With parallel termination of a transmission line, the line characteristic impedance \$Z0\$ is matched with the pull-down resistance \$Rt\$. But, in that case, shouldn't the receiver see half of the transmitted voltage?
In my understanding, with no reflection, the whole situation can be brought back to a voltage divider with equal resistances.
Obviously, I made an error somewhere. So, how can the receiver see the full voltage with parallel termination?
Line impedance is not the same as a series resistance which would cause a voltage drop. Line impedance, altough it happens to be measured in ohms, tells how the electric and magnetic fields are related outside the wires (=in the space where the wave actually travels, it's not in the metal). You can use the line impedance in calculations of what happens to a wave in line joints and terminations, but it has no use in ohms law.
In ordinary 2 wire cables, but not in waveguides, you can make most of the wave calculations by using the voltages and currents which they cause to the wires, but remember, that the wave and thus also the actual energy flow is outside the metal, it's only guided by the wires.
In your case the wave comes from the source, which obviously has very low internal series resistance. The wave meets the matched load, no reflection is caused. If the signal is DC, the rest can be calculated with ohms law. You get full voltage if the wire hasn't remarkable DC resistance.
The characteristic impedance of a cable is the value that should be taken into consideration when terminating to avoid reflections; it doesn't mean that it acts as that impedance in a potential divider sort of way.
Consider a really short 50 ohm coax terminated in 50 ohms driven by a zero ohm source at a low frequency - the full driving voltage will appear at the load both mathematically and intuitively (if you think about it).
I think you are confusing, case A:
simulate this circuit – Schematic created using CircuitLab
With, case B:
The difference between A and B is only at the source. The voltages I indicated is the amplitude of the signal at that point.
Indeed you're right that there is a 2:1 voltage division between the voltage at Vsource and the output voltage. That's because Rsource and Rload make a voltage divider.
However the voltage at Vsource isn't really the input voltage of the transmission line. That voltage is only present at the right side of Rsource.
In case B this is more obvious as I have replaces the voltage source and series resistor with their Thevenin equivalent, a current source and parallel resistor.
Note how the characteristic impedance of the transmission line is not actually present in the circuit. The characteristic impedance is the impedance with which you should feed in and extract power from the transmission line. If you don't do that you get signal reflections.
