The Rest of the Story
 

Hi Keith,

Thanks for the additional explaination. I am wondering if I misunderstood Cecil's original premise.

On Fri, 7 Mar 2008 06:26:46 -0800 (PST)
Keith Dysart wrote:

On Mar 7, 8:30*am, Roger Sparks wrote:
Hi Keith,

I must still not be "getting" something because while I now follow your numbers and trig identity, it looks to me like you used Cecil's premise "PRs = 50w + Pref * " to show that '50 W plus Pref' = 68 + 68cos(2wt-61..9degrees) watts, which would be correct for the 12.5 ohm case.

So, rather than disproving Cecil's premise, you successfully demonstrated that it was correct in the instantaneous case.

What am I missing?

The actual dissipation in the source resistor was computed using
circuit theory
to derive the voltage and current through the resistor and then
multiplying them
together to get the power dissipation:
Vrs(t) = 82.46 cos(wt -30.96 degrees)
Irs(t) = 1.649 cos(wt -30.96 degrees)
Prs.circuit(t) = Vrs(t) * Irs(t)
= 68 + 68 cos(2wt -61.92 degrees)

This was then shown not to be equal to the results using Cecil's
hypothesis
because
Prs.before(t) = 50 + 50 cos(2wt)
Pref(t) = 18 - 18 cos(2wt)
which would give, using Cecil's hypothesis
Prs.cecil(t) = 68 + 32 cos(2wt)

So I accept the circuit theory result of
Prs.circuit(t) = 68 + 68 cos(2wt -61.92 degrees)
and conclude that, since the results using Cecil's hypothesis are
different, Cecil's hypothesis must be incorrect.

That is, the power dissipated in the source resistor after the
reflection
returns is not the sum of the power dissipated in the resistor before
the
reflection returns plus the power in the reflected wave.

Now it does turn out that the average power dissipated in the source
resistor is the sum of the average power before the reflection returns
plus the average power in the reflected wave since
Prs.circuit.average = average( 68 + 68 cos(2wt -61.92 degrees) )
= 68
This does agree with Cecil's analysis using average powers. But energy
flows must balance on a moment by moment basis if energy is to be
conserved so when we do the instantaneous analysis we find that
Cecil's
hypothesis does not hold.

...Keith

PS: To compute Vrs(t) and Irs(t) using circuit theory:
The generator output voltage
Vg(t) = Vf.g(t) + Vr.g(t)
where Vf.g(t) is the line forward voltage at the generator
and Vr.g(t) is the line reflected voltage at the generator.

The generator output current
Ig(t) = If.g(t) + Ir.g(t)
where If.g(t) is the line forward current at the generator
and Ir.g(t) is the line reflected current at the generator.

Where Vs(t) is the source voltage
Vrs(t) = Vs(t) - Vg(t)
Irs(t) = Ig(t)
and the power is
Prs.circuit(t) = Vrs(t) * Irs(t)


--
73, Roger, W7WKB

The Rest of the Story
 

Cecil Moore wrote:
"The definition of irradiance is "the average energy per unit time. Any
deviation away---."

As energy per unit time is power, Cecil`s definition agrees with what my
dictionary says:
"Irradiance-The incident radiated power per unit area of a surface; the
radiometric counterpart of illumination, usually expressed in watts/cm,
squared."

This is different from Poynting which uses instantaneous values.

Best regards, Richard Harrison. KB5WZI

The Rest of the Story
 

Roger Sparks wrote:
On Fri, 07 Mar 2008 15:15:08 GMT
Cecil Moore wrote:
My formula applies *only to average power*.

I find myself surprised at your insistance that the instantaneous
case must be seperated from the average case in our example of waves
on a transmission line.

You have misunderstood what I am objecting to - which is:
Please don't say or imply that I have said or implied
anything about instantaneous values. I have NOT done so!
The formulas and concepts being used in the discussion
of instantaneous values are NOT mine! My objection is
to the false statements being made about what I have
posted.

Please go ahead and have your instantaneous discussion with
Keith but please don't say or imply that anything associated
with that discussion is about anything I have posted. Saying
or implying such a thing is simply false.

It is informative to look at each problem from many angles.

I agree - just stop saying that it is based on my assertions.
You or Keith may be right or wrong but either way, it is not
associated with anything I have posted. Please leave my name
out of any discussion concerning instantaneous values.

Did I misunderstand your premise, and you were really trying to say that
the inclusion of a 50 ohm source resistor would prevent the source from
ever 'seeing' anything but a 100 ohm load? I don't think that was your
intent.

Please add a dimension to your thinking. No matter what the
value of the load resistor, the source delivers 100 watts.
There are an infinite number of loads that the source could
"see" besides 100+j0 ohms that will make that condition true.

When the load is 0 ohms, the source "sees" 50+j50 ohms.
When the load is 12.5 ohms, the source "sees" 73.5+j44.1 ohms.
When the load is 25 ohms, the source "sees" 90+j30 ohms.
Only when the load is 50 ohms, does the source "see" 100+j0 ohms.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Roger Sparks wrote:
Thanks for the additional explaination. I am wondering if I misunderstood Cecil's original premise.

Roger, if you thought it involved any instantaneous
values then, yes, you misunderstood my premise.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Richard Harrison wrote:
Cecil Moore wrote:
"The definition of irradiance is "the average energy per unit time. Any
deviation away---."

Richard, your newsreader dropped part of what I
wrote which was:

The definition of irradiance is the "average energy
per unit area per unit time".
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 7, 8:17 am, Gene Fuller wrote:
Cecil Moore wrote:

Coherency, non-coherency, and interference is covered well
in "Optics" by Hecht and other textbooks. Optical physicists
have been tracking the EM energy flow for centuries. This
information may be new to you but it is old hat in physics.

Cecil,

You may or may not already know this, but a lot of detailed optical
analysis these days is done with full 3-D electromagnetic simulation,
starting from Maxwell equations and boundary conditions. Interference,
coherence, energy flow, and all of the other stuff you like to discuss
can be *output* from that analysis, but those items are not part of the
input. The "centuries old" optics simply does not get the job done. The
"centuries old" stuff may work in the (impossible) cases where
everything is completely lossless and ideal, but it doesn't give the
right answers in the real world.

73,
Gene
W4SZ

You can sure say that again...in fact, Maxwell doesn't really do it
either when you get to quantum mechanical effects. But that's a story
for another day.

Certainly, those who design and build FTIR spectrometers know
perfectly well that interference does not depend on a narrow-band
coherent source. Blackbody radiation works just fine, thank you. But
it doesn't take much beyond belief in linear systems to understand
that. I recall explaining to a company VP how it worked in terms of a
linear system, and it was very gratifying to see the virtual light
bulb lighting up in his head...he really got it.

Cheers,
Tom

The Rest of the Story
 

K7ITM wrote:
Certainly, those who design and build FTIR spectrometers know
perfectly well that interference does not depend on a narrow-band
coherent source.

How narrow-band? How coherent? In the irradiance (power
density) equation, Ptot = P1 + P2 + 2*sqrt(P1*P2)cos(A),
if the angle 'A' is varying rapidly, what value do you
use for cos(A)?

A constant average sustained level of destructive
interference cannot be maintained between two waves
unless they are coherent. If they are not coherent
the interference will average out to zero.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

K7ITM wrote:
... interference does not depend on a narrow-band coherent source.

OK Tom, here's a challenge for you. Given a system with
a constant steady-state destructive interference magnitude.
Exactly how can that constant steady-state destructive
interference magnitude be maintained if the two interfering
signals are not coherent?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

K7ITM wrote:
And exactly which part of "linear system" do you fail to understand?

I understand the meaning of your question now and
here is one for you:

Exactly which part of a constant, average, steady-state
condition of destructive interference do you fail to
understand?

Given two coherent signals interfering whose results are
10 watts of constant, average, steady-state destructive
interference, how do you propose to accomplish that
outcome when the signals are not coherent?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 7, 1:16*pm, Cecil Moore wrote:
Roger Sparks wrote:
Thanks for the additional explaination. I am wondering if I misunderstood Cecil's original premise.

Roger, if you thought it involved any instantaneous
values then, yes, you misunderstood my premise.

My understanding of your claim was that for the special case of
a 45 degree line supplied from a matched source, the energy
in the reflected wave is dissipated in the source resistor.

This sentence fragment from your document suggests this:
"reflected energy from the load is flowing through the source
resistor, RS, and is being dissipated there".

As "proof" of this, you computed average powers and showed that
the dissipation in the source resistor increased by the same
amount as the computed average power in the reflected wave.

But when an attempt it made to validate your claim using
instantaneous energy flows, the claim is proved false because
the dissipation in the source resistor does not occur at the
correct time to be absorbing the energy from the reflected
wave.

To prove your claim, I can see two paths:
- find some element in the circuit that stores the energy from
the reflected wave and releases it into the source resistor
at the correct time
- allow the violation of the principle of conservation of
energy

On the other hand, if you want to modify your claim to simply
be that the numerical value of the dissipation in the source
resistor has the same value as the pre-reflection dissipation
plus the numerical value of the energy in the reflected wave,
then the discrepancy is resolved. But then you can not claim
that the energy in the reflected wave is dissipated in the
source resistor.

...Keith