On Sun, 18 Nov 2007 16:10:57 -0500, "Stefan Wolfe"
wrote:

When integrating the actual equation at hand over its limits,
the value of the constant does not matter because it always subtracts out.

This is simply leaning on the Xerox to resuscitate a doomed method. We
only have to consider that if this "logic" were applied to the actual
equation (again, from you):
Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)
= (50 - 0) + (50 - 0) - 2SQRT P1*P2 sine(A)
That all terms would be rendered 0. After all, the two leading terms
are constants too.

Will we hear another chorus of "it doesn't matter?" :-)

You are simply compounding errors.

73's
Richard Clark, KB7QHC

On Sun, 18 Nov 2007 17:40:53 -0500, "AI4QJ" wrote:

This is funny.

A chorus indeed.

73's
Richard Clark, KB7QHC

On Sat, 17 Nov 2007 21:46:30 -0500, "Stefan Wolfe"
wrote:

Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Hi Dan,

Now let's return to this equation.

It is drawn from the classic optics formula for finding the Intensity
of radiation at a point in space that is illuminated by two sources.
The intensity is a function of illuminators and their relative (at
that point) phase.

Phase can also be thought of as distance (which returns us to the
point in space). As each path has a different length; then relative
phase, their difference in length, is simplified by casting out all
full cycles to leave only the remnant, or partial cycle. Of course,
keeping count of the complete angular distances can be preserved,
nothing will change in the result.

A is a single value (as a point in space is 1D) and is expressed as
that relative phase. A, thus in the world of all possible variations,
can be any single value between 0 and 2·Pi radians. It follows that
the cosine operation then renders a single value between +1 and -1.

If you were to integrate this over some portion of time, or for all
time; then it wouldn't change the answer one iota. The intensity of
interference at a fixed spot in space from fixed illuminators does not
change with time.

If you were to integrate this over some portion of space, or for all
space; then you would come up with different solutions. Across all
space the intensity would become the sum of the two sources'
illumination.

There is no necessity of changing cos(A) to -sin(A) + C at all. There
is no issue of unknown constants, the original intensities do not
disappear.

However, Integration does not resurrect Cecil's sucker punch. There
is no missing power (energy, calories, what-have-you) as the question
was tailored to an illusion that too many bought into.

Lest we go 'round the mulberry bush on that again, please respond to
my critique posted some time ago if you fault IT.

73's
Richard Clark, KB7QHC

On Sun, 18 Nov 2007 20:22:21 -0500, "AI4QJ" wrote:

point to me to where it is?
It is my first post to this thread, at level 2, responding directly to
Dave.

2. You are saying that A is not a function of time. OK, OF COURSE that
changes the appearance of the curve I was using. Still, the total energy
expended over one cycle (can be thought of as phase angle - distance, not
time), 0 to 2pi radians, is = zero

As always, engineering is done by strict language. Look at your own.
"Expended" energy is power. Integration of power over time will never
result in no power unless no energy was "expended."

(It is still valid to integrate and
cancel out C ;-)) Over the full cosine power cycle, energy is still
conserved, however the model is totally different.

The model never changed, the consideration of superposition (partial
solutions) may momentarily suspend us, but upon its completion the
complete solution resolves to exactly what I've posted. Hence, there
is no "missing" power (energy, etc.).

Why didn't you mention
that A was not a function of time before?

Consider the genesis of the formula. It informs us all that this is a
point location solution of total illumination from two remote sources.
There is no need for me to create a shopping list of all the things
this formula is NOT a function of.

I am only trying to show conservation of energy which must hold true for any
of these power formulas to be correct. And this equation still shows
conservation of energy being true, even if A is the phase angle.

If it is any different from what is not already in my post, feel free
to elaborate.

73's
Richard Clark, KB7QHC

On Sun, 18 Nov 2007 21:36:37 -0500, "AI4QJ" wrote:


"Richard Clark" wrote in message
.. .
On Sun, 18 Nov 2007 20:22:21 -0500, "AI4QJ" wrote:

point to me to where it is?
It is my first post to this thread, at level 2, responding directly to
Dave.

I can only handle one issue at a time (regarding what you said, it IS
possible to have a zero net energy expenditure as long as energy that was
sent out was received back again at a later time or at a different place;
that is what the sine/cosine functions essentially tell you, whether they
are functions of time or distance):

Hi Dan,

Well then, your concept of expended is quite different from most and
flexible to the point of not really meaning much. So what is the
point in using the word?

Regarding:

"Neither of these artificial conditions actually exist in the reality
of superposed waves, and that is the con. The group has been fixated
on the separate artificial environments with their partial solutions
as though they actually exist independent of the reality of the
superposed, complete solution."

This is a bit philosophical for me but let me say that I believe that a
square wave is definitely the sum of superimposed odd harmonics in
accordance with Fourier. I believe each of the superimposed waves exists
independently. If you do not believe me, just ask anyway who lives next door
to a CB'er with a linear amp and transmits over modulated square waves (due
to saturating his amplifier). The square waves exist but so do the component
waves. I am not sure if this is related to what you are saying but it seems
to be the same thing.

I believe that cupiditas is the root of evil; however, both of our
beliefs have nothing to do with partial solutions of Superposition
posing as real world entities.

73's
Richard Clark, KB7QHC

On Nov 18, 7:22 am, Cecil Moore wrote:
K7ITM wrote:
On Nov 17, 4:03 pm, Cecil Moore wrote:
The waves are launched by the external reflection from a
Z0-match and the internal reflection from the load.

So the waves are going opposite directions along the line??

No, all reflections travel toward the source and
therefore, are traveling in the same direction.
Their Poynting vectors are all toward the source.
Given the following Z0-match impedance discontinuity
in a transmission line with the source to the left
and the load to the right:

Z0-match
------Z01---+---Z02------
Pfor1-- Pfor2--
--Pref1=0 --Pref2

The power reflection coefficient is
rho^2 = [(Z02-Z01)/(Z02+Z01)]^2

Pref1 is a combination of two reflected waves

1. P1 = Pfor1(rho^2) "the external reflection from the
Z0-match"

2. P2 = Pref2(1-rho^2) "the internal reflection from
the load"

Pref1 = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Pref1 equals zero at a Z0-match so P1+P2 and A=180 deg.
--
73, Cecil http://www.w5dxp.com

Ah, finally you get around to telling us the setup.

So to get to the conditions in the original posting, we must have a
total power coming into that Z01:Z02 junction exactly equal to the
power leaving it. For example, if Z01 = 50 ohms as implied by your
numbers, and Z02 = 100 ohms, barring stupid math errors, I make out
that the left-to-right power on the Z01 line is 450 watts, and the
right-to-left power on the Z02 line is 56.25 watts, for a total of
506.25 watts. Since you've only accounted for 171 watts, the
remainder must be going off to the right from that junction. Change
the phases, and the power will split differently.

This seems to all agree with standard superpostion. So what the heck
was the point of the original posting in this thread?

Or, why do I even bother reading these things in the first place,
since they all turn out to be pretty boring?

Once again, we see that everything interesting going on in the system
is happening right at the discontinuity where waves arrive and are
reflected. Same in a Wilkinson combiner, same in a "magic T" (which I
suppose the Wilkinson is, if you look at it the right way), same as in
a resistive combining network (if you account for power dissipated in
the resistors), ...

AI4QJ wrote:
. . .
This is a bit philosophical for me but let me say that I believe that a
square wave is definitely the sum of superimposed odd harmonics in
accordance with Fourier. I believe each of the superimposed waves exists
independently. If you do not believe me, just ask anyway who lives next door
to a CB'er with a linear amp and transmits over modulated square waves (due
to saturating his amplifier). The square waves exist but so do the component
waves. I am not sure if this is related to what you are saying but it seems
to be the same thing.

This is indeed a philosophical question. A consequence of superposition
is that there's *no possible way* to tell if a particular square wave is
made from separately generated sinusoids, a single step, or combinations
of any of an infinite number of other possible periodic waveforms. With
the proper sorts of filters, you can take the square wave apart into any
of those infinite sets of functions, "proving" the "independent
existence" of each. Sinusoids are mathematically convenient, but they're
by far not the only choice.

Roy Lewallen, W7EL

Richard Clark wrote:
However, Integration does not resurrect Cecil's sucker punch. There
is no missing power (energy, calories, what-have-you) as the question
was tailored to an illusion that too many bought into.

There's no illusion and no missing power, just interference
at a point balanced by the opposite kind of interference
somewhere else - just following the conservation of energy
principle. It was a rhetorical question.
--
73, Cecil http://www.w5dxp.com

K7ITM wrote:
This seems to all agree with standard superpostion. So what the heck
was the point of the original posting in this thread?

Some posters deny that destructive interference is associated
with zero reflected energy toward the source when a Z0-match
exists. If you are not one of those posters, the thread was
not aimed at you.
--
73, Cecil http://www.w5dxp.com

Antonio Vernucci wrote:
- reflected power does not reach the transmitter, as it is fully
reflected back toward the load

The re-reflection is associated with destructive interference
toward the source and an equal magnitude of constructive
interference toward the load. The energy in the canceled
reflected waves is redistributed to a region that allows
constructive interference to occur, i.e. in the opposite
direction to the direction of reflected wave cancellation
toward the source.

Probably this is what you call destructive interference at the
trasmitter and constructive interference at the load.

At a Z0-match *point*, destructive interference *toward* the
transmitter and constructive interference *toward* the load.

It is akin to the passive elements of a Yagi causing
destructive interference to the rear and constructive
interference toward the front. What is the front/back
ratio of a Z0-match? :-)
--
73, Cecil http://www.w5dxp.com