what happens to reflected energy ?
 

On Jun 27, 1:38*pm, Keith Dysart wrote:
In any region, the energy flowing in (i.e. power) to the region minus
the energy flowing out (i.e. power) is equal to the additional energy
per unit time (i.e. power) being stored in the region. While not
called the "conservation of power law" it is an obvious corollary
to "conservation of energy".

I'm sorry, that is simply not true for power. The energy content of a
1us pulse containing one joule and the energy content of a one sec
pulse containing one joule are equal and that one joule is all that
must be conserved. The 1us pulse containing 1,000,000 watts can be
converted to a one second pulse containing 1 watt. Where did the other
999,999 watts go??? Energy has been conserved but the power changed
from 1,000,000 watts to 1 watt using exactly the same energy. Perhaps
this characteristic of power is what you are missing. Also, all the
energy can be conserved in reactance while power falls to absolute
zero. This often happens during a fraction of a cycle. That is what is
wrong with you trying to track instataneous power - it doesn't work
unless one standardizes to at least one cycle. Within each fraction of
a cycle, any principle of conservation of power will surely be
violated. If it appears that power is ever conserved, it is only by
accident. Such is the case with many megacycles/second where the
result of a fraction of a cycle will have a negligible effect on the
joules/sec.

The obvious alternative is that the computed energy in the reflected
wave is sometimes just a figment.

And God created the heavens and earth in six days and rested on the
seventh. I'm glad you are happy with your faith-based physics. In the
field of real-world physics, EM waves cannot exist without ExH energy.
The only way to win this argument is to prove to everyone that they
are not really detecting reflected waves containing energy when they
look at themselves in the mirror. Good luck on that one.

Question: How were the first three days measured before the creation
of the sun on the 4th day?

Not to mention that in your 1/8 wavelength example (http://www.w5dxp.com/nointfr.htm)
you do not explain where the energy is stored so that it can be
returned at a different time.

Energy is stored in the transmission line and delivered as needed to
satisfy the conservaton of energy principle. Years ago, I showed how
energy can flow *into the source* (negative power) during a fractional
part of a cycle in a conjugately matched system.

Such declarations do permit an easy out, despite not aligning with
reality.

If you can take one joule per microsecond (1 megawatt) and conserve
that one megawatt of power over a century, you can get rich selling
it. Let us know when you get your patent on conservation of power. :-)
Good Grief!

If that is the case, the whole concept of reflected energy seems
somewhat bogus. Over a whole cycle, the power delivered by the
generator is passed on towards the load. If that is all you want
to know, then there is no need at all for "reflected power".

But, as you can grok from the subject of this thread, that is not all
that is needed to know. The last gasp of the loser is that it didn't
matter anyhow. Reflected energy has always mattered to optical
physicists who know it obeys the laws of physics. Now it seems to
matter to some hams. If it doesn't matter to you, why do you continue
posting?

And to stop besmirching Hecht, it seems most probable that his
comment was in the context of optics. After all, the book had that
title.

Hint: RF waves are covered in every physics book whose title is
"Light". There is absolutely no difference, from a physics standpoint,
between a coherent light wave and a coherent RF wave except for
frequency. The both obey exactly the same laws of physics which you
seem to concede for visible light but not for light at RF frequencies.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 27, 2:23*pm, Keith Dysart wrote:
Example 1: Step function applied to a transmission line. After the
* * * * * *line settles, a forward and reflected voltage wave
* * * * * *continue on the line but no energy is being transferred.

As far as I am concerned, if Maxwell's equations don't work on an
example, it might as well be ignored. There is nothing during DC
steady-state that allows Maxwell's equations to work because there are
no EM waves during DC steady-state. Why don't you already know that?

I can take your approach and do you one better. Please prove that you
exist. If you cannot prove that you exist, then nothing you say is of
any consequence. See, I can do it also.

Example 2: On a line with infinite VSWR no energy crosses a
* * * * * *voltage minimum or maximum.

Completely false assumption. You are back to asserting that since the
north-bound traffic equals the south-bound traffic on the Golden Gate
Bridge that there is no traffic and no bridge maintenance is required.
When are you going to give up on that irrational wet dream of yours?
No *NET* energy crosses at a voltage zero or current zero point. That
doesn't make the north-bound energy equal to zero and doesn't make the
south-bound energy equal to zero. It just makes them equal. Just
because there is no NET traffic flow on the Golden Gate Bridge doesn't
mean there is zero traffic flow in both directions. Please stop
clowning around with such absurb notions.

Example 3: With the 1/8 wavelength line described in
* * * * * *http://www.w5dxp.com/nointfr.htmthe energy can not be
* * * * * *properly accounted for on a moment by moment basis..

There is no conservation of power principle. If you would track the RF
joules and the conversion of RF joules to heat instead of the joules/
second, everything would become clear to you. As it is, you are
laboring under some serious misconceptions about the laws of physics.
Power simply doesn't balance within a single cycle - because it
doesn't have to - because there is no conservation of power principle.

People who don't learn from their mistakes are doomed to commit the
same mistakes over and over. Keith, you seem to be all output and no
input. Please enable your input channels for a change.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 27, 2:20 pm, K1TTT wrote:
but the equivalent points out that your statements about it sourcing
constant power is incorrect.

You might like to actually test it. The two sources are always
delivering 100W.

With an open circuit load:
The current source is delivering 100 W to the parallel source
resistor.
The voltage source delivers nothing.
Total: 100W

With a shorted load:
The voltage source is delivering 100W to the series source resistor.
The current source delivers nothing.
Total: 100W

With a 50 ohm load:
The voltage source is delivering 50W.
The current source is delivering 50W.
Total: 100W
The load is receiving 50W.
Each generator resistor is dissipating 25W.

i have also pointed out that your statements about your 'step wave'
are obviously incorrect because you have applied assumptions that are
only valid in the sinusoidal steady state to a step function that can
never be in steady state.

I am not sure where you think there is an error. Perhaps you can
point them out in the following example:

Generator:
- 100V step in to an open circuit
- 50 ohm source impedance
Line:
- 50 ohm
- open circuit

Generator is commanded to produce a step.
This will produce 50 V and 1 A at the line input which will
propagate down the line.
The open end of the line has a reflection co-efficient of 1.0.
Just before the 50 V step reaches the end of the line, the
whole line will be at 50 V and 1 A will be flowing everywhere.
The 50 V step hits the end and is reflected, producing a 50 V
step (on top of the 50V already there) which propagates back
to the generator. In front of the 50 V step, the current is
still 1 A (which provides the charge necessary to produce
the reverse propagating 50 V step. Behind the step, the
current is 0.
When the reverse 50 V step (which is actually a step from
50V to 100V) reaches the generator, the source impedance
matches the line impedance so there is no further reflection.
The line state is now 100V and 0A all along its length.

The settling time was one round-trip.

The generator is still producing the step, so the forward
step voltage wave is still 'flowing' and being reflected so
there is still a reflected step voltage wave, each of 50 V.

Since the generator open circuit voltage is 100 V and the
line voltage is now 100 V, current is no longer flowing
from the generator to the line.

Does this agree with your understanding?

I have snipped the rest of your post since until the above
is agreed, there is no sense in proceeding further.

....Keith

what happens to reflected energy ?
 

On Jun 27, 2:26*pm, Cecil Moore wrote:
On Jun 27, 12:49*pm, Keith Dysart wrote:

Cecil simply sidesteps these little inconveniences by refusing to
consider anything other than sinusoidal RF excitation and by
refusing to consider any time based analysis.

That's simply false. Using Fourier analysis, I reduce anything other
than a sinusoid to multiple sinusoidal RF excitations, perform the
sinusoidal analysis, and then use superposition to find the answer.

You need to expand your solution space. Some problems are so much
easier to solve in the time domain.

I can't even imagine doing some problems in the frequency domain.
Let's see: I turn on my flashlight maybe once per week, so the
fundamental is 1.6e-6 Hz, and say the risetime is 1 millisecond
-- my head hurts already -- that's about 600,000,000 harmonics to
be computed. No wonder you give up on some problems so readily.

I also reject any example where Maxwell's equations do not work. Your
insistance that magical waves can somehow exist during DC steady-state
violates the known laws of physics.

It is not my insistence. It follows from the math.

Besides if you convert it to the frequency domain you should
be happy that they exist since they then align with your
understandings.

EM waves CANNOT exist during DC
steady-state because electrons are traveling at a constant velocity.

Are you sure you meant this? The electron velocity changes? Or did
you mean the wave velocity? Nope. That does not work either.

You can measure DC voltage with an AC voltmeter but that doesn't
change DC voltage to AC voltage.

And for especial fun... Why are you sure DC is so special?
v(t)=A cos(wt)
describes a sinusoid. It has the parameter w to specifiy the
frequency. Set it to 0, and voila: DC. It falls right out
of the same definition as is used for a sinusoid. It is a
sinusoid.

....Keith

what happens to reflected energy ?
 

On 27 jun, 17:39, Keith Dysart wrote:
On Jun 27, 2:20 pm, K1TTT wrote:

but the equivalent points out that your statements about it sourcing
constant power is incorrect.

You might like to actually test it. The two sources are always
delivering 100W.

With an open circuit load:
The current source is delivering 100 W to the parallel source
resistor.
The voltage source delivers nothing.
Total: 100W

With a shorted load:
The voltage source is delivering 100W to the series source resistor.
The current source delivers nothing.
Total: 100W

With a 50 ohm load:
The voltage source is delivering 50W.
The current source is delivering 50W.
Total: 100W
The load is receiving 50W.
Each generator resistor is dissipating 25W.

i have also pointed out that your statements about your 'step wave'
are obviously incorrect because you have applied assumptions that are
only valid in the sinusoidal steady state to a step function that can
never be in steady state.

I am not sure where you think there is an error. Perhaps you can
point them out in the following example:

Generator:
- 100V step in to an open circuit
- 50 ohm source impedance
Line:
- 50 ohm
- open circuit

Generator is commanded to produce a step.
This will produce 50 V and 1 A at the line input which will
propagate down the line.
The open end of the line has a reflection co-efficient of 1.0.
Just before the 50 V step reaches the end of the line, the
whole line will be at 50 V and 1 A will be flowing everywhere.
The 50 V step hits the end and is reflected, producing a 50 V
step (on top of the 50V already there) which propagates back
to the generator. In front of the 50 V step, the current is
still 1 A (which provides the charge necessary to produce
the reverse propagating 50 V step. Behind the step, the
current is 0.
When the reverse 50 V step (which is actually a step from
50V to 100V) reaches the generator, the source impedance
matches the line impedance so there is no further reflection.
The line state is now 100V and 0A all along its length.

The settling time was one round-trip.

The generator is still producing the step, so the forward
step voltage wave is still 'flowing' and being reflected so
there is still a reflected step voltage wave, each of 50 V.

Since the generator open circuit voltage is 100 V and the
line voltage is now 100 V, current is no longer flowing
from the generator to the line.

Does this agree with your understanding?

I have snipped the rest of your post since until the above
is agreed, there is no sense in proceeding further.

...Keith

Sorry, I ommited aknowledge to you I understoond your example and
what you mean with:

"What happens to Vfor1(rho) = 50v(0.7143) = 35.7v? "
"What happens to Pfor1(rho^2) = 50w(0.51) = 25.5w? "

(Last represent the cancellating (interference?) term of transmitted
power (1.Rho^2) towards generator from the reflected power from the
load to render VRef1=0 (doing the accounts with phasorial V-I math,
though I suppose will give similar results employing the eq-1 of your
World Radio article, but I am not sure if I'm catching it very well
yet. Have you P1, P2, P3 and P4, for your 100 W example, to clear
it? )

Miguel

what happens to reflected energy ?
 

On 27 jun, 19:05, lu6etj wrote:
On 27 jun, 17:39, Keith Dysart wrote:





On Jun 27, 2:20 pm, K1TTT wrote:

but the equivalent points out that your statements about it sourcing
constant power is incorrect.

You might like to actually test it. The two sources are always
delivering 100W.

With an open circuit load:
The current source is delivering 100 W to the parallel source
resistor.
The voltage source delivers nothing.
Total: 100W

With a shorted load:
The voltage source is delivering 100W to the series source resistor.
The current source delivers nothing.
Total: 100W

With a 50 ohm load:
The voltage source is delivering 50W.
The current source is delivering 50W.
Total: 100W
The load is receiving 50W.
Each generator resistor is dissipating 25W.

i have also pointed out that your statements about your 'step wave'
are obviously incorrect because you have applied assumptions that are
only valid in the sinusoidal steady state to a step function that can
never be in steady state.

I am not sure where you think there is an error. Perhaps you can
point them out in the following example:

Generator:
- 100V step in to an open circuit
- 50 ohm source impedance
Line:
- 50 ohm
- open circuit

Generator is commanded to produce a step.
This will produce 50 V and 1 A at the line input which will
propagate down the line.
The open end of the line has a reflection co-efficient of 1.0.
Just before the 50 V step reaches the end of the line, the
whole line will be at 50 V and 1 A will be flowing everywhere.
The 50 V step hits the end and is reflected, producing a 50 V
step (on top of the 50V already there) which propagates back
to the generator. In front of the 50 V step, the current is
still 1 A (which provides the charge necessary to produce
the reverse propagating 50 V step. Behind the step, the
current is 0.
When the reverse 50 V step (which is actually a step from
50V to 100V) reaches the generator, the source impedance
matches the line impedance so there is no further reflection.
The line state is now 100V and 0A all along its length.

The settling time was one round-trip.

The generator is still producing the step, so the forward
step voltage wave is still 'flowing' and being reflected so
there is still a reflected step voltage wave, each of 50 V.

Since the generator open circuit voltage is 100 V and the
line voltage is now 100 V, current is no longer flowing
from the generator to the line.

Does this agree with your understanding?

I have snipped the rest of your post since until the above
is agreed, there is no sense in proceeding further.

...Keith

Sorry, *I ommited aknowledge to you I understoond your example and
what you mean with:

"What happens to Vfor1(rho) = 50v(0.7143) = 35.7v? "
"What happens to Pfor1(rho^2) = 50w(0.51) = 25.5w? "

(Last represent the cancellating (interference?) term of transmitted
power (1.Rho^2) towards generator from the reflected power from the
load to render VRef1=0 (doing the accounts with phasorial V-I math,
though I suppose will give similar results employing the eq-1 of your
World Radio article, but I am not sure if I'm catching it very well
yet. Have you P1, P2, P3 and P4, for your 100 W example, to clear
it? *)

Miguel- Ocultar texto de la cita -

- Mostrar texto de la cita -

Or if you prefer, tell me if in your article: P1=48.98 W; P2=53.15 W;
P3=51.02 W; P4=51.02 W

what happens to reflected energy ?
 

On Jun 27, 4:01*pm, Cecil Moore wrote:
On Jun 27, 1:38*pm, Keith Dysart wrote:

In any region, the energy flowing in (i.e. power) to the region minus
the energy flowing out (i.e. power) is equal to the additional energy
per unit time (i.e. power) being stored in the region. While not
called the "conservation of power law" it is an obvious corollary
to "conservation of energy".

I'm sorry, that is simply not true for power. The energy content of a
1us pulse containing one joule and the energy content of a one sec
pulse containing one joule are equal and that one joule is all that
must be conserved. The 1us pulse containing 1,000,000 watts can be
converted to a one second pulse containing 1 watt.

Perhaps some of your difficulty is revealed in your phraseology. A
pulse does not 'contain' power. It can deliver energy at some rate.
If the pulse is rectangular, the rate will be constant for the
duration of the pulse. With some other profile, the rate will
vary over the duration of the pulse.

Perhaps a simple analogy would help. Near my house is a 50 m water
tower with a bunch of pipes connected to the bottom. The rate at
which water is added to the tower is always equal to the sum
of the rates flowing in on all the pipes (assume positive flow
raises the level in the tank, while negative flow reduces it).

Rephrased, for greater certainty: At any instant in time, the
rate at which water is being added to the tower is always equal
to the sum of the rates flowing in on all the pipes.

At any instant in time, all the water (and flows) can be
accounted for.

Same for energy (and energy flow).

snip

The obvious alternative is that the computed energy in the reflected
wave is sometimes just a figment.

And God created the heavens and earth in six days and rested on the
seventh.

Some do say, but this appears to be rather a non-sequitor.

I'm glad you are happy with your faith-based physics. In the
field of real-world physics, EM waves cannot exist without ExH energy.

Perhaps, then, you are simply arguing that these are not EM waves
since they do not have ExH energy?

The only way to win this argument is to prove to everyone that they
are not really detecting reflected waves containing energy when they
look at themselves in the mirror. Good luck on that one.

Question: How were the first three days measured before the creation
of the sun on the 4th day?

Continuing with non-sequitors?

Not to mention that in your 1/8 wavelength example (http://www.w5dxp.com/nointfr.htm)
you do not explain where the energy is stored so that it can be
returned at a different time.

Energy is stored in the transmission line and delivered as needed to
satisfy the conservaton of energy principle.

Nope. That also failed to account for the energy when observed in the
time domain. See http://sites.google.com/site/keithdysart/radio6.

Years ago, I showed how
energy can flow *into the source* (negative power) during a fractional
part of a cycle in a conjugately matched system.

Such declarations do permit an easy out, despite not aligning with
reality.

If you can take one joule per microsecond (1 megawatt) and conserve
that one megawatt of power over a century, you can get rich selling
it. Let us know when you get your patent on conservation of power. :-)
Good Grief!

If that is the case, the whole concept of reflected energy seems
somewhat bogus. Over a whole cycle, the power delivered by the
generator is passed on towards the load. If that is all you want
to know, then there is no need at all for "reflected power".

But, as you can grok from the subject of this thread, that is not all
that is needed to know. The last gasp of the loser is that it didn't
matter anyhow. Reflected energy has always mattered to optical
physicists who know it obeys the laws of physics. Now it seems to
matter to some hams. If it doesn't matter to you, why do you continue
posting?

Did I miss something? Was it not you who said "What happens over a
complete cycle is what is relevant."?

And to stop besmirching Hecht, it seems most probable that his
comment was in the context of optics. After all, the book had that
title.

Hint: RF waves are covered in every physics book whose title is
"Light". There is absolutely no difference, from a physics standpoint,
between a coherent light wave and a coherent RF wave except for
frequency. The both obey exactly the same laws of physics which you
seem to concede for visible light but not for light at RF frequencies.

Several differences:
- Transmission lines work down to DC
- At lower RF, it is possible to independantly measure voltage and
current

This allows a better understanding of the behaviour, not constrained
by the capabilities of the mearsuring instruments.

....Keith

what happens to reflected energy ?
 

On Jun 27, 4:27*pm, Cecil Moore wrote:
On Jun 27, 2:23*pm, Keith Dysart wrote:

Example 1: Step function applied to a transmission line. After the
* * * * * *line settles, a forward and reflected voltage wave
* * * * * *continue on the line but no energy is being transferred.

As far as I am concerned, if Maxwell's equations don't work on an
example, it might as well be ignored. There is nothing during DC
steady-state that allows Maxwell's equations to work because there are
no EM waves during DC steady-state. Why don't you already know that?

I always thought that Maxwell's equations were more complete than that
and worked all the way down to DC. Two of them do not even include
time
and nothing says that a derivative with respect to time can't be 0.

I can take your approach and do you one better. Please prove that you
exist. If you cannot prove that you exist, then nothing you say is of
any consequence. See, I can do it also.

From the above, you have proved that I exist. Thank you.

Example 2: On a line with infinite VSWR no energy crosses a
* * * * * *voltage minimum or maximum.

Completely false assumption. You are back to asserting that since the
north-bound traffic equals the south-bound traffic on the Golden Gate
Bridge that there is no traffic and no bridge maintenance is required.
When are you going to give up on that irrational wet dream of yours?
No *NET* energy crosses at a voltage zero or current zero point. That
doesn't make the north-bound energy equal to zero and doesn't make the
south-bound energy equal to zero. It just makes them equal. Just
because there is no NET traffic flow on the Golden Gate Bridge doesn't
mean there is zero traffic flow in both directions. Please stop
clowning around with such absurb notions.

I suppose, but then you have to give up on P(t)=V(t)*I(t), generally
considered to be a rather fundamental equation.

Example 3: With the 1/8 wavelength line described in
* * * * * *http://www.w5dxp.com/nointfr.htmtheenergy can not be
* * * * * *properly accounted for on a moment by moment basis.

There is no conservation of power principle.

There is no mention of power above; simply energy.

Are you saying that conservation of energy only applies some of
the time?

If you would track the RF
joules and the conversion of RF joules to heat instead of the joules/
second, everything would become clear to you. As it is, you are
laboring under some serious misconceptions about the laws of physics.
Power simply doesn't balance within a single cycle - because it
doesn't have to - because there is no conservation of power principle.

In your example, the RF energy does seem to disappear and re-appear,
when tracked on a moment by moment basis.

People who don't learn from their mistakes are doomed to commit the
same mistakes over and over. Keith, you seem to be all output and no
input. Please enable your input channels for a change.

Well, it would help if you could actually find and articulate a flaw
in http://sites.google.com/site/keithdysart/radio6.

....Keith

what happens to reflected energy ?
 

On Jun 27, 4:42*pm, Keith Dysart wrote:
It is not my insistence. It follows from the math.

Unfortunately for your arguments, math models do not dictate reality.
If the math model doesn't match reality, it is invalid. Your math
models obviously do not match reality.

Are you sure you meant this? The electron velocity changes? Or did
you mean the wave velocity? Nope. That does not work either.

Yes, acceleration and deceleration of electrons (in the conductor) is
required for EM waves to even exist. That's an obvious change in
electron velocity. Is that another fact of physics that shoots your
theory down? I repeat: EM waves are impossible during DC steady-state.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 27, 6:59*pm, Keith Dysart wrote:
From the above, you have proved that I exist. Thank you.

Nope, I believe you are only a figment of my imagination. Please prove
that you actually exist.

I suppose, but then you have to give up on P(t)=V(t)*I(t), generally
considered to be a rather fundamental equation.

I have absolutely no problem with giving up on the conservation of
power principle in which no rational technical person can possibly
believe.

Are you saying that conservation of energy only applies some of
the time?

No, I am saying that if you cannot balance the energy equation at all
times, you have made a mistake. You are not tracking joules. You are
attempting to track watts which can appear and disappear at any time.
The only condition where watts can be tracked is over an integer
multiple of complete cycles. That's why watts can be tracked when the
frequency is in the MHz. Trying to track instantaneous watts within a
fraction of a cycle is a moronic attempt at power superposition, a no-
no that we all learned in EE101.

In your example, the RF energy does seem to disappear and re-appear,
when tracked on a moment by moment basis.

No, the power can disappear and re-appear but the energy cannot. You
have not even come close to tracking the energy.

Well, it would help if you could actually find and articulate a flaw
inhttp://sites.google.com/site/keithdysart/radio6.

The flaw is your belief in a conservation of power principle that
doesn't exist. Instantaneous power is not required to obey any
conservation principle. What you are doing on that web page is
attempting to superpose powers apparently without a clue.

Superposition of power is a no-no. The power density equation allows
us to accomplish the addition of *average* powers taking interference
into effect. I know of no such mathematical equations for
instantaneous power and your instantaneous power superposition
technique is obviously invalid.
--
73, Cecil, w5dxp.com