The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

There are no reflections at the source so the reflected
energy flows through the source resistor. There is no
interference to redistribute any energy. There is no
other place for the reflected energy to go.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

On Mar 4, 10:27*pm, Cecil Moore wrote:
Keith Dysart wrote:
Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

There are no reflections at the source so the reflected
energy flows through the source resistor. There is no
interference to redistribute any energy. There is no
other place for the reflected energy to go.

That is the conundrum, isn't it?

And yet the analysis of instantaneous energy flows definitely
shows that the reflected energy is not the energy being dissipated
in the source resistor.

Encountering this conundrum, and not wanting to give up on
conservation of energy, is what helped me form my views on
the nature of reflected energy.

...Keith

The Rest of the Story
 

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

The Rest of the Story
 

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.

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


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.
--
73, Roger, W7WKB