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.
However, dBc/Hz is the power referenced to the carrier and I'm not sure what that is in this case.
I suspect the carrier in this case is the average optical power, which they may be thinking of as a many-terahertz carrier.
some authors present system noise floor measurements in units of dBc/Hz. Is this wrong since in this case there's no carrier?
It's not clear to me why somebody would choose those units for a noise floor. It may be wrong, but I'd want to see the context where you read it to say for sure.
I find that the trace on the RF spectrum analyzer shows harmonics as a series of peaks. The levels between peaks is at the same level as the background level (i.e. when there is no signal input). Can we therefore infer that the RIN at these points (i.e. if we integrate from 10 Hz, say, up to the 1st harmonic) is equal to or less than the system RIN?
In the RIN measurements I've seen, there are no measurable harmonics, just a single peak related to the laser's intrinsic relaxation oscillation frequency. Are you testing with a modulation signal applied to the laser? Most RIN measurement's I've seen were done with the laser operated CW, and I'd think the results are easier to interpret for a CW optical signal.
In general spectrum analyzers have a noise floor, but I wouldn't call it "RIN", because it is not "relative intensity" --- it doesn't change in proportion to the optical power. The measurement system noise is a fixed "floor" and you can't measure power spectral density below that floor. So whenever the trace is down at the noise floor, you're not measuring anything about the device you are testing, just the capabilities of the analyzer.
General comment
The RIN measurement is fairly difficult to do. Unless the laser has very bad performance you need a very low-noise detector, very low-noise preamplifier, and a very sensitive spectrum analyzer (with a low noise floor). You will want to test the noise floor of your whole receiver system (detector, preamplifier, spectrum analyzer) before measuring your laser to be sure you know when you're measuring the laser behavior and when you're just seeing instrument noise.
Edit
To follow up your questions in comments:
Sorry I'm not familiar with RIN measurements on pulsed lasers. But the units of dBc/Hz make a lot more sense now --- they're just talking about the fundamental of the pulse signal as the carrier.
The measurements I'm familiar with, you're most interested in the peak frequency in the RIN spectrum. I don't think you could do this with a pulsed laser because you'd have to pulse at a higher frequency than the RIN peak, which would also be beyond the modulation capabilities of the laser. But maybe there are tricks I'm not aware of.
I will suggest that for a pulsed RIN measurement, you don't need the bias tee, though you might want a blocking capacitor for the sake of your SA input. The peak of the fundamental of the pulse signal gives you the laser signal power that you'd be measuring the noise relative to.
is it fair to say then that the laser has equal or better noise performance?
I'd say it this way: if the laser noise is too small to measure on your detector/SA system, then the measurement system is not adequate to measure the noise of that laser.
how would you recommend characterising the system noise floor?
Typically, you turn on the photodetector and pre-amp, but don't apply any laser signal. Then take a sweep on the spectrum analyzer, using the exact settings you'll use for your measurement. This gives the combined floor for the detector plus the SA.
You should be able to display this for comparison to your laser RIN measurements by just using the save-trace features of the SA, without any need for calculations.
Best Answer
Noise Figure requires a given Input Noise Density, often 50 ohms.
You may be more interested in the Noise Voltage, so you can use 300 ohm antennas, or 600 ohm audio, or magnetic beacon sensors that are just inductors.
So let us compute theNoise Density (nanoVolts/square_root_Hertz). And the equivalent Rnoise. I use Rnoise as an easy design concept, to combine with other discrete resistors in a circuit and predict the total noise.
The datasheet gives 40 microVolts INPUT, over 150MHz bandwidth. That is, this is input_referred_noise (RTI).
Now scale that by sqrt(150,000,000). For easy math, use sqrt(100,000,000) which is 10,000.
So the internal broadband noise (per root_Hertz) is
Knowing 62 ohms produces 1 nanoVolt, and this math predicts 4nV, the
equivalent internal Rnoise of that first transistor is
Lots of transistors can achieve that noise floor.
I recall Hewlett Packard had long had access to ultra_small resolution photo-lithography machines, maybe even electron-beam machines.
I would not be surprised if HP produced special electronic systems for the US government. For national defense there really is no price limit on monitoring other nations' radio/radar emissions. Even back in the 1960s.
These systems simply are built, to avoid any surprises, at whatever cost.