Does the standard termination of 120 ohm really suppress reflections in recessive state in CAN?

Question

Suppose we have a multi-node CAN network with the ends of transmission line terminated with the standard 120 ohm. When the recessive state is on the transmission line, the voltage on both CAN_H and CAN_L will be 2.5V, so, when this state reaches the termination resistors at the end of the transmission line, the voltage at both ends of the termination resistor is 2.5V(voltage drop is 0V), so no current flows and voltage would see a high impedance state and should reflect back, but is not observed. What is the flaw in my explanation? Please explain in detail.

I don't clearly understand how this standard termination is actually suppressing reflections in recessive state?

Answer 1

When any signal is sent into an infinitely long transmission line, the signal sees the characteristic impedance all the time so there is no discontinuities and no reflections. If you have a bus of certain length, then the signal sees the bus to have the characteristic impedance until the signal hits the end which has infinite impedance - a discontinuity. Therefore the ends are terminated with a resistor matching the characteristic impedance so that when the signal reaches end of the wires, it will still see from that point on the resistor impedance which is no different from the impedance of the infinitely long transmission line. So correctly terminated transmission line has no reflections from the ends, no matter what signal is sent into it. The reflections do not depend on what signal is sent to the bus and at what source impedance - the dominant state drives the bus with extremely low impedance, and the recessive state stops driving the bus so it has extremely high impedance. Basically same as connecting a voltage source ( e.g. battery) to the bus, and disconnecting it. The CAN bus will look like a 60 ohm resistance, from any point.