Transmission line impedance question

[likephysics]

I am a bit confused on source impedance and Tline impedance, when both are equal.

Let's say source impedance is 50 ohms and Tline imepdance is also 50 ohms.
Why does the incident wave divide between the source impedance and tline impedance.
The impedance is same, so reflection co-efficient would be zero.

If the tline and load impedance are same, then the voltage does not get divided.

 

[Averagesupernova]

Maybe I am reading this wrong but it seems to me you are confused about dissipating half the power in the source when driving a load with the same impedance as the source impedance? In other words, you would have the same confusion if we eliminated the transmission line and drove the load directly?

 

[likephysics]

Maybe I am reading this wrong but it seems to me you are confused about dissipating half the power in the source when driving a load with the same impedance as the source impedance? In other words, you would have the same confusion if we eliminated the transmission line and drove the load directly?

You read it right.
why doesn't the half power dissipation happen when the 50 ohm tline is terminated by 50 ohm load.

Let's say we have a 1V source with 50 ohm source impedance driving a 50 ohm tline.
The voltage on the tline would be 0.5V.

Now if the tline is terminated by 50 ohm load, the equivalent circuit would be

0.5V source with source impedance of 50 ohms, terminated by a 50 ohm load. right?
what am I missing.

 

[Averagesupernova]

You read it right.
why doesn't the half power dissipation happen when the 50 ohm tline is terminated by 50 ohm load.

Let's say we have a 1V source with 50 ohm source impedance driving a 50 ohm tline.
The voltage on the tline would be 0.5V.

No, it would not. You are talking about a source driving a 50 ohm transmission line with the far end open right?

Now if the tline is terminated by 50 ohm load, the equivalent circuit would be

0.5V source with source impedance of 50 ohms, terminated by a 50 ohm load. right?
what am I missing.

Think of it this way. When the source and load are matched impedance-wise, the same amount of power that is dissipated in the load will also be dissipated in the source. If you match the line impedance to the source and load it is as if the line is not even there at all. Of course there is some delay and in the real world there is some slight loss. But that is it.
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Thought experiment: Take a perfect 50 ohm transmission line that has only one end. It goes on in one direction infinitely. If you hook an ohmmeter to this line, just a plain old analog ohmmeter with a d'arsonoval moving coil with a needle, the meter will read 50 ohms as long as you leave it hooked. This is one way to think about what the characteristic impedance of a transmission line is. The inductance and capacitance along the line cause this voltage to current ratio which in our thought experiment works out to 50 ohms.

 

[likephysics]

So if I terminate a 50 ohm source with a parallel 50 ohm followed by a matched tline and load.
The load will see half the voltage, right?

 

[Averagesupernova]

So if I terminate a 50 ohm source with a parallel 50 ohm followed by a matched tline and load.
The load will see half the voltage, right?

No. You terminate it right at the source and you will get half of unloaded voltage. If you terminate it at the end of a transmission line you will get half of unloaded voltage. Do you understand thevenin? If you terminate it a second time like you described you will get less than half.

 

[sophiecentaur]

why doesn't the half power dissipation happen when the 50 ohm tline is terminated by 50 ohm load.

There is no loss mechanism in an ideal (50Ω) transmission line so all the power traveling on the line must be fed to the load (50Ω termination). The source impedance (transmitter) will involve some loss but, assuming that the line is terminated well / perfectly by the load, there will be no reflected power back to the transmitter so the power sharing between transmitter and load will be determined only by the transmitter source R_s. The load will 'appear' right at the output terminals of the transmitter. A voltage source will have 'zero' source resistance so there will be no loss and the load will get all the available power.
However, the transmitter efficiency would also have to be high when operating as a voltage source so the DC input power could be 'anything', compared with the output RF power. So the above statement is not the whole story. It's just a theoretical situation.