>
>On Tuesday, October 05, 1999 4:09 AM, measures [SMTP:measures@vcnet.com]
>wrote:
>> >
>> >On Tuesday, October 05, 1999 1:53 AM, measures
>[SMTP:measures@vcnet.com]
>> >wrote:
>> >> >
>> >> >Since I'm wearing egg anyway, more can't hurt.
>> >> >
>> >> >1. Smith Chart says 1/2 wave cable brings you all the around back
>to
>> >> >the original load impedance (50 ohms, here). Therefore the black
>box of
>> >> >the cable/load looks like 50 ohms, so, all should be happy. I say
>the
>> >> >VSWR is 93/50 = 1.86:1 - this caused by the mismatch at the load
>end.
>> >>
>> >> ? You have 50 +/- j0 ohms at the position of the SWR meter, and it
>reads
>> >> 1.86:1? I don't see how, Mike.
>> >>
>> >
>> >Rich, this says that as long as I have a multiple-half-wave of cable,
>> >the impedance of the cable plays no part in the SWR.
>>
>> ok
>>
>> >If this were true, what happens to the standing waves that are
>created
>> >at the load end ( 93 ohm cable, 50 ohm load )?
>>
>> >Why would connecting an SWR meter to the near end eliminate them?
>>
>> It eliminates reflections only at the halfwave point.
>> >
>
>It doesn't eliminate the standing waves along the cable.
? True. . However, the meter is not moving along the cable. It is
fixed at the generator end of the halfwave transmission line.
>The reflections from the mismatched load/cable are still on the cable.
>What happened to them?
? Reflections exist only within the 93 ohm, half wavelength section.
- Rich..., 805.386.3734, www.vcnet.com/measures.
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