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When trying to solve a basic transmission line problem, I was making the mistake of using voltage divider across the input load. Why is this wrong?

What I mean is, assume there is a generator impedance Zg and a load ZL on a lossless transmission line. I can find the input impedance Zin at the generator, but to find the voltage across the load, I can't simply do

\$V_g\frac{Z_L}{Zg + Zin}\$.

Why is that wrong?

schematic

simulate this circuit – Schematic created using CircuitLab

Schematic isn't that good. Basically some impedance Zg and some impedance Zl on a transmission line.

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  • \$\begingroup\$ Schematic please. Not quite visualizing what you are talking about. \$\endgroup\$ – mkeith Jan 28 '17 at 1:51
  • \$\begingroup\$ yes Vg*ZL/(Zg + ZL)=Vout neglecting f and Zo of transmission line \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 28 '17 at 2:08
  • \$\begingroup\$ @TonyStewart.EEsince'75 What do you mean? Also I wrote Zg + Zin. Zin takes into account Zo and I don't think the frequency matters because we aren't dealing witha specific load. \$\endgroup\$ – Goldname Jan 28 '17 at 2:20
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    \$\begingroup\$ @mkeith I added a schematic. \$\endgroup\$ – Goldname Jan 28 '17 at 2:23
  • \$\begingroup\$ For "low enough frequency", your equation I guess should hold. \$\endgroup\$ – niki_t1 Jan 28 '17 at 3:02
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One way to calculate Zin as the input impedance to the line with load is $$ Z_{in} = Z_0 (\frac{Z_L + j Z_0 \tan{\beta l}}{Z_0 + j Z_L \tan{\beta l}}) $$ which is a non-linear function of \$Z_L\$ and \$Z_0\$, and wavelength and line length.

The whole transmission line plus the load can therefore be modeled as one lumped impedance as \$Z_{in}\$. But you cannot use this model yet isolate \$Z_L\$ out to form a voltage divider, which would require you to assume \$Z_{in} = Z_0 + Z_L\$ (which is wrong).

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You are somewhat on the right track but just a bit confused. For the steady state case, the voltage divider rule says that the voltage should be Vg * Zin/(Zin+Zg). ZL does not come into it at all, except that you may need to know ZL to calculate Zin.

Note that ALL of the impedances may have a frequency dependence. In particular, at DC, if the transmission line is lossless, then Zin = ZL. In this special case only, then, the divider formula above could be written as Vg * ZL/(ZL+Zg).

Hope this clarifies everything.

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Rewrote the answer:

Assuming that you really want to use the thevenin equivalent of the combination generator + transmission line, the T.E. should be derived properly. You should find the output voltage when the load is disconnected (=Eo) and the output current when the load is 0 Ohms. Let that be Io.

Then your equivalent has the source voltage Eo and serial impedance Eo/Io

Calculating Eo and Io should from the very beginning take into the account that it takes some time the signal to travel through the line. It causes phase difference. The whole calculation should be done in phasor domain by using complex numbers. The result would contain, how long is the delay on the line, what is the frequency and the line charasteristic impedance also should exist in the right formula. The line impedance is not a resistance . It's needed when calculating the voltage and current components of a propagating wave. At both ends also should the reflected wave exist in the calculation.

Maybe you have thought that the line can be considered as a serial element with the load. Unfortunately it's a two port (= quad pole). That, how the signal source sees the load through a two port,can't be calculated by adding two impedances - not even in complex phasor domain.

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  • \$\begingroup\$ It never assume a resistor, I can do a voltage divider to get the voltage across the thevenin impedance right? \$\endgroup\$ – Goldname Jan 28 '17 at 2:35
  • \$\begingroup\$ @Goldname Just rewrote a longer answer \$\endgroup\$ – user287001 Jan 28 '17 at 3:09
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Your formula is not wrong. For the transmission line to be lossless its input and output impedances must match the source and load, and if they are the same then the input and output voltages must be equal (ie. no voltage lost in the line). Therefore it is equivalent to having the load connected directly to the source, and the transmission line 'disappears' (just like in your schematic, where the transmission line is represented by a lossless wire).

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    \$\begingroup\$ That is not what it means for a transmission line to be lossless. \$\endgroup\$ – The Photon Jan 28 '17 at 5:23

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