Electrical – SNR and Attenuation Question

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.

"How is it even possible for a single fixed measurer to determine both RSSI and Noise?" - very good question. The noise they are talking about is receiver noise and not interfering signal. At very low powers, the noise is mostly the thermal noise of the receiver: ie, if you were to disconnect the antenna and replace it with a 50 Ohm load (most RF systems are 50 Ohm) you will measure a certain level of noise. So, even if you had all the ideal components, your noise power would be P = k*T*B*G, where k is the Boltzmann's constant, T is the temperature in K, B is the bandwidth in Hz, and G is the gain of your system. In reality, every component adds noise as specified by its noise figure (listed in the datasheet of every RF component). If you look again at the noise power equation, you will see that by reducing bandwidth, you also reduce the noise. However, high bandwidth is necessary for high data rates, which explains why you need good SNR for high data rates.

"Why both values are negative and measured in dBm" - 0 dBm means the power is 1 mW. -20 dbm means the power is .01 mW. The minus indicates the number of dB below 0 dBm. Without the minus, it would have been above 0 dBm

"But who radiate that power?" - in case of noise, it is internal, in case of signal, the transmitter. However, fundamentally it doesn't matter.

"But why is its value so small then?" - it comes from what is called Friis transmission formula. So, with several simplifications, imagine that my transmit antenna radiates power isotropically in all directions. So, your power is uniformly distributed on the surface of a sphere of radius r (and surface area 4*pi*r^2), where r is the distance from the transmit antenna. In Imagine, that your receive antenna is about 1 m^2 and it can capture all the radiation that hits its surface. Now, it can only capture 1/(4*pi*r^2) of all the radiation, making the receive power very tiny and RF engineering a complex field :). This is a very hand wavy explanation but I hope it makes sense