Owen Duffy wrote:
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.

Thanks very much, Owen. I used the Smith chart to get 85, 130,
and 500 above. That's about as good an accuracy as I ever need.
Also MicroSmith says the impedance value is j507.2 ohms.

The stub, which is 45+45 = 90 degrees physically, is electrically
130 degrees long. There is an ~80 degree shift in the Gamma angle
at the 600--50 ohm impedance discontinuity resulting in ~40 degrees
being added to the electrical length of the stub by the impedance
discontinuity.

Let's say we now want to turn that stub into an electrical 1/4WL
stub. If we made the two sections the same number of degrees,
how many degrees would they occupy?

| X deg | X deg |
source====Z01=========Z02====open 1/4WL stub

where Z01=600 ohms and Z02=50 ohms

I get 16.1 degrees for X. A stub that is physically ~1/12WL
long is 1/4WL resonant, i.e. a 90 degree phase shift from end
to end. Does this remind anyone of a base-loaded mobile antenna?
--
73, Cecil http://www.qsl.net/w5dxp