3
$\begingroup$

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...

Improve this question
$\endgroup$
4
  • $\begingroup$ By multiplying you, still maintain the same unit - in the autocorrelation, one serves as function not as signal.. $\endgroup$ Commented Jan 7, 2016 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. $\endgroup$ Commented Jan 7, 2016 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". $\endgroup$ Commented Jan 7, 2016 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. $\endgroup$ Commented Jan 13, 2016 at 14:41

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Commented Jan 7, 2016 at 18:37

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.