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 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

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

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

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

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

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

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

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?

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