This is not true. We are not dealing with a transmission line. This is a totally different problem.
We are dealing with a divider where one impedance is a 2 k resistor and the second is a 50 ohm resistor in parallel with about 150 pf of cable and scope capacitance.
The discussed "probe" divider consists of: 2k resistor with an impedance of 2000 + j0 (it is the probe tip as well as it is a part of the sdcard's divider) and 50ohm resistive impedance (50+j0) "as seen at the input side" of the transmission line (represented by the 1m long 50ohm coax) terminated (at its output) by the 50ohm termination resistor (50+j0).
There is no capacitor involved, I would say (not considering small reactancies of capacitive parasitics and inductive parasitic reactancies of any wires involved).
ScopeModel.png Assumes cable and scope capacitance is 150 pf and input is to a 50 ohm terminator.
The properly terminated coaxial cable (or other transmission lines properly terminated) does not behave as two wires with two capacitors at the near and far ends wired.
The equivalent model of a transmission line is complex and distributed, afaik it includes N capacitors, 2N inductors, X resistors and other parameters probably. So in order to simulate the stuff properly you need a model for the coax there.
The square wave turns into this because of the cable and scope capacitance.
The coax is a transmission line (when terminated properly) which cannot be considered similar to the Layden jar.
Your scope has got a channel A (my assumption) with a BNC connector on the front panel. Near the connector there is (most probably) an information depicted: 1M || 25pF (or 50pF). That says: "Dear user of our o'scope, this channel A input has an complex impedance equivalent to 1megaohm resistor with parallel capacity of 25pF (or 50pf) wired from inside at this BNC connector. The internal circuitry of the channel A (voltage dividers, amplifiers, etc.) has been compensated for that input complex impedance thus the AC response of the channel A is flat within guaranteed bandwidth."
It means you must not consider that capacitance for your simulation, nor for your design (it is compensated by the o'scope itself). The 1Meg internal resistance in parallel with 50ohm (the 50ohm BNC terminator plugged into the channel A BNC connector from outside) does not play any role too.. The termination impedance is considered (and designed) resistive only (50 + j0).
So it is my current understanding that you must not include the input capacitance of the o'scope into any simulation. The stray capacitance at both ends of the coax might be something like 1pF max. The parasitic parallel capacity of the lower resistor (probe tip) would be 0.5pF max with small axial resistor or less with SMD one.
The resistors (and capacitors and wires) have serial inductive reactance as well, let us assume we will neglect it for today. The coax parameters (like pF/m, uH/m, ohm/m..) must not be consider for the simulation too, as all that stuff is already included int the statement "it is the 50ohm coax cable with following model parameters", when properly terminated.
I played with this test on my scope and get the same result though noisy since the signal is only about 80 mv.
Try it based on my schematics above with the coax (terminated by 50ohm BNC terminator at the o'scope input). Your results will be much nicer.. Of course you have to cut the coax and solder a BNC connector, maybe you want to avoid that :)
Probing High-Speed Digital Designs by Dr. Howard Johnson. First publ. in Electronic Design, March, 1997:
A report on the suggested "probe" in my above schematics for your divider problem measurement setup - see the "Figure 3—Resistive-Input Style Probe (Z0 Probe)"
Your probe will be the "2k" tip probe, usually "1k" tip is used by experts, but it does not matter for your purposes, moreover the 2k fits your sdcard's divider problem as explained above:
On the last page there is a detailed guide for you: "How to Make A 1K-Ohm Probe". They recommend RG-174 for 1GHz bandwidth (but you may go with RG58 as well, the 174 is thinner). You will use 2k instead 1k as it is the part of your sdcard divider. See the details on parasitics.
BTW, the suggested probe in my above schematics has been rated extremely well in comparison with a passive 10:1 probe and an 1pF FET active probe:
"..The resistive-input probe is cheap, it has a terrific bandwidth, and it is more tolerant of long ground wires than the other probes.. ..In high-speed system developments, the ubiquitous 10-pF 10:1 capacitive-input probe is no longer adequate. The two alternatives are the FET-input probe and the resistive-input probe. Of the two, the resistive-input probe is cheaper, it has as good or better bandwidth, and it is more tolerant of long ground wires. These advantages come at the cost of higher IOH required in your digital circuits in order to drive the 1K resistor. In modern high-speed systems, because the extra drive current is almost always readily available, the resistive-input probe makes a lot of sense. As we go higher in frequency, the FET-input probes will run into increasing difficulties. At signaling rates faster than about 300-pS rise-fall, only a resistive-input style probe can maintain a high enough input impedance to remain useful."
PS: another reading for you - a comparison of the suggested "probe" with 7 GHz Agilent Infiniimax $10k probe (while measuring a 100ps fast rising edge): http://koti.kapsi.fi/jahonen/Electronics/DIY%201k%20probe/