Superposition
 

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

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

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

*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?
--
73, Cecil http://www.w5dxp.com

Superposition
 

"Cecil Moore" wrote in message
et...
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

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

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

*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?
--
73, Cecil http://www.w5dxp.com

Cecil

Congratulations, you appear to have solved the world's energy problems for
ever! Electricity will now be too cheap to meter, just like they promised
when nuclear power plants were first built.

Mike G0ULI

Superposition
 

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

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

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

*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?

The peak intensity of a standing wave will always be greater than the
simple sum of the two waves. Itot = 4*I*cos^2(deg/2). But I think
nature will somehow conspire against you if you try to make use of
more than the 100 watts input. ;-)

ac6xg

Superposition
 

On Nov 16, 12:34 pm, 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

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

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

*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?
--
73, Cecil http://www.w5dxp.com

Nice "when are you going to stop beating your mother" sort of
question. And what was your reply?

Superposition
 

"Cecil Moore" wrote in message
et...
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:

Can we take this to mean that this is not the steady state condition?

Tam/WB2TT

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

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

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

*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?
--
73, Cecil http://www.w5dxp.com

Superposition
 

Mike Kaliski wrote:

"Cecil Moore" wrote in message
*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?

Congratulations, you appear to have solved the world's energy problems
for ever! Electricity will now be too cheap to meter, just like they
promised when nuclear power plants were first built.

Sorry, unlike others on this newsgroup, I don't support
a violation of the conservation of energy principle. To
answer my own question above, the extra 71 joules/sec
of constructive interference comes from 71 joules/sec
of destructive interference occurring somewhere else.
--
73, Cecil http://www.w5dxp.com

Superposition
 

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.
The average value of the energy in the standing waves *always*
equals the average values of the energy components of the
forward and reflected waves added together. If there are X joules
in the standing waves, there will be X joules in the sum of the
forward and reflected waves.

But I think
nature will somehow conspire against you if you try to make use of more
than the 100 watts input. ;-)

It was a rhetorical question for people who say, "Just do
a vector analysis and the energy will take care of itself."
Understanding exactly how the energy takes care of itself
is the point of this thread. Hint: 171 joules/sec from two
50 joules/sec waves requires an additional source of energy.
In the case of this example, the 71 joules/sec of constructive
interference requires 71 joules/sec of destructive interference
energy occurring somewhere else.
--
73, Cecil http://www.w5dxp.com

Superposition
 

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

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

50 joules/s are carried by Wave#1 if alone. The same applies to Wave#2

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

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

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.

73

Tony I0JX

Superposition
 

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

Superposition
 

Tam/WB2TT wrote:
"Cecil Moore" wrote:
Given two coherent EM waves superposed in a 50 ohm environment
at considerable distance from any source:

Can we take this to mean that this is not the steady state condition?

Sorry if I somehow gave you that idea. It is a steady-state
problem.
--
73, Cecil http://www.w5dxp.com

Superposition
 

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

Superposition
 

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

Superposition
 

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

Superposition
 

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.

Superposition
 

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

Superposition
 

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

Superposition
 

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

Superposition
 

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

Superposition
 

"Cecil Moore" wrote in message
. net...
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

Superposition
 

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.

Superposition
 

Dave wrote:
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.


Sorry, 0 or 2A. Then the power is 0 or 200W as I said.

Superposition
 

Tam/WB2TT wrote:
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.

Of course, the sources simply deliver the extra power.
But in our previous examples, the source is nowhere around
the point of interference and therefore cannot be the source
of the extra power. Away from any source, the energy required
by constructive interference *always* comes from destructive
interference somewhere else.

In a Z0-matched transmission line with reflections, the
constructive interference is toward the load at the Z0-match
point and the destructive interference is toward the source
at the Z0-match point.

An understanding of the constructive and destructive
interference at a Z0-match point is a necessary and
sufficient condition for understanding where the
reflected energy goes which is the whole purpose
of this thread.
--
73, Cecil http://www.w5dxp.com

Superposition
 

I really meesed up that last paragraph!

Dave wrote:

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
consisting of a 50 volt voltage source, not a 50 Ohm voltage source!
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.
As I said before, 0 or 2A, which gives 0 or 200 W.

Superposition
 

On Nov 17, 12:15 am, Cecil Moore wrote:
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

In typical Cecil fashion, you trimmed out the only part I really cared
about having you answer: "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." Depending on
how _I_ do that, I can get various answers, since some power goes
elsewhere in some of the methods, but I _never_ get more power out of
a steady-state system than I put in. Barring stupid math mistakes,
anyway.

Adios,
Tom

Superposition
 

Ian White GM3SEK wrote:
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.

Do you think Eugene Hecht of "Optics" fame is wrong?
The unit of irradiance is "watts per unit area" and is
NOT a "physical absurdity". Hecht uses "watts per unit
area of destructive interference" quite often in his
classic textbook. He says the spacial average of all
interference must be zero so that the watts per unit
area of constructive interference must be balanced by
the watts per unit area of destructive interference
elsewhere in order to satisfy the conservation of
energy principle.

Nothing is wrong, Ian, you are simply ignorant. I suggest
you read the chapter on interference in "Optics" and try
to comprehend it. It might do you good to learn something new.

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

By all means don't try to learn and understand anything new.
Newsgroup gurus apparently already know all there is to know
and are therefore incapable of additional learning.

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.

And that is exactly why you don't understand reflected energy.
An understanding of of interference can be had from a voltage
analysis but you obviously have never performed such. It is
common knowledge that V1^2+V2^2 is not equal to (V1+V2)^2.
The question as to why they are not equal has been avoided
even though it is easy to answer. If (V1+V2)^2 V1^2+V2^2
then the superposition of voltages has resulted in constructive
interference. If (V1+V2)^2 V1^2+V2^2 then the superposition
of voltages has resulted in destructive interference. Away
from any source, constructive interference must always equal
destructive interference to avoid violating the conservation
of energy principle. At a Z0-match point, the reflected energy
is redistributed back toward the load by constructive interference.
An equal magnitude of destructive interference occurs toward
the source thus eliminating reflected energy toward the source.
It is the same way that thin-film non-reflective glass works.
--
73, Cecil http://www.w5dxp.com

Superposition
 

"Tam/WB2TT" wrote in message
. ..

"Cecil Moore" wrote in message
. net...
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

Wouldn't you have to double the voltage to get the 1 amp from each source to
flow ?

Superposition
 

On Sat, 17 Nov 2007 12:02:26 -0500, "Ralph Mowery"
wrote:

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

Wouldn't you have to double the voltage to get the 1 amp from each source to
flow ?

Hi Ralph,

That is one of the usual properties of a current source.

73's
Richard Clark, KB7QHC

Superposition
 

"K7ITM" wrote
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.
__________

The physics of EM radiation.

As an example, consider an array comprised of two, identical radiators on
the same vertical axis, in the same physical orientation, with a vertical
separation of 1 wavelength, each driven with equal r-f power and relative
phase by the same r-f source.

The fields from the two radiators are generated and radiated separately, but
once well past the near-field boundary of the array, the EM field existing
at every point in free space will be the vector sum of those separate
fields.

When the net field at the radiation peak of the array is measured in the far
field, there will be no way to determine from that measurement whether the
field was generated using a single radiator with X power input, or the
described 2-element array having about 1/2 that power input.

RF

Superposition
 

Cecil Moore wrote:
Ian White GM3SEK wrote:
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.

Do you think Eugene Hecht of "Optics" fame is wrong?
The unit of irradiance is "watts per unit area" and is
NOT a "physical absurdity". Hecht uses "watts per unit
area of destructive interference" quite often in his
classic textbook. He says the spacial average of all
interference must be zero so that the watts per unit
area of constructive interference must be balanced by
the watts per unit area of destructive interference
elsewhere in order to satisfy the conservation of
energy principle.

Nothing is wrong, Ian, you are simply ignorant. I suggest
you read the chapter on interference in "Optics" and try
to comprehend it. It might do you good to learn something new.

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

By all means don't try to learn and understand anything new.
Newsgroup gurus apparently already know all there is to know
and are therefore incapable of additional learning.

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.

And that is exactly why you don't understand reflected energy.
An understanding of of interference can be had from a voltage
analysis but you obviously have never performed such. It is
common knowledge that V1^2+V2^2 is not equal to (V1+V2)^2.
The question as to why they are not equal has been avoided
even though it is easy to answer. If (V1+V2)^2 V1^2+V2^2
then the superposition of voltages has resulted in constructive
interference. If (V1+V2)^2 V1^2+V2^2 then the superposition
of voltages has resulted in destructive interference. Away
from any source, constructive interference must always equal
destructive interference to avoid violating the conservation
of energy principle. At a Z0-match point, the reflected energy
is redistributed back toward the load by constructive interference.
An equal magnitude of destructive interference occurs toward
the source thus eliminating reflected energy toward the source.
It is the same way that thin-film non-reflective glass works.

Hiding behind authority again, Cecil? Using a few carefully edited
quotes from Hecht doesn't prove anything. Ian hit the nail on the
head: Vague philosophical arguments using second and third order
abstractions that you can't prove to have any connection to reality
aren't going to convince anyone.
73,
Tom Donaly, KA6RUH

Superposition
 

Richard Fry wrote:
"K7ITM" wrote
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.
__________

The physics of EM radiation.

As an example, consider an array comprised of two, identical radiators
on the same vertical axis, in the same physical orientation, with a
vertical separation of 1 wavelength, each driven with equal r-f power
and relative phase by the same r-f source.

The fields from the two radiators are generated and radiated separately,
but once well past the near-field boundary of the array, the EM field
existing at every point in free space will be the vector sum of those
separate fields.

When the net field at the radiation peak of the array is measured in the
far field, there will be no way to determine from that measurement
whether the field was generated using a single radiator with X power
input, or the described 2-element array having about 1/2 that power input.

RF

So in the limit, as the number of radiators is increased to infinity,
the amount of power it would take to produce the measured sum would
go to zero. Nice logic.
73,
Tom Donaly, KA6RUH

Superposition
 

Hi Dave,

Your analysis, or critique rather, has been misdirected. In fact,
everyone has been suckered. Not unusual given the problem was crafted
to be disingenuous. It is, after all, this group's form of "Three
Card Monte." Can you pick the card that is the one-eyed Jack?

With a little re-ordering here....

On Sat, 17 Nov 2007 16:08:04 +0000, Dave wrote:

Wave#3 = Wave#1 superposed with Wave#2
Wave#3 is the only thing real here.

These two waves superpose to V = 92.38v and I = 1.85a
Note: P = 171 joules/sec
Superposition is a process that gives us a solution to a system that
exists, not a figurative one. However, the process of superposition
requires the suspension of reality to perform computation, and to
render the solution that is real.

yes, lets say source S1 supplies a voltage V1 into a load L1, where L1
is a pure 50 Ohm resistance.
Dave, herein lies everyone's presumption, and one that has been
"suggested" originally. What you "interpret" does not exist
independently. In fact, S1 has no independent existence (neither does
the other source). This is the artificial contrivance of partially
solving a superposition problem.

Let's simply look at these "suggestions:"
Wave#1: V = 50v at 0 deg, I = 1.0a at 0 deg, P = 50 joules/sec
In the reality of two waves, this artificial condition is arrived at
only through removing the second wave from the reality. And like
wise:

Wave#2: V = 50v at 45 deg, I = 1.0a at 45 deg, P = 50 joules/sec
In the reality of two waves, this artificial condition is arrived at
only through removing the first wave from the reality.

Neither of these artificial conditions actually exist in the reality
of superposed waves, and that is the con. The group has been fixated
on the separate artificial environments with their partial solutions
as though they actually exist independent of the reality of the
superposed, complete solution.

You have changed the circuit.

Of course he has. That is the allowed method of computing
superposition (he has in fact not done the full method of
superposition analysis - but that is immaterial to the discussion
except only to note that Cecil's posts are often rife with error).

The egregious error is found he
*During each second*, Wave#1 supplies 50 joules of energy
Wave #1 is not independent in reality, so the statement is wrong. What
is provided (50 joules) is only a partial solution in the method of
computing the superposition which for that computation, suspends
reality to examine the separate constituents in an artificial
environment.

73's
Richard Clark, KB7QHC

Superposition
 

Tom Donaly wrote:

So in the limit, as the number of radiators is increased to infinity,
the amount of power it would take to produce the measured sum would
go to zero. Nice logic.
73,
Tom Donaly, KA6RUH


Mathematica 6.0 for Sun Solaris SPARC (64-bit)
Copyright 1988-2007 Wolfram Research, Inc.

In[1]:= 0 Infinity

Infinity::indet: Indeterminate expression 0 Infinity encountered.

Out[1]= Indeterminate

Superposition
 

Dave wrote:
Tom Donaly wrote:

So in the limit, as the number of radiators is increased to infinity,
the amount of power it would take to produce the measured sum would
go to zero. Nice logic.
73,
Tom Donaly, KA6RUH


Mathematica 6.0 for Sun Solaris SPARC (64-bit)
Copyright 1988-2007 Wolfram Research, Inc.

In[1]:= 0 Infinity

Infinity::indet: Indeterminate expression 0 Infinity encountered.

Out[1]= Indeterminate
0 Infinity
is interpreted as zero times infinity

Superposition
 

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

But these are not "sources" per se, they are EM
waves with a fixed constant energy content. EM waves
simply cannot, willy-nilly, increase their energy
content. The actual source could conceivably be
light-years away so it would take light-years to
increase the energy content of the EM waves.

So please tell us how a 50 watts/unit-area wave can
increase its energy content to 85.5 watts/unit-area
while being light-years away from any source.
--
73, Cecil http://www.w5dxp.com

Superposition
 

K7ITM wrote:
In typical Cecil fashion, you trimmed out the only part I really cared
about having you answer: "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."

They were confined to a transmission line. Coherent
waves traveling in the same direction in a transmission
line are forced to also be collinear. Unlike space, a
transmission line forces collinearity upon the EM waves.
But the same thing happens at a 1/4WL thin-film non-
reflective coating on glass. Assuming one brings the
Poynting vectors into collinearity, one can see what
is happening at http://www.w5dxp.com/thinfilm.gig
--
73, Cecil http://www.w5dxp.com

Superposition
 

Richard Fry wrote:
The physics of EM radiation.

It seems strange to me that people who know that
antenna gain in one direction comes at the expense
of gain in another direction cannot carry that
concept over to transmission lines. Constructive
interference in one direction means destructive
interference in another direction. That's what
creates the radiation pattern for antennas. It
is also exactly the same thing that routes
energy toward the antenna instead of toward the
source in a Z0-matched transmission line.

--
73, Cecil http://www.w5dxp.com

Superposition
 

Tom Donaly wrote:
Hiding behind authority again, Cecil? Using a few carefully edited
quotes from Hecht doesn't prove anything. Ian hit the nail on the
head: Vague philosophical arguments using second and third order
abstractions that you can't prove to have any connection to reality
aren't going to convince anyone.

The void technical content of your objection is noted, Tom.
Why don't you present some theory and math that prove me
wrong instead of just waving your hands and uttering ad
hominem attacks?
--
73, Cecil http://www.w5dxp.com

Superposition
 

"Richard Clark" wrote in message
...
On Sat, 17 Nov 2007 12:02:26 -0500, "Ralph Mowery"
wrote:

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

Wouldn't you have to double the voltage to get the 1 amp from each source
to
flow ?

Hi Ralph,

That is one of the usual properties of a current source.

73's
Richard Clark, KB7QHC

If you want to drive 1 amp into a 100K resistor, make sure the current
source can develop 100,000 Volts!

Tam

Superposition
 

On Sat, 17 Nov 2007 13:52:54 -0500, "Tam/WB2TT"
wrote:

Wouldn't you have to double the voltage to get the 1 amp from each source
to
flow ?

Hi Ralph,

That is one of the usual properties of a current source.

73's
Richard Clark, KB7QHC

If you want to drive 1 amp into a 100K resistor, make sure the current
source can develop 100,000 Volts!

Hi All,

It is with some reflected amusement that I pause here to relate a
story. I was once tasked to calibrate an HP precision current source.
I had faithfully connected my standard shunt and voltmeter to do just
this, and the source performed exactly as specified (HP equipment that
wasn't broke, always did). I then disconnected the leads and was
immediately bit. The source performed exactly as specified!

73's
Richard Clark, KB7QHC

Superposition
 

Tom Donaly wrote
So in the limit, as the number of radiators is increased to infinity,
the amount of power it would take to produce the measured sum would go to
zero.
____________

As the number of radiators in a given array never can reach infinity,
neither will the input power for a given peak ERP from that array ever go to
zero. Obviously there are practical limits as well.

But this does not change the realities that...

1) other things equal, the greater the number of discrete radiators in an
array, the less input power is needed for that array to produce a given peak
ERP, and

2) the peak free-space, far field produced by a given ERP is the same for
all combinations of antenna gain and antenna input power producing that ERP.

This has been proven in commercial FM and TV broadcast systems for many
decades.

RF