On Wed, 5 Mar 2008 21:37:06 -0800 (PST)
Keith Dysart wrote:

On Mar 6, 12:04*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 06:06:04 -0800 (PST)

Keith Dysart wrote:
On Mar 5, 8:12*am, Roger Sparks wrote:
On Tue, 4 Mar 2008 17:00:31 -0800 (PST)

Keith Dysart wrote:
On Mar 4, 3:36*pm, Cecil Moore wrote:
After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

Thanks Keith. I see what you are doing now, although I still don't understand your logic in faulting Cecil on the instantaneous values. I agree with you that the instantaneous values can be tracked, but don't see a fault in Cecil's presentation.

Roy and I went around a few times on whether the source reflects in a case like this. The source reflection controls whether the 50 ohm source resistor acts like 50 ohms to the reflected wave, or acts like a short circuit in parallel with the 50 ohm source resistor.

--
73, Roger, W7WKB

Roger Sparks wrote:
Thanks Keith. I see what you are doing now, although I still
don't understand your logic in faulting Cecil on the instantaneous
values. I agree with you that the instantaneous values can be
tracked, but don't see a fault in Cecil's presentation.

I made no assertions about instantaneous power at all and
have added a note in my article to that effect. My assertions
about instantaneous power cannot possibly be wrong because
I didn't make any. :-) Nothing in my article applies to or
is concerned with instantaneous power.

For the power density/irradiance/interference equation to
be applicable, certain conditions must be met. One of those
conditions is that all component powers must be *average*
powers resembling power density/irradiance from optics.
Another condition is that the phase angle between the two
waves being superposed must be constant and therefore the
two associated waves must be coherent (phase-locked) with
each other.

Roy and I went around a few times on whether the source reflects
in a case like this. The source reflection controls whether the
50 ohm source resistor acts like 50 ohms to the reflected wave,
or acts like a short circuit in parallel with the 50 ohm source
resistor.

What Roy (and others) are missing is that there is more than
one mechanism in physics that can cause a redistribution of
reflected energy back toward the load. An ordinary reflection
is not the only cause. In a one-dimensional environment,
e.g. a transmission line, there is an additional mechanism
present that can redirect and redistribute the reflected
energy back toward the load.

1. Reflection - what happens when a *single wave* encounters
an impedance discontinuity. Some (or all) of the reflected
energy reverses direction.

2. Wave interaction - what happens when *two waves* superpose,
interact, AND effect a redistribution of their energy components
as described on the FSU web page at:

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that
are 180-degrees ... out of phase with each other meet, they
are not actually annihilated, ... All of the photon energy
present in these waves must somehow be recovered or
redistributed in a new direction, according to the law of
energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive
interference, so the effect should be considered as a
*redistribution of light waves and photon energy* ..."

In the simple ideal voltage source described in my article,
there are no reflections because the source resistance equals
the characteristic impedance of the transmission line.

In Part 1 of the article, there is also no wave interaction
because the forward wave and reflected wave are 90 degrees
out of phase. So for that special case, none of the reflected
energy is redistributed back toward the load. Therefore, for
that special case, all of the reflected energy is dissipated
in the source resistor because all conditions for a redistribution
of the reflected energy have been eliminated.

In the special case described in Part 1, because of the 90
degree phase difference, the forward wave and reflected
wave are completely independent of each other almost as if
they were not coherent. The result in that special case is
that the power components can simply be added because in
that special case, (V1^2 + V2^2) = (V1 + V2)^2, something
that is obviously NOT true in the general case.

Part 2 of the article will describe what happens when
the forward wave and reflected wave interact at the source
resistor and effect a redistribution of reflected energy
back toward the load *even when there are no reflections*.
This is the key concept, understood for many decades in the
field of optical physics, that most RF people seem to be
missing.
--
73, Cecil http://www.w5dxp.com

On Mar 6, 1:17*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 21:37:06 -0800 (PST)
[snip]
Thanks Keith. *I see what you are doing now, although I still don't understand your logic in faulting Cecil on the instantaneous values. *I agree with you that the instantaneous values can be tracked, but don't see a fault in Cecil's presentation. *

For the special situation described in
http://www.w5dxp.com/nointfr.htm
Cecil is attempting to show that the reflected energy is dissipated in
the source resistor.

The logic he employs is:
- before the reflection arrives back at the generator, the source
resistor
is dissipating X watts.
- the reflected wave has an energy flow of Y watts.
- after the reflection arrives back at the generator, the source
resistor
is dissipating Z watts.
- since Z is equal to X + Y, the energy in the reflected wave is being
dissipated in the source resistor.

In other words, since the dissipation in the source resistor increases
by the same amount as the power in the reflected wave, the energy in
the reflected wave must be being dissipated in the source resistor.

Cecil analyzes the circuit for a number of load resistances and
suggests
that the equality holds for any load resistance.

For example, with a load resistance of 12.5 ohms, the original
dissipation
in the source resistor is 50 W which increases to 68 W when the 18 W
reflected wave arrives back at the generator. That is, X = 50, Y = 18
and Z = 68, so Z is equal to X + Y.

Cecil does all of this analysis using average powers.

But we know that the power dissipation varies as a function of time
and
that the power in the reflected wave is a function of time. It is my
contention that if it is the energy in the reflected wave that is
increasing the dissipation in the source resistor, the dissipation in
source resistor should occur at the same time that the reflected wave
delivers the energy.

In other words, not only should Z.average = X.average + Y.average,
but Z.instantaneous should equal X.instantaneous + Y.instantaneous
for if the dissipation in the source resistor is not tracking the
energy in the reflected wave, it can not be the energy in the
reflected wave that is heating the resistor.

So using the same 12.5 ohm example,
X.inst = 50 + 50 cos(2wt)
Y.inst = 18 + 18 cos(2wt)
X.inst + Y.inst = 68 + 68 cos(2wt)
but
Z.inst = 68 + 68 cos(2wt - 61.9degrees)

So Z.inst is not equal to Y.inst + X.inst.

This means that the dissipation in the resistor is not happening
at the same time as the energy is being delivered by the reflected
wave, which must mean that it is not the energy from the reflected
wave that is heating the source resistor.

So while analyzing average power dissipations suggests that the
energy from the reflected wave is dissipated in the source resistor,
analysis of the instantaneous power shows that it is not.

...Keith

Keith Dysart wrote:
Cecil does all of this analysis using average powers.

That's because the power density/irradiance/interference
equation only works for average powers. One condition
for valid use of that equation is that the powers must
have been averaged over one complete cycle and usually
over many, many cycles. How many cycles does it take
for a bright interference ring to register on the human
retina?

My example is set up such that the average interference
is zero over each complete cycle. I cannot think of a
way to eliminate interference internal to a cycle. Both
destructive and constructive interference occur during a
cycle, which is what you are seeing, but interference
averages out to zero over each complete cycle.

In other words, not only should Z.average = X.average + Y.average,
but Z.instantaneous should equal X.instantaneous + Y.instantaneous
for if the dissipation in the source resistor is not tracking the
energy in the reflected wave, it can not be the energy in the
reflected wave that is heating the resistor.

Destructive interference from one part of the cycle is
being delivered as constructive interference in another
part of the cycle. That is completely normal.

The same thing happens with an ideal standing wave. The
average power in a standing wave equals zero although
some instantaneous power can be calculated as existing
in the standing wave. You have discovered why Eugene
Hecht says that instantaneous power is "of limited utility".

In fact, what we are dealing with in the example is a
standing wave because the forward wave and reflected
wave are flowing in opposite directions through the
source resistor. I hope your math took that into account.

This means that the dissipation in the resistor is not happening
at the same time as the energy is being delivered by the reflected
wave, which must mean that it is not the energy from the reflected
wave that is heating the source resistor.

All it means is that there must be some localized interference
during each cycle for which you have failed to account.
In other words, you are superposing powers which is a
known invalid thing to do when interference is present.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
So using the same 12.5 ohm example,
X.inst = 50 + 50 cos(2wt)
Y.inst = 18 + 18 cos(2wt)
X.inst + Y.inst = 68 + 68 cos(2wt)

Where do you take into account that the forward wave
is 90 degrees out of phase with the reflected wave?

Z.inst = 68 + 68 cos(2wt - 61.9degrees)

Where does the 61.9 degrees come from?
--
73, Cecil http://www.w5dxp.com

On Mar 6, 11:12*am, Cecil Moore wrote:
Keith Dysart wrote:
So using the same 12.5 ohm example,
X.inst = 50 + 50 cos(2wt)
Y.inst = 18 + 18 cos(2wt)
X.inst + Y.inst = 68 + 68 cos(2wt)

Where do you take into account that the forward wave
is 90 degrees out of phase with the reflected wave?

Z.inst = 68 + 68 cos(2wt - 61.9degrees)

Where does the 61.9 degrees come from?

The computation of all of these can be seen in the spreadsheet at
http://keith.dysart.googlepages.com/...d%2Creflection

Recall that: cos(a)cos(b) = 0.5( cos(a+b) + cos(a-b) )

The power in the resistor is computed by multiplying the voltage
Vrs(t) = 82.46 cos(wt -30.96 degrees)
by the current
Irs(t) = 1.649 cos(wt -30.96 degrees)
yielding
Prs(t) = 68 + 68 cos(2wt -61.92 degrees)

For reflected power at the generator
Vr.g(t) = 42.42 cos(wt +90 degrees)
Ir.g(t) = 0.8485 cos(wt -90 degrees)
Pr.g(t) = Vr.g(t) * Ir.g(t)
= -18 + 18 cos(2wt)

Note that I previously made a transcription error in the sign
of one of the terms.

So X.inst + Y.inst is actually 68 + 32 cos(2wt).

...Keith

Keith Dysart wrote:
The computation of all of these can be seen in the spreadsheet at
http://keith.dysart.googlepages.com/...d%2Creflection

The error in your calculations has been diagnosed.
Since interference exists at the instantaneous level,
it is invalid to add the instantaneous powers directly
as you have done.

Hint: if (V1^2 + V2^2) is not equal to (V1 + V2)^2
then it is invalid to add powers directly. Everyone
should have learned that fact in EE201 when the
professor said: "Thou shalt not superpose powers".
--
73, Cecil http://www.w5dxp.com

On Thu, 6 Mar 2008 06:10:26 -0800 (PST)
Keith Dysart wrote:

On Mar 6, 1:17*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 21:37:06 -0800 (PST)
[snip]
Thanks Keith. *I see what you are doing now, although I still don't understand your logic in faulting Cecil on the instantaneous values. *I agree with you that the instantaneous values can be tracked, but don't see a fault in Cecil's presentation. *

For the special situation described in
http://www.w5dxp.com/nointfr.htm
Cecil is attempting to show that the reflected energy is dissipated in
the source resistor.

The logic he employs is:
- before the reflection arrives back at the generator, the source
resistor
is dissipating X watts.
- the reflected wave has an energy flow of Y watts.
- after the reflection arrives back at the generator, the source
resistor
is dissipating Z watts.
- since Z is equal to X + Y, the energy in the reflected wave is being
dissipated in the source resistor.

In other words, since the dissipation in the source resistor increases
by the same amount as the power in the reflected wave, the energy in
the reflected wave must be being dissipated in the source resistor.

Your logic and conclusion seem correct to me.


Cecil analyzes the circuit for a number of load resistances and
suggests
that the equality holds for any load resistance.

For example, with a load resistance of 12.5 ohms, the original
dissipation
in the source resistor is 50 W which increases to 68 W when the 18 W
reflected wave arrives back at the generator. That is, X = 50, Y = 18
and Z = 68, so Z is equal to X + Y.

Cecil does all of this analysis using average powers.

But we know that the power dissipation varies as a function of time
and
that the power in the reflected wave is a function of time. It is my
contention that if it is the energy in the reflected wave that is
increasing the dissipation in the source resistor, the dissipation in
source resistor should occur at the same time that the reflected wave
delivers the energy.

In other words, not only should Z.average = X.average + Y.average,
but Z.instantaneous should equal X.instantaneous + Y.instantaneous
for if the dissipation in the source resistor is not tracking the
energy in the reflected wave, it can not be the energy in the
reflected wave that is heating the resistor.

Hard to follow this long sentence/paragraph. The dissipation in the source resistor should be the sum of instantaneous energy flows from both source(forward) and reflected energy flows. We would not expect that the instantaneous energy flow would be equal from both source and reflection because of the sine wave nature of the energy flow.


So using the same 12.5 ohm example,
X.inst = 50 + 50 cos(2wt)
Y.inst = 18 + 18 cos(2wt)
X.inst + Y.inst = 68 + 68 cos(2wt)
but
Z.inst = 68 + 68 cos(2wt - 61.9degrees)

So Z.inst is not equal to Y.inst + X.inst.

I think you are mixing average energy flow (the 50 and 18 watts) with instantaneous flow (50 cos(2wt) and 18 cos(2wt)).

I think Z.inst = x.inst + y.inst = 50 cos(2wt) + 18 cos(2wt). The two energy components arrive 90 degrees out of phase (in this example) so we should write z.inst = 50 cos(2wt) + 18 cos(2wt + 90).


This means that the dissipation in the resistor is not happening
at the same time as the energy is being delivered by the reflected
wave, which must mean that it is not the energy from the reflected
wave that is heating the source resistor.

So while analyzing average power dissipations suggests that the
energy from the reflected wave is dissipated in the source resistor,
analysis of the instantaneous power shows that it is not.

...Keith

Either your math or mine is not correct. Which is incorrect?

--
73, Roger, W7WKB

Roger Sparks wrote:
The dissipation in the source resistor should be the
sum of instantaneous energy flows from both source
(forward) and reflected energy flows. We would not
expect that the instantaneous energy flow would be
equal from both source and reflection because of the
sine wave nature of the energy flow.

The key question: Is the square of the sum of the
two voltages equal to the sum of the squares of
the two voltages? If yes, there is no interference
and it is valid to add the powers directly as
Keith has done.

If no, interference exists and it is *INVALID* to
add the powers directly as Keith has done. Every
EE was warned about superposing powers at the
sophomore level if not before. This is why.

So what we need to know is:

Is [Vfor(t) + Vref(t)]^2 equal to
Vfor(t)^2 + Vref(t)^2 ???

Does [70.7v*cos(wt) + 42.4v*cos(wt+90)]^2 equal
[70.7v*cos(wt)]^2 + [42.4v*cos(wt+90)]^2 ???

Is 2[70.7v*cos(wt)*42.4v*cos(wt+90)] always zero???

The answer is obviously 'NO' so Keith's direct addition
of powers, i.e. superposition of powers, is invalid as
it always is when interference is present.

When the interference term is properly taken into
account, the instantaneous dissipation in the source
resistor will no doubt equal the dissipation from
the forward wave plus the dissipation from the
reflected wave plus the interference term which is
minus for destructive interference and plus for
constructive interference. The interference will
average out to zero over each single complete cycle.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Is 2[70.7v*cos(wt)*42.4v*cos(wt+90)] always zero???

This is a test to see if interference exists. It turns
out that constructive interference exists for the
first 90 degrees and third 90 degrees of the forward
wave cycle. Destructive interference exists for the
second and fourth 90 degrees of the cycle. The magnitude
of the destructive interference exactly equals the
magnitude of the constructive interference as expected
so the net interference is zero as expected.
--
73, Cecil http://www.w5dxp.com