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Old March 5th 08, 07:25 PM posted to rec.radio.amateur.antenna
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First recorded activity by RadioBanter: Mar 2007
Posts: 3,521
Default The Rest of the Story

Keith Dysart wrote:
You used your tool to attempt to show that the reflected power
is dissipated in Rs.


The tool proves that the reflected average power is dissipated
in Rs because it has no where else to go. If instantaneous
reflected power were relevant, why don't we read about it in
any of the technical textbooks under the wave reflection model?

I did a finer grained analysis using instantaneous power to show
that it is not.


My tool is known not to work for instantaneous power and
was never intended to work for instantaneous power. So your
argument is just a straw man diversion. Eugene Hecht explains
why average power density (irradiance) must used instead of
instantaneous power.

"Furthermore, since the power
arriving cannot be measured instantaneously, the detector
must integrate the energy flux over some finite time, 'T'.
If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

The use of averages in analysis can be misleading.


The misuse of a tool, designed to be used only with
averages, can be even more misleading. When you measure
an open-circuit using a DC ohm-meter on a dipole, are
you really going to argue that the DC ohm-meter is not
working properly? That's exactly what you are arguing here.
When one misuses a tool, as you are doing, one will get
invalid results. There's no mystery about that at all.

You are saying that the energy model, designed to be
used with average powers, does not work for instantaneous
values. When you try to use it for instantaneous values,
you are committing a well understood error. Why do you
insist on committing that error?

5. ______________________________________________.


Now you have got the issue. Since the reflected power is
not dissipated in Rs, the answer must be one of 1 to 5.

5. is probably the best choice.


Until you fill in the blank for number 5, you are just
firing blanks. :-) Exactly what laws of physics are you
intending to violate with your explanation?

And that is why it became necessary to rethink the nature
of energy in reflected waves.


Nope, it's not. Reflected waves obey the laws of superposition
and reflection physics. That's all you need to understand.
Now new laws of physics or logical diversions required.
--
73, Cecil http://www.w5dxp.com
 
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