>
>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.
Where on the coax?
>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!
? they do.
- Rich..., 805.386.3734, www.vcnet.com/measures.
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