[AMPS] Re: The Worldwide, No Holds Barred, SWR Quiz.

[Vic Rosenthal]

Jon Ogden wrote:
> 
> At a half wavelength of coax, you rotate completely around the VSWR circle and thus see the impedance of the load at the input of the coax. 

Correct.

> In this case, the load is 50 Ohms.  At the half wavelength of coax, the load is 50 Ohms as well.  

Correct.

> The SWR in a 50 Ohm system then is 1:1.  In a 93 Ohm system you would have the SWR mismatch of 1.86:1, but in a 50 Ohm system, our VSWR is 1:1.

Not correct!  A generator would see a 50 ohm load at this point, and would be
able to match it as well as in the case of an arbitrary length of 50-ohm coax
feeding a 50-ohm load.  However, the SWR on the coax is still 1.86 to 1.  SWR is
*defined* as the ratio of maximum to minimum voltage on the line; if the
characteristic impedance of the line is different from the impedance of the
load, the voltage and current will vary (in opposite directions) as you move
along the line.  The 1/2 wavelength point is the point at which the voltage and
current are the same as at the load, so the impedance is the same.  But you
still have the variation in voltage and current!

Jon, if you were right, SWR meters would be useless, since readings would change
as cable lengths changed!  You may be misled by the fact that in practical
systems, common-mode ('antenna') currents on the outside of the coax often
disrupt SWR readings.
 
> So the bottom line is that if you have a 50 Ohm load and a 1/2 wavelength of coax that it matters not what the impedance of that coax is!  It's inherently narrow band, but that's the case.

It matters not from the point of view of matching to the generator.  It does
matter from the point of view of additional losses on the line due to SWR, or
coax voltage ratings, etc.

73,
Vic, K2VCO
Fresno CA

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