(Confused) RF circuit concepts- Voltage division?

[nomisme]

how could you calculate the input voltage at an input impedance from a voltage source?

I saw they use voltage division law to find the voltage at the input impedance(Zin) from a voltage source(Vs).
They do it like Vs * Zin/(Zin+Zs).
The problem came... We all know that we can determine the voltage only after considering all resistances in a circuit. They still have the transmission line and load impedance. If I denote them as Zt and Zl, wouldn't the voltage to the input impedance be Vs * Zin/(Zin+Zs+Zt+Zl) instead?

Because if you simply consider Vs * Zin/(Zin+Zs), the voltage wave will be zero after going through the input impedance before even entering the transmission line connected with a load on the other end. This doesn't make any sense.

Help.. If you don't understand what I am saying, please let me know.

 

[Simon Bridge]

The input impedence to what?

The usual situation is that you have a black box with some sort of network inside and two terminals sticking out of it. You have no access to the internals of the box. All is not lost: this can be modeled as an ideal voltage source in series with a reactance. The Thevinin equivalent.

You get the value of the ideal voltage source by measuring the open-circuit voltage between the terminals.
The internal reactance (input impedance - whatever) you can find by putting a load between the terminals, measuring the new voltage across the terminals, and doing a bit of math.

You'll notice that this approach combines the effects of all the reactances except for the load, and includes them as part of the "internal reactance" of the black box. If you happened to have a very long coil of wire between the external load and the black box, then that will throw the calculation off - unless you treat that as part of the load. It's outside the box so you can, in principle, measure it.

If you want to use a different model, then the calculation is going to be different.
I think you need to provide a specific example - a link to a circuit of the kind you have in mind for instance.

 

[nomisme]

The input impedence to what?

The usual situation is that you have a black box with some sort of network inside and two terminals sticking out of it. You have no access to the internals of the box. All is not lost: this can be modeled as an ideal voltage source in series with a reactance. The Thevinin equivalent.

You get the value of the ideal voltage source by measuring the open-circuit voltage between the terminals.
The internal reactance (input impedance - whatever) you can find by putting a load between the terminals, measuring the new voltage across the terminals, and doing a bit of math.

You'll notice that this approach combines the effects of all the reactances except for the load, and includes them as part of the "internal reactance" of the black box. If you happened to have a very long coil of wire between the external load and the black box, then that will throw the calculation off - unless you treat that as part of the load. It's outside the box so you can, in principle, measure it.

If you want to use a different model, then the calculation is going to be different.
I think you need to provide a specific example - a link to a circuit of the kind you have in mind for instance.

Hi Simon,

Thanks for your explanation in advance.

The input impedance I mentioned refers to an input resistance to a transmission line on the left. (The load is on the far right and the voltage source is on the far left and the transmission line is in the middle.)

The voltage from the source will experience an input impedance before entering the transmission line. We want to first find out its voltage at the input impedance.

The problem is that if we model the input impedance as a resistor, wouldn't the voltage wave becomes zero after passing through it? Because according to the voltage division rule provided above, it only considers Rs and Rin so the others resistances in the system are ignored. So after passing the voltage through both Rs and Rin, the voltage should be zero.If this is the case, there should be no wave going through the transmission line and load. This is a very fundamental concept that I don't understand. How could you determine the voltage at the input without considering the resistance of the transmission and the load.

I am really confused..

 

[meBigGuy]

A transmission line has an input impedance. For example a 50 ohm line appears initially as 50 ohms, but if there is no load at the other end it eventually becomes an open circuit. This load is effectively across the transmission line input terminals.

You will connect an output load. If you connect 50 ohms to the output, then the 50 ohm input load will remain 50 ohms because there are no reflections.

You will drive with a source that has an output impedance. The output impedance is in series with the 50 ohms input impedance. This causes a voltage drop when you drive the transmission line.

The input impedance of the transmission line can be modeled as a resistor across the transmission line. So if the source is 50 ohms and the line is 50 ohms, then Vs/2 initially appears across the transmission line.

 

[nomisme]

A transmission line has an input impedance. For example a 50 ohm line appears initially as 50 ohms, but if there is no load at the other end it eventually becomes an open circuit. This load is effectively across the transmission line input terminals.

You will connect an output load. If you connect 50 ohms to the output, then the 50 ohm input load will remain 50 ohms because there are no reflections.

You will drive with a source that has an output impedance. The output impedance is in series with the 50 ohms input impedance. This causes a voltage drop when you drive the transmission line.

The input impedance of the transmission line can be modeled as a resistor across the transmission line. So if the source is 50 ohms and the line is 50 ohms, then Vs/2 initially appears across the transmission line.

Thank you both of you.
I think I ve got it now.

 

[Simon Bridge]

Great - well done.