RadioBanter - The Rest of the Story

RadioBanter (https://www.radiobanter.com/)

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- - The Rest of the Story (https://www.radiobanter.com/antenna/131062-rest-story.html)

The Rest of the Story

On Mar 14, 6:00*pm, Cecil Moore wrote:
Keith Dysart wrote:
Still handwaving. Show the expressions and the numbers
that make it balance. Otherwise, just handwaving.

Show the expressions and the numbers for how many angles
can dance on the head of a pin???? Shirley, you jest.
This is not the proper venue to try to establish your
new religion.

Are you really going to let me be the last man standing
this time?

No, you died in action a few weeks ago and don't realize
it. It happened the first time you superposed powers when
(V1^2 + V2^2) is not equal to (V1 + V2)^2. Too bad you
are incapable of comprehending exactly what that means.
You can easily work it out for yourself but you haven't
yet attempted to do so. Hopefully, one of the resident
gurus whom you trust will explain it to you.

Until you correct your ignorance on that subject, *you* are
just handwaving. I suggest you contact someone who is more
knowledgeable than you on the subject. Ask Richard C. to
prove his assertion that the reflections from a non-reflective
thin-film surface are brighter than the sun. If you detect
what is wrong with his argument, you will know what is wrong
with yours.
--
73, Cecil *http://www.w5dxp.com

The Rest of the Story

On Mar 14, 5:29*pm, Cecil Moore wrote:
Keith Dysart wrote:
If there were two transmission lines, then I could see
why you might want two pipes in an analogy.

But since there is only one transmission line, an
analogy with one pipe makes more sense.

Unfortunately for your argument, molecular water and
EM waves are considerably different animals. Water
energy traveling in opposite directions in a pipe
interact. EM waves traveling in opposite directions
in a transmission line interact only at an impedance
discontinuity or at an impedor. As long as only a
constant Z0 environment exists, the forward wave and
reflected wave pass like ships in the night. For you
to prove otherwise, you are going to have to define
the position and momentum of a single photon. Good
luck on that one.

I accept your retraction of your analogy.

...Keith

The Rest of the Story

On Mar 14, 5:51*pm, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 14, 7:59 am, Cecil Moore wrote:
P(t).reactance = [V(t).reactance][I(t).reactance]
Where is that term in your equations?

It is unnecessary. But if you believe me wrong, show me
where it goes, compute the values, and show how it
accounts for the energy that is not dissipated in the
source resistor.

It is unnecessary to account for all of the instantaneous
power???? Your problem is greater than just a simple
misunderstanding of the laws of physics by which we must
all abide.

The DC energy is stored in your vehicle's battery until
it is needed to start your vehicle. That delay between
stored energy and needed energy is related to the
(undefined) wavelength. Think about it.

In an AC circuit, the reactance has no say as to when
to store the energy and when it is delivered back to the
system. It is also related to wavelength which is defined.

When the source voltage is zero at its zero crossing
point/time, the instantaneous power dissipation in
the source resistor is NOT zero! Doesn't that give
you pause to wonder where the instantaneous power
is coming from when the instantaneous power delivered
by the source is zero????

Previously explained, as you may have forgotten, with
the energy cyclically returned from the line.

For your convenience, I copy from a previous post:
----
Recall that
Pg(t) = 32 + 68cos(2wt)

For some of the cycle, energy flow is from the line
towards the resistor and the voltage source.

But this is not the energy in the reflected wave
which has the function
Pr.g(t) = -18 + cos(2wt)
and only flows in one direction, towards the source.
And it is this supposed energy that can not be
accounted for in the dissipation of the source
resistor.
----

And, as expected
Ps(t) = Prs(t) + Pg(t)

Energy is conserved.

...Keith

The Rest of the Story

"Keith Dysart" wrote in message
...
On Mar 14, 5:51 pm, Cecil Moore wrote:
Energy is conserved.

qed.

now, can we get back to antennas?? its much more fun mocking art than the
endless discussions of electrons and holes sloshing back and forth.

The Rest of the Story

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?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story

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

The Rest of the Story

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

The Rest of the Story

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

The Rest of the Story

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