measures wrote:
>>Not if the impedance of your transmission line is the same is the
>>impedance of your source. Draw it out on a Smith Chart, Rich.
>>
>? In this case, the impedance of the source/generator is zero ohms, the
>characteristic impedance of the half wavelength transmission line is 93
>ohms, and the termination is 50-ohms.
>
>>You move along a constant VSWR circle with 50 Ohms as the center of that
>>circle. The radius of the circle is the magnitude of the VSWR. This
>>assumes a 50 Ohms sytem and a Smith Chart normalized to 50 Ohms.
>>
>"This is assuredly not a 50 ohm system due to the presence of the half
>wavelength 93-ohm coax."
Rich,
MUST you debate everything? Even when somebody agrees with you?
It IS a 50 Ohm system because you are defining the measuring device as 50
Ohms. Therefore, you want to see what the SWR with a 50 Ohm device is.
So you use the 50 Ohms as your characteristic impedance.
Now, you are correct in that it matters not what the impedance of your
source is. I thought more about this. All that is being done in your
example is using a 50 Ohm instrument to measure the VSWR. And at the 1/2
wavepoint in the 93 Ohm coax we have a 50 Ohm impedance, hence the
instrument reads 1:1 SWR. It matters not if the meter and the source are
matched or not since the meter is only interested in what is after itself.
Please though, Rich. I agree with you. Enough debating on this!
73,
Jon
KE9NA
-------------------------------------
Jon Ogden
KE9NA
http://www.qsl.net/ke9na <--- CHECK IT OUT! It's been updated!!!!!
"A life lived in fear is a life half lived."
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