Cecil Moore wrote:
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. ______________________________________________.


Gentlemen;

What I was taught long ago was that the phenomenon that we call
reflected energy is radiated on the return. That portion of energy that
is not radiated is re-reflected and is radiated. This continues until
the level of energy no longer supports radiation. Resistance also
dissipates a portion of the transmitted energy. That resistance includes
the resistor under discussion above and that resistance found in the
coax conductors.

Of course my Elmer could have been wrong.


Dave WD9BDZ

On Mar 5, 11:06*am, Cecil Moore wrote:
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.

You used your tool to attempt to show that the reflected power
is dissipated in Rs.

I did a finer grained analysis using instantaneous power to show
that it is not.

The use of averages in analysis can be misleading.

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. ______________________________________________.

Now you have got the issue. Since the reflected power is
not dissipated in Rs, the answer must be one of 1 to 5.

5. is probably the best choice.

And that is why it became necessary to rethink the nature
of energy in reflected waves.

...Keith

On Mar 5, 8:06 am, Cecil Moore wrote:
....blah, blah...

So consider the case of a section of lossless uniform transmission
line of characteristic impedance R0, which I write as R instead of Z
since it of course must be real-valued, connected between two sources
S1 at end 1 and S2 at end 2. These sources each have source impedance
R0: they are perfectly matched to the characteristic impedance of the
line. The line is long enough that we can observe any standing waves
that may be on it. (For believers in directional couplers, that can
be short indeed, but it does not need to be short.) Source S1 is set
to output a sinusoidal signal of amplitude A1 into a matched load, on
frequency f1. Similarly S2 outputs a sinusoidal signal A2 into a
matched load at frequency f2, which is distinct from f1.

It is easy to show mathematically, and to measure in practice, that
the amplitude of the frequency f1 is constant along the line, and
similarly that the amplitude of the frequency f2 is constant along the
line. That is to say, there is no standing wave at either frequency.
Energy at f1 travels on the line only in the direction from S1 to S2,
and vice-versa for f2.

That says to me that the energy on the line at f1 is absorbed entirely
by source S2, and the energy at f2 is absorbed entirely by S1, with no
reflection at the boundaries between S1 and the line, and the line and
S2.

At this point, I leave it as an exercise for the reader to interpret
or explain exactly what is meant by "absorbed by." This may involve
understanding that in a Thevenin or Norton simple model of each
source, the energy delivered by the voltage or current source at any
moment in time may not equal that which it would deliver into a
matched load at the same point in the cycle...

Cheers,
Tom

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

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

Keith Dysart wrote:
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.

Your analysis seems to be flawed. You are adding average
power terms to instantaneous power terms which is mixing
apples and oranges.
--
73, Cecil http://www.w5dxp.com

On Mar 5, 11:11*am, Cecil Moore wrote:
Keith Dysart wrote:
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.

Your analysis seems to be flawed. You are adding average
power terms to instantaneous power terms which is mixing
apples and oranges.

I do not think that is the case.

The expression for instantaneous power in Rs before the reflection
(or, if you prefer, when a 50 ohm load is used), is

Prs(t) = 50 + 50cos(2wt)

It is trivial to compute the average of this since the average
of a sine wave is 0, but that does not make the expression the
sum of an average and an instantaneous power.

As an exercise, compute the power in a 50 ohm resistor that has
a 100 volt sine wave across at, that is
V(t) = 100 cos(wt)

You will find the result is of the form shown above.

So when you add the instantaneous power in Rs before the reflection
arrives with the instantaneous power from the reflection it will
not sum to the instantaneous power dissipated in Rs after the
reflection returns.

Thus conveniently showing that for this example, the reflected
power is not dissipated in Rs.

...Keith

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

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.

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

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