"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

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

"Tam/WB2TT" wrote in message
. ..

"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

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

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

"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

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

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.

I see Cecil is still superposing his waves of average power.

I have an example that's more fun yet.

Take a 10 volt source and connect it through a 10 ohm resistor to
another 10 volt source. The positive terminals of the sources are
connected to the ends of the resistor, and the negative terminals are
connected together -- "grounded", if you prefer. Turn on one source,
leaving the other off. (An "off" voltage source is a short circuit.)
Result: 10 watts of dissipation in the resistor. Turn off the first
source and turn on the second. Result: 10 watts of dissipation in the
resistor. Now turn both sources on. Result: An exercise for the reader.

This is a linear circuit for which superposition holds.

Roy Lewallen, W7EL

Roy Lewallen wrote:
I see Cecil is still superposing his waves of average power.

That's at least unfair and at most unethical, Roy. I am not
superposing power. I am using the accepted irradiance equations
from optical physics to predict the energy result of superposing
EM waves, something that was being done by physicists before you
were born.

I have an example that's more fun yet.
Take a 10 volt source ...

It may be more fun but irrelevant. Please explain how the sun
can adjust its energy output depending upon what might or might
not be happening on earth. If you can do that, I will retract
everything I have said about this subject.
--
73, Cecil http://www.w5dxp.com

I reinforce what Roy has said with different words: Superposition and
linearity are one and the same. If a circuit or process is linear, then
superposition gives correct results. If superposition works, then the
circuit or process is linear. Power is not a linear process. Power
involves multiplication.

When investigating the results of multiple voltage and/or current sources
(including E or H from an antenna) all at the same frequency, one performs
an addition (vector addition). Then, and only then, one may (note the
permissive form) use the resultant voltage or current or impedance (two of
the three) to calculate (complex) power. (In the antenna case, recall that
Z is near 377 ohms only in the far field.)

73, Mac N8TT

--
J. McLaughlin; Michigan, USA
Home:
"Roy Lewallen" wrote in message
snip

I see Cecil is still superposing his waves of average power.

I have an example that's more fun yet.

Take a 10 volt source and connect it through a 10 ohm resistor to another
10 volt source. The positive terminals of the sources are connected to the
ends of the resistor, and the negative terminals are connected together --
"grounded", if you prefer. Turn on one source, leaving the other off. (An
"off" voltage source is a short circuit.) Result: 10 watts of dissipation
in the resistor. Turn off the first source and turn on the second. Result:
10 watts of dissipation in the resistor. Now turn both sources on. Result:
An exercise for the reader.

This is a linear circuit for which superposition holds.

Roy Lewallen, W7EL

J. Mc Laughlin wrote:
Power is not a linear process. Power involves multiplication.

Nobody has said that power is a linear process.
The equation for adding EM powers has been known
for decades and if it didn't work, it would have
been discarded. As it is, one can find the power
density equation in any appropriate physics textbook.
The method of adding powers associated with two
coherent collinear EM waves is:

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

where 'A' is the phase angle between the E-fields
of the two waves.
--
73, Cecil http://www.w5dxp.com