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