For architectural reasons, we have a need to take a twisted pair (of an RS485 bus) into a PCB and then out. The PCB traces would add about 2 cm (just less than an inch) of length to the bus. The stubs connected to this differential trace pair in the PCB are very short (<1 cm).
Question: How to match the PCB differential trace pair impedance to that of the twisted pair cable (~120 Ohms)?
Given the low bit frequency (about 250kbits/s), the first answer that comes to mind is that the trace pair impedance does not really matter, since the shortest wavelength within the relevant bandwidth is so much larger than the trace pair on the PCB (say, a couple of meters vs. a couple of centimeters). This also appears to be true for CAN buses entering PCBs, given some of the discussions that I have seen in this site.
Having acknowledged that, if we still wanted to get as close to 120 Ohms differential impedance with the trace pair on the PCB, how can we compute the trace width and separation? Edge coupled micro strip calculations that one can find on web sites such as eeweb.com incorporate a reference ground plane on which the common-mode currents can return. In our case, it is just a twisted pair (no ground wire or shield) entering the PCB (i.e., no common-mode current return path on the wire), so should we even use a ground plane under the trace pair?