One other note: At 30:50 in the video you mentioned these were "real little things happening in the circuit". I'll assume you know exactly why your getting the knee in the signal. That is the transmission line and the knee in the signal is telling you you probed closer to the source side of the transmission line as opposed to the destination. Surely you recall your t-line fundamentals from school?
Of course, and that's was what I was alluding to. Those reflections are present in the circuit at the probed point, that's actually happening to the signal on the board at that point, it's not the probe causing that.
I wasn't going to go down that rabbit hole, the video was already way longer than I wanted it to be.
Alas, I didn't learn about t-lines at school, but did come across this myself a couple of weeks ago.. I wrote the following for another forum, which may be of interest to some.
I was looking at similarly distorted waveform on the output of a 22.5MHz oscillator on a PCB I'd knocked up:
Here's a photo showing the oscillator and the traces to the ICs it fans out to and where on the header, close to the oscillator, I was probing.
Thinking that at just 22.5MHz, it's a relatively low frequency signal, I didn't pay too much attention to its routing. I was more interested in having modular PCBs so I could experiment with different parts. These are pretty small PCBs, so the total distance from the oscillator output pin, through the header and to the furthest input pin on the AND gate IC is just 35mm. That didn't seem too far.
Alas, my intuition is often wrong. So let's sim it. To begin with, we need to understand the geometry of the PCB and the trace. This is a two layer PCB, with a ground plane at the bottom and the signal trace on the top - a microstrip. Here's a cross section:
The propagation speed of an electrical signal through a top-layer PCB microstrip, can be calculated with:
Here, c is the speed of light, and Er is the relative permittivity (or dielectric constant) of the PCB substrate. For this PCB, I'm using FR-4 material for the substrate, so this value is around 4.5. The scaling of Er by 0.475 and adding 0.67, accounts for the fact that the FR-4 substrate is only below the microstrip and we have air above. If this were a four layer PCB and our microstrip was on one of the inner layers, we'd need a different formula.
er=4.5; % relative permittivity of dielectric for FR4
eeff = 0.475*er+0.67; % effective permittivity for top layer microstrip (substrate material is only on one side)
c=3e8; % speed of light
vp_mps=c/sqrt(eeff); % speed in m/s
vp_mmpns=1e3*vp_mps*1e-9 % speed in mm/ns
tpd=1000/vp_mmpns % propagation delay in ps/mm
So in 1ns, a signal will travel around 180mm. This means it takes 35/180=195ps for a signal to travel from the oscillator to the AND gate IC.
If the clock were a 22.5MHz sine wave, then its rise time would be 1/22.5e6/2=22ns. However, it is actually closer to a square wave with a rise time of 600ps, so approximately .35/.6 = 600MHz bandwidth. In the former case, the rise time is much greater (112x) than the propagation delay, in the latter, it's fairly close (3x). Let's run a spice sim to see why this relative difference is important.
In order to tell our spice simulator how to simulate this, we need to create a transmission line model, based on the above microstrip geometry:
.model top_trace u level=3 plev=1 elev=1 nl=1
+ th=1.378mil ht=63mil kd=4.5 dlev=1
+ wd=12mil
.end
We can then create a schematic in spice that consists of a voltage source, followed by the transmission line model, and then a 4pF capacitor to simulate the input pin capacitance of the IC at the end of microstrip. I've split the transmission line model in to 8 equal parts, so we can probe at different points to see what is going on along the line.
Rather than immediately trying to model the oscillator's actual output and real trace length, we can start with a simplified example. We can look at how a single 600MHz sine wave pulse (the highest frequency component in the square wave) would propagate down an 800mm microstrip (a very big PCB!).
We can see the voltages at different points along the line in the top 8 waveforms, with the bottom 8 waveforms showing the current at the same points. We can see that because we have a capacitor at the end of the microstrip, the pulse is reflected back. When the reflection initially occurs, we can see the current reverse direction (it becomes negative), then we can see it lead voltage. The simulated round trip time is around 9.6ns. The speed was therefore 2*800mm/9.6n = 167mm/ns, which is close to the estimated value above (The formulas used by spice will be more accurate than the approximation above).
If we decrease the length of the microstrip to 400mm, we can see that the wave is reflected back sooner, as it has less distance to travel.
If we decrease the length again to 80mm (so that the round trip time is 2*80/167=0.95ns), we can see the reflected pulse is constructively interfering with the initial pulse. The pulse shape observed depends upon where exactly along the trace you probe:
If we reduce the length to 8mm (with a roundtrip time of 2*8/167=95ps), the reflections are so fast, that they don't distort the waveform:
So we can see that if the round trip time is greater than about 1/6th of the rise time, the reflections are going to distort what we see with a probe. And the distortion observed will depend where along the trace we probe.
We can then change the voltage source in the schematic to create a square wave closer to what the oscillator outputs, to see if we can see something similar as was seen with the real oscilloscope, in order to validate our model.
And that looks pretty similar. If we then shorten the microstrip length, the distortion disappears.
So, as most probably already knew, we need to start worrying about reflections when the round trip time is greater than about 1/8th of rise time and keep the traces as short as possible.