Gilmer, Mike wrote:
> The SWR in the cable is set by the load/cable mismatch.
>PERIOD.
No. Not correct. If so, then please explain the operation of a quarter
wave transformer.
The VSWR is only dependent on the load/cable mismatch PROVIDED THE CABLE
IS THE SAME IMPEDANCE AS THE SOURCE!
If the source impedance is different from the cable impedance is
different than the load impedance all, you have a different ball game.
You need to understand the frame of reference from where you operate.
>
>If this isn't true, then I haven't learned a thing :(
>
>If it is, I'll continue:
>
>Remove the SWR meter and the generator. Cable and load only. What is
>this black box?
>It's 1.86:1 VSWR regardless of line length. The cable could be 2 meters
>or two miles (yes, ignoring attenuation).
>Why would putting a SWR meter at one end change this?
>
>In general you normalize to the cable, in this case 93 ohms.
No. That's not correct. You normalize to your system impedance. In
this case it is 50 Ohms since Rich specified a 50 Ohm VSWR meter. The 50
Ohm load is not 1.86:1 in reference to the 50 Ohm source. That's where
you are mixed up.
>
>
>> 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.
>>
>
>Yes, to transform the impedance back to "itself" it matters not what the
>cable Z is.
>The SWR is a different matter.
If the source sees 50 Ohms at the cable interface then it sees no VSWR as
there is no mismatch.
73,
Jon
KE9NA
--------------------------------------------------------------------------
The Second Amendment is NOT about duck hunting!
Jon Ogden
jono@enteract.com
www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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