>
>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!
>
? Let me see if I have this straight. You agree that the 50-ohm SWR
meter will read 1:1 at the ends of the 1/2 wavelength section, and that
it will read 1.86:1 at the 1/4 wavelength point?
- thanks.
- Rich..., 805.386.3734, www.vcnet.com/measures.
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