Totally - I have been wondering about that. I was measuring on the AC coupling caps since they are so easy to probe relative to the QFN's they terminate into. The only way to get the bandwidth out of the probe is to hand place it on the test points, the solder-in probe heads attenuate way too much. I will try to see if I can get the probe directly on the QFN pads. I would expect the period to change with the different lengths of the transmission line and presumably different impedances from the geometry differences - although I am a beginner in wave theory.
It's definitely worth reading up on how waves propagate up and down transmission lines, and how different kinds of termination (series vs parallel) actually work. The very short version goes something like this.
Suppose you have:
- a driver with low output impedance (let's assume 0R for the sake of simplicity), and zero rise time.
- a long 50 Ohm transmission line, with no termination, and a high impedance receiver at the far end
- logic levels denoted by 0V for a logic 0, and 1V for a logic 1
- initial condition is that the entire transmission line is settled at 0V, and we consider what happens when there's a rising edge at the driver.
At t = 0, the voltage at the driver output instantly changes from 0V to 1V. The wave begins to propagate away from the driver toward the receiver, at a rate determined by the inductance and capacitance per unit length. Energy is stored in the inductance and capacitance of the transmission line, and while the wave front is propagating, current is drawn from the driver. It's a 50 Ohm line, which means the current being drawn is simply given by Ohm's law, so I = V / R = 1 / 50 = 20mA.
At any impedance discontinuity, part of the wave will pass through and the rest is reflected back. Conservation of energy is, of course, maintained.
The wave reaches the receiver, where the 50 Ohm transmission line ends, and the impedance of the receiver input is effectively infinite. This means the coefficient of reflection is 1, so all the energy is reflected back toward the driver. The voltage at the receiver becomes 2V, which may even damage the receiver. Current continues to be drawn from the driver. The receiver's input impedance may change, since it's now seeing an input voltage that well exceeds the usual logic '1' level.
The wave then reaches the driver, and this time it sees an impedance of 0R. This again means that 100% of the wave is reflected, but the sign also changes, and the negative going edge now propagates back towards the receiver. At the receiver, the voltage drops from 2V down to 0V. The wave reverses direction and continues to travel up and down the transmission line until it eventually dissipates, and the entire transmission line eventually settles at 1V.
This, clearly, isn't going to work, and it may even damage the receiver. Bad idea.
Now consider what happens if you put a 50 Ohm resistor across the receiver input.
At t = 0, the voltage at the driver jumps to 1V and the wave begins to propagate. When it reaches the receiver, the impedance it now sees is equal to the characteristic impedance of the transmission line, so the coefficient of reflection is 0. All the energy in the wave front is absorbed by the resistor and turned into heat, and the driver sources 20mA indefinitely. The voltage at the receiver transitions cleanly from 0V to 1V and all is well.
Note also that the voltage at every point on the transmission line switches cleanly from 0V to 1V. You can probe it anywhere and you'll see the same thing.
As an alternative, now imagine a 50 Ohm resistor placed in series with the driver, located close to the driver end of the transmission line, effectively making the driver's output impedance 50 Ohms. (Some high speed driver ICs include this resistor, and may allow a choice of values to match different transmission lines).
The voltage at the driver output, therefore, is divided down by the potential divider: 50 Ohms output impedance, and the transmission line itself looks like 50 Ohms to ground. So a 0.5V step propagates from driver toward receiver.
At the receiver, the incoming wave sees a high impedance, so as in the un-terminated case, it doubles in amplitude and reverses direction. The voltage at the receiver switches cleanly from 0V to 1V, which is good. The wave propagates back towards the driver, where it sees 50 Ohms, and 100% of the wave is absorbed and turned into heat in the series resistor.
In this case, current is only drawn from the driver while the wave is propagating. In the steady state (after 2 transit times - there, and back again), no current is drawn, as the receiver is high impedance. So, series termination means less dissipation in the driver. I generally tend to prefer series termination for this reason.
But: the waveform you see depends on where you probe. You only get a clean 0-1 transition at the receiver. Everywhere else, the signal goes from 0V to 0.5V, then from 0.5V to 1V a moment later. How long it sits at 0.5V depends on the distance from the receiver.
That's the ideal case. Real drivers have finite rise time, real transmission lines are lossy, and even the most expensive probes will distort the signal - so what you see on your scope may not bear much relation to what I've described.
Series termination tends to be the better choice for point-to-point connections, where one end is always the driver and one end is always the receiver. Parallel termination is preferred on bidirectional links, because you can simply put a resistor to GND next to each transceiver, and the line will be correctly terminated for both directions. Parallel termination (resistor to GND at both ends) is also better for multi-drop buses which may be driven from any point on the bus, and received at any other point.
It might be worth fabricating a PCB with a driver, receiver, a very long transmission line, and sites for series and parallel termination resistors. Drive a square wave, and probe at different places with different resistors fitted to see the effect.
Read up on time domain reflectometry (TDR); it works by detecting reflected signals, so any good description of how it works should cover impedance discontinuities and their effect on forward vs reflected waves.