The Rest of the Story
 

K7ITM wrote:
Now, exactly what part of "linear system" do you fail to understand?

Have you stopped beating your wife?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Gene Fuller wrote:
Did you ever wonder why all of the basic phenomena, both optical and RF,
were known to the "ancients", yet you are the first one to pull
everything together in this miraculous new version of a reflection model?

I am the one quoting the wisdom of those ancients
which seems to have somehow fallen by the wayside
and been replaced by some pseudo scientific religion.
What is it about the conservation of energy principle
that you disagree with? What is it about the wave
reflection model that you disagree with? What is it
about the principle of superposition that you disagree
with?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

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.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

How do you justify this conclusion? *It seems to me that until the reflection arrives, the source will have a load of Rs = 50 ohms plus line = 50 ohms for a total of 100 ohms. *Prs.before = 50 watts.

The voltage on Rs, before the reflection returns is
Vrs(t) = 70.7cos(wt)
Prs(t) = Vrs(t)**2/50
= 50 + 50cos(2wt)

Prs.before.average = average(Prs(t))
= 50
since the average of cos is 0.

For the equation Vrs(t) = 70.7cos(wt), are you finding the 70.7 from sqrt(50^2 + 50^2)?

If so, how do you justify that proceedure before the reflection returns? I think the voltage across Rs is 50v until the reflection returns. I also think the current would be 1 amp, for power of 50w, until the reflection returns.


My apologies for leaving out the "(t)" everywhere which would have
made it clearer.

There would be no reflected power at the source until the reflection returns, making the following statements incorrect.

I should have been more clear. The below applies after the reflected
wave returns.
And I should have included the "(t)" for greater clarity.

The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

Rs is shown as a resistance on only one side of the line. *It would simplify and focus the discussion if Rs were broken into two resistors, each placed on one side of the source to make the circuit balanced. *

I am not sure how this would help. It would make the arithmetic
somewhat more
complex.

...Keith

Well, as drawn, the circuit is unbalanced. Reflections are about time and distance traveled during that time, so we really don't know where the far center point is located for finding instantaneous values. We are making an assumption that the length of the source and resistor rs is zero which is OK, but we need to be aware that we are making that assumption. By splitting the resistor into two parts, and then adding them together to make the calculations, we can see that a balanced circuit is intended, but the source resistors are combined for ease of calculation.

--
73, Roger, W7WKB

The Rest of the Story
 

Cecil Moore wrote:
K7ITM wrote:
Now, exactly what part of "linear system" do you fail to understand?

Have you stopped beating your wife?

(I might also ask why you're going to so much trouble to be
disagreeable with something that agrees with what you were
posting...but I think I already know the answer to that one.)

Tom, we were getting along quite well before you asked
your leading question, obviously designed to elicit
anger, not just once but twice.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
So the energy flows that should add up, do add up.

How did you take care of the fact that the forward wave
and reflected wave are flowing in opposite directions
through the resistor?

How did you take care of the 90 degree phase difference
between the forward wave and the reflected wave through
the resistor?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

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.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

How do you justify this conclusion? *It seems to me that until the reflection arrives, the source will have a load of Rs = 50 ohms plus line = 50 ohms for a total of 100 ohms. *Prs.before = 50 watts.

The voltage on Rs, before the reflection returns is
Vrs(t) = 70.7cos(wt)
Prs(t) = Vrs(t)**2/50
* * * *= 50 + 50cos(2wt)

Prs.before.average = average(Prs(t))
* * * * * * * * * *= 50
since the average of cos is 0.

For the equation Vrs(t) = 70.7cos(wt), are you finding the 70.7 from sqrt(50^2 + 50^2)?

No. Cecil's circuit has a 100 V RMS source. For this source

Vs(t) = 141.4 cos(wt)

Before the reflection returns, the line acts as 50 ohms. This, in
series with
the 50 source resistor means that Vs(t)/2 is across Rs and Vs(t)/2
appears
across the line.

Vrs(t) = Vf.g(t) = Vs(t)/2 = 70.7 cos(wt)

The power dissipated in the resistor is

Prs(t) = Vrs(t)**2 / 50
= ((70.7 * 70.7) / 50) * cos(wt) * cos(wt)
= 100 * 0.5 * (cos(wt+wt) + cos(wt-wt))
= 50 (cos(2wt) + 1)
= 50 cos(2wt) + 50

If so, how do you justify that proceedure before the reflection returns? *I think the voltage across Rs is 50v until the reflection returns. *I also think the current would be 1 amp, for power of 50w, until the reflection returns.

That is the average power. But since the voltage is a sinusoid, the
instantaneous power as a function of time is
Prs(t) = 50 + 50 cos(2wt)
which does average to 50 W, thus agreeing with your calculations.


My apologies for leaving out the "(t)" everywhere which would have
made it clearer.

There would be no reflected power at the source until the reflection returns, making the following statements incorrect.

I should have been more clear. The below applies after the reflected
wave returns.
And I should have included the "(t)" for greater clarity.

The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

Rs is shown as a resistance on only one side of the line. *It would simplify and focus the discussion if Rs were broken into two resistors, each placed on one side of the source to make the circuit balanced. *

I am not sure how this would help. It would make the arithmetic
somewhat more
complex.

...Keith

Well, as drawn, the circuit is unbalanced. *Reflections are about time and distance traveled during that time, so we really don't know where the far center point is located for finding instantaneous values. *We are making an assumption that the length of the source and resistor rs is zero which is OK, but we need to be aware that we are making that assumption. *

I find it valuable to understand how such an ideal circuit would
operate, then
extend the solution if necessary.

In practice, many circuits are small compared to the wavelength and
these assumptions
do not materially affect the answer.

Of course, it is important to know when they do, at which time more
complex analysis
becomes necessary.

By splitting the resistor into two parts, and then adding them together to make the calculations, we can see that a balanced circuit is intended, but the source resistors are combined for ease of calculation.

...Keith

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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