The only thing you got wrong is the capitalization of "dB" and "dBW". The "d" is lower case because it represnts "deci-", the SI prefix for 1/10. Meaning that a decibel is 1/10 of a *bel*, a unit which is almost never used in practice.

Everything else you wrote is correct. 20 dBW = 100 W. "20 dB" tells you nothing about the power until you know the reference level.

Reading the question and the comments, there may be a conceptual misunderstanding : the attenuator WILL attenuate any noise presented on its input (even from just a 50 ohm source impedance), to the same extent it attenuates the signal.

However it also generates noise of its own, which may be represented as the noise from a perfect resistor equal to its own output impedance, and this is added at the output to the (attenuated) input signal and noise. So if input and output Z are both 50 ohms, the net result is attenuated signal + marginally increased noise (i.e. NF = attenuation).

But if its output impedance is lower, the added noise is also lower, thus improving the noise voltage as Andy states.

So represent the attenuator as a perfect attenuator (attenuating noise) in series with a Johnson noise voltage source equal to the output impedance. The rest is just applying the formulae.

EDIT: re: updated question.

(1) There is nothing special about 290K except that it's a realistic temperature for the operation of a passive circuit. The reason they chose it is that the article quotes a noise floor ( -174dBm/Hz) which is correct for a specific temperature : yes, 290k.

(2) While any resistance in the attenuator will contribute noise, I realise that it is not a satisfactory explanation as to why you get the same noise out of an attenuator, because (as Andy says) you could make a capacitive attenuator which is not a Johnson noise generator. So we have to look a little deeper, and remember these noise sources are the statistics of the individual electrons that make up the current.

So, let's say we build a (50 ohm in, 50 ohm out) attenuator, and attempt to cheat Johnson by using a capacitive divider. That implies a node within the attenuator which conducts some of the input current to ground. At this node, we have two current paths; a fraction of the current flows to output, the rest to ground. What determines which path an individual electron will take? Essentially, chance. Collectively? Statistics. So this is a noise source.

Or let's just add series capacitance to provide enough attenuation : we thereby avoid dividing the current flow and eliminate the noise source, right? At the cost of reducing the signal current; our statistics now operate with a smaller sample size and consequently greater variance : more noise.

These results are the best you can do, there is no way round them.

## Best Answer

Deci-Bels always express a

powerratio. Specifically, dB is defined as 10Log_{10}(pwr2/pwr1). Therefore "20 dB" isexactly the same thingas "100 times as much power". This power ratio is never ambiguous, but sometimes what it applies to can be. However, this is no different than for a statement like "100 times as much power". It may be ambiguous what has 100 times more power than what else, but the ratio itself is clear.dB are sometimes used to specify gains of amplifiers that work on voltages. Since dB always specifies a power ratio, and power is proportional to the square of the voltage, dB in this context can be thought of as 10Log

_{10}((V2/V1)²), which is the same thing as 20Log_{10}(V2/V1).Power depends not only on the voltage but also the impedance that voltage is driving. Sometimes those impedances aren't know, or the system works inherently with voltages and the actual power isn't relevant, so the simplification is made that the power ratio is the square of the voltage ratio. This is often the case in audio circuits. In other applications, like RF, the impedances are known and important, so then they are taken into account and dB represents the actual power ratio.