> "Adding a matching circuit at the transmitter (tuner) or adjusting the
> length of the feedline such that a 50 ohm resistive impedence is presented
> to the source does not change the VSWR in the feedline. Yes, the VSWR
> between the source and the tuner will be 1:1, but the VSWR in the feedline
> will remain unchanged."
Sometimes VSWR does change significantly along a line -- and not always for
reasons of loss nor common-mode RF current on the outside of a coaxial line.
Consider this example:
At the operating frequency, a dipole at its input terminals is exactly 50
ohms resistive (50+j0). Let's feed the dipole with an electrical half-wave
of low-loss 600-ohm open line. VSWR on the line is 12:1. Correct?
Finally, let's connect a random length of 50 ohm, low-loss coax to the input
of the 600 ohm line.
What is the VSWR on the 50 ohm section of line?
Possible Answers:
A. Is it still about 12:1 because VSWR does not change on a low-loss line.
B. Not enough information because you didn't state the coax length.
C It's now about 6:1
D. It is now 1:1
E. You can't terminate a coaxial line into a 600-ohm balanced line without a
current balun and get an answer.
F. None of the above.
Bonus question: What is the impedance at the input to the 600-ohm line
section?
So, here we have one transmission line composed of two types. Ignoring
loss, is the VSWR really the same at all points on the transmission line?
Paul, W9AC
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