Why is there no voltage drop along a feedline

Question

This page shows a graph of voltage along a transmission line by distance. Considering the case of a simple purely restive feedline of 50 ohms matched to a load of 50 ohms, I would expect the voltage at the source end to be twice that of the voltage at the load end of the transmission line (line and load act like voltage divider), so the VSWR would be 2:1, not 1:1. It seems to me that this transmission line, unterminated, would achieve that 1:1 ratio. Why is the SWR/VSWR considered to be infinite for an unterminated or shorted line? Are these measurements built on the assumption that reactive components to the impedance dominate over purely real (resistive) components?

Answer 1

It is important to have your definitions straight. The source voltage is measured after the source's output impedance, as shown in below image:

^{simulate this circuit – Schematic created using CircuitLab}

You are right that V1 in the image is double amplitude with respect to "Source Voltage", but that voltage is not part of the equation.

So with an ideal cable and matched in- and output impedances, the source voltage is equal to the voltage at the load end and with that reflection coefficient is 0:

Reflection coefficient

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{50 - 50 }{50 +50} = 0\$

VSWR

VSWR for a properly terminated line is defined as:

\$VSWR = \dfrac{V_{MAX}}{V_{MIN}} = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+0}{1-0} = 1\$

VSWR for an shorted line

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{0 - 50}{0 + 50} = -1\$

\$VSWR = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+1}{1-1} = \dfrac{2}{0} = \infty\$

Which basically means the voltage along the line varies between 2 and 0 times the source voltage.

Voltage standing wave ratio is the ratio between the largest amplitude and the lowest amplitude found along the line. If the lowest amplitude nears zero, then VSWR will go to infinity.

VSWR for an open line

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{\infty - 50}{\infty + 50} = 1\$

\$VSWR = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+1}{1-1} = \dfrac{2}{0} = \infty\$

Which again basically means the voltage along the line varies between 2 and 0 times the source voltage.

Answer 2

Like Paul R says in his comment, it looks like you are confusing the characteristic impedance of a transmission line with the properties of a resistor.

The characteristic impedance of a transmission line is the ratio between the voltage signal and the current signal of a travelling wave on the line. Since it is the ratio of a voltage to a current, it has the units of impedance. And, since the current and voltage are in phase (to a reasonable approximation) on many types of line, the impedance value is real, but that does not mean that it causes power to be lost from the circuit like a resistor does.

The power represented by this voltage and current isn't lost as heat, it's propagated down the line towards the load.