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

Keith Dysart wrote:
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.

There are no limitations. If zero interference exists,
then 100% of the reflected energy is dissipated in
the source resistor.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
If you wish to have your equalities apply at only selected
points within the cycle, that works for me.

Not only at selected points within the cycle but
also for average values. If zero average interference
exists then 100% of the average reflected energy is
dissipated in the source resistor which is the subject
of my Part 1 article. If the instantaneous interference
is zero, 100% of the instantaneous reflected power is
dissipated in the source resistor.

I repeat: When zero interference exists, 100% of the
reflected energy is dissipated in the source resistor.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
But I notice that you have not yet indicated which
energy equation I may have written that was unbalanced.

Why should I waste my time finding your conservation of
energy violations?

I repeat: When there exists zero interference, 100%
of the reflected energy is dissipated in the source
resistor. Since you think you provided an example where
that statement is not true, your example violates the
conservation of energy principle.
--
73, Cecil http://www.w5dxp.com

I have consolidated three replies below...

On Mar 20, 12:29*am, Cecil Moore wrote:
Keith Dysart wrote:
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.

There are no limitations. If zero interference exists,
then 100% of the reflected energy is dissipated in
the source resistor.

Sentence one says "no limitations". Sentence two specifies a
limitation. But your paper did provide that limitation and
indicated that circuit (Fig 1-1) with a 45 degree line was
an example which satisfied that limitation.

But in subsequent discussion you have waffled about whether,
for the circuit in Fig 1-1, "100% of the reflected energy is
dissipated in the source resistor" is applicable for all
time or only for those instances when the source voltage is
equal to 0.

Could you clarify whether your claim for the circuit of
Fig 1-1 applies to all time, or just to those instances
when the source voltage is 0.

On Mar 20, 12:34 am, Cecil Moore wrote:
Keith Dysart wrote:
If you wish to have your equalities apply at only selected
points within the cycle, that works for me.

Not only at selected points within the cycle but
also for average values. If zero average interference
exists then 100% of the average reflected energy is
dissipated in the source resistor which is the subject
of my Part 1 article. If the instantaneous interference
is zero, 100% of the instantaneous reflected power is
dissipated in the source resistor.

Perhaps this has clarified. So you are only claiming that
reflected energy is dissipated in the source resistor at
those instances when the source voltage is zero. Good.

Now as to averages: Averaging is a mathematical operation
applied to the signal which reduces information. I do agree
that the increase in the average dissipation in the source
resistor is numerically equal to the average value of the
reflected power. But this is just numerical equivalency.
It does not prove that the energy in the reflected wave
is dissipated in the source resistor. To prove the latter,
one must show that the energy in the reflected wave, on
an instance by instance basis is dissipated in the source
resistor because conservation of energy applies at the
instantaneous level.

And I have shown in an evaluation of the instantaneous
energy flows that the energy dissipated in the source
resistor is not the energy from the reflected wave.

I repeat: When zero interference exists, 100% of the
reflected energy is dissipated in the source resistor.

But only at those instances where the source voltage is
zero.

On Mar 20, 12:50 am, Cecil Moore wrote:
Keith Dysart wrote:
But I notice that you have not yet indicated which
energy equation I may have written that was unbalanced.

Why should I waste my time finding your conservation of
energy violations?

Mostly to prove that my analysis has an error.

I repeat: When there exists zero interference, 100%
of the reflected energy is dissipated in the source
resistor. Since you think you provided an example where
that statement is not true, your example violates the
conservation of energy principle.

But if there is no error in my analysis (and you have not
found one), then perhaps you should examine whether the
clause "When there exists zero interference, 100% of the
reflected energy is dissipated in the source resistor"
is in error. Sometimes one has to re-examine one's deeply
held beliefs in the light of new evidence. It is the only
rational thing to do.

...Keith

Keith Dysart wrote:
Sentence one says "no limitations". Sentence two specifies a
limitation.

Semantic games. There are no limitations within the
stated boundary conditions just like any number of
other concepts.

Could you clarify whether your claim for the circuit of
Fig 1-1 applies to all time, or just to those instances
when the source voltage is 0.

There are no claims regarding instantaneous power in
Fig 1-1 or anywhere else in my web article. All the
claims in my web article refer to *average* powers and
the article states exactly that. For the purpose and
subject of the web article, the subject of instantaneous
power is just an irrelevant diversion. No other author
on the subject has ever mentioned instantaneous power.
Given average values, time doesn't even appear in any
of their equations. Apparently, those authors agree
with Eugene Hecht that instantaneous power is "of
limited utility".

Here's my claim made in the article: When the *average*
interference at the source resistor is zero, the *average*
reflected power is 100% dissipated in the source resistor.
I gave enough examples to prove that claim to be true.
Since the instantaneous interference averages out to
zero, this claim about *average* power is valid.

When Tom, K7ITM, asserted that the same concepts work
for instantaneous power, I took a look and realized that
he was right. One can make the same claim about instantaneous
power although I do not make that claim in my web article.

When the instantaneous interference at the source resistor
is zero, the instantaneous reflected power is 100% dissipated
in the source resistor.

Perhaps this has clarified. So you are only claiming that
reflected energy is dissipated in the source resistor at
those instances when the source voltage is zero. Good.

I am claiming no such thing. Please cease and desist with
the mind fornication, Keith. You cannot win the argument
by being unethical.

Mostly to prove that my analysis has an error.

I have pointed out your error multiple times before, Keith,
and you simply ignore what I say. Why should I waste any
more time on someone who refuses to listen?

One more time:
Over and over, you use the equation Ptot = P1 + P2 even
though every sophomore EE student knows the equation is
(usually) invalid. The valid method for adding AC powers is:

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)

The last term is called the interference term which you
have completely ignored in your analysis. Therefore, your
analysis is obviously in error. When you redo your math
to include the interference term, your conceptual blunders
will disappear. Until then, you are just blowing smoke.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
On Mar 16, 10:21 am, Cecil Moore wrote:
snip
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?


For the first 90 degrees of time, the circuit can be represented as

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| /
Vs \ 50 ohm resistor
100v RMS /
| \
| |
+--------------+----------------------+
gnd

After 90 degrees of time has passed, the circuit can be represented as

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| |
Vs ---
100v RMS --- 50 ohm inductive
| |
| |
+--------------+----------------------+
gnd

The sudden switch in circuit design at time 90 degrees is not unique to
start up, but is true for any adjustment made to Vs and any returning
wave from the shorted stub,. As a result, a true stable circuit will
never be found unless some voltage adjustment is allowed for the 90
timing shift caused by the shorted stub. Keith (in his analysis of the
circuit) recognizes that Vs drives into a reactive circuit.

If we want to understand how constructive and destructive interference
act to cause a 50 ohm resistor to evolve into a 50 ohm capacitor, then
we need to examine how traveling waves might do this. It would be nice
to have a formula or wave sequence that fully addressed this evolution.

From circuit theory, we have the inductive reactance of a
short-circuited line less than 1/4 wavelength long is

XL = Zo * tan (length degrees)

= 50 * tan(45)

= 50 ohms


From traveling wave theory, we would have the applied wave from the
source arriving 90 degrees late to the stub side of resistor Rs. This
ignores the fact that current must already be passing through resistor
Rs because voltage has been applied to Rs from the Vs side 90 degrees
earlier. Whoa! Things are not adding up correctly this way!

We need to treat the wave going down the 50 ohm line as a single wave
front. The wave reverses at the short-circuit, reversing both direction
of travel and sign of voltage. When the wave front reaches the input
end, 90 degrees after entering (for this case), the voltage/current
ratio is identical to the starting ratio (the line was 45 degrees long,
tan(45) = 1), and the returning voltage directly adds to the voltage
applied from Vs. As a result, the current flowing through Rs will
increase, and Vs will see a changed (decreased) impedance.

After 90 degrees of signal application, we should be able to express the
voltage across Rs as

Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

Substitute so

Vrs(t) = Vs(wt + 90) - Vs(wt + 90)/2 + Vs(wt)/2

= Vs(wt + 90)/2 + Vs(wt)/2

Allow Vs to be represented by a sine wave, we have

2Vrs(t) = Vs*sin(wt + 90) + Vs*sin(wt)
= 2Vs*(sin(wt + 45)(cos(45))
Vrs(t) = Vs(sin(wt + 45)(cos(45))

Vs is defined as 100v RMS, which equals 100 * 1.414 = 141.4v Peak. The
maximum voltage would occur when the sin term was 90 degrees and equals
1, which would occur at wt = 45 degrees. We would have

Vrs(45) = 141.42 * sin(90)(cos(45)
= 141.42 * 1 * 0.7071
= 100v

Now consider the current. After the same 90 degrees of signal
application, we should be able to express the current through Rs as

Irs(t) = Is(t) + Iref(t)

Is(t) = Is(wt + 90), Iref(t) = Is(wt)

The reflected current has been shifted by 90 degrees due to the
reflection so we must rewrite Iref(t) to read

Iref(t) = Is(wt + 90)

Substitute,

Irs(t) = Is(wt + 90) + Is(wt)

Allow Is to be represented by a sine wave, we have

Irs(t) = Is*sin(wt + 90) + Is*sin(wt)

= 2*Is(sin(wt + 45)(cos(45))

How do we find Is? Is is the initial current found by dividing the
applied voltage at peak (141.42v) by the initial resistance (100 ohms).


Is = 141.42/100 = 1.4142a

The maximum current would occur when the sin term was 90 degrees and
equals 1, which would occur at wt = 45 degrees. We would have

Irs(t) = 2*Is(sin(wt + 45)(cos(45))

= 2 * 1.4142 * 1 * 0.7071

= 2a

These results agree with the results from Keith and from circuit theory.

We have a theory and at least the peaks found from the theory agree with
the results from others. How about Cecil's initial question which is
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?

We will use the equation

Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

The challenge looks at the time when Vs(t) = Vs*sin(wt + 90) = 0, which
occurs when wt = -90. When wt = -90, all the power to the source
resistor Rs is from the reflected portion described as Vref(t) =
Vs(wt)/2. The voltage across Rs would be

Vrs(-90) = Vref(-90)
= Vs*sin(-90)/2
= 141.4/2
= 70.7v

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. Both paths are available at all times.
Power flows through both paths to Rs at all times, but because of the
time differential in arrival timing, at some point Rs will receive power
only from Vs, and at another point, receive power only from Vref.

73, Roger, W7WKB

Roger Sparks wrote:
Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Good job, Roger. Let's simplify the equation through
substitution.

Let V1(t) = [Vs(t) - Vg(t)]

Let V2(t) = Vref(t)

Vrs(t) = V1(t) + V2(t)

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

Yes, because when Vs(t)=0, there is zero instantaneous
interference.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. Both paths are available at all times. Power
flows through both paths to Rs at all times, but because of the time
differential in arrival timing, at some point Rs will receive power only
from Vs, and at another point, receive power only from Vref.

What Keith is missing is that:

Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2

In his math, Keith is asserting that since

Prs(t) NOT= P1(t) + P2(t), then the reflected power is not
being dissipated in the source resistor. But every sophomore
EE knows NOT to try to superpose powers like that. Roger,
I'll bet you know not to try to superpose powers?

Since Keith doesn't listen to me, would you pass that
technical fact on to him?

When Keith uses the correct equation:

Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)]

he will see that the equation balances and therefore
100% of the reflected energy is dissipated in the
source resistor since the interference term averages
out to zero over each cycle.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Roger Sparks wrote:
Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Good job, Roger. Let's simplify the equation through
substitution.

Let V1(t) = [Vs(t) - Vg(t)]

Let V2(t) = Vref(t)

Vrs(t) = V1(t) + V2(t)

This works. It certainly helps to link my work with yours.

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

Yes, because when Vs(t)=0, there is zero instantaneous
interference.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). The short path is the
series path of two resistors composed of the source resistor Rs and
the input to the 50 ohm transmission line measured by Vg. The long
path is the series path of one resistor Rs and one capacitor composed
of the shorted transmission line. Both paths are available at all
times. Power flows through both paths to Rs at all times, but because
of the time differential in arrival timing, at some point Rs will
receive power only from Vs, and at another point, receive power only
from Vref.

What Keith is missing is that:

Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2

In his math, Keith is asserting that since

Prs(t) NOT= P1(t) + P2(t), then the reflected power is not
being dissipated in the source resistor. But every sophomore
EE knows NOT to try to superpose powers like that. Roger,
I'll bet you know not to try to superpose powers?

Using circuit theory, at the peak under steady conditions, we have
141.4v applied to 70.7 ohms which gives a current of 2a. The 70.7 ohms
is sqrt(50^2 + 50^2). Two amps flowing through Rs = 50 ohms, the power
to Rs is (2^2) * 50 = 4 * 50 = 200w.

Using your simplified equation for the voltage across Rsj

Vrs(t) = V1(t) + V2(t)

Vrs(t) = V1*sin(wt + 90) + V2*sin(wt)

In our case, V1 = V2 because both voltages are developed over a 50 ohm
load. As a result, in our case, 100w will be delivered to Rs both at wt
= 0 and wt = -90, two points 90 degrees apart. If we are looking for
the total power delivered to Rs, then it seems to me like we SHOULD add
the two powers in this case. This recognizes that power is delivered to
Rs via two paths, each carrying 100w. Alternatively, we could add the
two voltages together to find the peak voltage, and then square that
number and divide by the resistance of Rs. Both methods should give the
same result.

So it looks to me like Keith is right in his method, at least in this
case.


Since Keith doesn't listen to me, would you pass that
technical fact on to him?

When Keith uses the correct equation:

Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)]

he will see that the equation balances and therefore
100% of the reflected energy is dissipated in the
source resistor since the interference term averages
out to zero over each cycle.

I am still uncomfortable with my summary statement that both paths, the
long and short, are available and active at all times. On the other
hand, it is consistent with your advice that the waves never reflect
except at a discontinuity. The conclusion was that 100w is delivered to
Rs via each path, with the path peaks occurring 90 degrees apart in
time. Surprisingly, 100 percent of the reflected energy is ALWAYS
absorbed by Rs. There is no further reflection from Vs or Rs.

This is counter intuitive to me. I like to resolve the circuit into one
path, one wave form. We do that with circuit analysis. With traveling
waves, we frequently have two or more paths the exist independently, so
we must add the powers carried on each path, just like we add the
voltages or currents. (But watch out and avoid using standing wave
voltages or currents to calculate power!)

Most surprising to me is the observation that my beginning statement
(from my previous post) about the circuit evolving from a circuit with
two resistive loads, into a circuit with a resistive load and
capacitive load, is really incorrect. Once steady state is reached,
BOTH circuits are active at the same time, forming the two paths
bringing power to RS. Using that assumption, we can use the traveling
waves to analyze the circuit.

73, Roger, W7WKB

On Mar 20, 10:02*am, Cecil Moore wrote:
Keith Dysart wrote:
Sentence one says "no limitations". Sentence two specifies a
limitation.

Semantic games.

The whole question here revolves around the meaning of the
limitations on your claim. That is semantics. Not games.
And it is key to the discussion.

There are no limitations within the
stated boundary conditions just like any number of
other concepts.

But you need to clearly state your limitations and stop
flip flopping.

Could you clarify whether your claim for the circuit of
Fig 1-1 applies to all time, or just to those instances
when the source voltage is 0.

There are no claims regarding instantaneous power in
Fig 1-1 or anywhere else in my web article. All the
claims in my web article refer to *average* powers and
the article states exactly that. For the purpose and
subject of the web article, the subject of instantaneous
power is just an irrelevant diversion. No other author
on the subject has ever mentioned instantaneous power.

I am surprised, this being 2008, that I could actually be
offering a new way to study the question, but if you insist,
I accept the accolade.

Given average values, time doesn't even appear in any
of their equations. Apparently, those authors agree
with Eugene Hecht that instantaneous power is "of
limited utility".

Here's my claim made in the article: When the *average*
interference at the source resistor is zero, the *average*
reflected power is 100% dissipated in the source resistor.
I gave enough examples to prove that claim to be true.
Since the instantaneous interference averages out to
zero, this claim about *average* power is valid.

Without prejudice to the accuracy of the above, let us
explore a bit.

We know that conservation of energy requires that the
energy flows balance at all times, which means that at
any instance, the flows must account for all the energy.

Analysis has shown that when examined with fine granularity,
that for the circuit of Fig 1-1, the energy in the reflected
wave is not always dissipated in the source resistor. For
example, some of the time it is absorbed in the source.

Now when the instantaneous flows are averaged, it is true
that the increase in dissipation is numerically equal to
the average power for the reflected wave. But this does
not mean that the energy in the reflecte wave is dissipated
in the source resistor, merely that the averages are equal.

Now you qualify your claim with the term "*average* power".
You say "the *average* reflected power is 100% dissipated in
the source resistor."

But the actual energy in the reflected wave is not dissipated
in the source resistor. So what does it mean to say that the
*average* is?

When Tom, K7ITM, asserted that the same concepts work
for instantaneous power, I took a look and realized that
he was right. One can make the same claim about instantaneous
power although I do not make that claim in my web article.

When the instantaneous interference at the source resistor
is zero, the instantaneous reflected power is 100% dissipated
in the source resistor.

Perhaps this has clarified. So you are only claiming that
reflected energy is dissipated in the source resistor at
those instances when the source voltage is zero. Good.

I am claiming no such thing. Please cease and desist with
the mind fornication, Keith. You cannot win the argument
by being unethical.

Unfortunately you struck the sentence which I paraphrased
and then failed to explain which parts of it I may have
misinterpreted. That does not help. It would have been
more valuable for you to rewrite the original sentence
to increase its clarity.

Mostly to prove that my analysis has an error.

I have pointed out your error multiple times before, Keith,
and you simply ignore what I say. Why should I waste any
more time on someone who refuses to listen?

One more time:
Over and over,

I only have two expression involving power.

you use the equation Ptot = P1 + P2 even
though every sophomore EE student knows the equation is
(usually) invalid. The valid method for adding AC powers is:

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Which of the two needs the 'cos' term?

Ps(t) = Prs(t) + Pg(t)
or
Pg(t) = Pf.g(t) + Pr.g(t)

In fact neither do. They both stand quite well on their own.
Both are so correct, that they apply for any voltage function
provided by the source. What would you propose to use for cos(A)
when the source voltage function is aperiodic pulses? Fortunately,
the 'cos' term is not needed so the question is completely moot.

Non-the-less do feel free to offer corrected expression that include
the 'cos(A)' term.

The last term is called the interference term which you
have completely ignored in your analysis. Therefore, your
analysis is obviously in error. When you redo your math
to include the interference term, your conceptual blunders
will disappear. Until then, you are just blowing smoke.

The math holds as it is. But I invite you to offer an alternative
analysis that includes cos(A) terms. We can see how it holds up.

If you can't, then you should definitely reconsider who is
'blowing smoke'.

...Keith