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We have a source (Z_s) and load (Z_load) connected with a transmission line. It is known that the connection is matched, that means Z_s=Z_0 and Z_0=Z_load. If the voltage from source is 1V, what will be the voltage on load?

It may look as a textbook exercise, but it's not. I'm having hard time understanding what is going on if we place a matched transmission line between source and load. All the books I have focus on reflections and physical details, but not much about more simplistic real scenarios.

If the connection is matched as described can the transmission line be ignored (considering it's loseless) and voltage on load be calculated from standard voltage divider? Thus, getting 0.5V on load, or do we need to consider the line also and it's 1/3V on load?

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Z_s is the internal resistance of the source. So you will see 1V after this resistance, at the output of the source. Knowing this, you can omit the Z_s in your calculations. Now it's just a matter of calculating the voltage divider and since Z_o=Z_load, V_load will be half of the source voltage. Thus V_load = 0.5V

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  • \$\begingroup\$ And if we put another Z_s1 after the source, which is also matched (Z_s1=Z_o), then all the parts in line will have 1/3 of the source voltage? \$\endgroup\$
    – so_user
    Commented Nov 5, 2019 at 8:55
  • \$\begingroup\$ Yes. However, one will add Z_s1 tot Z_o to make it easier. So the line resistance will now be twice as much as the load resistance. So V_load = V_s / 3 indeed \$\endgroup\$
    – Swedgin
    Commented Nov 5, 2019 at 9:05

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