Roy Lewallen wrote:
It turns out that we're saved -- For the forward traveling pulse, Ip1 =
Vp1/Z0. For the reverse traveling pulse, Ip2 = -Vp2/Z0. So when the
appropriate substitutions are made, we find that 2*Vp1*Vp2 +
2*Z0*Ip1*Ip2 = 0, so the energy in the sum of the pulses is equal to the
sum of the energies of the pulses. And this is true regardless of the
values of Vp1, Vp2, Ip1, and Ip2. That is, it's true for any two pulses,
for any overlap length. _Provided they're traveling in opposite
directions._

Yes, signals traveling in opposite directions don't interfere.

What happens when one pulse is the inverse of the other, that is, one is
positive and the other negative? Don't they cancel?

No, they don't. In the overlap region, the voltage is indeed zero. But
the current is twice that of each original pulse. The energy is simply
all stored in the magnetic field (line inductance) during the overlap.
The above equations still hold.

Yes, signals traveling in opposite directions don't interfere.

The conclusion I reach is that yes, a specific amount of energy
accompanies a pulse on a transmission line having purely real Z0, and is
confined to the pulse width. Although it can swap between E and H
fields, the energy in the confines of the pulse stays constant in value,
and simply adding when pulses overlap.

This is simply not true for coherent, collinear waves traveling
in the same direction. "Optics", by Hecht has an entire chapter
on "Interference". He says: "Briefly then, interference
corresponds to the interaction of two or more lightwaves yielding
a resultant irradiance that deviates from the sum of the component
irradiances." Irradiance is the power density of a lightwave, i.e.
watts per unit-area. Paraphrasing Hecht: Interference corresponds
to the interaction of two RF waves in a transmission line yielding
a resultant total power that deviates from the sum of the component
powers. If the total power is less than the sum of the component
powers, destructive interference has taken place (normally toward
the source). If the total power is greater than the sum of the
component powers, constructive interference has taken place
(normally toward the load). It is the goal of amateur radio
operators to cause *total destructive interference* toward the
source and *total constructive interference* toward the antenna.
These terms are defined in "Optics", by Hecht, 4th edition on
page 388. Quoting Hecht:

"In the case of *total constructive interference*, the phase
difference between the two waves is an integer multiple of
2*pi and the disturbances are in-phase."

When the phase angle is an odd multiple of of pi, "it is
referred to as *total destructive interference*.

If anyone works out the phase angles between the voltages, one
will discover that they match Hecht's definitions above.

Every text on EM wave interference that you can find will explain
how the bright interference rings are four times the intensity of
the dark interference rings so the average intensity is two times
the intensity of each equal-magnitude wave. Of course, that outcome
honors the conservation of energy principle. Using 'P' for power
density, the equation that governs such interference phenomena
in EM waves is:

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

where 'A' is the angle between the two electric fields. Every
textbook on optical physics contains that irradiance equation.
If Ptot is ever zero while P1 and P2 are not zero, one can be
absolutely certain that the "lost" energy has headed in the
opposite direction in a transmission line because there is
no other possibility. Energy is *never* lost.

RF waves in a transmission line obey the same laws of physics as
do light waves in free space. Coherent, collinear waves traveling
in the same direction do indeed interfere with each other.
Sometimes the interference is permanent as it is at an ideal
1/4WL anti-reflective thin-film coating on glass.

Sine waves are another problem -- there, we can easily have overlapping
waves traveling in the same direction, so we'll run into trouble if
we're not careful. I haven't worked the problem yet, but when I do, the
energy will all be accounted for. Either the energy ends up spread out
beyond the overlap region, or the energy lost during reflections will
account for the apparent energy difference between the sum of the
energies and the energy of the sum. You can count on it!

There is no problem. Optical physicists figured it out long
before any of us were born.

www.mellesgriot.com/products/optics/oc_2_1.htm

"If the two [out-of-phase] reflections are of equal amplitude,
then this amplitude (and hence intensity) minimum will be
zero."

This applies to reflections toward the source at a Z0-match
in a transmission line.

"... the principle of
conservation of energy indicates all 'lost' reflected intensity
[in the reflected waves] will appear as enhanced intensity in
the transmitted [forward wave] beam."

i.e. All the energy seemingly "lost" during the cancellation
of reflected waves toward the source at a Z0-match in a
transmission line, is recovered in the forward wave toward
the load.

That is exactly what happens when we match our systems. We
cause destructive interference toward the source in order
to eliminate reflections toward the source. The "lost"
energy joins the forward wave toward the load making the
forward power greater than the source power.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Every text on EM wave interference that you can find will explain
how the bright interference rings are four times the intensity of
the dark interference rings so the average intensity is two times
the intensity of each equal-magnitude wave.

I certainly misspoke there. The bright interference rings
are four times the intensity of one of the two equal
waves. The dark interference rings are, of course,
zero intensity.

If the intensity of one wave is P, the intensity of the
bright rings will be 4P and the intensity of the dark
rings will be zero. The average intensity will, of course,
be 2P, the sum of the two wave intensities.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

The bright interference rings
are four times the intensity of one of the two equal
waves. The dark interference rings are, of course,
zero intensity.

If the intensity of one wave is P, the intensity of the
bright rings will be 4P and the intensity of the dark
rings will be zero.

That's right. And we know that intensity is proportional to the
square of the EM field, so if P=9 then field=3. When there are two
such EM fields superposed, then we have 3+3 squared which is four
times greater than 3 squared. And owing to this supposed
'inequality', we have the sophomoric (literally) notion that there is
"extra" energy which must come from somewhere else.

ac6xg

Jim Kelley wrote:
Cecil Moore wrote:
If the intensity of one wave is P, the intensity of the
bright rings will be 4P and the intensity of the dark
rings will be zero.

That's right. And we know that intensity is proportional to the square
of the EM field, so if P=9 then field=3. When there are two such EM
fields superposed, then we have 3+3 squared which is four times greater
than 3 squared. And owing to this supposed 'inequality', we have the
sophomoric (literally) notion that there is "extra" energy which must
come from somewhere else.

The intensity is watts/unit-area, i.e. real energy.
If the intensity of the bright rings is 4P there is
indeed greater than average energy which requires a
zero P dark ring somewhere else in order to
average out to 2P. The "extra" energy in the bright
rings comes from the dark rings. The conservation of
energy principle allows nothing else. It is not a
sophomoric notion. It is the laws of physics in action.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

If the intensity of the bright rings is 4P there is
indeed greater than average energy which requires a
zero P dark ring somewhere else in order to
average out to 2P.

I guess you didn't notice the thing about the square of the sum of two
numbers being four times as great as the square of one of the numbers.
It simply means, for example, that a doubling in voltage is a
quadrupling in power. It doesn't mean there's "extra" energy anywhere.

The "extra" energy in the bright
rings comes from the dark rings.

There really is no extra energy.

The conservation of
energy principle allows nothing else.

Conservation of energy doesn't allow for there to *be* extra energy in
the first place! That's the most fundamental principle. Putting the
word 'extra' in quotes doesn't change that.

It is not a
sophomoric notion.

The notion was apparently contrived so that sophomores wouldn't feel
the need to ask as many questions.

It is the laws of physics in action.

As far as you know.

ac6xg

Jim Kelley wrote:
I guess you didn't notice the thing about the square of the sum of two
numbers being four times as great as the square of one of the numbers.
It simply means, for example, that a doubling in voltage is a
quadrupling in power. It doesn't mean there's "extra" energy anywhere.

On the contrary, constructive interference cannot happen
without that "extra" energy sucked up from destructive
interference (or from a local source). That's what you
are completely missing. Constructive interference between
two waves requires more energy than exists in those two
component waves. In the absence of a local source, another
two waves *must* be engaging in destructive interference
at the same time in order to supply the constructive
interference.

Conservation of energy doesn't allow for there to *be* extra energy in
the first place!

Therefore, those bright rings are just an illusion and
don't contain more than the average energy in the two
superposed waves. I guess that's your story and you
are sticking to it.
--
73, Cecil http://www.w5dxp.com

The idiot savant Cecil Moore wrote:

Therefore, those bright rings are just an illusion and
don't contain more than the average energy in the two
superposed waves.

You couldn't be more wrong in your interpretation.

Perhaps you will understand it this way: the bright rings have an
intensity which is equal in energy to the sum of the energy in the two
superposed waves. Energy doesn't have to come from anywhere else to
balance the equation.

The error you're making is the assumption that since the intensities
don't add up, something has to come from somewhere else. But that's a
wrong assumption to make. One shouldn't expect intensity to add like
that. Obviously it doesn't add like that. The fields are what add and
subtract, not intensity (or power). Power is the time derivative of
energy - that's the reason a squared term comes in. But power and
intensity don't propagate, therefore they don't superpose, and
shouldn't be added algebraically. Fields, voltage, current, energy -
it makes sense to sum those things algebraically.

When you double something and then square the sum, you get a factor of
four. But you're still only doubling the 'thing' of interest.
Depositing two paychecks and squaring the sum doesn't make your bank
balance go up by a factor of four. But it is certainly true that you
would have to borrow money from someone to make that happens. The
point is that it's not a realistic expectation.

I assume you know that the meter face on your Bird wattmeter is
calibrated exponentially. So, when your Bird power meter reading goes
up by a factor of four, what do you think the voltage across its meter
movement has actually increased by? The thing is, the Bird company
understands that when power reading quadruples, the circuit voltage
has only doubled. You'll note they don't claim that since the reading
quadrupled instead of doubling, the "extra" power had to come from
somewhere else.

If you can't understand this, that's fine. But it's not an excuse to
get so bloody belligerent with people about it.

ac6xg

Cecil Moore wrote:


Yes, signals traveling in opposite directions don't interfere.



Yes, signals traveling in opposite directions don't interfere.



This is a distinction with no technical value. Waves in the same
location are subject to the usual rules of linear superposition of the
fields. Whether you want to call this "interference" is simply a
philosophical choice. There is a whole gamut of results resulting from
the superposition, ranging from zero field to a maximum of all the field
magnitudes combined. The terms "destructive" and "constructive" are
sometimes used to denote the extreme cases, but those terms are not so
well defined for the more intermediate cases.

There is utterly no scientific distinction that applies to "signals
traveling in opposite directions." The mathematical results may look
special in the opposite direction case, but the same basic equations
apply in all cases.

73,
Gene
W4SZ

Gene Fuller wrote:
Cecil Moore wrote:
Yes, signals traveling in opposite directions don't interfere.

This is a distinction with no technical value. Waves in the same
location are subject to the usual rules of linear superposition of the
fields. Whether you want to call this "interference" is simply a
philosophical choice.

Not so. Here's what Eugene Hecht says: "... optical
interference corresponds to the interaction of two
or more [plane] light waves yielding a resultant
irradiance that deviates from the sum of the component
irradiances."

Superposition can occur with or without interference. If
P1 and P2 are the power densities for two plane waves:

If Ptot = P1 + P2, there is no interference because the
resultant power density does not deviate from the sum of
the component power densities.

If Ptot P1 + P2, there exists interference because
the resultant irradiance does deviate from the sum of
the component power densities.

There is utterly no scientific distinction that applies to "signals
traveling in opposite directions."

Interference only occurs when coherent, collinear waves
are traveling in the same direction. When they are
traveling in opposite directions, standing waves are
the result. Let's limit our discussion to plane waves.

The mathematical results may look
special in the opposite direction case, but the same basic equations
apply in all cases.

Yes, but boundary conditions apply. The phasors of the plane
waves traveling toward each other are rotating in opposite
directions so interference is impossible. Here is a slide
show about interference which only occurs when the waves are
traveling in the same direction.

http://astro.gmu.edu/classes/a10594/...8/l08s025.html
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
Cecil Moore wrote:
Yes, signals traveling in opposite directions don't interfere.

This is a distinction with no technical value. Waves in the same
location are subject to the usual rules of linear superposition of the
fields. Whether you want to call this "interference" is simply a
philosophical choice.

Not so. Here's what Eugene Hecht says: "... optical
interference corresponds to the interaction of two
or more [plane] light waves yielding a resultant
irradiance that deviates from the sum of the component
irradiances."

Superposition can occur with or without interference. If
P1 and P2 are the power densities for two plane waves:

Why do you attribute such magic to the word "interference"? Do you think
that Hecht's "interaction" is any different than superposition?

What if the waves are not quite anti-parallel, say at an angle of 179
degrees? Is interference now possible?

Suppose the waves are only 1 degree from parallel. Does that negate the
interference?

Repeating: This is a distinction with no technical value.

73,
Gene
W4SZ