Alternatively, I suspect the scope isn't 50ohm input but is really 50ohm//15pF. In that case you will see the effect stated in the P6156 manual, viz:
No; I have used a 3GHz scope to do the test. I'm damn sure it has proper 50 Ohm inputs! Please check the specs before making these kind of statements.
Here is evidence that you
don't have a 50ohm input.
The pictures below are with my Tek 485 and HP10020a 1.5GHz Z0 probe. The key point about the 485 is that it has
two input attenuators: one is the traditional 1Mohm scope attenuator, the other is a traditional RF 50ohm attenuator. An RF relay connects the input to
either one or the other attenuators.
The first picture shows the source: the Tek485 cal out (<1ns risetime) correctly loaded with a 50ohm through terminator that ought to be good to a GHz or so. The probe (500ohm input) is directly connected to the source, without leads. Timebase is 5ns/div.
The second picture shows the trace when the scope's (real) 50ohm attenuator is used. I think you will agree that is a nice enough waveform.
The third picture shows the trace when using the scope's 1Mohm input with an inline 50ohm through terminator. Hence the input "seen" by the cable is really 50ohm//20pF. You will note the dip you see in your traces. Where do you think that has come from?
The fourth picture is used to highlight the probe cable length, by switching to the 50ohm attenuator so the input impedance is 25ohms. You will note the position of the step coincides with the dip in the third picture, thus demonstrating that the dip is due to the 20pF input capacitance.
Incidentally, those familiar with using TDRs will realise that a dip in a trace is due to capactance, and a peak is due to inductance.
BTW, a Z0 probe does
not use a lossy transmission line; it uses standard coax.