On Mon, 14 Aug 2006 23:04:30 GMT, Owen Duffy wrote:



PS: I think the problem you have given can be solved with simple trig:
find the reactance of the Z02 section using one trig term,

Z=-j50*cot(45)=-j50

find the
length of Z01 that would deliver that reactance using one trig term,

l=acot(50/600)=85.2

add that length and the actual length of Z01 section, find the

Z01'=85.2+45=130.2

reactance of the Z01 section using one trig term. I could do that in a

X=-j600*cot(130.2)=j507.7

flash with a scientific hand calculator while you were sharpening your
pencil.

It is a trivial problem either way, and can only ever be an
approximation of a practical problem.

But you wouldn't get that accuracy from the Smith chart.

Owen
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