Problem is understanding relationship between voltage, power, and power spectral density. You have to pay close attention to whether you are talking about the voltage of a signal, it's power density as a function of frequency, or the total power over a bandwidth.
If by 10mV, you mean the total power in the noise signal over 100MHz, you should get in the habit of writing "volts squared" to be clear that you mean power. The power spectral density is expressed as 1uV/rtHz to allow us to view it with the same regard as we would view a deterministic signal, so that we may quickly use the standard dB formula for power calculation, so if the signal is 1V (that's 1 volt) and noise is 1uV, then
SNR = 20log10 (1V/1uV) = 120dB.
But if you intend to find the total power in the noise over the band, you have to take the integral of the other representation of the power spectral density, which, in a sense, is the true form of the expression, which is volts-squared-per-Hertz, or watts per hertz, assuming the signal is being fed into a 1-ohm load. As you recall, you are allowed to assume a 1-ohm load because you are interested in ratios, and each signal would have to drive the same load no matter what the value of the load, so the true value of the load is not really relevant. Calculating the integral, the total power in the noise would be:
(1uV/rtHz)^2 * 100Mz = (10^-12 V^2/Hz) * 10^8Hz = 10^-4 watts.
Note that the total power over the band has to be expressed as volts-sqared or watts(again assuming a 1-ohm load).
Doing the same for signal, you would get (1V^2/Hz* 10^8 Hz) = 10^8V^2 =
10^8 wattsThe SNR over the entire band is then (10^8*V^2)/(10^-4*V^2)= 10^12 = 12 bels = 120dB.
120 dB.Another way to think about this is that SNR really is the ratio of powers, P1/P2. But if you've already calculated powers for each of the signal and noise, why use another power formula to calculate their ratio? Just take the ratio. Also, if I may digress, this is a good opportunity to restate the principle behind the 20log10(V1/V2) formula since understanding it facilitates understanding why PSD is expressed in the V/rtHz format.
We all know that power is V-squared-over-R. This comes from the definitions of charge, force, fields, energy, time, and power.
P = V^2/R
If two signals V1 and V2 drive the same R, and you want to calculate their ratio of powers (which is what we are interested in), then whatever the values of V1 and V2 are, they will be each squared in the power ratio formula to calculate their P1 and P2:
P1/P2 = [(V1^2)/R]/[V2^2/R] = (V1/V2)^2
We use logarithms of ratios in all branches of science to keep us from going mad when the ratios vary from ultra-tiny to ultra-huge on the same sheet of paper, noting of course, the ratio is no longer the true ratio, but the log-base-10 of the ratio.
power ratio (using logs) = log10[ (V1/V2)^2].
But in many cases like amplifiers, we are calculating a gain of output power over input power, **given their output voltages and input voltages**, respectively. This gain is often less than 10, and unless we do something, our paper will end up with a bunch of numbers that start with 0 followed by decimals:
0.3456 (bels)How annoying this would be.
We'd rather have at least one whole number, followed by a decimal, followed by some numbers. We fudge by saying whatever log value we get, just lie and say it is 10 times as much to shift over the decimal one spot, but take care to keep in mind that we lied:
3.456 (no longer bels but decibels, hence the deci which means 1/10th)3.456 decibels = 0.3456 bels
So we say power ratio, in decibels, is
10log10(P1/P2).*However*, most times we are not given P1 and P2 (or earthquake Richter power 1 and Richter power 2), but the voltages. This is where the 2 comes from. Popping it out from the circuit power ratio expression above:
power ratio in decibels for an electronic circuit with V1 and V2, is
2*10*log10(V1/V2).And this is why you should not apply the standard log formula to calculate the ratios of two values that are already measure power themselves, because the whole point of the factor of 2 in the standard expression is to account for the presumption that the two values that form the ratio argument to the log are "voltages". The V/rtHz format for PSD is purposely specified in that way (even though square root of a Hertz doesn't make any sense to anybody) in anticipation of how it will most likely be used.
Total noise power in sub-channel would be
(1uV/rtHz)^2*1MHz = 10^-12V^2*Hz * 10^6Hz = 10^-6V^2 = 10^-6 watts.
Total signal power in sub-channel would be (treating signal as being random with PSD):
(1V/rtHZ)^2*1MHz = 1V^2*10^6Hz = 10^6 V^2 = 10^6 watts
SNR = 10^6 watts / 10 ^-6 watts = 10^12 = 12 bels = 120 dB.
Yeah, if we could only increase SNR so easily. ;)
-Le Chaud Lapin-