Electronic – Why is a lower Noise Figure considered better than a higher Noise Figure

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.

I think that you do have to compare powers and also in a certain bandwith. The noise is a power in a certain bandwidth, if you choose a smaller BW the noise will be less !

If your signal is only 1 MHz wide, you don't need your the 100 MHz bandwidth (the LNA could be 100 MHz but after mixing you would filter to a 1 MHz bandwidth and get rid of the extra noise).

So you cannot compare your 50 mV signal to the noise power (in a specified BW). You need a signal power (in a specified BW) and compare that to the noise (in the same BW).

Think of it like: if your signal is only a 50 mV sinewave it will use a very small BW (it is only one frequency). Now compare that to for example a WiFi signal which uses many frequencies in a certain BW. Combined they can also peak at 50 mV but they would contain a lot more power (and information!).