On 15 mayo, 22:42, Cecil Moore wrote:
On May 14, 1:56*pm, Wimpie wrote:

Now you can forget the whole reflected power story.

Which is exactly what most people do - just forget the problem and
hope it will go away. Some have said just to calculate the voltage/
current and the energy (or power) will take care of itself. What this
requires is ignorance of the well-known laws of physics from the field
of optics. When one understands the laws of physics that effect the
boundary conditions for RF waves including the role of interference,
all of the component energies can be easily tracked through a system
in the same manner the track irradiance through an optical system
because they cannot measure the voltages and currents. Method#1 is the
way most RF engineers analyze a problem. Method#2 is the way optical
physicists are forced to solve the problem.

Method#1: Calculate the voltages and currents and superpose them. (One
way is using s-parameter voltage equations.) Calculate power at the
end of this process.

Method#2: Calculate the irradiances (power densities) and merge them
together using the irradiance equations. The irradiance equations are
what one gets when one squares the s-parameter normalized voltage
equations.

When you say "forget reflected power", you are asking people to forget
the ExH power contained in every EM wave. You are asking people to
forget that EM waves must necessarily travel at the speed of light in
the medium, i.e. component EM wave energy cannot stand still in
standing waves. You are asking people to forget the conservation of
energy principle.

Earlier I suggested that we concentrate on a simpler lossless example
that doesn't involve the source impedance at all. It doesn't matter
what the source impedance is. All that matters is that the source is
delivering 70.707 volts to the 50 ohm feedline. Here it is again:

100W--50ohm--+--1/4WL 100ohm--200 ohm load

During steady-state, Pfwd is 100w and Pref is 0 watts on the 50 ohm
line. Pfwd is 112.5w and Pref is 12.5w on the 100 ohm line. Exactly
what happens at the Z0-match point '+' *to reverse the flow of that
12.5w reflected power wave? (There are no RF source or DC supply
arguments behind which to hide.) The reflected voltage reflection
coefficient at point '+' is -0.3333, i.e. the negative of the forward
voltage reflection coefficient at the load.

The ExH power (Poynting vector) in the reflected wave from the load is
12.5 watts. It is necessarily traveling toward the source at the speed
of light in the Z0=100 ohm medium. It necessarily contains measurable
real-world energy and momentum. Exactly what happens at point '+' to
reverse the momentum and energy flow of that reflected EM wave? All
the arguments in the world about what happens in the source has not
answered that question and this example doesn't involve the source at
all.

I do generally not agree on this one, please convert your Thevenin
source to a Norton source.

Sorry, that particular source was not my choice but was instead W7EL's
choice in his food-for-thought series. He specifically states he is
NOT talking about any equivalent circuit but is instead talking about
a good approximation to a real-world source with a 50 ohm source
resistance.

I do not see how the above energy flow problem/example can be solved
without an understanding of the principles of interference (covered in
section 4.3 of "Reflections" by Walter Maxwell. If you can adequately
answer the questions without any direct or implied reference to
interference, you will be the first I know of on ham newsgroups to
have done so. Hint: squaring the s-parameter equations yields those
interference power terms. If you choose not to respond to this simpler
example, I will understand your reasons based on "forgetting reflected
power".
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Hello Cecil,

I'm rather familiar with S-parameters and unbounded EM phenomena. We
are discussing coaxial cables with very small diameter with respect to
wavelength, therefore full TEM condition is applicable, even some
small distance left or right from a discontinuity. So you can fully
replace E cross H by a voltage and current approach (as is used in any
vector network analyser).

Regarding impedance, reflection coefficient and interference issues;
when you know one of the two (and Z0), you can calculate the other
(of course you need to consider phase). All interference issues are
accounted for in Z, given a transmission line impedance (Z0).

Transmission line theory is not a goal, but just a means to reach your
goal. It is fully acceptable to convert Ref.Coef. to Z (and vice
versa) to efficiently solve your problem. So when you convert an RC to
a Z, you can forget all transmission line calculations that you used
to find that Z. As Tom mentioned, we are discussing narrow band
systems, not wide band transmission systems.

I am not ignoring a problem (as you suggested), I am just using the
right tool to solve a problem. I remember from several people, the
best Engineer is the one that knows what things he has to take into
account and what things he can ignore. You introduced optics here, are
you going to introduce QM also to show that I (and some others) are
completely wrong w.r.t to PA output impedance?

Cecil, I think you have sufficient knowledge to form an opinion
without hiding behind others. You also have the equipment to figure
out some things yourself, and I gave some hints to help you. The only
question is, are you willing to do this?

With kind regards,


Wim
PA3DJS
www.tetech.nl