A noise figure of 3dB, translates to an equivalent noise temperature equal to the room temperature. What is the significance of this relation? Is there a physical interpretation for this?
Electronic – 3 dB Noise Figure
noiseRFsignal-to-noise
Related Solutions
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.
Where does 8.28e-19 come from?
Thermal noise of a 50 ohm resistor in a 1 Hz bandwidth at 27 degC is 9.1e-10 volts
To convert this to an equivalent power it needs squaring and this produces a number of 8.28e-19.
The formula also reduces to 20 log\$_{10}(\frac{V_{NOISE}}{9.1e-10}\$) i.e. it compares actual RMS noise against 1Hz-limited voltage noise from a 50 ohm resistor.
Best Answer
Noise figure (NF) is the decibel equivalent of noise factor (F) and noise factor is the amount an amplifier degrades the signal to noise ratio of a given input signal. The only significance this has with temperature is that the stated figure/factor of an amplifier will be at 290 kelvin (room temperature).
See this wiki page.