That's great! The schematic really helps.

Yep! your reasoning seems sound provided that you calculated the voltage drop across the 50 ohm resistor by measuring the signal on each end of it with respect to ground and taking the difference, and thre is no "funny stuff" -like added noise, and that the amplitude was taken in two separate measurements to preclude phase shift cause a problem with a differential measurement.

The output having twice the amplitude of voltage across the 50 ohm resistor seems to be the key.

 

I measured just like that. With two oscilloscope probes. One across both the resistor and SA-input and one across just the SA-input. I did not get twice the voltage across the 50 ohm resistor and the SA as I got across just the SA. I measured the following ratio:

(50ohm resistor and SA voltage)/(SA voltage) = 3/2 = 1,5

Then you have definitely confirmed that the analyzer input is damaged. It's very odd that it is exactly twice the impedance that it should be though! If the frequency response is still good, you can make the analyzer usable by putting a 100 ohm feedthrough load on its input connector.

 

Thank you for all the help. I appreciate it.

 

I measured just like that. With two oscilloscope probes. One across both the resistor and SA-input and one across just the SA-input. I did not get twice the voltage across the 50 ohm resistor and the SA as I got across just the SA. I measured the following ratio:
(50ohm resistor and SA voltage)/(SA voltage) = 3/2 = 1,5

Seems your SA has standard input impedance 75Ω:


ADDED:
I changed my opinion - if impedances are not equal, because of additional 50Ω resistor,
then level voltage on input SA is not predictable as long as reflections of signal are possible.