On 5/15/2011 3:42 PM, Cecil Moore wrote:
On May 14, 1:56 pm, 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

It is a 100 ohm feedline. The impedance at the + point looking toward
the load is 50 ohms. The 100 W source will see the first 50 ohms and
then the impedance at the + point. The total impedance seen by the
source is therefore 100 ohms. A 100 W source will deliver 100V to this
100 ohm combination and 50V of it will appear at the + point. Where does
the 70.707V come from?

John