This is related to the propagation speed of the signal, which is 6in/ns in the copper cables on the FR4 board. Simply put, as long as the round-trip time of the signal on the trace is greater than the rise time of the signal, the trace on the PCB should be handled as a transmission cable.

Let's see what happens when a signal travels over a long trace. Suppose there is a 60-inch long PCB trace, the return path is the ground plane on the inner layer of the PCB board close to the signal cable, and the signal cable and the ground plane are open at the far end.

Thesignal propagates forward on this trace, it takes 10ns to transmit to the end of the trace, and 10ns to return to the source, so the total round-trip time is 20ns. If the signal round-trip path above is regarded as an ordinary current loop, there should be no current on the return path because it is open at the far end. This is not the case, however, the return path has current for the initial period of time after the signal is on.

Add a signal with a rise time of 1ns to this cable. In the first 1ns time, the signal only traveled 6 inches on the cable. I don’t know whether the far end is open or short, so how much impedance does the signal feel? OK? If the signal round-trip path is regarded as an ordinary current loop, there will be a contradiction, so it must be treated as a transmission cable.

Actually, there is parasitic capacitance between the signal trace and the return ground plane. When the signal propagates forward, the voltage at point A does not change continuously. For parasitic capacitance, the changing voltage means current is generated. Therefore, the impedance felt by the signal is the impedance presented by the capacitor, and the parasitic capacitance constitutes the path for the current to flow back. At each point the signal travels forward, it experiences an impedance resulting from the application of a changing voltage to the parasitic capacitance, commonly referred to as the transient impedance of the transmission cable.

When the signal reaches the far end, the voltage at the far end rises to the final voltage of the signal, and the voltage does not change any more. Although the parasitic capacitance still exists, but there is no voltage change, the capacitance is equivalent to an open circuit, which corresponds to the DC situation.

Therefore, the short-term performance of this signal path is not the same as the long-term performance. In the initial short period, the performance is the transmission cable. Even with an open circuit at the far end of the transmission cable, during signal transitions, the front end of the transmission cable will behave like a finite-value resistor.