On Mar 16, 10:21*am, Cecil Moore wrote:
Keith Dysart wrote:
And, as expected
Ps(t) = Prs(t) + Pg(t)

Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

* * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted
* * *100v RMS * * * * * * *50 ohm line * * * * | Stub
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+
* * * * gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?

Vs(t) = 1.414cos(wt)

After settling, from some circuit analysis...

Vg(t) = 100 cos(wt+45degrees)
Ig(t) = 2 cos(wt-45degrees)

Vrs(t) = Vs(t) - Vg(t)
= 100 cos(wt-45degrees)
Irs(t) = Ig(t)
= 2 cos(wt-45degrees)

Is(t) = Ig(t)
= 2 cos(wt-45degrees)

Ps(t) = Vs(t) * Is(t)
= 100 + 141.4213562 cos(2wt-45degrees)

Prs(t) = Vrs(t) * Irs(t)
= 100 + 100 cos(2wt-90degrees)

Pg(t) = Vg(t) * Ig(t)
= 0 + 100 cos(2wt)

For confirmation of conservation of energy, the above
is in agreement with
Ps(t) = Prs(t) + Pg(t)

Ps(t) = 100 + 141.4213562 cos(2wt-45degrees)
so, Ps(t) = 0, occurs whenever
141.4213562 cos(2wt-45degrees)
is equal to
-100
which, for example, would happen when 2wt = 180 degrees
or wt = 90 degrees.

At this time, again for example, the energy being
dissipated in the source resistor
Prs(t) = 100 + 100 cos(2wt-90degrees)
= 100 + 0
= 100

Since no energy is being delivered from the source,
(Ps(t) is 0), then given
Ps(t) = Prs(t) + Pg(t)
the energy must be coming from the line. Let us
check the energy flow at the point Vg at this
time
Pg(t) = 0 + 100 cos(2wt)
= -100
as required.

So the energy being dissipated in the source
resistor at this time is being returned from
the line.

Just for completeness we can compute the line
state at the point Vg in terms of forward and
reverse waves...

Vf.g(t) = 70.71067812 cos(wt)
Vr.g(t) = 70.71067812 cos(wt+90degrees)
Vg(t) = Vf.g(t) + Vr.g(t)
= 100 cos(wt+45degrees)

If.g(t) = 1.414213562 cos(wt)
Ir.g(t) = 1.414213562 cos(wt-90degrees)
Ig(t) = If.g(t) + Ir.g(t)
= 2 cos(wt-45degrees)

Pf.g(t) = Vf.g(t) * If.g(t)
= 50 + 50 cos(2wt)

Pr.g(t) = Vr.g(t) * Ir.g(t)
= -50 + 50 cos(2wt)

And since
Pg(t) = Pf.g(t) + Pr.g(t)
= 0 + 100 cos(2wt)
confirming the previously computed values.

It is valuable to examine Pr.g(t) at the time when
Ps(t) is zero. Substituting wt = 90degrees
into Pr.g(t)...

Pr.g(t) = -50 + 50 cos(2wt)
= -50 -50
= -100
which would appear to be the 100 watts needed to
heat the source resistor. This is misleading.
When all the values of t are examined it will be
seen that only the sum of Pf.g(t) and Pr.g(t),
that is, Pg(t), provides the energy not provided
from the source that heats the source resistor.
wt=90 is a special case in that Pf.g(t) is 0 at
this particular time.

The line input impedance does have a reactive
component and it is this reactive component that
can store and return energy. The energy flow
into and out of this impedance (pure reactance
for the example under consideration) is described
by Pg(t).

In summary,
Ps(t) = Prs(t) + Pg(t)

and by substitution, if the solution is preferred
in terms of Pf and Pr,
Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t)

Ps and Pr alone are insufficient to explain the
heating of the source resistor.

...Keith

Keith Dysart wrote:
Pr.g(t) = -50 + 50 cos(2wt)
= -50 -50
= -100
which would appear to be the 100 watts needed to
heat the source resistor.

Which, contrary to your previous assertions, agrees
with what I have said previously. This is the
destructive interference energy stored in the
feedline 90 degrees earlier and being returned
as constructive interference to the source
resistor. It's impossible to sweep it under the
rug when the source power is zero, huh?

All power comes from the source. Since power is
being delivered to the source resistor at a time
when the source is delivering zero power, it must
have been previously been stored in the reactance
of the transmission line.
--
73, Cecil http://www.w5dxp.com

On Mar 17, 1:06*am, Cecil Moore wrote:
Keith Dysart wrote:
Pr.g(t) = -50 + 50 *cos(2wt)
* * * * = -50 -50
* * * * = -100
which would appear to be the 100 watts needed to
heat the source resistor.

Which, contrary to your previous assertions, agrees
with what I have said previously. This is the
destructive interference energy stored in the
feedline 90 degrees earlier and being returned
as constructive interference to the source
resistor. It's impossible to sweep it under the
rug when the source power is zero, huh?

The intriguing question is:
1. Do you just stop reading as soon as you find a snippet
that aligns with your claim?
2. Do you keep reading, but do not understand because you
are so gleeful at finding a snippet that aligns with
your claim?
3. Or do you read, understand, but choose to disingenuously
ignore that which follows the snippet that aligns with
your claim?

Regardless, if you use the snippet above to support
your claim, you have effectively modified your claim.
You are now saying that the reflected energy is
dissipated in the source resistor only at particular
times, such as when the source voltage is 0, and that
you are not interested in what happens during the rest
of the cycle.

For example, if you were to do the same analysis,
except do it for t such that wt equals 100 degrees,
instead of 90, you would find that you need more terms
than just Pr.g to make the energy flows balance.

When wt equals 100 degrees
Ps = -28.170
Prs = 65.798
Pg = -93.96
so, as expected
Ps = Prs + Pg

And
Pr.g = -96.985
Pf.g = 3.015
Ooooppppss, no way to make those add to 65.798.

But
Ps = Prs + Pf.g + Pr.g
since
Pg = Pf.g + Pr.g

So all is well with world.

All power comes from the source. Since power is
being delivered to the source resistor at a time
when the source is delivering zero power, it must
have been previously been stored in the reactance
of the transmission line.

This is true. But it is, of course, Pg(t) that
describes the energy flow in and out of the line,
not Pr.g(t).

So are you now agreeing that it is not the energy
in the reflected wave that accounts for the
difference in the heating of the source resistor
but, rather, the energy stored and returned from
the line, i.e. Pg(t)?

Note also, that when wt is 100 degrees, not only is
the source resistor being heated by energy being
returned from the line, the source is also absorbing
some of the energy being returned from the line
(Ps = -28.170).

...Keith

Keith Dysart wrote:
Regardless, if you use the snippet above to support
your claim, you have effectively modified your claim.

False. My claim is what it has always been which is:
An amateur radio antenna system obeys the conservation
of energy principle and abides by the principles of
superposition (including interference) and the wave
reflection model. Everything I have claimed falls out
from those principles.

Your claims, however, are in direct violation of the
principles of superposition and of the wave reflection
model, e.g. waves smart enough to decide to be reflected
when the physical reflection coefficient is 0.0. Your
claims even violate the principles of AC circuit theory,
e.g. a reactance doesn't store energy and deliver it back
to the system at a later time in the same cycle.

You are now saying that the reflected energy is
dissipated in the source resistor only at particular
times, such as when the source voltage is 0, and that
you are not interested in what happens during the rest
of the cycle.

You haven't read my article yet, have you? Here's a quote:
"For this *special case*, it is obvious that the reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there. But remember, we chose a special
case (resistive RL and 1/8 wavelength feedline) in order to
make that statement true and it is *usually not true* in the
general case."

If there is one case where your assertion is wrong, then
your assertion is false. I found that special case when
the source voltage is zero that makes your assertions
false.

For example, if you were to do the same analysis,
except do it for t such that wt equals 100 degrees,
instead of 90, you would find that you need more terms
than just Pr.g to make the energy flows balance.

Yes, you have realized that destructive and constructive
interference energy must be accounted for to balance the
energy equations. I have been telling you that for weeks.

I repeat: The *only time* that reflected energy is 100%
dissipated in the source resistor is when the two component
voltages satisfy the condition: (V1^2 + V2^2) = (V1 + V2)^2.
None of your examples have satisfied that necessary condition.
All it takes is one case to prove the following assertion false:

"Reflected energy is *always* re-reflected from the
source and redistributed back toward the load."

You appear to think that if you can find many cases where
an assertion is true, then you can simply ignore the cases
where it is not true. I have presented some special cases
where it is not true. It may be true for 99.9% of cases,
but that nagging 0.1% makes the statement false overall.

Equally false is the assertion: "Reflected energy is
always dissipated in the source resistor." The amount
of reflected energy dissipated in the source resistor
can vary from 0% to 100% depending upon network
conditions. That statement has been in my article from
the beginning.

So are you now agreeing that it is not the energy
in the reflected wave that accounts for the
difference in the heating of the source resistor
but, rather, the energy stored and returned from
the line, i.e. Pg(t)?

I have already presented a case where there is *zero*
power dissipated in the source resistor in the presence
of reflected energy so your statement is obviously just
false innuendo, something I have come to expect from
you when you lose an argument.
--
73, Cecil http://www.w5dxp.com

On Mar 17, 10:05*am, Cecil Moore wrote:
Keith Dysart wrote:
Regardless, if you use the snippet above to support
your claim, you have effectively modified your claim.

False. My claim is what it has always been which is:
An amateur radio antenna system obeys the conservation
of energy principle and abides by the principles of
superposition (including interference) and the wave
reflection model.

Well who could argue with that. But don't all systems
obey? And why limit it to amateur?

Everything I have claimed falls out from those principles.

But the question then becomes "Have the prinicples been
correctly applied?"

Your claims, however, are in direct violation of the
principles of superposition and of the wave reflection
model, e.g. waves smart enough to decide to be reflected
when the physical reflection coefficient is 0.0. Your
claims even violate the principles of AC circuit theory,
e.g. a reactance doesn't store energy and deliver it back
to the system at a later time in the same cycle.

An intriguing set of assertions. It would be good if you
could point out the equations that are in violation.

Is there an error in either
Ps(t) = Prs(t) + Pg(t)
or
Pg(t) = Pf.g(t) + Pr.g(t)

Both by derivation and by example, these seem to be true.

Do you disagree?

You are now saying that the reflected energy is
dissipated in the source resistor only at particular
times, such as when the source voltage is 0, and that
you are not interested in what happens during the rest
of the cycle.

You haven't read my article yet, have you? Here's a quote:
"For this *special case*, it is obvious that the reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there.

I would suggest that it is not 'obvious'. Some times the
energy from the reflected wave is being absorbed in the
source.

'Obvious' is often an excuse for an absence of rigour.

But remember, we chose a special
case (resistive RL and 1/8 wavelength feedline) in order to
make that statement true and it is *usually not true* in the
general case."

If there is one case where your assertion is wrong, then
your assertion is false. I found that special case when
the source voltage is zero that makes your assertions
false.

Which assertion is wrong?
Ps(t) = Prs(t) + Pg(t) ?
Pg(t) = Pf.g(t) + Pr.g(t) ?

For example, if you were to do the same analysis,
except do it for t such that wt equals 100 degrees,
instead of 90, you would find that you need more terms
than just Pr.g to make the energy flows balance.

Yes, you have realized that destructive and constructive
interference energy must be accounted for to balance the
energy equations. I have been telling you that for weeks.

You do use the words, but do not offer any equations that
describe the behaviour. The claim is thus quite weak.

I repeat: The *only time* that reflected energy is 100%
dissipated in the source resistor is when the two component
voltages satisfy the condition: (V1^2 + V2^2) = (V1 + V2)^2.
None of your examples have satisfied that necessary condition.
All it takes is one case to prove the following assertion false:

"Reflected energy is *always* re-reflected from the
source and redistributed back toward the load."

I have never made that claim. So if that is your concern, I agree
completely that it is false.

My claim is that "the energy in the reflected wave can not usually be
accounted for in the source resistor dissipation, and that this is
especially so for the example you have offerred".

You appear to think that if you can find many cases where
an assertion is true, then you can simply ignore the cases
where it is not true. I have presented some special cases
where it is not true. It may be true for 99.9% of cases,
but that nagging 0.1% makes the statement false overall.

Having never claimed the truth of the statement, I have
never attempted to prove it.

Equally false is the assertion: "Reflected energy is
always dissipated in the source resistor." The amount
of reflected energy dissipated in the source resistor
can vary from 0% to 100% depending upon network
conditions. That statement has been in my article from
the beginning.

But you have claimed special cases where the energy
*is* dissipated in the source resistor.

Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)
demonstrate this to be false for the example you have offerred.

So are you now agreeing that it is not the energy
in the reflected wave that accounts for the
difference in the heating of the source resistor
but, rather, the energy stored and returned from
the line, i.e. Pg(t)?

I have already presented a case where there is *zero*
power dissipated in the source resistor in the presence
of reflected energy so your statement is obviously just
false innuendo, something I have come to expect from
you when you lose an argument.

Since it was a question, you were being invited to clarify
your position. But it begs the question:
When zero energy is being dissipated in the source resistor
in the presence of reflected energy, where does that reflected
energy go?

Recalling that by substitution
Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t)
so that when the resistor dissipation is 0 for certain values of t
Ps(t) = 0 + Pf.g(t) + Pr.g(t)
= Pf.g(t) + Pr.g(t)
Pr.g(t) = Ps(t) - Pf.g(t)

Please provide equations that describe how the reflected energy is
split between the source and the forward wave.
For completeness, these equations should generalize to describe
the splitting of the energy for all values of t, not just those
where the dissipation in the source resistor is 0.

...Keith

Keith Dysart wrote:
My claim is that "the energy in the reflected wave can not usually be
accounted for in the source resistor dissipation, and that this is
especially so for the example you have offerred".

The instantaneous example to which you are referring does
not meet my special case requirement of *ZERO INTERFERENCE*
so your above statement is just another straw man and is
thus irrelevant to my claim. But you already know that.

I agree with you that "the energy in the reflected wave can
not usually be accounted for in the source resistor dissipation."
I have stated over and over that the reflected energy
dissipation in the source resistor can range from 0% to 100%.

My claim is that when the special case of *ZERO INTERFERENCE*
exists between the two voltages superposed at the source resistor,
then 100% of the reflected energy is dissipated in the source
resistor. You have not provided a single example which proves
my claim false.

The test for *ZERO INTERFERENCE* is when, given the two
voltages, V1 and V2, superposed at the source resistor,

(V1^2 + V2^2) = (V1 + V2)^2

None of your examples have satisfied that special case
condition. My average reflected power example does NOT
satisfy that condition for instantaneous power!

Here's the procedure for proving my claim to be false.
Write the equations for V1(t) and V2(t), the two instantaneous
voltages superposed at the source resistor. Find the time
when [V1(t)^2 + V2(t)^2] = [V1(t) + V2(t)]^2. Calculate
the instantaneous powers *at that time*. You will find that,
just as I have asserted, 100% of the instantaneous reflected
energy is dissipated in the source resistor.

Until you satisfy my previously stated special case condition
of *ZERO INTERFERENCE*, you cannot prove my assertions to be
false.
--
73, Cecil http://www.w5dxp.com

On Mar 19, 11:47*am, Cecil Moore wrote:
Keith Dysart wrote:
My claim is that "the energy in the reflected wave can not usually be
accounted for in the source resistor dissipation, and that this is
especially so for the example you have offerred".

The instantaneous example to which you are referring does
not meet my special case requirement of *ZERO INTERFERENCE*
so your above statement is just another straw man and is
thus irrelevant to my claim. But you already know that.

I agree with you that "the energy in the reflected wave can
not usually be accounted for in the source resistor dissipation."
I have stated over and over that the reflected energy
dissipation in the source resistor can range from 0% to 100%.

My claim is that when the special case of *ZERO INTERFERENCE*
exists between the two voltages superposed at the source resistor,
then 100% of the reflected energy is dissipated in the source
resistor. You have not provided a single example which proves
my claim false.

The test for *ZERO INTERFERENCE* is when, given the two
voltages, V1 and V2, superposed at the source resistor,

(V1^2 + V2^2) = (V1 + V2)^2

None of your examples have satisfied that special case
condition. My average reflected power example does NOT
satisfy that condition for instantaneous power!

Here's the procedure for proving my claim to be false.
Write the equations for V1(t) and V2(t), the two instantaneous
voltages superposed at the source resistor. Find the time
when [V1(t)^2 + V2(t)^2] = [V1(t) + V2(t)]^2. Calculate
the instantaneous powers *at that time*. You will find that,
just as I have asserted, 100% of the instantaneous reflected
energy is dissipated in the source resistor.

Until you satisfy my previously stated special case condition
of *ZERO INTERFERENCE*, you cannot prove my assertions to be
false.
--
73, Cecil *http://www.w5dxp.com

In a previous iteration, I had accepted that your claim only
applied when there was zero interference. I converted that to
the particular values of wt for which your claim aaplied,
e.g. 90 degrees.

I then restated your claim as applying only at those particular
times and not at other points in the cycle, but you were
unhappy with that limitation.

So you have to choose...
Does it only apply at those particular points in the cycle
where wt is equal to 90 degrees or appropriate multiples?
(A restatement of only applying when there is no interference.)
or
Does it apply at all times in the cycle?

...Keith

Keith Dysart wrote:
On Mar 17, 10:05 am, Cecil Moore wrote:
My claim is what it has always been which is:
An amateur radio antenna system obeys the conservation
of energy principle and abides by the principles of
superposition (including interference) and the wave
reflection model.

Well who could argue with that.

Well, of course, you do when you argue with me. For
instance, you believe that reflections can occur when
the reflections see a reflection coefficient of 0.0,
i.e. a source resistance equal to the characteristic
impedance of the transmission line. This obviously
flies in the face of the wave reflection model. When
you cannot balance the energy equations, you are
arguing with the conservation of energy principle.
--
73, Cecil http://www.w5dxp.com

On Mar 19, 3:16*pm, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 17, 10:05 am, Cecil Moore wrote:
My claim is what it has always been which is:
An amateur radio antenna system obeys the conservation
of energy principle and abides by the principles of
superposition (including interference) and the wave
reflection model.

Well who could argue with that.

Well, of course, you do when you argue with me.

Now there is an ego. Anyone arguing with you is
definitely against conservation of energy. Amusing.

For
instance, you believe that reflections can occur when
the reflections see a reflection coefficient of 0.0,
i.e. a source resistance equal to the characteristic
impedance of the transmission line.

You won't find anywhere that I said that. In fact, I
was quite pleased that I had helped you relearn this
tidbit from your early education which previous posts
made clear you had forgotten. Did you eventually look up "reflection",
"lattice" or "bounce diagram"?

This obviously
flies in the face of the wave reflection model. When
you cannot balance the energy equations, you are
arguing with the conservation of energy principle.

But I notice that you have not yet indicated which
energy equation I may have written that was unbalanced.
You do have a nack with unfounded assertions, don't
you?

...Keith

On Mon, 17 Mar 2008 03:00:10 -0700 (PDT)
Keith Dysart wrote:

On Mar 17, 1:06*am, Cecil Moore wrote:
Keith Dysart wrote:
Pr.g(t) = -50 + 50 *cos(2wt)
* * * * = -50 -50
* * * * = -100
which would appear to be the 100 watts needed to
heat the source resistor.


Thanks for a comprehensive analysis Keith. It was helpful to me to see how you step by step analyzed the circuit.

--
73, Roger, W7WKB