Hi Dallas.
> Thanks to all of those who responded to my initial question.
> I guess I need to clarify the question. I know that SWR should
> be Z(load) divided by Z(line) when load is > than line and vice versa,
> but then the formula for finding Z(load) must be tainted. The only ref I have
> found is Z=squareroot (R^2+J^2) When using this product, the above
> formula does not
> apply. Am I using the wrong formula to determine Z or the wrong formula
> to determine SWR?
> example: 1-j50 is essentially Z=50, but it does not provide 1:1 SWR.
> of course 50-j0 does. A Smith Chart or X/Y vector graph with SWR circles is a
> simple
> graphical solution, but I was looking for a mathematical one.
No, these formulas are incorrect. It's little longer :-)
1+SQRT(SQR(R-Z)+SQR(X))/SQRT(SQR(R+Z)+SQR(X))
SWR = ---------------------------------------------
1-SQRT(SQR(R+Z)+SQR(X))/SQRT(SQR(R-Z)+SQR(X))
where R and X is antenna resistance and reactance, Z - line impedance
And this is not full formula - it's for lines without losses, but quite right.
SWR in your case is 100.01 by this formula. This is correct for antenna end.
For TX end real SWR will be slightly lower because of losses in line.
For Ed K4SB:
I'm sorry, but you absolutely don't understand how it works.
Insert SWR-meter in center of your half wave 600 Om coax :-)
May be after this you'll understand that:
1.SWR-meter on TX end don't show us real SWR in line.
You can read about this in manual for MFJ-259 SWR Analyzer.
2.Line always transforms impedances. When it half wave it's 1:1,
when it quarter it's maximum, all other lengths is between these two.
3.SWR depends on line and load impedances and not depends on line length.
You can see it from above formula.
(Yes it depends on line length, but we can ignore this in real system
- it slowly falls down from load to TX because of losses in line).
73 de Serge UT2IO.
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