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

Stand by for the other three articles.
--
73, Cecil http://www.w5dxp.com

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

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

On Mar 4, 8:00*pm, 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

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

I should have mentioned that there are some energy flows that do
add, as expected.

The energy delivered by the generator to the line (or equivalently,
the energy flowing in the line at the generator end) is the sum
of the forward and reverse energy flows...

Pf.g = 50 + 50cos(2wt)
Pr.g = -18 + 18cos(2wt)
Pline.g = 32 + 68cos(2wt)

And the energy delivered by the source is always equal to the
energy being dissipated in the resistor plus the energy being
delived to the line...

Prs = 68 + 68cos(2wt-61.9degrees)
Rline.g = 32 + 68cos(2wt)

Psource = 100 + 116.6cos(2wt-30.96degrees)

Psource is equal to Prs + Pline.g

So the energy flows that should add up, do add up.

...Keith

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

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

That is irrelevant. The power (irradiance) model
doesn't apply to instantaneous energy and power.
Hecht says as much in "Optics".

Nobody has ever claimed that the energy/power
analysis applies to instantaneous values. The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Keith Dysart wrote:
When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

That is irrelevant. The power (irradiance) model
doesn't apply to instantaneous energy and power.
Hecht says as much in "Optics".

Nobody has ever claimed that the energy/power
analysis applies to instantaneous values. The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

Interesting.

Do you also use only the RMS phase and RMS interference to come up with
your RMS answers?

8-)

73,
Gene
W4SZ

Gene Fuller wrote:
Cecil Moore wrote:
Keith Dysart wrote:
When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

That is irrelevant. The power (irradiance) model
doesn't apply to instantaneous energy and power.
Hecht says as much in "Optics".

Nobody has ever claimed that the energy/power
analysis applies to instantaneous values. The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

Interesting.

What is interesting is that in the formula for power
dissipated in the source resistor, the 50 watts is
an average power. It is an invalid procedure to try
to add instantaneous power to an average power.

Do you also use only the RMS phase and RMS interference to come up with
your RMS answers?

I didn't say anything about "RMS phase and RMS interference".
The phase angle used in Hecht's irradiance equation is the
phase between the electric fields of the two waves. The
magnitude of the interference is an average magnitude based
on the RMS values of voltage and current.

I do exactly what Eugene Hecht did in "Optics". He said:

"Furthermore, since the power
arriving cannot be measured instantaneously, the detector
must integrate the energy flux over some finite time, 'T'.
If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

i.e. The irradiance/interference equation does not work
for instantaneous powers which are "of limited utility".
--
73, Cecil http://www.w5dxp.com

On Mar 4, 10:25*pm, Cecil Moore wrote:
Keith Dysart wrote:
When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

That is irrelevant. The power (irradiance) model
doesn't apply to instantaneous energy and power.
Hecht says as much in "Optics".

Nobody has ever claimed that the energy/power
analysis applies to instantaneous values.

Are you saying that conservation of energy does NOT apply to
instantaneous values?

The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

I understand the analysis technique you are proposing. That
it leads to the wrong conclusion can be easily seen when the
instantaneous energy flows are studied. This merely demonstrates
that the analysis technique has its limitations.

The bottom line remains that the reflected energy is not
dissipated in the source resistor, even for the special cases
under discussion.

...Keith

Keith Dysart wrote:
....
The bottom line remains that the reflected energy is not
dissipated in the source resistor, even for the special cases
under discussion.

...Keith

What if the source resistor is of finite length
Alan

Keith Dysart wrote:
On Mar 4, 10:25 pm, Cecil Moore wrote:
Nobody has ever claimed that the energy/power
analysis applies to instantaneous values.

Are you saying that conservation of energy does NOT apply to
instantaneous values?

Of course not. I am saying that the 50 watts in the
source resistor power equation is an average power.
It is invalid to try to add instantaneous power to
an average power, as you tried to do.

The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

I understand the analysis technique you are proposing. That
it leads to the wrong conclusion can be easily seen when the
instantaneous energy flows are studied. This merely demonstrates
that the analysis technique has its limitations.

The proposed analysis technique is a tool. Trying to apply
that tool to instantaneous powers is like trying to use a
DC ohm-meter to measure the feedpoint impedance of an antenna.
Only a fool would attempt such a thing.

The bottom line remains that the reflected energy is not
dissipated in the source resistor, even for the special cases
under discussion.

Saying it doesn't make it so, Keith. There is nothing to
keep the reflected energy from being dissipated in the
source resistor.

If the reflected energy is not dissipated in the source
resistor, where does it go? Please be specific because it
is obvious to me that there is nowhere else for it to go
in the special zero interference case presented.

Here are some of your choices:

1. Reflected energy flows through the resistor and into
the ground without being dissipated.

2. The 50 ohm resistor re-reflects the reflected energy
back toward the load. (Please explain how a 50 ohm load
on 50 ohm coax can cause a reflection.)

3. There's no such thing as reflected energy.

4. Reflected waves exist without energy.

5. ______________________________________________.
--
73, Cecil http://www.w5dxp.com