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A Subtle Detail of Reflection Coefficients (but important to know)
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August 21st 03, 10:01 AM
Dr. Slick
Posts: n/a
(Tom Bruhns) wrote in message om...
So, when are you going to start using your own head and not depending
on others? When will you go through the simple set of equations to
find that Vr/Vf = (Zload-Zo)/(Zload+Zo), and there IS NO conjugate on
the Zo in the numerator? I went through all that long ago (several
years) when someone questioned it, and I trust my calcs. It happens
that they agree with those of many, many others.
Who agrees with you? Les Besser is an established authority, as is
the ARRL, and Pozar... i use their examples because I am not yet an
authority myself. Please do some more research before you type.
Hey, we are all standing on the shoulders of giants!
I certainly wouldn't want to stand on YOUR shoulders, though...
Which of the following do you not believe? Because if you believe it
all, then it's a couple VERY SIMPLE algebraic steps to get to the
equation for Vr/Vf.
1. On a TEM line, Vf/If = -Vr/Ir = Zo, which may be reactive.
2. On a TEM line, V=Vf+Vr and I=If+Ir.
3. Where a TEM line connects to a load, the equations in (1) and (2)
hold, and V=Zload*I -- because the net line voltage is the same as the
voltage across the load, and the net line current is the same as the
load current.
Please do us a favor and go do the algebra, and if you get what Besser
teaches, show us the steps, and we'll resolve the difference from what
we get. If you get what we get, then go take it up with Besser.
Cheers,
Tom
I've checked my results with MIMP, and the reflection coefficients
matched the equation:
"For passive circuits, 0=[rho]=1,
And strictly speaking: Reflection Coefficient
=(Zload-Zo*)/(Zload-Zo)
Where * indicates conjugate."
A bit angry aren't we? Typical of one who has absolutely lost an
argument...
When you can show us your free energy device, we would love to see
it!
Slick
"Your Rage Has Imbalanced You!" - Lancelot
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