Cecil Moore wrote:
Jim Kelley wrote:

The peak intensity of a standing wave will always be greater than the
simple sum of the two waves.


No reference to peaks here, Jim. Everything is average values.

Evidently then you haven't adequately familiarized yourself with the
nature of the equations that you use.

73, ac6xg

Antonio Vernucci wrote:
But if both waves are sumultaneouly present, the power carried by each
wave when alone is no longer a meaningful number.

Why is the ExB Poynting vector of each wave no longer
proportional to the energy content? Why does the energy
content of the component waves have to change when they
superpose? Where does that energy change go? Do the
necessary joules disappear and/or appear from thin air?

As a matter of fact
when superposing two coherent waves (same frequency, fixed phase
relationship), one MUST first sum voltages (or currents) and then
calculate power.

That's what I did and the result was 171 joules/sec.
The Poynting vector for each of the two source waves
is 50 joules/sec. Why is the energy content of the
component waves not a meaningful number?

Summing wave powers could only be done in case of incoherent waves.

No, there is a special equation to be used for summing coherent
waves, i.e. the irradiance equation from optical physics. For
power density:

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

where 'A' is the angle between the two E-fields.

In conclusion, the answer to your question is that the apparent extra 71
joules/s come front the fact that 100 joules/s taken as reference is a
number having no physical meaning.

For every second that passes, 50 + 50 = 100 joules has no
physical meaning? Are you saying that an EM wave is not
associated with ExB joules/sec?
--
73, Cecil http://www.w5dxp.com

Why is the ExB Poynting vector of each wave no longer
proportional to the energy content? Why does the energy
content of the component waves have to change when they
superpose? Where does that energy change go? Do the
necessary joules disappear and/or appear from thin air?

Joules do not disappear, they just get distributed over the free space in a
non-uniform manner.

In certain regions of the space the two waves add up (apparently creating extra
power), in other regions they cancel out (apparently destroying power). The
integral of total radiated power does not change.

In your example you considered a location where the two waves have a 45 deg.
shift. At another location, where the two waves have a zero deg. shift, you
would observe an even higher apparent power creation. Conversely, at locations
where the two waves have a 180 deg. shift you would observe absence of power.

The principle causing the apparent power creation at your location is the same
principle by which an antenna formed by two stacked dipoles features a gain of
up to 3 dB with respect to a single dipole, and can then deliver up to twice the
power to a receiver placed at the maximum radiation heading (and zero power at a
receiver placed at 90 degrees from that heading). .

That's what I did and the result was 171 joules/sec.
The Poynting vector for each of the two source waves
is 50 joules/sec. Why is the energy content of the
component waves not a meaningful number?

Each wave produces 50 joules/s when alone. When the two waves are superimposed,
each wave produces not only its 50 joules/s but also 35.5 more joules that it
"robs" from other regions of the space. If you would plainly sum the power of
two components (i.e. 50 + 50), you would neglect the fact that coherent waves
necessarily interfere with each other in the space, in constructive or
destructive manner depending on the receiver location.

Summing wave powers could only be done in case of incoherent waves.

No, there is a special equation to be used for summing coherent
waves, i.e. the irradiance equation from optical physics. For
power density:

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

where 'A' is the angle between the two E-fields.

Please re-read my sentence more carefully. My statement was that summing powers
(that is. Ptotal = P1 + P2) would only be correct for incoherent waves.

For coherent waves, plainly summing powers would generally be incorrect (apart
from one particular phase angle), and one must nstead use the equation you have
shown.

For every second that passes, 50 + 50 = 100 joules has no
physical meaning? Are you saying that an EM wave is not
associated with ExB joules/sec?

see previous remarks.

73

Tony I0JX

On Nov 16, 3:10 pm, Cecil Moore wrote:
K7ITM wrote:
Nice "when are you going to stop beating your mother" sort of
question. And what was your reply?

It's a rhetorical question, Tom. What is your reply?
When someone (besides Eugene Hecht) explains it to
my satisfaction I will stop beating that dead horse.
--
73, Cecil http://www.w5dxp.com

From your original posting, "The following is from an email to which I
replied today." There was no indication than anything following that
was your reply, and I was curious what your reply was. You're welcome
to beat dead horses as much as you like, but that doesn't mean I need
to.

Assuming the two "waves" existed independently at some points in
space, you'll have to first tell us _exactly_ what was done to combine
them into one wave. Right now I'm not accepting that you will be able
to combine two independent waves carrying 50 watts each into a single
wave carrying more than 100 watts. Thus, it's a "when are you going
to stop beating your mother" problem, as posed. There's really
nothing interesting except at the point at which the waves combine.
But then that's already been explained more than once.

Jim Kelley wrote:
Evidently then you haven't adequately familiarized yourself with the
nature of the equations that you use.

The last term in the following power density equation
is known as the "interference term". If it is positive,
the interference is constructive. If it is negative,
the interference is destructive.

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

Ptotal = 50w + 50w + 2*SQRT(2500)cos(45)

Ptotal = 100w + 100w(0.7071) = 170.71w

The interference term is 70.71 watts of constructive
interference indicating that there must exist 70.71 watts
of destructive interference elsewhere in the system.
If the constructive interference happens at an impedance
discontinuity in a transmission line in the direction of
the load then there must be an equal magnitude of
destructive interference toward the source.
--
73, Cecil http://www.w5dxp.com

Antonio Vernucci wrote:
Each wave produces 50 joules/s when alone. When the two waves are
superimposed, each wave produces not only its 50 joules/s but also 35.5
more joules that it "robs" from other regions of the space.

My point exactly! The same thing is true when it happens
in a transmission line at a Z0-match point. The region of
constructive interference toward the load (forward energy
wave) "robs" energy from the region of destructive
interference toward the source (reflected energy waves).
That's how antenna tuners work.
--
73, Cecil http://www.w5dxp.com

K7ITM wrote:
From your original posting, "The following is from an email to which I
replied today." There was no indication than anything following that
was your reply, and I was curious what your reply was.

The posting was my reply to that original email.

Right now I'm not accepting that you will be able
to combine two independent waves carrying 50 watts each into a single
wave carrying more than 100 watts.

It happens all the time at a Z0-match in a transmission
line. Please reference Dr. Best's article in the Nov/Dec
2001 QEX. He combines a 75 joule/sec wave with an 8.33
joule/sec wave to get a 133.33 joule/sec wave.

Ptotal = 75 + 8.33 + 2*SQRT(75*8.33) = 133.33 joules/sec

Dr. Best's article was the first time I had ever seen
the power density irradiance equations from the field
of optical physics used on RF waves.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Jim Kelley wrote:
Evidently then you haven't adequately familiarized yourself with the
nature of the equations that you use.

The last term in the following power density equation
is known as the "interference term". If it is positive,
the interference is constructive. If it is negative,
the interference is destructive.

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

Ptotal = 50w + 50w + 2*SQRT(2500)cos(45)

Ptotal = 100w + 100w(0.7071) = 170.71w

The interference term is 70.71 watts of constructive
interference indicating that there must exist 70.71 watts
of destructive interference elsewhere in the system.
If the constructive interference happens at an impedance
discontinuity in a transmission line in the direction of
the load then there must be an equal magnitude of
destructive interference toward the source.



I share Tom B's suspicions. Since Cecil's analysis is leading to
physical absurdities such as "watts of destructive interference" and
vagueries such as "elsewhere in the system", it means that something is
wrong. It could be either in his statement of the problem, the
suitability of his chosen method of analysis, or the way Cecil is
applying that method; or any combination of the above.

Either way, it is Cecil's tarbaby, and nobody else needs to get stuck to
it.

The rest of us can continue to use the methods that have existed for a
hundred years to account for the voltages, currents and phases at any
location along a transmission line, and at any moment in time.


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

"Cecil Moore" wrote in message
news
Antonio Vernucci wrote:
Each wave produces 50 joules/s when alone. When the two waves are
superimposed, each wave produces not only its 50 joules/s but also 35.5
more joules that it "robs" from other regions of the space.

My point exactly! The same thing is true when it happens
in a transmission line at a Z0-match point. The region of
constructive interference toward the load (forward energy
wave) "robs" energy from the region of destructive
interference toward the source (reflected energy waves).
That's how antenna tuners work.
--
73, Cecil http://www.w5dxp.com

You can come up with a lot simpler example that at first might look like a
paradox. Consider two DC current sources of 1 amp each. Each current source
will deliver 50 W to a 50 Ohm resistor. Now connect the two current sources
in parallel, and the resultant 2 amps will deliver 200W to the same 50 Ohm
resistor. There is nothing wrong here.

Tam

Cecil Moore wrote:
The following is from an email to which I replied today.

Given two coherent EM waves superposed in a 50 ohm environment
at considerable distance from any source:

Wave#3 = Wave#1 superposed with Wave#2

Wave#1: V = 50v at 0 deg, I = 1.0a at 0 deg, P = 50 joules/sec

yes, lets say source S1 supplies a voltage V1 into a load L1, where L1
is a pure 50 Ohm resistance.


Wave#2: V = 50v at 45 deg, I = 1.0a at 45 deg, P = 50 joules/sec

Source S2 supplies a voltage V2 puts into a load L2, where L2 is a pure
50 Ohm resistance.

These two waves superpose to V = 92.38v and I = 1.85a
Note: P = 171 joules/sec

You have changed the circuit.

Source S1 is no longer connected to a load L1 consisting of a 50 Ohm
load. It is connected to a load L3, consisting of a pure resistance in
series with a voltage source.

Since you have changed the circuit source 1 is connected to, you should
not be surprised it supplies a different power.

Move the phase difference to 180 degrees, and source S1 would supply no
power at all.


*During each second*, Wave#1 supplies 50 joules of energy and
Wave#2 supplies 50 joules of energy for a total of 100 joules of
energy being supplied *every second* to the superposition process.
Yet the results of that superposition process yields 171 joules
of energy *during each second*, 71 joules more than is being
supplied to the process. Where are the extra 71 joules per second
coming from?

35.5 J/s (Watts) comes from the source S1 and 35.5 J/s (Watts) comes
from the source S2. It's different to the first case, as they are
connected to a different circuit.

You can do the same with DC - you don't need to use AC at all. Put a 50
V battery in series with the pure 50 Ohm load and it supplies 50 W. Put
it in series with another load, consisting of a 50 Ohm voltage source in
series with a 50 Ohm load, and it is no surprise it delivers a different
power. Depending on what way you connect the two batteries, the current
would be 0 A or 4 A, and so the power 0 or 200W.