Revisiting the Power Explanation
 

Cecil Moore wrote:
Gene Fuller wrote:
OK, we are getting somewhere. According to the equation, we know that
interference has the same units as irradiance. Interesting. Does that
mean that interference doesn't work for fields? Are there multiple
definitions for interference?

My IEEE Dictionary is 130 miles away so I cannot look
up the technical definition for interference. The units
of irradiance are watts per unit area. In a transmission
line, we can consider the unit area to be constant and
simply drop the units of area. The resultant "irradiance"
equation for transmission line power is:

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

2*SQRT(P1*P2)cos(A) is the interference term in watts
where A is the angle between the two interfering fields.

This is all explained at: http://www.w5dxp.com/energy.htm
If the above equation is followed, adding powers is easier
than superposing voltages and then calculating power.


Cecil,

This is a perfect example of how the messages get so confused around
here. You cannot simply ignore dimensionality, units, vector vs. scalar,
etc. That leads to really silly conclusions like expressing interference
in the same units as irradiance.

I don't know if dimensional analysis is still taught these days, but way
back when I was taking science classes it was often stated that the very
first thing to do when attempting to solve any problem was to make sure
that all of the dimensions and units matched up. The numbers mean
nothing if the equations are wrong. I suspect a few others in this
newsgroup had similar training.

73,
Gene
W4SZ

Revisiting the Power Explanation
 

Gene Fuller wrote:
This is a perfect example of how the messages get so confused around
here. You cannot simply ignore dimensionality, units, vector vs. scalar,
etc. That leads to really silly conclusions like expressing interference
in the same units as irradiance.

Just remember when you are standing at the Pearly Gates
outside looking in, it was you and not me who called
Eugene Hecht's conclusions "really silly". :-)
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Richard Clark wrote:
Cecil no doubt hesitates to trod this path that leads to the
destruction of his other favorite, failed example of anti-glare glass.

I found out later that it wasn't anti-glare glass.
It was anti-reflective glass. Out of ignorance, I
just chose the wrong word. However, everything I
said was true about anti-reflective glass. A 1/4WL
thin film acts virtually the same way a 1/4WL series
matching section acts and reflections are eliminated
through wave cancellation. Incidentally, your attempt
at superposing powers was why your reflections turned
out to be brighter than the surface of the sun. If
you had instead used the irradiance equation, your
result and mine would have been identical - reflections
eliminated.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Richard Clark wrote:
Cecil Moore wrote:

reflections eliminated.

As I said, you lack of experience in the field of Optics is glaring.

I have some anti-reflective glass, Richard. It
does a reasonable job of wave cancellation even
under less than ideal conditions with ambient
light.

Our previous example involved an ideal laser and
an ideal 1/4WL coating of thin film. The index
of refraction of the thin film material was
exactly the square root of the index of refraction
of the glass. Those are ideal conditions for 100%
wave cancellation of reflections as anyone in
the field of optics should know.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

On Sat, 31 Mar 2007 11:11:13 GMT, "Dave" wrote:

and hence my recommendation to abandon power as a useful method of analyzing
waves in transmission lines. power is a scalar, you can't add power without
resorting to digging out the phase angles that come from the current or
voltage waves.

Hi Dave,

This is not a very compelling reason for method that has a wide
application in optics; Cecil has no real experience in this field, and
it shows frequently, but the math is quite common. It relates,
peripherally, to Keith's advice to investigate lattice diagrams, but
Cecil no doubt hesitates to trod this path that leads to the
destruction of his other favorite, failed example of anti-glare glass.

73's
Richard Clark, KB7QHC

Revisiting the Power Explanation
 

On Sat, 31 Mar 2007 12:21:00 -0500, Cecil Moore
wrote:

reflections eliminated.

As I said, you lack of experience in the field of Optics is glaring.

Revisiting the Power Explanation
 

Richard Clark wrote:
Cecil Moore wrote:
Your own model showed the falsity of this supposed 100% cancellation.

Actually, it showed perfectly the 100% cancellation
by two waves each of equal magnitude and opposite phase.
It was your superposition of powers that was at fault
and you apparently don't even realize your mistake.
Do you want to go through it again?

Have you tried the brain teaser I posted? It is a
lot like non-reflective thin films.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

On Fri, 30 Mar 2007 11:46:15 -0800, Richard Clark wrote:

On Fri, 30 Mar 2007 18:19:20 GMT, Walter Maxwell
wrote:

Oh, 'cmon Richard, are you saying that if a load reflected wave

Is not the same statement as your earlier one:
On Fri, 30 Mar 2007 14:45:56 GMT, Walter Maxwell wrote:

The part I feel is contradicted is that when total re-reflection
is caused without a total discontinuity such as a physical
short or open circuit

What is in your "without a total discontinuity" that is now found in
your "load reflected wave?"

Are we to now parse "total discontinuity" as being wholly different
from "partial discontinuity" such that waves suddenly mix from that
difference?

My example of the classic AT/ATR tube evidences EVERY observation you
offer, except it is a necessary load without which those observations
would never appear. If I were to replace its "total discontinuity"
with a weak tube (it exhibits less than total short); it too would
exhibit EVERY observation you offer EXCEPT they would be imperfect or
"partial discontinuities" repeated every quarter wave. It is obvious
that the effect follows the physical load, not the waves (they haven't
changed when the tube went bad). The physical load is the principle
in the process of interference.

Richard, I don't consider the AT/ATR tube relevant to the discussion of re-reflection of reflected power
incident on the output of a pi-network in an RF TX.

There is absolutely no example of interference that does not rely on a
load to reveal it.

If what you just said is true, then how do you explain nulls in a radiation pattern of an antenna having more
than one radiator, for example two verticals spaced 1/4 wl and fed in quadrature, thus creating a cardioid
pattern with a total null in one direction in azimuth? If the null was not created by interference between the
radiations from the two verticals, then how do you explain the formation of the null?

73's
Richard Clark, KB7QHC

Walt, W2DU

Revisiting the Power Explanation
 

On Sat, 31 Mar 2007 17:40:45 GMT, Cecil Moore
wrote:

Those are ideal conditions for 100%
wave cancellation of reflections as anyone in
the field of optics should know.

Amusing. You don't really know anyone of them, do you? Certainly
none who have to live with the invalidity of this sophomoric
treatment.

Your own model showed the falsity of this supposed 100% cancellation.
As I said, your visiting the topic of Lattice diagrams is not even on
the horizon, and behind it, Fresnel's equations are even more remote.
(I can well imagine you flipping through the index to find those pages
to presume their understanding, your Xeroxing other's work is
apparent.)

Like I said, your are quite ignorant to the matters of Optics. RF
runs a poor second, as your professional life in binary design has
corrupted your thinking.

Revisiting the Power Explanation
 

Richard Clark wrote:
On Sat, 31 Mar 2007 18:57:02 GMT, Cecil Moore
wrote:

Do you want to go through it again?

Do I need to see you crawl again?

Here it is again. Reflections are eliminated at the
air to thin-film surface.

In the following fixed font diagram, IR is the Index
of Refraction.

air | 1/4WL thin-film | Glass
1W Laser---IR=1.0---|----IR=1.222-----|--IR=1.493---...
Ifor=1W | Ifor=1.0101W | Ifor=1W
Iref=0W | Iref=0.0101W | Iref=0

Note that I is "irradiance", not current.

Given: The irradiance reflection coefficient is 0.01
at both interfaces. The irradiance transmission coefficient
is 0.99 at both interfaces.

Have you tried the brain teaser I posted? It is a
lot like non-reflective thin films.

proves my point adequately (your fumbling with patch-work proofs like
bracing the SQRT with absolutes is funny only once however).

Well, Hecht apparently didn't have to deal with nit-pickers
as exist on this newsgroup. If such had any importance at
all, I'm sure he would have mentioned it.
--
73, Cecil http://www.w5dxp.com