Jim Kelley wrote:

Cecil Moore wrote:
Given the
line, the unit area term can be dropped without error.

In the engineering profession, it would probably mean without job; in
science, without publication.

The number of watts inside a coaxial transmission line
is understood by any any rational person to be distributed
over the area of the coax. With a fixed-given unit-area,
the Poynting Vector is customarily given in watts, not
watts/unit-area. The same thing applies to watts within
a laser beam.

You are probably right about published white papers. You
are wrong about the engineering profession. All engineers
need to do is get close enough.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
Please try again after adding 1 more wavelength of 450
Ohm line between the generator and the 75 Ohm line you
added. Kindly explain where the 'reflected power' on
this new section of 450 Ohm line goes.

It doesn't leave the 450 ohm line as long as the generator
sees 75 ohms as a load. Here is a similar example:

source---75 ohm line--+--1/2WL 450 ohm line---75 ohm load

Where does the reflected energy on the 450 ohm line go?
Since there is a 75 ohm Z0-match at point '+', it circulates
between the load and point '+'. Decrease the length of the
75 ohm line by one inch until it doesn't exist anymore.
The same conditions continue to exist all during that time.
The reflections at point '+' disappear in the process of
wave cancellation which is a type of permanent interference.

Now consider that the 75 ohm line can be one foot
long and everything is the same as the 1WL of 75
ohm line (except the delays).

This would be quite incorrect.

No, this would be 100% correct. One foot of 75 ohm coax
is enough to establish a 75 ohm environment. A 75 ohm
load on the source is enough to establish a 75 ohm
environment. Please see:

http://www.w2du.com/r3ch19a.pdf
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:
When you think of interference as being the instantaneous sum of waves
at a given position and time, then there is really only one kind of
interference to be had - though there are a variety of results which can
be obtained from it.

If, as you say, interference is only a result and
not a cause, how can there possibly be "a variety
of results which can be obtained from (interference)"?
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:
Electromagnetic waves reflect only from real physical boundaries.

Is the V/I ratio at the heart of a source a "real
physical boundary"? If not, why not?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Jim Kelley wrote:

Electromagnetic waves reflect only from real physical boundaries.

Is the V/I ratio at the heart of a source a "real
physical boundary"? If not, why not?

Exactly two reasons: mu, and epsilon.

A physical boundary is neither defined by, nor affected by voltage,
current, irradiance, luminosity, power, wealth, energy, phase of the
moon, sun spots, kharma, or even gravitas.

Has this become a new point of contention, Cecil? You've said that
you always claimed that waves only reflect from physical boundaries.
And that's all I'm saying.

73, ac6xg

Jim Kelley wrote:
Has this become a new point of contention, Cecil? You've said that you
always claimed that waves only reflect from physical boundaries. And
that's all I'm saying.

When it comes to a source, I seem to have been wrong
about that. A source seems to create its own physical
boundary.
--
73, Cecil http://www.w5dxp.com

On Apr 4, 9:25 pm, Cecil Moore wrote:
When it comes to a source, I seem to have been wrong
about that. A source seems to create its own physical
boundary.

Truly, you have constructed a world view that is much more
complicated than necessary.

Consider the 75 Ohm resistor at the right hand end of a 75 Ohm
transmission line. The load is matched to the line and there
is no discontinuity and (dare I say it?) no reflection.

Want a generator at that end? Put an ideal current source
in parallel with the 75 Ohm resistor. What do you have but a
generator with a 75 Ohm output impedance. And no discontinuity.

Do all the above with a 50 Ohm resistor. As a load, the 50
Ohm resistor is a discontinuity with reflections. As a
generator it has a 50 Ohm output impedance and there is a
discontinuity.

Is not the symmetry rather enticing? And simple?

And using superposition, you can analyze the incident wave
and its reflection, and along with the generated wave, sum
them to obtain the total system response.

Truly elegant. And all so simple.

....Keith

On 4 Apr 2007 19:02:43 -0700, "Keith Dysart" wrote:

On Apr 4, 9:25 pm, Cecil Moore wrote:
When it comes to a source, I seem to have been wrong
about that. A source seems to create its own physical
boundary.

Truly, you have constructed a world view that is much more
complicated than necessary.

Hi Keith,

That isn't the half of it (without going into your further treatment)
our Cecileo will simply twist this admission of error into glowing
self validation - "and yet it moves."

73's
Richard Clark, KB7QHC

Keith Dysart wrote:
Consider the 75 Ohm resistor at the right hand end of a 75 Ohm
transmission line. The load is matched to the line and there
is no discontinuity and (dare I say it?) no reflection.

Yet that violates the convention that reflected energy
absorbed by the source was never sourced. Why do you
think that convention was adopted in the first place?

Is not the symmetry rather enticing? And simple?

Apparently it has enticed you to ignore reality. This
argument has been raging for a good 20 years now. Some
of the brightest engineers in the world still disagree.
Your simplistic theories are easily disproved by a
bench experiment. Why you cling to them is strange.

And all so simple.

Make that simple-minded.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
That isn't the half of it (without going into your further treatment)
our Cecileo will simply twist this admission of error into glowing
self validation - "and yet it moves."

When I find myself in an error, Richard, I correct it.
What do you do?
--
73, Cecil http://www.w5dxp.com