In the worst case scenario, a CAN message will contain alternating bits (...01010101...), which will produce a square wave with a fundamental frequency of HALF the baud rate:
This ignores the higher harmonics to produce a clean square wave, however.
See this post about CAN circuitry: https://resources.altium.com/p/can-bus-designing-can-bus-circuitry
In my case, the network should have a 1mbit/s baud rate. Assuming the worst-case scenario, when the network is transmitting a sequence of alternating bits (01010101) the signal will be a square wave of frequency 500kHz, or equal to half the baud rate.
We know the resistance at 60 ohms, we can thus calculate the capacitor.
This post seems to suggest that it's not actually important to place the cut-off frequency above the baud rate anyway: https://e2e.ti.com/blogs_/b/industrial_strength/posts/the-importance-of-termination-networks-in-can-transceivers
Typical concerns I have heard with this termination technique include, “Will this filter my CAN bus signals?” and “Do I need to place the corner frequency above my data rate?” The simple answer to both questions is no. Since the capacitor does not place a direct current (DC) load on the differential bus signal �� it only filters the alternating current (AC) signal and the common-mode signal – and the differential signal is what determines the bus state, you do not need to set the corner frequency of the filter above the data rate.
