Electronic – How may noise voltage and noise figure be compared

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.

Here is a page that describes some basic techniques for noise figure measurement: https://www.maximintegrated.com/en/app-notes/index.mvp/id/2875

The simplest technique, mentioned by The Photon, is called the 'gain method'. In this method, you need to know the gain of the amplifier (which you can measure with a spectrum analyzer and a signal generator) and the output noise power when the input is terminated. The 'magic number' for noise power is -174 dBm/Hz at room temperature. Take the difference between the measured noise power and the room temperature noise power (-174 dBm/Hz) and subtract the gain to get the noise figure. It should be possible to add some more gain after the amplifier in question to get well above the spectrum analyzer noise floor, then figure out what the noise figure of the first state is. Essentially, you build a relatively high gain amplifier, measure the gain and noise figure of that, then stick your amplifier on in front of that, measure the gain and noise figure of the combination, then use Friis' formula to get the noise figure of the first stage (F = F1 + (F2-1)/G1 + ...).