Constructive interference in radiowave propagation
 

Gene Fuller wrote:
It is easy to give examples where the waves survive the superposition,
because they always do. It is rather strange that you are making this
argument after all the back and forth about traveling waves and standing
waves. Do we now have multiple flavors of EM waves? Some that obey
superposition and some that don't?

They all obey superposition which can occur with or
without interference. And you are wrong about all
waves surviving superposition. Canceled waves do
not survive wave cancellation in the direction that
they are traveling. Access this web page and set
the two waves to equal frequencies, equal magnitudes,
and opposite phases, i.e. 0 and 180 degrees.

http://micro.magnet.fsu.edu/primer/j...ons/index.html

When you do that, the waves are canceled in their
original direction of travel. The energy in those
canceled waves certainly survives, but those two
original waves cease to exist never to be seen
again.

I must have missed class the day they went over the theory of
"cancellation".

You must have. Please run the above java application
and alleviate your ignorance about what you missed. Why
do the waves disappear when they are of equal magnitude
and opposite phase?

I stand 100% behind my two messages to Walt. If you actually read them
you would note that I said for most cases it makes no difference whether
the waves interfere forever or whether they interact and "cancel".

Of course it makes all the difference in the world. That's
what the entire argument is all about. You simply cannot
sweep the truth under the "does not matter" rug. And until
you can say "all cases" instead of "most cases" your
argument is irrelevant. If it doesn't work for all, it
doesn't work at all.

The bottom line is that EM waves do not interact in free space.

It is indeed difficult to get two beams of light collinear
in space space. But it is not difficult at all to get two
RF waves collinear in a transmission line. It happens every
time someone adjusts his antenna tuner for a Z0-match.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Keith Dysart wrote:
Do Born and Wolf offer crisp definitions of the boundaries
between coherent, partially coherent, and mutually incoherent?
Or is it a continuum arbitrarily divided into 3 regions for
the purposes of discussion?

The IEEE Dictionary has an interesting definition for
"degree of coherence". Imax means intensity maximum
and Imin means intensity minimum.

Visibility = (Imax-Imin)/(Imax+Imin)

Light is considered "highly coherent" when Visibility
exceeds 0.88 and "partially coherent" when Visibility
is less than 0.88. Incoherent for "very small values"
of Visibility.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

On Apr 8, 7:09 pm, Cecil Moore wrote:
Keith Dysart wrote:
Do Born and Wolf offer crisp definitions of the boundaries
between coherent, partially coherent, and mutually incoherent?
Or is it a continuum arbitrarily divided into 3 regions for
the purposes of discussion?

The IEEE Dictionary has an interesting definition for
"degree of coherence". Imax means intensity maximum
and Imin means intensity minimum.

Visibility = (Imax-Imin)/(Imax+Imin)

Light is considered "highly coherent" when Visibility
exceeds 0.88 and "partially coherent" when Visibility
is less than 0.88. Incoherent for "very small values"
of Visibility.

This is good; a continuum with high coherence at one end, low
coherence at the other and medium in the middle, and, of course,
since the ends are infinitely small, no such thing as perfect
coherence or "NO" coherence (at least in the real world).

This then takes us back to the original point; there is always
some interference, though it may be small enough that an
engineer does not find it of interest.

....Keith

Constructive interference in radiowave propagation
 

Cecil Moore wrote:
Gene Fuller wrote:
It is easy to give examples where the waves survive the superposition,
because they always do. It is rather strange that you are making this
argument after all the back and forth about traveling waves and
standing waves. Do we now have multiple flavors of EM waves? Some that
obey superposition and some that don't?

They all obey superposition which can occur with or
without interference. And you are wrong about all
waves surviving superposition. Canceled waves do
not survive wave cancellation in the direction that
they are traveling. Access this web page and set
the two waves to equal frequencies, equal magnitudes,
and opposite phases, i.e. 0 and 180 degrees.

http://micro.magnet.fsu.edu/primer/j...ons/index.html


When you do that, the waves are canceled in their
original direction of travel. The energy in those
canceled waves certainly survives, but those two
original waves cease to exist never to be seen
again.

I must have missed class the day they went over the theory of
"cancellation".

You must have. Please run the above java application
and alleviate your ignorance about what you missed. Why
do the waves disappear when they are of equal magnitude
and opposite phase?


[snip]


Cecil,

That's really funny. A grad student and a programmer put together a
simply java applet to try to illustrate the concept of interference, and
you treat it as a new bible. I bet the authors would be appalled by your
interpretation.

By the way, did you look beyond the pretty pictures and read the section
where the authors said,

"All of the wave examples presented in Figure 1 portray waves
propagating in the same direction, but in many cases, light waves
traveling in different directions can briefly meet and undergo
interference. After the waves have passed each other, however, they will
resume their original course, having the same amplitude, wavelength, and
phase that they had before meeting."


Hmmm, I think that is exactly what I said in this thread on RRAA.



73,
Gene
W4SZ

Constructive interference in radiowave propagation
 

Keith Dysart wrote:
This is good; a continuum with high coherence at one end, low
coherence at the other and medium in the middle, and, of course,
since the ends are infinitely small, no such thing as perfect
coherence or "NO" coherence (at least in the real world).

But remember that definition is for fiber optics
sources, not amateur radio sources. Coherency
in amateur radio systems can get as close as
a zero reading on a reflected power meter.

Still there are those nagging assertions of Born
and Wolf that for two equal magnitude signals,
the total intensity possible for incoherent
signals is double the intensity of one signal.
The total intensity possible for coherent
signals is four times the intensity of one
signal.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Gene Fuller wrote:
That's really funny. A grad student and a programmer put together a
simply java applet to try to illustrate the concept of interference, and
you treat it as a new bible. I bet the authors would be appalled by your
interpretation.

One more example of an ignorant person making fun of something
he doesn't understand. One of those signals is s11(a1). The
other is s12(a2). Added together they equal zero. That's the
S-Parameter equation for reflections toward the source.

b1 = s11(a1) + s12(a2) = 0

If s11, a1, s12, and a2 are all not zero, the above equation
describes wave cancellation, something you say never happens.

By the way, did you look beyond the pretty pictures and read the section
where the authors said,

"All of the wave examples presented in Figure 1 portray waves
propagating in the same direction, but in many cases, light waves
traveling in different directions can briefly meet and undergo
interference. After the waves have passed each other, however, they will
resume their original course, having the same amplitude, wavelength, and
phase that they had before meeting."

Yes, that happens "in many cases" but NOT IN ALL CASES. You
apparently missed the point which is the part where they said:

"... when two waves of equal amplitude and wavelength that are
180-degrees ... out of phase with each other meet, they are not
actually annihilated, ... All of the photon energy present in
these waves must somehow be recovered or redistributed in a
new direction, according to the law of energy conservation ...
Instead, upon meeting, the photons are redistributed to regions
that permit constructive interference, so the effect should be
considered as a redistribution of light waves and photon energy
rather than the spontaneous construction or destruction of light."

Hmmm, I think that is exactly what I said in this thread on RRAA.

No, what you have said on RRAA is that wave cancellation never
happens because wave cancellation doesn't occur in many
cases. That is obviously faulty logic and all it takes to prove
you wrong is one case of wave cancellation. That case happens
every time a ham adjusts his antenna tuner for zero reflected
power. If we consider the java ap as the reflected waves flowing
toward the source, setting them to 0 and 180 degrees is exactly
what happens at the antenna tuner.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

On Mon, 09 Apr 2007 02:45:52 GMT, Cecil Moore
wrote:

All of the photon energy present in
these waves must somehow be recovered or redistributed in a
new direction, according to the law of energy conservation

New heights of sheer stupidity. This sounds like bingo night in the
church basement where no one actually loses any money, it just gets
shuffled around.

"Recovered OR redistributed" ... this certainly qualifies for next
year's Oscar for chuckles. Luckily the Nobel committee doesn't follow
Hollywood or they would have awarded Mighty Mouse the Physics award
for antigravity (certainly 50 million children's admiration couldn't
lead them astray on this choice!).

Constructive interference in radiowave propagation
 

I'd vowed that I wouldn't hit this tarbaby yet again. But here I go.

Among the junk science being bandied about here is the following
supposition:

Suppose you have beams from two identical coherent lasers which, by a
system of (presumably partially reflective and partially transmissive)
mirrors, are made to shine in exactly the same direction from the same
point (which I'll call the "summing point"). Further, suppose that the
paths from the two lasers to this summing point differ by an odd number
of half wavelengths. So beyond the summing point, where the laser beams
exactly overlie each other, there is no beam because the two exactly
cancel. Or, in other words, the sum of the two superposed fields is
zero. The recurring argument is that because each laser is producing
energy and yet there is no net field and therefore no energy in the
summed beams, something strange has happened at the summing point (or
"virtual short circuit"), and creative explanations are necessary to
account for the "missing energy". One such proposed explanation is that
the mere meeting of the two beams is the cause of some kind of a
reflection of energy, and that each wave somehow detects and interacts
with the other.

Well, here's what I think. I think that no one will be able to draw a
diagram of such a summing system which doesn't also produce, due solely
to the reflection and transmission of the mirrors, a beam or beams
containing exactly the amount of energy "missing" from the summed beam.
No interaction(*) of the two beams at or beyond the summing point is
necessary to account for the "missing" energy -- you'll find it all at
other places in the system. Just as you do in a phased antenna array,
where the regions of cancelled field are always accompanied by
complementary regions of reinforced field. Somewhere, in some bounce
from a mirror or pass through it, the beams will end up reinforcing each
other is some other direction. My challenge is this: Sketch a system
which will produce this summation of out-of-phase beams, showing the
reflectivity and transmissivity of each mirror, and showing the beams
and their phases going in all directions from the interactions from each
mirror. Then show that simple interaction of the beams with the mirrors
is insufficient to account for the final distribution of energy.

Next, do the same for a transmission line. Show how two coherent
traveling waves can be produced which will propagate together in the
same direction but out of phase with each other, resulting in a net zero
field at all points beyond some summing point. But also calculate the
field from waves reflected at the summing point and elsewhere in the
system due to simple impedance changes. Show that this simple analysis,
assuming no interaction between the traveling waves, is insufficient to
account for all the energy. A single case will do.

Until someone is able to do this, I'll stand firm with the unanimous
findings of countless mathematical and practical analyses which show
superposition of and no interaction between waves or fields in a linear
medium.

(*) By "interaction" I mean that one beam or wave has an effect on the
other, altering it in some way -- for example, causing it to change
amplitude, phase, orientation, or direction. I'm not including
superposition, that is the fact that the net field of the two waves is
the sum of the two, in the meaning of "interaction".

Roy Lewallen, W7EL

Constructive interference in radiowave propagation
 

Richard Clark wrote:
Cecil Moore wrote:
All of the photon energy present in
these waves must somehow be recovered or redistributed in a
new direction, according to the law of energy conservation

New heights of sheer stupidity. This sounds like bingo night in the
church basement where no one actually loses any money, it just gets
shuffled around.

One more example of an ignorant person making fun of
things he doesn't understand. The principle of the
conservation of energy indeed states that energy is
not gained or lost - it just gets shuffled around.
When EM wave cancellation occurs, the energy remains
in EM wave form and simply gets redistributed.

As we tune our antenna tuners while watching the
reflected power indication, we are varying the
magnitude and phase of the two component reflected
waves at the tuner input. When those two component
waves are adjusted to equal magnitudes and opposite
phase, reflections toward the source are obviously
canceled since they can be measured as going to zero
in the direction of the source. The following S-Parameter
equation describes the final outcome toward the source.

b1 = s11(a1) + s12(a2) = 0, total destructive interference

Since energy cannot be destroyed by an antenna tuner
and since there are only two directions available, in
a lossless system, the energy in waves canceled toward
the source must be redistributed (re-reflected) to waves
traveling toward the load as constructive interference.
That's the other S-Parameter equation.

b2 = s21(a1) + s22(a2) = Vfor/SQRT(Z0)

Square those equations and you get the power equations.
|b1|^2 = net reflected power = 0
|b2|^2 = net forward power = all the available power
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

"Cecil Moore"
Still there are those nagging assertions of Born
and Wolf that for two equal magnitude signals,
the total intensity possible for incoherent
signals is double the intensity of one signal.
The total intensity possible for coherent
signals is four times the intensity of one signal.
________

It is a fairly common practice in broadcast designs to combine the outputs
of two r-f amplifiers of equal power rating, using a 4-port, 3 dB coaxial
hybrid. The two amplifiers are driven by a single exciter through a
suitable splitter. The antenna connects to one output port of the hybrid,
and the other output port is connected to a dummy load.

When the relative r-f phases of the two txs are suitably set, the antenna
connection of the hybrid receives the total output power of the two txs, and
the dummy load port receives zero. When the relative r-f phases of the txs
are changed by 90 degrees from that setting, then the conditions at the
output ports are reversed.

The total average power available at the hybrid output for both of these
conditions is twice that of a single tx without the hybrid.

Does the quote from Born and Wolf support this?

RF

Constructive interference in radiowave propagation
 

Roy Lewallen wrote:
Next, do the same for a transmission line. Show how two coherent
traveling waves can be produced which will propagate together in the
same direction but out of phase with each other, resulting in a net zero
field at all points beyond some summing point.

Roy, it is done all the time represented by the
S-Parameter equation for the reflected wave
toward the source. Assuming an impedance discontinuity
in a transmission line:

b1 = s11(a1) + s12(a2) = 0

If b1=0, then s11(a1) and s12(a2) are indeed two coherent
traveling waves propagating together in the same direction
but out of phase with each other. The net reflected wave
field is zero.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Cecil Moore wrote:


One more example of an ignorant person making fun of something
he doesn't understand.

Cecil,

Sorry, I simply cannot keep up with you. We seem to be switching from RF
to optics in order to explain something or other, and now we are
switching from optics to S-parameters to explain the explanation.

You accuse me of saying that waves never "cancel", right after I told
you that "cancel" is not a good description in detailed technical
analysis. Do you actually read anything here?

Going back to your oft-stated line about Galileo, you seem to be on the
other side now. You keep insisting on a Cecil-centric universe, with the
consequent requirement to add more and more crystalline spheres to
explain all of your "creative" ideas.

Stick to the basic EM understanding that has stood for more than 100
years. No need to keep inventing new principles. The old ones work just
fine.

(You must be getting desperate; the ad hominem attacks have started.) 8-)

73,
Gene
W4SZ

Constructive interference in radiowave propagation
 

Richard Fry wrote:
The total average power available at the hybrid output for both of these
conditions is twice that of a single tx without the hybrid.

Does the quote from Born and Wolf support this?

Yes, there is obviously no interference between the
two transmitters if the powers simply add together.
You have doubled the current capability without
doubling the voltage capability - that's not
interference.

For interference to occur, both the E-fields and
the H-fields must be superposed at the same time
such that both fields increase or decrease by the
same percentage. So how do we double the voltage
and double the current in the 50 ohm transmission
line to the antenna?

Put a transformer on the output of each transmitter.
Wire the secondaries in series. We have doubled the
voltage output. Now drive the 50 ohm load with that
doubled voltage and see what happens. This is equivalent
to constructive interference. Hope you are running
the transmitters at half power. :-)
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Gene Fuller wrote:
Sorry, I simply cannot keep up with you. We seem to be switching from RF
to optics in order to explain something or other, and now we are
switching from optics to S-parameters to explain the explanation.

So are you saying the physics concepts from the field
of optics are wrong? Are you saying the S-Parameter
equations are invalid? If not, seems you are having
a hard time defending your concepts against those
valid concepts. That should tell you something about
your (simplified short cut) concepts.

I'm going to keep it up until you give up on the notion
that you already know everything and therefore reality
obeys your every whim. You cannot dismiss wave cancellation
simply because you find that part of reality distasteful.

In analyzing an impedance
discontinuity in a transmission line, the S-Parameter
equations are accepted as valid. b1 is the normalized
reflected voltage toward the source.

b1 = s11(a1) + s12(a2) = 0

When b1 is zero, waves s11(a1) and s12(a2) have been
canceled because they are of equal magnitude and opposite
phase. Simple cause and effect - not rocket science.

If you can prove those concepts from the field of optics
are invalid and that an S-Parameter analysis is invalid,
now would be the time.

You accuse me of saying that waves never "cancel", right after I told
you that "cancel" is not a good description in detailed technical
analysis. Do you actually read anything here?

Wave cancellation occurs all the time, Gene. Every time
a ham tunes his antenna tuner for zero reflected power,
he has caused two reflected waves to cancel. I am amazed
at how many otherwise intelligent posters to this newsgroup
attempt to engage in the copout of sweeping under the rug
anything that they do not understand and/or don't want to
deal with.

Stick to the basic EM understanding that has stood for more than 100
years.

That's exactly what I am doing. You seem to be relatively
ignorant of those century old concepts. I don't know when
interference was first explained but it was long before
you and I were born. I am repeating principles that have
been around for a century including the wave reflection
model.

What has actually happened is that seductive short cuts
have left many ignorant of the basic principles, e.g.
Standing waves can exist without reverse traveling waves.
Reverse traveling waves exist without a source of energy.
Waves never interact. etc. etc. etc.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Gene Fuller wrote:
We seem to be switching from RF
to optics in order to explain something or other, ...

Gene, I forgot to ask. At exactly what EM frequency
do the RF waves stop obeying the century old laws of
physics for visible light?
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

On Apr 8, 10:20 pm, Cecil Moore wrote:
Keith Dysart wrote:
This is good; a continuum with high coherence at one end, low
coherence at the other and medium in the middle, and, of course,
since the ends are infinitely small, no such thing as perfect
coherence or "NO" coherence (at least in the real world).

But remember that definition is for fiber optics
sources, not amateur radio sources. Coherency
in amateur radio systems can get as close as
a zero reading on a reflected power meter.

Still there are those nagging assertions of Born
and Wolf that for two equal magnitude signals,
the total intensity possible for incoherent
signals is double the intensity of one signal.
The total intensity possible for coherent
signals is four times the intensity of one
signal.

I take this to mean that with largely incoherent sources
the intensity is doubled everywhere.

With largely coherent sources, the average intensity is
doubled everywhere, but there is also a spatial distribution
where the peak intensity is 4 times, but the minimum is zero
(thus the same average of two).

....Keith

Constructive interference in radiowave propagation
 

On Apr 9, 9:04 am, Cecil Moore wrote:
Richard Fry wrote:
The total average power available at the hybrid output for both of these
conditions is twice that of a single tx without the hybrid.

Does the quote from Born and Wolf support this?

Yes, there is obviously no interference between the
two transmitters if the powers simply add together.
You have doubled the current capability without
doubling the voltage capability - that's not
interference.

For interference to occur, both the E-fields and
the H-fields must be superposed at the same time
such that both fields increase or decrease by the
same percentage. So how do we double the voltage
and double the current in the 50 ohm transmission
line to the antenna?

Except that Richard's description sure seems to meet the
requirements of coherency. Can you offer a way for use
to know whether two signals are coherent?

Secondly, I am at a complete loss to understand how you
can be arguing that when two signals of a particular
power interfere, the result is 4 times the power. This
sure seems like you're getting something from nothing.

What happened to the staunch acceptance of 'conservation of
energy'?

....Keith

Constructive interference in radiowave propagation
 

On Apr 9, 3:35 am, Roy Lewallen wrote:
I'd vowed that I wouldn't hit this tarbaby yet again. But here I go.

Among the junk science being bandied about here is the following
supposition:

Suppose you have beams from two identical coherent lasers which, by a
system of (presumably partially reflective and partially transmissive)
mirrors, are made to shine in exactly the same direction from the same
point (which I'll call the "summing point"). Further, suppose that the
paths from the two lasers to this summing point differ by an odd number
of half wavelengths. So beyond the summing point, where the laser beams
exactly overlie each other, there is no beam because the two exactly
cancel. Or, in other words, the sum of the two superposed fields is
zero. The recurring argument is that because each laser is producing
energy and yet there is no net field and therefore no energy in the
summed beams, something strange has happened at the summing point (or
"virtual short circuit"), and creative explanations are necessary to
account for the "missing energy". One such proposed explanation is that
the mere meeting of the two beams is the cause of some kind of a
reflection of energy, and that each wave somehow detects and interacts
with the other.

Well, here's what I think. I think that no one will be able to draw a
diagram of such a summing system which doesn't also produce, due solely
to the reflection and transmission of the mirrors, a beam or beams
containing exactly the amount of energy "missing" from the summed beam.
No interaction(*) of the two beams at or beyond the summing point is
necessary to account for the "missing" energy -- you'll find it all at
other places in the system. Just as you do in a phased antenna array,
where the regions of cancelled field are always accompanied by
complementary regions of reinforced field. Somewhere, in some bounce
from a mirror or pass through it, the beams will end up reinforcing each
other is some other direction. My challenge is this: Sketch a system
which will produce this summation of out-of-phase beams, showing the
reflectivity and transmissivity of each mirror, and showing the beams
and their phases going in all directions from the interactions from each
mirror. Then show that simple interaction of the beams with the mirrors
is insufficient to account for the final distribution of energy.

Next, do the same for a transmission line. Show how two coherent
traveling waves can be produced which will propagate together in the
same direction but out of phase with each other, resulting in a net zero
field at all points beyond some summing point. But also calculate the
field from waves reflected at the summing point and elsewhere in the
system due to simple impedance changes. Show that this simple analysis,
assuming no interaction between the traveling waves, is insufficient to
account for all the energy. A single case will do.

Until someone is able to do this, I'll stand firm with the unanimous
findings of countless mathematical and practical analyses which show
superposition of and no interaction between waves or fields in a linear
medium.

(*) By "interaction" I mean that one beam or wave has an effect on the
other, altering it in some way -- for example, causing it to change
amplitude, phase, orientation, or direction. I'm not including
superposition, that is the fact that the net field of the two waves is
the sum of the two, in the meaning of "interaction".

Roy Lewallen, W7EL

Constructive interference in radiowave propagation
 

Keith Dysart wrote:
I take this to mean that with largely incoherent sources
the intensity is doubled everywhere.

Being a little more precise:
With mutually incoherent equal-magnitude sources,
the maximum possible peak intensity is double the
intensity of a single wave. There's no interference.

For "largely incoherent sources", the peak intensity
would be slightly more than double.

With largely coherent sources, the average intensity is
doubled everywhere, but there is also a spatial distribution
where the peak intensity is 4 times, but the minimum is zero
(thus the same average of two).

With mutually coherent equal-magnitude sources,
the maximum possible peak intensity is four times
the intensity of a single wave, i.e. there is total
constructive interference. (This can happen
at a Z0-match in an RF transmission line.)

For "largely coherent sources" the peak intensity
would be slightly less than four times.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

On Apr 9, 3:35 am, Roy Lewallen wrote:
I'd vowed that I wouldn't hit this tarbaby yet again. But here I go.

Thanks for doing so.

The junk science is often presented with very rational sounding
arguments and it can be difficult to detect the flaws. This example
was a case for me and you expose the flaw nicely.

What I have difficulty with is deciding on the value of the junk
science. On the one hand it misleads many; on the other, debunking
provides opportunity to develop deeper understanding.

When I first started lurking on this group many, many years ago,
I didn't even know that there was a question about "where does the
reflected power go". Following the debates and iterating to the
correct answers has been extremely educational, much more so than
just accepting the correct explanations without question.

The promoters of junk science fulfill an important role in this
process and I can't decide if their net effect is good or bad.
The bad effects are, of course, when they successfully lead
others astray.

On the whole, good or bad? I haven't decided.

....Keith

Constructive interference in radiowave propagation
 

"Cecil Moore"
Being a little more precise:
With mutually incoherent equal-magnitude sources,
the maximum possible peak intensity is double the
intensity of a single wave. There's no interference.
____________

BUT, the two equal-power signals in my scenario are exactly coherent at the
output port where they combine to produce twice average output power of
either tx.

Please explain, in light of your concepts?

RF

Constructive interference in radiowave propagation
 

Keith Dysart wrote:
Except that Richard's description sure seems to meet the
requirements of coherency. Can you offer a way for use
to know whether two signals are coherent?

If we can cause them to interfere, that is proof that
they are coherent. If their reflected waves superpose
with destructive or constructive interference present,
then they are coherent.

Secondly, I am at a complete loss to understand how you
can be arguing that when two signals of a particular
power interfere, the result is 4 times the power. This
sure seems like you're getting something from nothing.

Maybe it seems that way to you but it's because of
confusion. If the intensity of each of two coherent
waves is one watt/unit-area and total destructive
interference occurs at one point, two watts/unit-area
have seemingly disappeared. Since energy cannot disappear,
those two watts/unit-area must appear as constructive
interference somewhere else. If total constructive
interference appears somewhere else, the total intensity
is one watt/unit-area from the first wave, one watt/unit-area
from the second wave, plus the two watts/unit-area from
the total destructive interference. Adding all those
intensities up gives us four watts/unit-area at the point
of total constructive interference. The intensity at
the total destructive interference point plus the
intensity at the total constructive interference point
still averages out to two watts/unit-area, exactly the
intensities in the original waves and exactly what it
takes to satisfy the conservation of energy principle.

Your "something for nothing" doesn't exist. In the absence
of a local source, any constructive interference must be
offset by an equal magnitude of destructive interference.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Richard Fry wrote:
"Cecil Moore"
Being a little more precise:
With mutually incoherent equal-magnitude sources,
the maximum possible peak intensity is double the
intensity of a single wave. There's no interference.
____________

BUT, the two equal-power signals in my scenario are exactly coherent at
the output port where they combine to produce twice average output power
of either tx.

Please explain, in light of your concepts?

I thought I did. You have combined those outputs
without producing any interference. If there is
no interference, as far as power goes, it doesn't
matter if they are coherent or not.

Assuming P1 is the power output of the first
transmitter and P2 is the power output of the
second transmitter, if Ptotal = P1 + P2 + 0,
the interference (last) term is obviously zero,
i.e. it's prima facie - no interference.

If you double the H-field (current) while keeping
the E-field (voltage) constant, you have not
produced any interference.

If you double the E-field (voltage) while keeping
the H-field (current) constant, you have not
produced any interference.

If you double both the E-field (voltage) and the
H-field (current) at the same time, you have produced
total constructive interference.

If you zero both the E-field (voltage) and the
H-field (current) at the same time, you have produced
total destructive interference.

For interference to exist, both the E-field and
H-field must be changed by the same percentage
thus keeping their Z0 ratio constant.

If both fields decrease, extra energy is available
during that destructive interference event. If both
fields increase, extra energy is required by that
constructive interference event. In the absence
of a local source, |destructive interference| must
always equal |constructive interference|.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Keith Dysart wrote:
The junk science is often presented with very rational sounding
arguments and it can be difficult to detect the flaws. This example
was a case for me and you expose the flaw nicely.

Hint to omniscient gurus: One cannot use ignorance
for exposing flaws. Roy says in his Food for Thought
article:

I personally don't have a compulsion to understand where this power "goes".

Seemingly, that feeling of his is supposed to be enough
incentive to discourage the rest of us to give up on our
quest for tracking the energy through the system.

Roy has ploinked me for disagreeing with him. What
does that say about his inability to technically
defend his concepts?

The S-Parameter equations completely debunk what Roy
posted.

b1 = s11(a1) + s12(a2) = 0

|b1|^2, the reflected power, equals zero because
of wave cancellation involving those components
of a1 (forward normalized voltage) and a2 (reflected
normalized voltage).

If s11, a1, s12, and a2 are all non-zero, then wave
cancellation has occurred between s11(a1) and s12(a2)
proving Roy's statements to be false. The above wave
cancellation happens every time a ham adjusts his
antenna tuner for zero reflected power.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

On 9 Apr 2007 07:08:56 -0700, "Keith Dysart" wrote:

On Apr 9, 9:04 am, Cecil Moore wrote:
Yes, there is obviously no interference between the
two transmitters if the powers simply add together.

is more than slightly disconnected from:

Secondly, I am at a complete loss to understand how you
can be arguing that when two signals of a particular
power interfere, the result is 4 times the power. This
sure seems like you're getting something from nothing.

What happened to the staunch acceptance of 'conservation of
energy'?

Called discarding the baby with the bath water. Such is the fate of
observed reality in the face of novel theories.

This specious argument (posing quadruple powers to confound
expectations) demands the slight of hand rhetoric in that when clear
minds consider the entire field of radiation is summed, then you get
the average of reinforced energy and canceled energy which is:
2

This merely reinforces yet another observed (but rhetorically
dismissed) reality:
Richard Fry wrote:
The total average power available at the hybrid output for both of these
conditions is twice that of a single tx without the hybrid.

73's
Richard Clark, KB7QHC

Constructive interference in radiowave propagation
 

Richard Clark wrote:
This specious argument (posing quadruple powers to confound
expectations) demands the slight of hand rhetoric ...

Do you really think that Born and Wolf engage in
"slight of hand rhetoric"? Quoting them speaking
of the combined intensity of two equal magnitude
waves, I1 and I2.

"... and the intensity varies between a maximum
value Imax = 4*I1, and a minimum value Imin = 0
(Fig. 7.1)."

Fig. 7.1 is a sinusoid with a varying amplitude
from 0 to 4*I1 on the Y axis and relative phase
angle plotted on the X axis.

The average intensity is, of course, 2*I1 in
accordance with the conservation of energy
principle. Such is the nature of constructive
and destructive interference.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Cecil Moore wrote:
Richard Fry wrote:
"Cecil Moore"
Being a little more precise:
With mutually incoherent equal-magnitude sources,
the maximum possible peak intensity is double the
intensity of a single wave. There's no interference.
____________

BUT, the two equal-power signals in my scenario are exactly coherent
at the output port where they combine to produce twice average output
power of either tx.

Please explain, in light of your concepts?

I thought I did. You have combined those outputs
without producing any interference. If there is
no interference, as far as power goes, it doesn't
matter if they are coherent or not.

Assuming P1 is the power output of the first
transmitter and P2 is the power output of the
second transmitter, if Ptotal = P1 + P2 + 0,
the interference (last) term is obviously zero,
i.e. it's prima facie - no interference.

If you double the H-field (current) while keeping
the E-field (voltage) constant, you have not
produced any interference.

If you double the E-field (voltage) while keeping
the H-field (current) constant, you have not
produced any interference.

If you double both the E-field (voltage) and the
H-field (current) at the same time, you have produced
total constructive interference.

If you zero both the E-field (voltage) and the
H-field (current) at the same time, you have produced
total destructive interference.

For interference to exist, both the E-field and
H-field must be changed by the same percentage
thus keeping their Z0 ratio constant.

If both fields decrease, extra energy is available
during that destructive interference event. If both
fields increase, extra energy is required by that
constructive interference event. In the absence
of a local source, |destructive interference| must
always equal |constructive interference|.

Cecil,

You are quite prolific at manufacturing even more crystalline spheres.

8-)


73,
Gene
W4SZ

Constructive interference in radiowave propagation
 

Cecil Moore wrote:
Gene Fuller wrote:
We seem to be switching from RF to optics in order to explain
something or other, ...

Gene, I forgot to ask. At exactly what EM frequency
do the RF waves stop obeying the century old laws of
physics for visible light?

Cecil,

It is interesting that you rarely address the technical points that I
make. It is usually something along the lines of the above throwaway
comments.

What are you trying to hide?

I am still waiting to learn the technical details of "cancellation",
including the proper units and the characteristic equations. There has
been deafening silence in response to my similar query about
"interference".

I have been called lots of things in my life, mostly deserved, but
rarely have I been called ignorant. It is obvious that this thread has
long outlived any chance for a meaningful discussion.

You win! Keep on building more spheres.

73,
Gene
W4SZ

Constructive interference in radiowave propagation
 

Gene Fuller wrote:
You are quite prolific at manufacturing even more crystalline spheres.
8-)

I just love the technical content of your postings. :-)
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

On Mon, 09 Apr 2007 17:17:40 GMT, Cecil Moore
wrote:

The average intensity is, of course, 2*I1

Which satisfies (after a flurry of reference shuffling, oblique
jargon, obscure rhetoric, and ordinary BAFFLEGAB) the simple inquiry:
The total average power available at the hybrid output for both of these
conditions is twice that of a single tx without the hybrid.

Does the quote from Born and Wolf support this?

Constructive interference in radiowave propagation
 

On Mon, 09 Apr 2007 18:01:05 GMT, Gene Fuller
wrote:

You are quite prolific at manufacturing even more crystalline spheres.

Hi Gene,

Wait until they are embroidered with epicycles.

73's
Richard Clark, KB7QHC

Constructive interference in radiowave propagation
 

Richard Fry wrote:

"Cecil Moore"

Still there are those nagging assertions of Born
and Wolf that for two equal magnitude signals,
the total intensity possible for incoherent
signals is double the intensity of one signal.
The total intensity possible for coherent
signals is four times the intensity of one signal.

________

It is a fairly common practice in broadcast designs to combine the
outputs of two r-f amplifiers of equal power rating, using a 4-port, 3
dB coaxial hybrid. The two amplifiers are driven by a single exciter
through a suitable splitter. The antenna connects to one output port of
the hybrid, and the other output port is connected to a dummy load.

When the relative r-f phases of the two txs are suitably set, the
antenna connection of the hybrid receives the total output power of the
two txs, and the dummy load port receives zero. When the relative r-f
phases of the txs are changed by 90 degrees from that setting, then the
conditions at the output ports are reversed.

The total average power available at the hybrid output for both of these
conditions is twice that of a single tx without the hybrid.

Does the quote from Born and Wolf support this?

RF

The quote from Born and Wolf that Cecil cites supports the 'profound'
notion that (E1 + E2)^2 / (E1 + E2) = 4.

73, Jim AC6XG

Constructive interference in radiowave propagation
 

Gene Fuller wrote:
I am still waiting to learn the technical details of "cancellation",
including the proper units and the characteristic equations. There has
been deafening silence in response to my similar query about
"interference".

I have posted the details of cancellation more than
once and you haven't disagreed except in an ad
hominem way with zero technical content.

Once again here is the S-Parameter equation for
the reflected wave at a Z0-match impedance
discontinuity in a transmission line:

b1 = s11(a1) + s12(a2) = 0

b1 is the normalized reflected voltage at the
Z0-match. |b1|^2 is the reflected power in
the direction of the source.

s11(a1) is the normalized external reflected
voltage at the Z0-match.

s12(a2) is the normalized internal reflected
voltage making it through the Z0-match back
from the mismatched load.

Since b1 = 0, s11(a1) and s12(a2) have canceled.
s11(a1) and s12(a2) are of equal magnitudes and
opposite phases. Those waves are canceled toward
the source. Their combined energy components join
the forward wave toward the load.

Wave cancellation is what happens when a ham tunes
his antenna tuner for zero reflected power. I dare
say you have engaged in that very behavior leading
up to wave cancellation of reflections in the
direction of the source.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Jim Kelley wrote:
The quote from Born and Wolf that Cecil cites supports the 'profound'
notion that (E1 + E2)^2 / (E1 + E2) = 4.

There sure appears to be something wrong with that
equation, Jim. If E1 = E2 = 1 volt, then you have
4 / 2 = 4 which seems a bit wrong, if you don't
mind me saying so. What dimensions does your '4'
above have? Seems it would have to be 4 volts.

Assuming E1 = E2, I think what you meant to say was:

(E1 + E2)^2 / E1^2 = 4 (dimensionless)

which is what Born and Wolf say in equation (17)
chapter 7. Please note that is total constructive
interference as defined by Hecht in "Optics"

Also please note that if those signals are the opposite
phase:

(E1 - E2)^2 / E1^2 = 0

That's total destructive interference as defined
by Hecht.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

Cecil Moore wrote:
Richard Clark wrote:

Walter Maxwell wrote:

It is true, however, that two non-coherent fields from two different
sources would just plow through each
other with no effect on either.


Does the binary transition from a one micro-degree longer
cycle (non-coherent) to 0 (coherence) same length cycle really
plunge us into a new physical reality of waves colliding with
rebounds and caroms where formerly there was absolutely no
interaction before?


Of course, you are being facetious but the answer is simple.
If the two signals are mutually incoherent, they don't
interfere. Permanent wave cancellation is impossible
between two waves that are not coherent. Hecht in "Optics"
devotes an entire chapter to the "Basics of Coherence Theory".
So do Born and Wolf in "Principles of Optics". Here is what
Walt was obviously saying except in Born and Wolf's words:

"If the two beams originate in the same source, the fluctuations
in the two beams are in general correlated, and the beams are
said to be completely or partially *coherent* depending on
whether the correlation is complete or partial. In beams from
different sources, the fluctuations are completely uncorrelated,
and the beams are said to be mutually *incoherent*. When such
beams from different sources are superposed, no interference is
observed under ordinary experimental conditions, the total intensity
being everywhere the sum of the intensities of the individual beams."

In case you missed it, that says *NO INTERFERENCE* between mutually
incoherent waves. Seems reasonable to say that "no interference"
means the same thing as "no effect".

The last paragraph above is a perfect example of your tendancy to
misinterpret these texts, Cecil. Born and Wolf does not (and would
not) assert that there is "no effect" when mutually incoherent waves
are superposed. It's not reasonable to say that. There is certainly
an effect. In fact Walt and I use the effect whenever we tune our basses.

An illustration can be viewed at:

http://www.kettering.edu/~drussell/D...rposition.html

73, Jim AC6XG

Constructive interference in radiowave propagation
 

Cecil Moore wrote:

With mutually coherent equal-magnitude sources,
the maximum possible peak intensity is four times
the intensity of a single wave, i.e. there is total
constructive interference. (This can happen
at a Z0-match in an RF transmission line.)

For "largely coherent sources" the peak intensity
would be slightly less than four times.

So according to your theory I can take a 1 watt laser, split the beam
into two coherent beams, recombine the beams in-phase together along
the same path thus creating constructive interference, and obtain 2
watts of laser power. Or would it be 4 watts?

:-)

ac6xg

Constructive interference in radiowave propagation
 

Cecil Moore wrote:

Jim Kelley wrote:

The quote from Born and Wolf that Cecil cites supports the 'profound'
notion that (E1 + E2)^2 / (E1 + E2) = 4.


There sure appears to be something wrong with that
equation, Jim. If E1 = E2 = 1 volt, then you have
4 / 2 = 4 which seems a bit wrong, if you don't
mind me saying so. What dimensions does your '4'
above have? Seems it would have to be 4 volts.

Assuming E1 = E2, I think what you meant to say was:

(E1 + E2)^2 / E1^2 = 4 (dimensionless)

which is what Born and Wolf say in equation (17)
chapter 7. Please note that is total constructive
interference as defined by Hecht in "Optics"

Also please note that if those signals are the opposite
phase:

(E1 - E2)^2 / E1^2 = 0

That's total destructive interference as defined
by Hecht.

Right. So do you get the point, or not?

73, ac6xg

Constructive interference in radiowave propagation
 

Jim Kelley wrote:
The last paragraph above is a perfect example of your tendancy to
misinterpret these texts, Cecil. Born and Wolf does not (and would not)
assert that there is "no effect" when mutually incoherent waves are
superposed. It's not reasonable to say that. There is certainly an
effect. In fact Walt and I use the effect whenever we tune our basses.

One wave has no effect on the other wave, Jim.
Please pay attention.
--
73, Cecil http://www.w5dxp.com

Constructive interference in radiowave propagation
 

"Keith Dysart" wrote
The promoters of junk science fulfill an important role in this
process and I can't decide if their net effect is good or bad.
The bad effects are, of course, when they successfully lead
others astray.

On the whole, good or bad? I haven't decided.

Keith, I really like when junk science gurus exibit
constructive interference, the bafflegab sums to
four times the normal power!

Mike W5CHR

Constructive interference in radiowave propagation
 

Jim Kelley wrote:
So according to your theory I can take a 1 watt laser, split the beam
into two coherent beams, recombine the beams in-phase together along the
same path thus creating constructive interference, and obtain 2 watts of
laser power. Or would it be 4 watts?

If it were total constructive interference, two 1/2W
beams would yield an intensity of 2 watts. Of course,
at another location, total destructive interference
would have to occur where the intensity was zero.

Put another way, the bright interference rings would
contain two watts per unit area while the flat black
rings would contain zero watts per unit area thus
averaging out to the original total of one watt.

As I said before, this ain't rocket science.
--
73, Cecil http://www.w5dxp.com