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Re: [TowerTalk] Elevated vertical

To: "Al Kozakiewicz" <>, <>
Subject: Re: [TowerTalk] Elevated vertical
From: "Paul Christensen" <>
Date: Sat, 24 Dec 2011 10:38:20 -0500
List-post: <">>
> "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 

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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