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signal loss on a perfectly matched transmission line

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ykishore

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If a transmission line is terminated perfectly with a load resistance equal to its characteristic impedance, does this mean only half of the input signal reaches the load? I mean like if we consider the characteristic impedance as a lumped resistance, say Z0, now the load Rs=Z0, as its perfectly matched; so applying a simple voltage divider for series resistances, now does this mean anytime we send a signal on a transmission line, say a PCB trace, only half the signal is going to reach the load? i.e., if we send a 5Vp-p square wave on the line, the signal received at the load is going to be 2.5Vp-p and we need plan our source signals accordingly? or am I missing something here?

I believe the power dissipated in the resistor is related to the duty cycle of the square wave so as far as I can remember it doesn't drop all the way to 2.5V, but you will have to double check. Sorry I cant be more precise off top of my head.

If a transmission line is terminated perfectly with a load resistance equal to its characteristic impedance, does this mean only half of the input signal reaches the load?

Yes, for a controlled impedance system, the voltage with proper load is half the open circuit voltage.

I mean like if we consider the characteristic impedance as a lumped resistance, say Z0, now the load Rs=Z0, as its perfectly matched; so applying a simple voltage divider for series resistances, now does this mean anytime we send a signal on a transmission line, say a PCB trace, only half the signal is going to reach the load? i.e., if we send a 5Vp-p square wave on the line, the signal received at the load is going to be 2.5Vp-p and we need plan our source signals accordingly?

Correct!

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related to the duty cycle of the square wave so as far as I can remember it doesn't drop all the way to 2.5V

No, that's not correct.
If we have 50 Ohm source impedance and 50 Ohm load, we have half the source voltage at the load, for any signal any frequency. It's indeed simple voltage divide math.

You can also think of it as .. with no load you get the reflected wave that gives twice the initial t=0 voltage compared to matched circuit impedance.

All matched impedance source and load with no reflections in transmission line, the initial peak wave voltage is reduced 50% by impedance ratio and stays there.

No, that's not correct.
If we have 50 Ohm source impedance and 50 Ohm load, we have half the source voltage at the load, for any signal any frequency. It's indeed simple voltage divide math.

I did wonder...

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