Where's the energy? (long)
 

Let's take a look at the energy in pulses and sine waves. At the end of
the day, the energy all has to be accounted for, whether it superposes
or not. It's really not all that difficult to do the analysis, as long
as we're careful not to fall into the traps which seem to have tripped
up quite a few others.

First a rectangular pulse. The energy E it takes to launch a pulse of
voltage Vp and duration T (seconds) on a transmission line of
characteristic impedance (assumed purely real) Z0 is Pp * T = Vp * Ip *
T where

Pp is the constant power applied as the pulse as created
T is the length of time the power was applied
Vp and Ip are the voltage and current of the traveling pulse

The pulse is a traveling wave, so for a forward traveling pulse, Ip = Vp
/ Z0; consequently, E = Vp^2 * T / Z0 = Ip^2 * T * Z0. Note that it's
essential to assume a purely resistive Z0 for this simple time-domain
analysis, since a reactive Z0 would cause a distortion of the pulse shape.

Once launched onto the line, we don't have any guarantee that all the
energy will stay within the spatial boundaries of the pulse -- all we
know for sure is how much total energy we've put into the line. But we
can conceptually freeze the pulse at any instant and see where the
energy is. Let's do that.

The obvious way to determine the energy in the pulse is to integrate the
power, which we can easily calculate. This is, after all, what we did to
find the energy we put into the line in the first place. But we're
interested in the energy distribution as a function of physical position
at an instant of time, so we can't find it by integrating the power.
(This is a mistake that seems to be commonly made.) Why not? Well, first
of all, energy is the *time* integral of the power. If we integrate the
power over a time interval of zero, the result is zero. We could look at
a single position on the line and integrate the power during the time it
takes for the wave to move by, to get the amount of energy which went by
during the time interval. But that's an indirect way of seeing where the
energy is on the line at a given time, and can easily lead to invalid
results. There are at least two potential problems with integrating the
power over a period of time to get the energy which passes a point. The
first is that we assign a sign to power, negative when energy is
traveling one way and positive when traveling the other. Consequently,
the result of the integral can be positive or negative. Although the
concept of negative potential energy is a valid one, I don't believe it
really applies to this situation, so one would have to be very careful
in interpreting and dealing with the sign resulting from the
integration. The second potential problem is that an integral never
produces a unique answer, but only an answer that's correct to within a
constant which has to be separately determined. Careless evaluation of
the constant or ignoring it altogether can produce invalid results.

So what I'm going to do is to evaluate the stored energy *per unit
length* of the line at each position along the line. The meaning of this
is that if we were to choose some sufficiently short segment length, the
amount of energy stored on each segment will be proportional to the
energy per unit length evaluated at that segment. In other words, I'll
evaluate the energy density as a function of position, or the energy
distribution along the line. This tells us where the energy is at the
instant of evalulation. I'm going to use the convention that the stored
or potential energy of a discharged line (V and I = 0) is zero.

The energy per unit length stored in the electric field, or line
capacitance, is C'V^2/2, where C' is the capacitance per unit length and
V is the voltage on the segment of line being evaluated. V is assumed to
not vary significantly over the segment length. We can let the segment
length approach zero as a limit, and say that the energy per unit length
is this value at any particular point along the line, where V is the
voltage at that point.

Likewise, the energy per unit length stored in the magnetic field, or
line inductance, is L'I^2/2 where L' is the inductance per unit length.
The total energy stored per unit length at any point is

E' = (C'V^2 + L'I^2)/2

On our line with purely real Z0, Z0 = sqrt(L'/C'), so L' = Z0^2 * C' and

E' = C'(V^2 + (Z0*I)^2)/2

where
E' is the total stored energy per unit length (or energy density) at
some point
V is the voltage at that point
I is the current at that point
Z0 is the (purely real) line characteristic impedance
C' is the capacitance per unit length

Length units for E' and C' can be anything as long as they're the same
for both.

Now let's look at a traveling pulse. We'll freeze it at some instant
while it's traveling down the line.

At any point to the left or right of the pulse, V and I are zero, so the
energy density is zero except where the pulse is. Where the pulse is, V
is the pulse voltage Vp and I the pulse current Ip, so

E' = C'(Vp^2 + (Z0*Ip)^2)/2

For a traveling wave, I = V/Z0, so

E' = C'Vp^2

This energy density is constant over the whole length of the pulse,
since Vp and Ip are constant over that distance. The total energy in the
pulse is then

E = E' * len = C'Vp^2 * len where len is the length of the pulse in
the same length units as C' and E'.

Because the energy density beyond the pulse in both directions is zero,
this is also the total energy in the line, which must equal the amount
we put in originally. So

E = C'Vp^2 * len = Vp^2 * T / Z0

from which we can calculate C' = T / (Z0 * len). Some manipulation of
this gives

T / len = sqrt(L'C') which relates line delay to L' and C', a result
which can be derived by other means.

All the energy in the line is accounted for -- it's traveling along with
the pulse, confined to the width of the pulse as we'd expect.

Ok, now let's fire another pulse at it from the other end of the line,
and see what happens when they completely overlap. Call the pulse 1 and
2 voltages Vp1 and Vp2, and currents Ip1 and Ip2. Assume that both have
the same duration T and therefore the same length len.

Voltages and currents (or E and H fields) add in the overlap region, so
the total V and I are the sum of the individual pulses' V and I. The
energy density in the overlap region is then:

E' = C'((Vp1 + Vp2)^2 + (Z0*(Ip1 + Ip2))^2)/2 * len

= C'(Vp1^2 + Vp2^2 + 2*Vp1*Vp2 + Z0(Ip1^2 + Ip2^2 + 2*Ip1*Ip2))/2

But what's the simple sum of the energy densities of the two pulses?

E1' + E2' = C'(Vp1^2 + Vp2^2 + Z0*(Ip1^2 + Ip2^2))/2

Oops! The energy density of the sum of the two pulses isn't the same as
the sum of the energy densities of the two pulses! And Since the overlap
region length is the same as the single pulse length, the same holds
true for the total energy. The problem is the two additional terms in
the total energy density 2*Vp1*Vp2 and 2*Z0*Ip1*Ip2.

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._

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.

Well, we ducked that bullet. But what if the two pulses are traveling in
the same direction? What then? The two troublesome terms don't cancel,
so some energy ends up getting created or destroyed. But before worrying
too much about that, try to imagine how you'd accomplish it. The
propagation speed is the same for all pulses, so there' no way one can
catch up with another if both are fired from the input. I believe you
can contrive a situation where two pulses can be generated, one from
each end which, if long enough, will partially overlap when going the
same direction, after reflection. But the overlap and energy
calculations will be different than for this example, and I'm sure the
energy of the summed pulses will equal the total energy in the line. I'd
appreciate seeing an analysis from anyone who thinks he can show
differently. Remember that this analysis assumed that no other pulse was
present at the input while the pulse was being generated. If one is, the
amount of energy going into the pulse will be different. It also assumed
a constant line Z0 and velocity factor (constant L' and C'), so a
different analysis would have to be used if that condition is violated.

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.

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!

As always, I appreciate any corrections to either the methodology or the
calculations.

Roy Lewallen, W7EL

Where's the energy? (long)
 

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

Where's the energy? (long)
 

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

Where's the energy? (long)
 

Roy Lewallen wrote:
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!

An example from optics will make the situation clear.

http://www.w5dxp.com/thinfilm.GIF

At t3, when the 0.009801 watt internal reflection arrives
to interfere with the 0.01 watt external reflection, what
is the resulting reflected power toward the source?

Anyone who can answer that simple question from the field
of optics will understand what happens to the energy in
a transmission line.

Hint: the reflected power is *not* 0.01w - 0.009801w.
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

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

Where's the energy? (long)
 

On Jan 23, 8:35*am, Cecil Moore wrote:
Yes, signals traveling in opposite directions don't interfere.
Call this assertion A.

Consider two antennas several wavelengths apart and driven with
the same frequency. Exploring the field strength far from the
antennas we find regions with zero field strength (nulls) and
regions with increased field strength. This variation in field
strength is usually ascribed to interference and the pattern
of variation is often called an interference pattern.

Similar results can be observed with light (google "two slit
experiment").

Locate one of these nulls far from the antennas and follow it
back towards the antennas. Eventually you will be on a line
between the two antennas.

From assertion A above, is it your contention that far from
the antennas it is "interference" that causes the variation
in field strength, but that on the line drawn between the two
antennas some other mechanism is responsible?

If so, what is the other mechanism? And does it only work
exactly on the line, or does it start working when you get
close to the line? How close?

Now I suggest that interference works just as well on the
line drawn between the antennas as it does every where
else and the conditions along that line are not a special
case.

That said, when we look at the two slit experiment, it is
generally agreed that the photons are redistributed such
that there are no photons in dark regions and more photons
in the bright regions.

On the line drawn between the two antennas, there are dark
regions and bright regions (the standing wave). By analogy,
there are no photons in the dark regions and more in the
bright regions. But the photons from the two sources were
travelling towards each other. What is the mechanism that
redistributes the photons such that there are none in the
dark regions? Do the photons stop and not enter the dark
region? Or do they turn into 'dark photons' as they
transit the dark regions? What are 'dark photons'?

...Keith

Where's the energy? (long)
 

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

Where's the energy? (long)
 

Keith Dysart wrote:
From assertion A above, is it your contention that far from
the antennas it is "interference" that causes the variation
in field strength, but that on the line drawn between the two
antennas some other mechanism is responsible?

Of course not - please don't be ridiculous. If the two
antenna elements were isotropic point sources, on a
line drawn between them, there could be no interference
and there would be only standing waves in free space
along that line assuming no reflections from nearby
objects, etc.

Everywhere else there are components of waves traveling
in the same direction so interference is possible anywhere
except on that line between the point sources. When the
sources are not a point, seems to me, interference could
occur at any and all points in space.

My "assertion A above" was about transmission lines,
an essentially one-dimensional context. Two waves in
a transmission line are either traveling in opposite
directions or in the same direction.

Incidentally, I came across another interesting quote
from one of my college textbooks, "Electrical Communication",
by Albert. "Such a plot of voltage is usually referred to
as a *voltage standing wave* or as a *stationary wave*. Neither
of these terms is particularly descriptive of the phenomenon.
A *plot* of the effective values of voltage ... is *not a wave*
in the usual sense. However, the term "standing wave" is in
wide-spread use." [Emphasis is the author's]
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

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

Where's the energy? (long)
 

On Jan 23, 1:12*pm, Cecil Moore wrote:
Keith Dysart wrote:
From assertion A above, is it your contention that far from
the antennas it is "interference" that causes the variation
in field strength, but that on the line drawn between the two
antennas some other mechanism is responsible?

Of course not - please don't be ridiculous. If the two
antenna elements were isotropic point sources, on a
line drawn between them, there could be no interference
and there would be only standing waves in free space
along that line assuming no reflections from nearby
objects, etc.

Everywhere else there are components of waves traveling
in the same direction so interference is possible anywhere
except on that line between the point sources. When the
sources are not a point, seems to me, interference could
occur at any and all points in space.

OK. So it is your contention that "far from the antennas
it is "interference" that causes the variation in field
strength, but that on the line drawn between the two
antennas some other mechanism is responsible".

But why do you say "Of course not" and then proceed to
paraphrase my statement?

When the mechanism abruptly changes from interference
when off the line to "standing wave" when EXACTLY (how
exact?) on the line, is there any discontinuity in
the observed field strengths?

...Keith

Where's the energy? (long)
 

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

It is not magic. "Interference" and "superposition" simply
have different definitions.

Interference is a subset of superposition, i.e. interference
cannot occur without superposition but superposition can occur
without interference. This subject is covered in every optics
text that I have ever seen, including Born and Wolf. Given two
waves of equal power densities (irradiances) if the resultant
irradiance is not equal to the sum of the two irradiances, then
interference has occurred.

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

Impossible in a transmission line which is the context.
In free space, I would guess that interference is possible
in their common direction of travel.

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

For coherent waves in free space, that would ensure interference
until the beams diverged. It should result in the usual light
and dark interference rings.

Repeating: This is a distinction with no technical value.

Maybe it would help if you published a video of you waving
your hands as you scream that assertion at the top of your
lungs? :-)
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

Keith Dysart wrote:
OK. So it is your contention that "far from the antennas
it is "interference" that causes the variation in field
strength, but that on the line drawn between the two
antennas some other mechanism is responsible".

It is unethical to bear false witness about what I said.
What I said was:

On a line drawn between two *isotropic point sources*,
when there are no reflections anywhere around, along
that line, interference is impossible. The only thing
existing along that line would be standing waves.
There is no point along that line where the power
density is not equal to the sum of the two sources,
i.e. there is superposition but no interference along
that line.

If the elements are not point sources, interference
is obviously possible at each and every point. I assume
your example elements are not point sources so what you
claimed was my contention was a false statement. If
you can't win the arguments without making false
statements about what I said, you lose anyway.
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

"Cecil Moore" wrote
Keith Dysart wrote:
From assertion A above, is it your contention that far from
the antennas it is "interference" that causes the variation
in field strength, but that on the line drawn between the two
antennas some other mechanism is responsible?

Of course not - please don't be ridiculous. If the two
antenna elements were isotropic point sources, on a
line drawn between them, there could be no interference
and there would be only standing waves in free space
along that line assuming no reflections from nearby
objects, etc.
______________

Cecil, hopefully you understand that even isotropic radiators near each
other and excited on the same frequency with the same amount of power will
generate far-field pattern nulls.

Maybe I'm misunderstanding you.

RF

Where's the energy? (long)
 

Richard Fry wrote:
Cecil, hopefully you understand that even isotropic radiators near each
other and excited on the same frequency with the same amount of power will
generate far-field pattern nulls.

Maybe I'm misunderstanding you.

I'm trying to understand how a line drawn between two
"isotropic radiators near each other" could ever be
in the far field.
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

"Cecil Moore wrote
Richard Fry wrote:
Cecil, hopefully you understand that even isotropic radiators near each
other and excited on the same frequency with the same amount of power
will generate far-field pattern nulls.

Maybe I'm misunderstanding you.

I'm trying to understand how a line drawn between two
"isotropic radiators near each other" could ever be
in the far field.
_______________

Everywhere it exceeds 2*D^2/lambda in length, where D is the greatest
dimension of the array.

RF

Where's the energy? (long)
 

On Jan 23, 1:46*pm, Cecil Moore wrote:
Keith Dysart wrote:
OK. So it is your contention that "far from the antennas
it is "interference" that causes the variation in field
strength, but that on the line drawn between the two
antennas some other mechanism is responsible".

It is unethical to bear false witness about what I said.
What I said was:

On a line drawn between two *isotropic point sources*,
when there are no reflections anywhere around, along
that line, interference is impossible. The only thing
existing along that line would be standing waves.

And you also wrote:
Everywhere else there are components of waves traveling
in the same direction so interference is possible anywhere
except on that line between the point sources.

I am having great difficulty finding any difference
between my writing:

So it is your contention that "far from the antennas
it is "interference" that causes the variation in field
strength, but that on the line drawn between the two
antennas some other mechanism is responsible".

and your paraphrase.

...Keith

Where's the energy? (long)
 

Richard Fry wrote:
"Cecil Moore wrote
Richard Fry wrote:
Cecil, hopefully you understand that even isotropic radiators near each
other and excited on the same frequency with the same amount of power
will generate far-field pattern nulls.

Maybe I'm misunderstanding you.

I'm trying to understand how a line drawn between two
"isotropic radiators near each other" could ever be
in the far field.

Everywhere it exceeds 2*D^2/lambda in length, where D is the greatest
dimension of the array.

Are you saying that the sources that are "near each other"
are far enough apart to be in each other's far field?
How could that be if the two sources are D apart?
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

Keith Dysart wrote:
I am having great difficulty finding any difference
between my writing:

So it is your contention that "far from the antennas
it is "interference" that causes the variation in field
strength, but that on the line drawn between the two
antennas some other mechanism is responsible".

and your paraphrase.

The difference is that your example contained elements
that are not zero dimensions. My assertions covered only
antenna elements of zero dimensions. I repeat:

On a line drawn between two coherent isotropic radiators,
in the absence of any reflections, interference along
that line is impossible because the average total
power density all along that line is constant.
There is no interference in standing waves given
"interference" as defined by Eugene Hecht in "Optics".
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

"Cecil Moore"
Everywhere it exceeds 2*D^2/lambda in length, where D is the greatest
dimension of the array.

Are you saying that the sources that are "near each other"
are far enough apart to be in each other's far field?
How could that be if the two sources are D apart?
_________

It isn't necessary for the radiators to be in each other's far field.

Here is a link to a plot of the far-field elevation pattern of a linear
array of three isotropic sources at one lambda vertical spacing.

http://i62.photobucket.com/albums/h8...picSources.gif

RF

Where's the energy? (long)
 

On Jan 23, 2:21*pm, Cecil Moore wrote:
Keith Dysart wrote:
I am having great difficulty finding any difference
between my writing:

* So it is your contention that "far from the antennas
* it is "interference" that causes the variation in field
* strength, but that on the line drawn between the two
* antennas some other mechanism is responsible".

and your paraphrase.

The difference is that your example contained elements
that are not zero dimensions. My assertions covered only
antenna elements of zero dimensions. I repeat:

On a line drawn between two coherent isotropic radiators,
in the absence of any reflections, interference along
that line is impossible because the average total
power density all along that line is constant.
There is no interference in standing waves given
"interference" as defined by Eugene Hecht in "Optics".

So then, for "two coherent isotropic radiator",
it is your contention that "far from the antennas
it is "interference" that causes the variation in field
strength, but that on the line drawn between the two
antennas some other mechanism is responsible".

...Keith

Where's the energy? (long)
 

Richard Fry wrote:
It isn't necessary for the radiators to be in each other's far field.

I'm just trying to understand your point. You said there
are nulls in the far field. If the radiators are not in
each other's far fields, how could the line drawn between
them be in the far field?
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

Keith Dysart wrote:
So then, for "two coherent isotropic radiator",
it is your contention that "far from the antennas
it is "interference" that causes the variation in field
strength, but that on the line drawn between the two
antennas some other mechanism is responsible".

Please define "field strength". The total average
power density along a line drawn between the two
point sources is constant, i.e. the average sum
of the energy in the E-field and H-field is
constant. If you are defining "field strength"
as only the E-field, of course standing waves
are the cause, not interference, as defined by
Hecht in "Optics".
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

"Cecil Moore" wrote
I'm just trying to understand your point. You said there
are nulls in the far field. If the radiators are not in
each other's far fields, how could the line drawn between
them be in the far field?


X Iso source 1

=================== to far field, where nulls will exist

X Iso source 2,
1 lambda from
Iso source 1



RF

Where's the energy? (long)
 

Richard Fry wrote:
"Cecil Moore" wrote
I'm just trying to understand your point. You said there
are nulls in the far field. If the radiators are not in
each other's far fields, how could the line drawn between
them be in the far field?


X Iso source 1

=================== to far field, where nulls will exist

X Iso source 2,
1 lambda from
Iso source 1

No, no, no, Richard. The line is drawn from one source to
the other source. Your line is not drawn from either
source to the other source.
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

"Cecil Moore" wrote
No, no, no, Richard. The line is drawn from one source to
the other source. Your line is not drawn from either
source to the other source.
__________

Obviously we are not talking about the same net radiations patterns.

Carry on.

RF

Where's the energy? (long)
 

Richard Fry wrote:
"Cecil Moore" wrote
No, no, no, Richard. The line is drawn from one source to
the other source. Your line is not drawn from either
source to the other source.

Obviously we are not talking about the same net radiations patterns.

Sorry about that. English semantics strikes again. Your
line was indeed "between" the two sources but not the
"from - to" line that I had in mind when I said "between". :-)
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

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

It is not magic. "Interference" and "superposition" simply
have different definitions.

Interference is a subset of superposition, i.e. interference
cannot occur without superposition but superposition can occur
without interference. This subject is covered in every optics
text that I have ever seen, including Born and Wolf. Given two
waves of equal power densities (irradiances) if the resultant
irradiance is not equal to the sum of the two irradiances, then
interference has occurred.

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

Impossible in a transmission line which is the context.
In free space, I would guess that interference is possible
in their common direction of travel.

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

For coherent waves in free space, that would ensure interference
until the beams diverged. It should result in the usual light
and dark interference rings.

Repeating: This is a distinction with no technical value.

Maybe it would help if you published a video of you waving
your hands as you scream that assertion at the top of your
lungs? :-)

Cecil,

Many people, myself included, treat the term "interference" in a
qualitative manner. The general meaning is that two entities somehow
interact in a noticeable way, and the result has some signature of that
interaction.

You appear to use a very precise, quantitative definition of
"interference." I do not recall ever seeing such a quantitative
definition. Could you please give us a reference or an exact quote from
some reasonably reputable source that defines "interference" in a
quantitative and unambiguous manner?

You imply that some interactions lead to "interference" and some do not.
How can the unwashed among us know when the magic occurs and when it
does not?

73,
Gene
W4SZ

Where's the energy? (long)
 

Gene Fuller wrote:

...
You imply that some interactions lead to "interference" and some do not.
How can the unwashed among us know when the magic occurs and when it
does not?

73,
Gene
W4SZ

You mean if I just wash it will increase my ability to understand? D*mn
man, I would NEVER have thought it possible. Indeed, if most were to
suggest that, I would laugh. But, given it is you, ... chuckle

And please, take this as a friendly joke! (albeit a poor one) I tire of
the religiously devout crying "blasphemy" and posting stones and
pitchforks! ROFLOL

Warm regards,
JS

Where's the energy? (long)
 

Roy Lewallen wrote:
[... very nice explanation]

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!

As always, I appreciate any corrections to either the methodology or
the calculations.

Roy Lewallen, W7EL

How about analyzing a vibrating string? If you play guitar, there's a very
nice note you can make by plucking a high string, then putting your finger
at exactly the correct spot and removing it quickly. The note will jump to
a much higher frequency and give a much purer sound. Clearly, the
mechanical energy has split into two waves that cancel at the node.

In principle, you could show the node is stationary, thus contains no
energy. But there is energy travelling on both sides of the null point -
you can hear it.

You can also create other notes by touching different spots on the
vibrating string. These create standing waves with energy travelling in
both directions, but cancelling at the null points. Very similar to
transmission lines.

Regards,

Mike Monett

Where's the energy? (long)
 

Mike Monett wrote:

How about analyzing a vibrating string? If you play guitar, there's a very
nice note you can make by plucking a high string, then putting your finger
at exactly the correct spot and removing it quickly. The note will jump to
a much higher frequency and give a much purer sound. Clearly, the
mechanical energy has split into two waves that cancel at the node.

In principle, you could show the node is stationary, thus contains no
energy. But there is energy travelling on both sides of the null point -
you can hear it.

You can also create other notes by touching different spots on the
vibrating string. These create standing waves with energy travelling in
both directions, but cancelling at the null points. Very similar to
transmission lines.

Regards,

Mike Monett

Sounds like a great idea. I'll look forward to seeing your analysis.

Roy Lewallen, W7EL

Where's the energy? (long)
 

Mike Monett wrote:
Roy Lewallen wrote:
[... very nice explanation]

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!

As always, I appreciate any corrections to either the methodology or
the calculations.

Roy Lewallen, W7EL

How about analyzing a vibrating string? If you play guitar, there's a very
nice note you can make by plucking a high string, then putting your finger
at exactly the correct spot and removing it quickly. The note will jump to
a much higher frequency and give a much purer sound. Clearly, the
mechanical energy has split into two waves that cancel at the node.

In principle, you could show the node is stationary, thus contains no
energy. But there is energy travelling on both sides of the null point -
you can hear it.

You can also create other notes by touching different spots on the
vibrating string. These create standing waves with energy travelling in
both directions, but cancelling at the null points. Very similar to
transmission lines.

Regards,

Mike Monett


Most undergraduate physics texts have, or should have, discussions of
vibrating strings. There's a good treatment of the subject in
William C. Elmore's and Mark A. Heald's book _Physics of Waves_
published by Dover. If you wanted to get in an argument you could
say that the energy on both sides of the node isn't traveling, but is
merely alternating between potential and kinetic. Such strings have loss
(or you wouldn't be able to hear them). Loss is a taboo subject on this
newsgroup because it makes wave behavior too hard to understand for the
savants posting here.
73,
Tom Donaly, KA6RUH

Where's the energy? (long)
 

Roy Lewallen wrote:

Mike Monett wrote:

[...]

Sounds like a great idea. I'll look forward to seeing your analysis.

Roy Lewallen, W7EL

LOL! I stopped playing guitar years ago!

Regards,

Mike Monett

Where's the energy? (long)
 

"Tom Donaly" wrote:

[...]

Most undergraduate physics texts have, or should have, discussions
of vibrating strings. There's a good treatment of the subject in
William C. Elmore's and Mark A. Heald's book _Physics of Waves_
published by Dover.

If you wanted to get in an argument you could say that the energy
on both sides of the node isn't traveling, but is merely
alternating between potential and kinetic.

Yes, I thought about that a bit before posting. It seems logical a
plucked string sends a wave in both directions, where it is
reflected and returns to create a standing wave.

When it forms a standing wave, it seems reasonable to say the energy
is alternating between potential and kinetic. But isn't that similar
to what happens on a transmission line that is exactly some multiple
of a quarter wavelength long?

Such strings have loss (or you wouldn't be able to hear them).

Loss is a taboo subject on this newsgroup because it makes wave
behavior too hard to understand for the savants posting here.

73,
Tom Donaly, KA6RUH

Regards,

Mike Monett

Where's the energy? (long)
 

Mike Monett wrote:
"Tom Donaly" wrote:

[...]

Most undergraduate physics texts have, or should have, discussions
of vibrating strings. There's a good treatment of the subject in
William C. Elmore's and Mark A. Heald's book _Physics of Waves_
published by Dover.

If you wanted to get in an argument you could say that the energy
on both sides of the node isn't traveling, but is merely
alternating between potential and kinetic.

Yes, I thought about that a bit before posting. It seems logical a
plucked string sends a wave in both directions, where it is
reflected and returns to create a standing wave.

When it forms a standing wave, it seems reasonable to say the energy
is alternating between potential and kinetic. But isn't that similar
to what happens on a transmission line that is exactly some multiple
of a quarter wavelength long?


Demo 4 of the TLVis1 program I posted reference to, shows that in a
transmission line with a pure standing wave (load reflection coefficient
magnitude of 1), the energy between nodes alternates between the
electric field (line capacitance) and magnetic field (line inductance).
This is true regardless of the line length or the source termination.

Roy Lewallen, W7EL

Where's the energy? (long)
 

Roy Lewallen wrote:

Demo 4 of the TLVis1 program I posted reference to, shows that in
a transmission line with a pure standing wave (load reflection
coefficient magnitude of 1), the energy between nodes alternates
between the electric field (line capacitance) and magnetic field
(line inductance).

This is true regardless of the line length or the source
termination.

Roy Lewallen, W7EL

Yes, this is a very nice demo. Thank you for posting it.

I have a question. In demo 4, the bottom window shows the Ee field
in green, Eh in red, and ETot in black.

When the demo starts, you can only see a green and a black trace.

If you pause it just as the wave hits the end, you can now see the
red trace, Eh. (This is an actual statement and has nothing to do
with the fact I am Canadian.)

What happened to the Eh trace as the wave is initally moving to the
right? Is it overlaid by the Ee trace in green? Or is it just not
plotted?

Then, when the wave hits the end and starts reflecting, the red
trace remains attached to ground, and the green trace moves up and
connects with the black trace. (Sorry for the confusing description
- you have to try it yourself to see.)

Now, as you single step, the green trace and the red trace appear to
be 180 degrees out of phase.

My problem here is someone wrote a web page that claims the electric
and magnetic fields are orthogonal:

http://www.play-hookey.com/optics/tr...etic_wave.html

I tried sending him an email to show if the fields were orthogonal
as he claims, it would look like a pure reactance, and no energy
would be transmitted. But he is stuck on his idea and won't budge.

Now my problem is figuring out exactly what happens at the
reflection, and why the Eh field behaves the way shown in your demo.

Regards,

Mike Monett

Where's the energy? (long)
 

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

Where's the energy? (long)
 

Mike Monett wrote:

Yes, this is a very nice demo. Thank you for posting it.

I have a question. In demo 4, the bottom window shows the Ee field
in green, Eh in red, and ETot in black.

When the demo starts, you can only see a green and a black trace.

If you pause it just as the wave hits the end, you can now see the
red trace, Eh. (This is an actual statement and has nothing to do
with the fact I am Canadian.)

What happened to the Eh trace as the wave is initally moving to the
right? Is it overlaid by the Ee trace in green? Or is it just not
plotted?

The traces are drawn in the order Eh, Ee, and total. During the initial
forward wave, Eh and Ee are equal, so the Ee overwrites the Eh trace.

Then, when the wave hits the end and starts reflecting, the red
trace remains attached to ground, and the green trace moves up and
connects with the black trace. (Sorry for the confusing description
- you have to try it yourself to see.)

Hopefully it'll all make sense once you think about how one trace will
always win when more than one have the same value.

Now, as you single step, the green trace and the red trace appear to
be 180 degrees out of phase.

My problem here is someone wrote a web page that claims the electric
and magnetic fields are orthogonal:

http://www.play-hookey.com/optics/tr...etic_wave.html

You're making the same error that Cecil often does, confusing time phase
with directional vector orientation. The orthogonality of E and H fields
refers to the field orientations of traveling plane TEM waves in
lossless 3D space or a lossless transmission line, at the same point and
time. The E and H fields of these traveling waves are always in time
phase, not in quadrature. The graphs show the magnitudes of the waves at
various points along the line. These represent neither the time phase
nor the spatial orientation of the E and H fields.

I tried sending him an email to show if the fields were orthogonal
as he claims, it would look like a pure reactance, and no energy
would be transmitted. But he is stuck on his idea and won't budge.

Good for him -- he's absolutely correct. If the E and H fields were in
time quadrature, you'd have a power problem. But they're not. They're in
phase in any medium or transmission line having a purely real Z0 (since
Z0 is the ratio of E to H of a traveling wave in that medium). This
includes all lossless media. But they're always physically oriented at
right angles to each other -- i.e., orthogonally, according to the right
hand rule.

Now my problem is figuring out exactly what happens at the
reflection, and why the Eh field behaves the way shown in your demo.

Go for it!

Roy Lewallen, W7EL

Where's the energy? (long)
 

Roy Lewallen wrote:

[...]

The traces are drawn in the order Eh, Ee, and total. During the
initial forward wave, Eh and Ee are equal, so the Ee overwrites
the Eh trace.

Good - thanks.

[...]

My problem here is someone wrote a web page that claims the
electric and magnetic fields are orthogonal:

http://www.play-hookey.com/optics/tr...etic_wave.html

You're making the same error that Cecil often does, confusing time
phase with directional vector orientation. The orthogonality of E
and H fields refers to the field orientations of traveling plane
TEM waves in lossless 3D space or a lossless transmission line, at
the same point and time.

Now you are confusing me with Cecil. I have no difficulty with the E
and H field orientation.

The E and H fields of these traveling waves are always in time
phase, not in quadrature.

Yes, that's what I tried to explain to him also.

The graphs show the magnitudes of the waves at various points
along the line. These represent neither the time phase nor the
spatial orientation of the E and H fields.

I tried sending him an email to show if the fields were
orthogonal as he claims, it would look like a pure reactance, and
no energy would be transmitted. But he is stuck on his idea and
won't budge.

Good for him - he's absolutely correct.

There is a bad mixup here. He claims:

"Note especially that the electric and magnetic fields are not in
phase with each other, but are rather 90 degrees out of phase. Most
books portray these two components of the total wave as being in
phase with each other, but I find myself disagreeing with that
interpretation, based on three fundamental laws of physics"

He claims the E and H fields are in quadrature. I claim he is wrong.

If the E and H fields were in time quadrature, you'd have a power
problem.

I believe that is what I tried to tell him. He bases his argument on
the following:

1. "The total energy in the waveform must remain constant at all
times."

Not true. It obviously goes to zero twice each cycle.

2. "A moving electric field creates a magnetic field. As an electric
field moves through space, it gives up its energy to a companion
magnetic field. The electric field loses energy as the magnetic
field gains energy."

Only if the environment is purely reactive. Not true with a pure
resistance.

3. "A moving magnetic field creates an electric field. This is
Faraday's Law, and is exactly similar to the Ampere-Maxwell law
listed above. A changing magnetic field will create and transfer its
energy gradually to a companion electric field."

Again, not true in a resistive environment.

But they're not. They're in phase in any medium or transmission
line having a purely real Z0 (since Z0 is the ratio of E to H of a
traveling wave in that medium). This includes all lossless media.

But they're always physically oriented at right angles to each
other - i.e., orthogonally, according to the right hand rule.

Yes, there is no confusion about this whatsoever.

[...]

Roy Lewallen, W7EL

Regards,

Mike Monett

Where's the energy? (long)
 

Gene Fuller wrote:
You appear to use a very precise, quantitative definition of
"interference." I do not recall ever seeing such a quantitative
definition. Could you please give us a reference or an exact quote from
some reasonably reputable source that defines "interference" in a
quantitative and unambiguous manner?

I've already posted what Eugene Hecht said about interference.

In the irradiance (power density) equation,
Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)
the last term is known as the "interference term", page 388 of
"Optics" by Hecht. Here's another reference:

http://en.wikipedia.org/wiki/Interference

A Google search for "electromagnetic wave interference" yielded
1,650,000 hits.

You imply that some interactions lead to "interference" and some do not.
How can the unwashed among us know when the magic occurs and when it
does not?

If the interference term in the above irradiance (power
density) equation is not zero, then interference is present.

In the s-parameter equation, b1 = s11*a1 + s12*a2, if b1
equals zero while s11, a1, s12, and a2 are not zero, then
total destructive interference is present.

Assume we superpose two coherent, collinear voltages, V1 and V2:

If (V1+V2)^2 V1^2+V2^2, then constructive interference is
present.

If (V1+V2)^2 V1^2+V2^2, then destructive interference is
present.
--
73, Cecil http://www.w5dxp.com

Where's the energy? (long)
 

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