What's the importance of source impedance termination?

The idea is that signals propagate at a finite speed, that is to say a certain signal takes t time to get from one end of the transmission line to the other line. The cable also has some intrinsic capacitance/inductance per unit length, which can be approximated with a characteristic impedance (assuming loss-less):

\begin{equation} Z_0 = \sqrt{\frac{L}{C}} \end{equation}

This is the impedance initially experienced by the source when the signal changes, with the signal level acting like a voltage divider circuit between R1 and Z0: \begin{equation} V_s = V_{in} \frac{Z_0}{R_1 + Z_0} \end{equation}

When the signal propogates to the end of the cable, it will realize that there's nothing to dump the signal energy into. The signal must go somewhere, so it bounces off the far end and returns to the source. When it reaches the source, the source voltage will be twice the original \$V_s\$, which will flow back through R1 to the source.

If \$R_1\$ = \$Z_0\$, \$V_S = V_{in}\$ and the entire transmission line has reached steady state because no more energy can be injected into or absorbed from the line. This is ideal because the line has reached steady state in ~2t (one t to get to the target, and one t to get back to the source).

If \$R_1\$ is too large, \$V_S\$ will still be larger than \$V_{in}\$ so the source will continue to dump energy into the transmission line, and the transmission line's voltage will slowly step up as the signal bounces back/forth.

If \$R_1\$ is too small, \$V_S\$ will overshoot when the signal gets back. In this case, a falling edge wave will propagate down the line because the source is trying to absorb the excess energy pumped into the line, and again the voltage will bounce back/forth till steady state is reached.

In the latter 2 cases the target voltage could bounce above/below a certain digital logic level multiple times so the receiver could get false data bits as a result. This could also be potentially damaging to the source because the reflected signal may rise induce excess stress on the source.

Now what happens if we attach something to the other side, like a resistor \$R_2\$?

Now the target can absorb energy, and only a fraction of the original signal is reflected. If \$R_2 = Z_0\$, we again have matched impedances and no signal is reflected.

If \$R_2\$ is too small/too big, we'll end up with similar reflection signals as above, except the signal is inverted.

Using \$R_1 = Z_0\$ can still be used to prevent repeated bouncing, but the steady state signal voltage will be the result of the voltage divider between R1 and R2. If \$R_2 = Z_0\$, there is no reflection so the value of R1 doesn't matter. We might as well pick R1 = 0 so the target voltage is the same as the source voltage. As supercat pointed out, you could also have the source drive a signal twice the magnitude the target is expecting and still use \$R_1 = R_2 = Z_0\$

I wrote an online transmission line simulator to play around with which demonstrates source termination. I found it useful for visualizing these signal propagation waves along the transmission line. Pick a large enough R2 and you can approximate an open, like the case you have. This only models loss-less transmission lines, but is usually accurate enough.

In terms of signal integrity (measured by step response at the receiver side) the three configurations are identical (Zsource - Zload):

1) 50 Ohm - infinity (source termination)
2) 0 Ohm - 50 Ohm (load termination)
3) 50 Ohm - 50 Ohm (termination at both ends)

However, in the 3rd variant there is a 50% decrease in amplitude. So, from practical standpoint the 3rd option should be avoided unless there is a compelling reason to do so.

Disclaimer: this covers single wire point-to-point ideal cable one direction communication between the source the receiver. If there is a junction on the way, then it may make sense to use dual termination - I didn't think about it.

Ok, here's the long but overly generalized description of what's going on...

The impedance of the transmission line (a.k.a. trace) is 50 ohms, which means that as the signal travels down the cable it looks like a 50 ohm load to the driver. When it hits the end of the trace, it reflects back and causes parts of the trace to temporarily reach a much higher/lower voltage than it should. We call this overshoot and undershoot.

With the 50 ohm source resistor, the resistor plus 50 ohm trace forms a voltage divider (div by 2). Just before the signal hits the end, the signal at that location is 50% of the needed amplitude. Just after the signal hits the end, the reflection combines with the 50% original signal and results in a perfect 100% amplitude signal. The reflection travels back to the source resistor where it is absorbed.

A receiver located at the very end of the trace will see a mostly perfect signal edge. But a receiver in the middle or near the resistor will first see a 50% signal and then a 100% signal. Because of this, source termination is only used when there is only one receiver and that receiver must be located at the end of the trace.

If the resistor does not match the impedance of the wire/trace/cable then the voltage divider is not 50%-- which results in an imperfect match and the reflection could cause issues.