I had a thought over the weekend and I really need input from experts. It may be a completely stupid question but I'm very curious and it's draining all of my brain energy.

Suppose I have a Linear Time Invariant Differential Amplifier in a test fixture and I use AFR to de-embed to get the 4-port s-parameters.

Let's pretend this amplifier has a differential noise figure of 0dB and produces no noise on its own.

The Differential Amplifier has two 50Ohm pure input termination resistors that are completely uncoupled with no parasitics and each have a PSD of 4KT*50 (V^2/Hz).

Also, the two traces of the package 4-port transmission line leading to these two pure input resistors are completely uncoupled and lossless and are 50Ohm.

Also, at the end of the line it's terminated with an impedance of 50Ohm and let's pretend that these terminations do NOT have any inherent noise aka PSD = 0 V^2/Hz

Now here's the question. Because there is a 50Ohm line in series with the noisy resistors, one can prove that the uncorrelated PSD of the differential input to the amplifier is 2*KT*50 (V^2/Hz). **Does this mean that the differential output PSD is (2*kT*50)*|SDD21|^2?**

The root of the question is does the amplifier's SDD21 act on the voltage across its differential input nodes or does it only act upon a forward travelling wave INTO the amplifier.

Now for Question 2 (please still answer question 1)

Assuming that the far end termination resistors DO have thermal noise and the channel is now lossy,

does this mean that the differential output PSD is (2*k*T*50)*(1+|SDD21_channel|^2)*|SDD21_amp|^2 [V^2/Hz]?This makes sense to me, but I'm skeptical and am desperate for input.

Of course assuming there are 0 reflections at all times at any point.