Initial condition: The advantage of symmetric end termination in transmission lines

The PCB (printed-circuit-board) transmission line in Figure 1a lacks an end termination. If you leave switch S₁ closed for a long time, the line comes to rest in a state with precisely 0V at all points. The transmission line in Figure 1b behaves similarly. With S₂ closed, it also comes to rest in a state with 0V at all points. The voltages in the two situations are the same, but the currents differ. In the first case, the line at rest carries no current. In the second case, the end termination supplies a substantial current as long as you hold the line in a low state. To make the numbers easy, assume 2V logic, a perfect switch at the source, and values of 100Ω for both R₁ and R₂. Those values produce a steady-state current of 20 mA for Figure 1b.

At time zero, with both lines in their respective steady-state conditions, open both switches. In the case of Figure 1a, just before you open the switch, no current flows through it. Opening S₂ therefore changes nothing; it has no effect on the circuit. Opening S₁ in Figure 1b has a different effect. In the steady-state condition, 20 mA spills continuously through the switch. When you interrupt that state of events by opening S₂, the current at the left end of the line changes from 20 mA to 0A. You can emulate that effect with a superposition of two linear-current sources, I_B and I_C, which connect (Figure 1c).

Current source I_B replaces the 20 mA of steady-state current flowing through S₂ in Figure 1b. It sets the initial conditions before your switching event, and it perpetually sinks 20 mA. At time zero, a 20-mA step of current from source I_C cancels the current from source I_B, bringing the net current to 0A. The combination of two sources duplicates the conditions at the left of Figure 1b the moment S₂ opens. The linear-current-source model clarifies the actions that occur at time zero. Directing a positive step of 20 mA into the line must create a positive-step-voltage waveform moving to the right with an amplitude of 20 mA×50Ω=1V. In a 2V system, that scenario makes a half-sized step.