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.
The signal-to-noise ratio needed for \$\mathrm{32\ kbit/s}\$ with a bandwith of \$\mathrm{100\ kHz}\$ is calculated in the following way:
\$\underbrace{C=B\cdot \log_2{\left(1+\dfrac SN\right)}}_{\textrm{Shannon's equation}}\longrightarrow \dfrac CB=\log_2{\left(1+\dfrac SN\right)}\longrightarrow 2^\frac{C}{B}=1+\frac SN\longrightarrow \frac SN=2^\frac CB - 1\$
If we substitute the values in the equation we get:
\$\dfrac SN=2^\frac{32\cdot 10^{3}\ \textrm{bits/s}}{10^5\ \textrm{Hz}} - 1=2^{0.32} - 1=0.2483\longrightarrow 10\cdot\log_{10}{(0.2483)}=-6.05\textrm{ dB}\$.
In Shannon's equation:
- C is the channel capacity in bits per second;
- B is the bandwidth of the channel in hertz;
- S is the average signal power received over the bandwidth in watts;
- N is the average noise power received over the bandwidth in watts;
- S/N is the signal-to-noise-ratio as a linear power ratio (not as decibels)
Sources:
https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem
https://www.st-andrews.ac.uk/~www_pa/Scots_Guide/iandm/part8/page1.html
Best Answer
Short answer: Yes. Sort of.
You don't mention what medium your signal is traveling on, but distance can effect both the signal and the noise and can even effect the bandwidth in some cases.
In a wireless system the biggest thing to look at is the free space losses in your signal. The noise will stay pretty much equal regardless of distance, but your signal power will be reduced as you get farther away. The exception to this is when you have to put higher power amps on your receiver that might induce more noise into your system.
In a wired system you have both your signal power decreasing, but also your noise level increasing. Also in a wired system you can have the non-ideal aspects of a wire cause a low pass or band pass filter that will change some as you go to longer distances that you will have to keep into account.