti Jan 7 '16 at 0:35
  • $\begingroup$ " RMS noise power" or any other "RMS power" is a misnomer. sure, RMS power can be calculated, which would likely be different than mean power (which is i think what you're groping for), but i don't think there is any useful meaning in RMS power. $\endgroup$ – robert bristow-johnson Jan 7 '16 at 0:57
  • $\begingroup$ I get your point on RMS power if you consider power transfer to be a scalar and directionless quantity. That's not always so, however (see Poynting's Theorem). I think it's also useful when thinking of power as the square of the average amplitude. In general it's just a useful way of saying "average magnitude, not average value". $\endgroup$ – Omegaman Jan 7 '16 at 7:11
  • 1
    $\begingroup$ As mentioned below you have to assign "meaning" to a reading based upon the process by which it was generated and your purposes. You are doing a time limited auto-correlation in your example; which will have both information and uncertainty; and residuals of time-limiting/filtering. I strongly suggest drawing and examing the "commutation" diagram cycling between signal- fourier/laplace representation-power spectral density-- and auto correlation. And examine the restrictions imposed by an attempt to "measure" any of these: i.e. bandwidth/averaging/time limiting. $\endgroup$ – rrogers Jan 13 '16 at 14:41

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.

  • $\begingroup$ Understanding units is key to understanding the physicality of the mathematics. In your example of $a\cdot e^x$, $x$ and thus $e^x$ is typically unitless, and the result takes the units of $a$. $V\cdot e^{-jwt}$ has the units of V times $cos(wt)-j\cdot sin(wt)$ where w is units of angle/sec and the trig results are unitless. In a mixer, typically the nonlinear component is a diode giving current times a unitless exponential of voltage divided by thermal voltage (Schockley eqn)-- thus voltage in, current out. Power is typically V*I, but I is V/R, so power is V^2/R as I show in my question. $\endgroup$ – Omegaman Jan 7 '16 at 18:37

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.