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I am having this basic doubt

In the basic crude basic block diagram, we have a source, transmission line and load, all having an impedance of 50Ohms. But according to maximum power transfer theorem, we need the source impedance to be equal to the load impedance for maximum power transfer. But there is a transmission line in the middle of 50Ohms.

This would mean that the source is looking into a 100ohms load (As the transmission line and load impedance get added as the resistances are in series). And the load is also looking into at a source impedance of 100Ohms.

So, how can we achieve maximum power transfer here? Or, is there something that I am missing?

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You are missing the fact that a transmission line is not a resistor. A 50 Ω line, terminated in a 50 Ω resistor, 'looks like' a 50 Ω load to the source driving it.

A 50 Ω line or load both define a ratio of voltage to current. However, for the line, it defines the ratio of the voltage wave to current wave that propagates along the line. The line itself has (ideally) no resistance.

There are several ways to make a source have an output impedance of 50 Ω, one way is to have a voltage source followed by a 50 Ω series resistor.

Let's say the 50 Ω source drives a 50 V step into a 50 Ω resistive load. That will need a 100 V step from the voltage source, because of voltage division between its series resistor and the load. The load current suddenly rises to 1 A, and the load voltage to 50 V.

Instead, let's put a length of 50 Ω transmission line between these. When the step occurs, the ratio of voltage wave in the line to the current wave will be 50 Ω. A voltage of 50 V will appear across the line, and a current of 1 A start flowing into it from the source. A 50 V voltage wave and a 1 A current wave set off down the line. When they reach the far end, they find a 50 Ω load. The load voltage now rises to 50 V and its current to 1 A. There has been no problem in meeting both the voltage and current from the line, they are in the correct ratio, so all the boundary conditions are met and no reflection is generated.

If instead the load had been something different, say 100 Ω, or an open circuit, or a short circuit, the current and voltage in the load would not have matched that arriving along the transmission line, and a reflected wave would have been generated to make up the differences.

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  • \$\begingroup\$ Sorry, I am not able to understand it. And yes indeed, but let us assume that the transmission line is long and it said to have a 50 Ohm resistance, in that case, what would happen? \$\endgroup\$ – Newbie May 16 at 14:56
  • \$\begingroup\$ Suppose if the transmission line has a characteristic impedance of 50Ohms, wont voltage division happen between the source & Line and between line & load? \$\endgroup\$ – Newbie May 16 at 14:58
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    \$\begingroup\$ @Newbie The Tx line does not have a 50 ohm resistance, it has a 50 ohm impedance. Voltage division does occur at the source, I'll update my answer. \$\endgroup\$ – Neil_UK May 16 at 15:46
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    \$\begingroup\$ The transmission line doesn't have 50 ohm impedance (an ideal lumped resistor can be described with a 50 ohm real impedance). It has 50 ohm characteristic impedance. \$\endgroup\$ – The Photon May 16 at 16:27
  • \$\begingroup\$ Thank you @Neil_UK for the answer. Helped me clear my doubts \$\endgroup\$ – Newbie May 16 at 16:38
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At the instant voltage \$V_{IN}\$ is applied to the sending end of the transmission line, the current taken by that line is \$V_{IN}/Z_0\$. It doesn't know there's a load at the far end so the current taken is defined by the cable or transmission line.

That continues until both voltage and current (speeding towards the load) reach the load end and, if the load has a resistance of \$Z_0\$, then all is well. End of story. There is a bigger story but, for this question, that's all that needs to be said.

So, how can we achieve maximum power transfer here?

As for maximum power transfer, you don't need to have a source impedance for this to work and you'll get maximum power without the source impedance.

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  • \$\begingroup\$ Suppose if the transmission line has a characteristic impedance of 50Ohms, wont voltage division happen between the source & Line and between line & load? \$\endgroup\$ – Newbie May 16 at 14:58
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    \$\begingroup\$ It will happen between source impedance and line but, if the load = \$Z_0\$ then all the power flowing down the line will be fully used by the load and there will be no potential division at the load. \$\endgroup\$ – Andy aka May 16 at 15:00
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    \$\begingroup\$ Imagine a 10 metre length of 50 ohm coax connected to a resistor of 50 ohm then, compare it with a 10 metre length of coax connected to an infinitely long piece of the same coax - there is no discrepancy here - the same voltage and current waves will continue into the infinitely long coax without a hiccup. That infinitely long coax presents itself as 50 ohms and is no different to an actual 50 ohm resistor. \$\endgroup\$ – Andy aka May 16 at 15:19
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You're missing a basic point. The source is matched perfectly to the transmission line, so you get maximum power transfer (100%) at that interface. And the transmission line is perfectly matched to the load, so maximum power transfer there also.

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  • \$\begingroup\$ Yes, but when you see from the source, it just doesn't see the 50ohm impedance of the transmission line alone, right? It seems the total 50+50=100ohm impedance of both the line and the load. So, voltage division won't happen? \$\endgroup\$ – Newbie May 16 at 15:01
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    \$\begingroup\$ No. Remember this is wave-front propagation phenomenon. Re-read what Andy said. At the time T0 when the edge of the wave first impinges upon the left hand edge of the transmission line, that's all the signal sees - the transmission line. It doesn't "know" anything about what's on the other end of the transmission line. \$\endgroup\$ – SteveSh May 16 at 16:38

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