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It is well known that the voltage sees different impedances while traveling down a tramission line, and this impedance is a function of the intrinsic impedance of the line Zo and tangent of wave frequency and the length away from the load according to:

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But my question is, since the voltage sees different impedances at different position along the line, why is it that power is always delivered to the load (i.e. an AC line that spans many Kms)? If voltage source truly sees different impedances down the line, then why would it deliver any power to the load at all since eventually somewhere on the line the impedance would be infinity?

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    \$\begingroup\$ What the heck is \$i\$? We use \$j\$ around here. \$\endgroup\$
    – Null
    Commented Oct 1, 2014 at 0:49
  • \$\begingroup\$ Why do you think "eventually somewhere on the line the impedance would be infinity"? \$\endgroup\$
    – The Photon
    Commented Oct 1, 2014 at 1:07
  • \$\begingroup\$ Suppose that the line has been shorted, then the expression of the line is given as Zin = jZotan(\$\beta\$*l), at various wave length, the line Zin could have infinite impedance, no impedance, a capacitive impedance or inductive impedance - but intuitively a short line will cause an instant current transfer, as if there was no impedance anywhere \$\endgroup\$
    – Fraïssé
    Commented Oct 1, 2014 at 1:11
  • \$\begingroup\$ IF the line is shorted, then you are right, power will not be delivered to the load. \$\endgroup\$
    – The Photon
    Commented Oct 1, 2014 at 2:05

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Think about the signal on the transmission line as being made up of a forward-travelling wave and a reverse-travelling wave.

At any given point on the line, the ratio of voltage to current in each of these two independent waves is determined by the characteristic impedance \$Z_0\$ of the line.

The forward travelling wave was initially created by the power source driving the line.

The reverse-travelling wave only exists because something caused reflections of the forward wave to return back up the line. Reflections only occur when there's some change in the line geometry, a circuit attached to the line, or a mismatched termination.

If voltage source truly sees different impedances down the line,

Your formula for \$Z_{in}\$ applies to one specific situation: There is a mismatched termination on an otherwise ideal line and you want to know the ratio of voltage and current (due to the combination of forward and reverse waves) at some location up the line from the termination. It accounts for the fact that the phase of the forward wave is different at this point on the line then where it encounters the load, and that the phase of the reverse wave has also changed as it traveled back down the line to where you're measuring.

It doesn't mean that the characteristic impedance of the line is changed in any way: The voltage/current ratio in the forward wave is still determined by \$Z_0\$ and the voltage/current ratio in the reverse wave is still determined by \$Z_0\$. And if there isn't a discontinuity or geometry change at this point in the line, there won't be any additional reflection created because \$Z_{in} \ne Z_0\$ at this point on the line.

... then why would it deliver any power to the load at all since eventually somewhere on the line the impedance would be infinity?

The apparent input impedance will only go to infinity if the load has reflected 100% of the energy in the signal (\$|\Gamma_L|=1\$). In this case, in fact there can't be any energy delivered to the load, because all of the energy is reflected back in the reverse wave.

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