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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

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Roger Sparks 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.

Note that his analysis proves that destructive interference
energy is indeed stored in the reactance of the network and
dissipated later in the cycle as constructive interference
in the source resistor at a time when the instantaneous
source power equals zero.
--
73, Cecil http://www.w5dxp.com

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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

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On Mar 18, 10:58*am, Roger Sparks wrote:
Thanks for a comprehensive analysis Keith. *It was helpful to me to see how you step by step analyzed the circuit.

You are most welcome.

...Keith

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On Mar 18, 12:16*pm, Cecil Moore wrote:
Roger Sparks 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.

Note that his analysis proves that destructive interference
energy is indeed stored in the reactance of the network and
dissipated later in the cycle as constructive interference
in the source resistor at a time when the instantaneous
source power equals zero.

I've made no claims about destructive or other interference
so I am intrigued by what you say my analysis proves.

Could you expand on how my analysis "proves" that
"destructive interference energy is indeed stored in the
reactance of the network and dissipated later in the cycle
as constructive interference in the source resistor at a
time when the instantaneous source power equals zero."

I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".
Are you saying that at other times, an explanation other
than interference is needed?
Or does interference explain the flows of the energy
in the reflected waves for all times?

It would be good if you could provide the equations that
describe these flows, especially the ones describing the
storage of some of the energy from the reflected wave in
the reactance of the network.

...Keith

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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

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Keith Dysart wrote:
I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".

:-) :-) That was my exact reaction when you wanted to
focus on the infinitesimally small times associated
with instantaneous powers. :-) :-) You are to blame
for that focus, not I.

It only takes one example to prove your claims wrong.
I didn't want to focus on instantaneous power at all
but you insisted. Now that your own techniques are
being used to prove you wrong, you are objecting.

At the time when the instantaneous source power is zero,
the only other sources of energy in the entire network
is reflected energy and interference energy. That's
a fact of physics.

You have gotten caught superposing powers, something
that every sophomore EE knows not to do. I repeat:
The only time that power can be directly added is
when there is *ZERO INTERFERENCE*. The test for
zero interference is: (V1^2 + V2^2) = (V1 + V2)^2
--
73, Cecil http://www.w5dxp.com

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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

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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

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On Mar 19, 12:00*pm, Cecil Moore wrote:
Keith Dysart wrote:
I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".

:-) :-) That was my exact reaction when you wanted to
focus on the infinitesimally small times associated
with instantaneous powers. :-) :-) You are to blame
for that focus, not I.

It only takes one example to prove your claims wrong.

Could you state the claim that you think you are proving
wrong?

I didn't want to focus on instantaneous power at all
but you insisted. Now that your own techniques are
being used to prove you wrong, you are objecting.

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

It only fails when you make the assertion that
for the circuit under consideration, the energy in
the reflected wave is dissipated in the source
resistor.

At the time when the instantaneous source power is zero,
the only other sources of energy in the entire network
is reflected energy and interference energy. That's
a fact of physics.

You have gotten caught superposing powers, something
that every sophomore EE knows not to do.

Could you kindly show me where?

The two expressions in which I have summed power are
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)

You have not previously indicated that either of these
are incorrect. Please take this opportunity.

I repeat:
The only time that power can be directly added is
when there is *ZERO INTERFERENCE*. The test for
zero interference is: (V1^2 + V2^2) = (V1 + V2)^2

Surely you overstate. Powers can be directly added
whenever you have a system with ports adding and
removing energy from the system. Conservation of
energy says that the sum of the energy being stored
in the system is equal to the sums of all the ins
and outs at the ports. There are no conditions on
the ports adding or removing energy.

...Keith

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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

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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

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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

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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

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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

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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

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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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

On Mar 20, 1:07*pm, Roger Sparks wrote:
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)

Just to ensure clarity, you are using a different convention for terms
and
signs than I do.

For me, Vg(t) is the actual voltage at point Vg. In terms of forward
and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at
point
g is sum of the forward and reverse voltage at point g.

In my terms, this leads to
Vrs(t) = Vs(t) - Vg(t)
= Vs(t) - Vf.g(t) - Vr.g(t)

Your Vg seems to be same as my Vf.g and you seem to be using a
different
sign for Vref than I do for Vr.

You must be carrying these differences through the equations correctly
because the final results seem to agree.

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

A small aside. It is conventional to use cos for sinusoidal waves
because
it maps more readily into phasor notation.

* * 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). *

This is not quite complete. The power to Rs does come from two places
but they are the voltage source and the line.

Ps(t) = Prs(t) + Pg(t)

where Pg(t) is the power at point g flowing out of the generator into
the line and Ps(t) is the power flowing from the source to the source
resistor and the line.

The power into the line can be seaparated into a forward component and
a reflected component:

Pg(t) = Pf(t) + Pr(t)

(Sometimes this is written as P = Pf - Pr, but that is just a question
of how the sign is used to represent the direction of energy flow.)

Substituting we have

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

Whenever Ps(t) equal 0, then Pf.g(t) also equals 0 so at those times,
this simplifies to
Prs = - Pr.g(t)
which is the same conclusion you came to.

But it is wise not to forget the Pf.g(t) term because at times when
Ps(t) is not equal to zero, the Pf.g(t) term will be needed to
balance the energy flows.

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.

And at still other times, the power in the source resistor will depend
on
all of Ps(t), Pf.g(t) and Pr.g(t), or, more simply, it will depend on
Ps(t) and Pg(t).

...Keith

The Rest of the Story
 

On Mar 20, 2:07*pm, 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)

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.

It would be excellent if you would correct my exposition
using the correct formulae. We could then see if your
proposal actually provides accurate results.

...Keith

PS. It won't.

The Rest of the Story
 

On Fri, 21 Mar 2008 05:10:01 -0700 (PDT)
Keith Dysart wrote:

On Mar 20, 1:07*pm, Roger Sparks wrote:
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)

Just to ensure clarity, you are using a different convention for terms
and
signs than I do.

For me, Vg(t) is the actual voltage at point Vg. In terms of forward
and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at
point
g is sum of the forward and reverse voltage at point g.

In my terms, this leads to
Vrs(t) = Vs(t) - Vg(t)
= Vs(t) - Vf.g(t) - Vr.g(t)

Your Vg seems to be same as my Vf.g and you seem to be using a
different
sign for Vref than I do for Vr.

You must be carrying these differences through the equations correctly
because the final results seem to agree.

Thanks for looking at my posting very carefully, because I think you are drawing the same conclusions as I, at least so far.

I think you are correct in saying that
For me, Vg(t) is the actual voltage at point Vg. In terms of forward
and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at
point
g is sum of the forward and reverse voltage at point g.

but doesn't this describe the standing wave?

At point g, the forward wave has passed through resistor Rs, loosing part of the original wave energy, and is inroute beginning the long path to the resistor Rs. The reflected wave has arrived back to one side of Rs and the voltage is already in series with the source voltage acting on Rs. I think we need to subtract the two terms so that we can separate the apples and oranges.

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

A small aside. It is conventional to use cos for sinusoidal waves
because
it maps more readily into phasor notation.

This is an important observation. cos(0) is 1 so to me it seems like we are saying that we begin at time zero with peak voltage or peak current. I wanted to begin with zero current at time zero so I used the sine. Am I missing something important here?


* * 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). *

This is not quite complete. The power to Rs does come from two places
but they are the voltage source and the line.

Ps(t) = Prs(t) + Pg(t)

where Pg(t) is the power at point g flowing out of the generator into
the line and Ps(t) is the power flowing from the source to the source
resistor and the line.

The power into the line can be seaparated into a forward component and
a reflected component:

Pg(t) = Pf(t) + Pr(t)

(Sometimes this is written as P = Pf - Pr, but that is just a question
of how the sign is used to represent the direction of energy flow.)

Substituting we have

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

Whenever Ps(t) equal 0, then Pf.g(t) also equals 0 so at those times,
this simplifies to
Prs = - Pr.g(t)
which is the same conclusion you came to.

No, I think my Pr.g(t) was positive, not negative. The difference is important because of the timing of the energy flows. Here is where it is important to observe that the forward wave entering the transmission line will have no further effect on resistor Rs until it has traveled 90 degrees of time. On the other hand, the reflected wave Pr.g(t) is only an instant of time away from entering the resistor Rs, but has not actually entered it because you are measuring it as being present at Vg, not as voltage Vrs.


But it is wise not to forget the Pf.g(t) term because at times when
Ps(t) is not equal to zero, the Pf.g(t) term will be needed to
balance the energy flows.

Pf.g(t) is the remaining power after the wave generated by Ps(t) has passed through Rs and left behind Prs(t). Another Prs(t + x) will be generated by the returning reflected wave.


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.

And at still other times, the power in the source resistor will depend
on
all of Ps(t), Pf.g(t) and Pr.g(t), or, more simply, it will depend on
Ps(t) and Pg(t).

...Keith

We differ on how to utilize the term Pg(t). I think it describes the standing wave, not the voltage across Rs.
--
73, Roger, W7WKB

The Rest of the Story
 

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

Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
But you need to clearly state your limitations and stop
flip flopping.

What you are calling "flip flopping" is me correcting
my errors. Once I correct an error, I don't flip-flop
back. My error was in assuming that the power-density
(irradiance) equation only works on average powers.
K7ITM convinced me that it works on instantaneous powers
also.

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.

I'm sure you are not the first, just the first to think
there is anything valid to be learned by considering
instantaneous power to be important. Everyone except
you discarded that notion a long time ago.

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.

Yes, yes, yes, now you are starting to get it. When
interference is present, the energy in the reflected
wave is NOT dissipated in the source resistor. Those
facts will be covered in Part 2 & 3 of my web article.

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.

For instantaneous values of voltage, the phase angle is
either 0 or 180 degrees so the cosine term is either +1
or -1. The math is perfectly consistent. That you don't
recognize the sign of the instantaneous interference term
as the cosine of 0 or 180 is amazing.

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

I did and you ignored it. There is no negative sign in the
power equation yet you come up with negative signs. That
you don't recognize your negative sign as cos(180) is
unbelievable.

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.

It it unfortunate that you don't comprehend that the cos(0) is
+1 and the cos(180) is -1. The sign of the instantaneous
interference term is the cosine term. The math certainly does
hold as it is - you are just ignorant of the "is" part.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
In my terms, this leads to
Vrs(t) = Vs(t) - Vg(t)
= Vs(t) - Vf.g(t) - Vr.g(t)

How about expanding those equations for us?

Vs(t) = 141.4*cos(wt) ????
Vg(t) = ____*cos(wt+/-____) ????
Vf.g(t) = ____*cos(wt+/-____) ????
Vr.g(t) = ____*cos(wt+/-____) ????

If you ever did this before, I missed it.

Given the correct voltage equations, I can
prove what I am saying about destructive and
constructive interference averaging out to zero
over one cycle is a fact.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
It would be excellent if you would correct my exposition
using the correct formulae. We could then see if your
proposal actually provides accurate results.

Roger didn't understand your terms and subscripts and
neither do I. I doubt that anyone understands your
formulas well enough to discuss them.

What is the equation for the forward voltage component
dropped across Rs?

What is the equation for the reflected voltage
component dropped across Rs?

Given valid equations for those two voltages, I can
prove everything I have been saying.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Fri, 21 Mar 2008 11:22:22 -0500
Cecil Moore wrote:

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

Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?
--
73, Cecil http://www.w5dxp.com

No, I really don't.

I do understand that you had better not try to find the power flowing from superposed voltages because they may be the sum of apples and oranges. That is, the superposed voltages may be flowing in opposite directions so they will each be carrying power but the power will be available/reachable only for the direction toward which the wave is traveling. An SWR meter comes to mind here.

--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:

Cecil Moore wrote:
Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?

No, I really don't.

It is because (V1 + V2)^2 usually doesn't equal V1^2+V2^2
because of interference. Keith's addition of powers
without taking interference into account is exactly
the mistake that the EE201 professors were talking about.
One cannot validly just willy-nilly add powers. It is
an ignorant/sophomoric thing to do.

If we add two one watt coherent waves, do we get a two
watt wave? Only in a very special case. For the great
majority of cases, we do *NOT* get a two watt wave. In
fact, the resultant wave can be anywhere between zero
watts and four watts. The concepts behind Keith's
calculations are invalid. If you are also trying to
willy-nilly add powers associated with coherent waves,
your calculations are also invalid.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Fri, 21 Mar 2008 19:43:12 GMT
Cecil Moore wrote:

Roger Sparks wrote:

Cecil Moore wrote:
Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?

No, I really don't.

It is because (V1 + V2)^2 usually doesn't equal V1^2+V2^2
because of interference. Keith's addition of powers
without taking interference into account is exactly
the mistake that the EE201 professors were talking about.
One cannot validly just willy-nilly add powers. It is
an ignorant/sophomoric thing to do.

If we add two one watt coherent waves, do we get a two
watt wave? Only in a very special case. For the great
majority of cases, we do *NOT* get a two watt wave. In
fact, the resultant wave can be anywhere between zero
watts and four watts. The concepts behind Keith's
calculations are invalid. If you are also trying to
willy-nilly add powers associated with coherent waves,
your calculations are also invalid.
--
73, Cecil http://www.w5dxp.com

OK, yes, I agree. It is OK to add powers when you are adding the power used by light bulbs. It is not OK to willy nilly multiply the voltage or current by the number of bulbs to learn the power used. You must carefully consider the circuit that connects the bulbs before selecting the proper method of calculating power, especially the possibility that the bulbs may be connected to phased power as in 3 phase or in traveling waves.

--
73, Roger, W7WKB

The Rest of the Story
 

On Mar 21, 9:12*am, Roger Sparks wrote:
On Fri, 21 Mar 2008 05:10:01 -0700 (PDT)

Keith Dysart wrote:
On Mar 20, 1:07*pm, Roger Sparks wrote:
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)

Just to ensure clarity, you are using a different convention for terms
and signs than I do.

For me, Vg(t) is the actual voltage at point Vg. In terms of forward and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at point
g is sum of the forward and reverse voltage at point g.

In my terms, this leads to
* Vrs(t) = Vs(t) - Vg(t)
* * * * *= Vs(t) - Vf.g(t) - Vr.g(t)

Your Vg seems to be same as my Vf.g and you seem to be using a
different nsign for Vref than I do for Vr.

You must be carrying these differences through the equations correctly
because the final results seem to agree.

Thanks for looking at my posting very carefully, because I think you are drawing the same conclusions as I, at least so far. *

I think you are correct in saying that For me, Vg(t) is the actual voltage at point Vg. In terms of forward
and
reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at
point g is sum of the forward and reverse voltage at point g. *

Ooooppps. That should have been Vg(t) = Vf.g(t) + Vr.g(t)

but doesn't this describe the standing wave? *

Not by itself. It is simply the function describing the voltage at
point g.

If you were to compute Vx(g) = Vf.x(t) + Vr.x(t) for many points
x, then you might end up with something akin to a standing wave.
Note that Vg(t) = Vf.g(t) + Vr.g(t) holds true for any function
while standing waves only really appear when the excitation
function is sinusoidal.

At point g, the forward wave has passed through resistor Rs,

This is not a way that I would recommend thinking about what is
happening. The voltage source and source resistor are lumped
circuit elements and thinking that wave is passing through
them is likely to lead to confusion.

It might be valuable to study the circuit completely with a
lumped voltage source, source resistor and reactance. Once
that circuit is understood, replace the reactance with the
transmission line. When studying the circuit with the
transmission line instead of the reactance, always check
to see whether you would use the same words if the TL were
replaced with the reactance. If not, question why.

loosing part of the original wave energy, and is inroute beginning the long path to the resistor Rs. *The reflected wave has arrived back to one side of Rs and the voltage is already in series with the source voltage acting on Rs. *I think we need to subtract the two terms so that we can separate the apples and oranges.

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

A small aside. It is conventional to use cos for sinusoidal waves
because
it maps more readily into phasor notation.

This is an important observation. *cos(0) is 1 so to me it seems like we are saying that we begin at time zero with peak voltage or peak current. *I wanted to begin with zero current at time zero so I used the sine. *Am I missing something important here?

It does not alter the outcome, so is not that important.

* * 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). *

This is not quite complete. The power to Rs does come from two places
but they are the voltage source and the line.

* Ps(t) = Prs(t) + Pg(t)

where Pg(t) is the power at point g flowing out of the generator into
the line and Ps(t) is the power flowing from the source to the source
resistor and the line.

The power into the line can be seaparated into a forward component and
a reflected component:

* Pg(t) = Pf(t) + Pr(t)

(Sometimes this is written as P = Pf - Pr, but that is just a question
of how the sign is used to represent the direction of energy flow.)

Substituting we have

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

Whenever Ps(t) equal 0, then Pf.g(t) also equals 0 so at those times,
this simplifies to
* Prs = - Pr.g(t)
which is the same conclusion you came to.

No, I think my Pr.g(t) was positive, not negative. *

I think that is just because your reference direction was different.
I use the same reference direction for both Pf and Pr. It is often
done that the reference direction is different for Pf and Pr.

The difference is important because of the timing of the energy flows. *Here is where it is important to observe that the forward wave entering the transmission line will have no further effect on resistor Rs until it has traveled 90 degrees of time. *On the other hand, the reflected wave Pr.g(t) is only an instant of time away from entering the resistor Rs, but has not actually entered it because you are measuring it as being present at Vg, not as voltage Vrs.

But it is wise not to forget the Pf.g(t) term because at times when
Ps(t) is not equal to zero, the Pf.g(t) term will be needed to
balance the energy flows.

Pf.g(t) is the remaining power after the wave generated by Ps(t) has passed through Rs and left behind Prs(t). *Another Prs(t + x) will be generated by the returning reflected wave.

Again, I don't think it wise to think of the wave as passing through
Rs.

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.

And at still other times, the power in the source resistor will depend
on
all of Ps(t), Pf.g(t) and Pr.g(t), or, more simply, it will depend on
Ps(t) and Pg(t).

...Keith

We differ on how to utilize the term Pg(t). *I think it describes the standing wave, not the voltage across Rs. *

I describe Vg(t) as the actual voltage that would be measured by an
oscilloscope at the point g; i.e. the beginning of the transmission
line. It is certainly not the voltage across Rs. Perhaps it IS what
you mean by standing wave. In any case the voltage across the resistor
is
Vs(t) = Vrs(t) + Vg(t)
Vrs(t) = Vs(t) - Vg(t)

Pg(t) is the power being delivered to the transmission line at any
instance t.
Pg(t) = Vg(t) * Ig(t)

...Keith

The Rest of the Story
 

On Mar 21, 12:37*pm, Cecil Moore wrote:
Keith Dysart wrote:
But you need to clearly state your limitations and stop
flip flopping.

What you are calling "flip flopping" is me correcting
my errors. Once I correct an error, I don't flip-flop
back.

Actually the flip-flopping I was referring to was the
constant changes in your view of the limitations that
apply to your claim.

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.

I'm sure you are not the first, just the first to think
there is anything valid to be learned by considering
instantaneous power to be important. Everyone except
you discarded that notion a long time ago.

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.

Yes, yes, yes, now you are starting to get it.

Is this a flop or a flip? Are you now agreeing that the
energy in the reflected wave is only dissipated in the
source resistor for those instances when Vs is 0?

When
interference is present, the energy in the reflected
wave is NOT dissipated in the source resistor. Those
facts will be covered in Part 2 & 3 of my web article.

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.

For instantaneous values of voltage, the phase angle is
either 0 or 180 degrees so the cosine term is either +1
or -1. The math is perfectly consistent.
[gratuitous insult snipped]

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

I did and you ignored it.

I could not find them in the archive. Could you kindly provide
them again, showing where the 'cos(A)' term fits in the
equations:
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)

There is no negative sign in the
power equation yet you come up with negative signs.
[gratuitous insult snipped]

Negative signs also arise when one rearranges equations.

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.
[gratuitous insult snipped]

...Keith

The Rest of the Story
 

On Mar 21, 12:53*pm, Cecil Moore wrote:
Keith Dysart wrote:
In my terms, this leads to
* Vrs(t) = Vs(t) - Vg(t)
* * * * *= Vs(t) - Vf.g(t) - Vr.g(t)

How about expanding those equations for us?

Vs(t) = 141.4*cos(wt) ????
Vg(t) = ____*cos(wt+/-____) ????
Vf.g(t) = ____*cos(wt+/-____) ????
Vr.g(t) = ____*cos(wt+/-____) ????

If you ever did this before, I missed it.

I did. And they are also conveniently in the spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection

For your convenience, in the example of Fig 1-1, 100 Vrms
sinusoidal source, 50 ohm source resistor, 45 degrees of
50 ohm line, 12.5 ohm load, after the reflection returns...

Vs(t) = 141.42135623731*cos(wt)
Vg(t) = 82.46211251*cos(wt+30.96375653)
Vf.g(t) = 70.71067812*cos(wt)
Vr.g(t) = 42.42640687*cos(wt+90)

And for completeness...

Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

Given the correct voltage equations, I can
prove what I am saying about destructive and
constructive interference averaging out to zero
over one cycle is a fact.

Not that anyone has disputed that. But it would
be good to see anyway.

...Keith

The Rest of the Story
 

On Mar 21, 1:00*pm, Cecil Moore wrote:
Keith Dysart wrote:
It would be excellent if you would correct my exposition
using the correct formulae. We could then see if your
proposal actually provides accurate results.

Roger didn't understand your terms and subscripts and
neither do I. I doubt that anyone understands your
formulas well enough to discuss them.

What is the equation for the forward voltage component
dropped across Rs?

What is the equation for the reflected voltage
component dropped across Rs?

Given valid equations for those two voltages, I can
prove everything I have been saying.

Since I am not sure exactly what your are looking for,
and you could not be sure that I did them right,
it would probably be best if you were to compute these
equations which you need.

...Keith

The Rest of the Story
 

On Mar 21, 5:03*pm, Roger Sparks wrote:
On Fri, 21 Mar 2008 19:43:12 GMT

Cecil Moore wrote:
Roger Sparks wrote:

Cecil Moore wrote:
Roger, do you understand why EE201 professors admonish
their students not to try to superpose powers?

No, I really don't. *

It is because (V1 + V2)^2 usually doesn't equal V1^2+V2^2
because of interference. Keith's addition of powers
without taking interference into account is exactly
the mistake that the EE201 professors were talking about.
One cannot validly just willy-nilly add powers. It is
an ignorant/sophomoric thing to do.

If we add two one watt coherent waves, do we get a two
watt wave? Only in a very special case. For the great
majority of cases, we do *NOT* get a two watt wave. In
fact, the resultant wave can be anywhere between zero
watts and four watts. The concepts behind Keith's
calculations are invalid. If you are also trying to
willy-nilly add powers associated with coherent waves,
your calculations are also invalid.
--
73, Cecil *http://www.w5dxp.com

OK, yes, I agree. *It is OK to add powers when you are adding the power used by light bulbs. *It is not OK to willy nilly multiply the voltage or current by the number of bulbs to learn the power used. *You must carefully consider the circuit that connects the bulbs before selecting the proper method of calculating power, especially the possibility that the bulbs may be connected to phased power as in 3 phase or in traveling waves.

My analysis used voltages, currents and impedances to compute
all the voltages and currents within the circuit. Some were
derived using superposition of voltages and currents but most
were derived using basic circuit theory (E=IR, Ztot=Z1+Z2, etc.)

Having done that, the powers for the three components (the
voltage source, resistor, and entrance to the transmission line)
in the circuit were computed. These powers were not derived using
superposition but by multiplying the current through the component
by the voltage across it. This is universally accepted as a valid
operation.

Having the power functions for each of the component, we can
then turn to the conservation of energy principle: The energy
in a closed system is conserved.

This is the basis for the equation
Ps(t) = Prs(t) + Pg(t)
This equation says that for the system under consideration
(Fig 1-1), the energy delivered by the source is equal to
the energy dissipated in the resistor plus the energy
delivered to the line. This is extremely basic and satisfies
the conservation of energy principle. This is not
superposition and any inclusion of cos(theta) terms would
be incorrect.

The equation
Pg(t) = Pf.g(t) + Pr.g(t)
is more interesting. The basis for this is superposition.
The forward and reverse voltage and current are superposed
to derive the actual voltage and current. It would seem
invalid to also sum the powers. To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

But by starting with the equations used to derive forward
and reverse voltage and current, it can be easily shown
with appropriate substitution that Ptot is always equal
to Pforward + Preverse (Or Pf - Pr if you use the other
convention for the direction of the energy flow)s. It
simply falls out from the way that Vf and Vr are derived
from Vactual and Iactual.

So
Pg(t) = Pf.g(t) + Pr.g(t)
is always true. For any arbitrary waveforms. Inclusion
of cos(theta) terms would be incorrect.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
. . .
The equation
Pg(t) = Pf.g(t) + Pr.g(t)
is more interesting. The basis for this is superposition.
The forward and reverse voltage and current are superposed
to derive the actual voltage and current. It would seem
invalid to also sum the powers. To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

But by starting with the equations used to derive forward
and reverse voltage and current, it can be easily shown
with appropriate substitution that Ptot is always equal
to Pforward + Preverse (Or Pf - Pr if you use the other
convention for the direction of the energy flow)s. It
simply falls out from the way that Vf and Vr are derived
from Vactual and Iactual.

It only holds true when Z0 is purely real. Of course, when it isn't,
time domain analysis becomes very much more cumbersome. But it's not
hard to show the problem using steady state sinusoidal analysis, and
that's where the cos term appears and is appropriate.


So
Pg(t) = Pf.g(t) + Pr.g(t)
is always true. For any arbitrary waveforms. Inclusion
of cos(theta) terms would be incorrect.

...Keith

Roy Lewallen, W7EL

The Rest of the Story
 

On Mar 21, 9:32*pm, Roy Lewallen wrote:
Keith Dysart wrote:
. . .
The equation
* Pg(t) = Pf.g(t) + Pr.g(t)
is more interesting. The basis for this is superposition.
The forward and reverse voltage and current are superposed
to derive the actual voltage and current. It would seem
invalid to also sum the powers. To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

But by starting with the equations used to derive forward
and reverse voltage and current, it can be easily shown
with appropriate substitution that Ptot is always equal
to Pforward + Preverse (Or Pf - Pr if you use the other
convention for the direction of the energy flow)s. It
simply falls out from the way that Vf and Vr are derived
from Vactual and Iactual.

It only holds true when Z0 is purely real. Of course, when it isn't,
time domain analysis becomes very much more cumbersome. But it's not
hard to show the problem using steady state sinusoidal analysis, and
that's where the cos term appears and is appropriate.

*
* So
* * Pg(t) = Pf.g(t) + Pr.g(t)
* is always true. For any arbitrary waveforms. Inclusion
* of cos(theta) terms would be incorrect.

Thanks for providing the limitation.

But I am having difficulty articulating where the math in the
following
derivation fails.

Starting by measuring the actual voltage and current at a single
point on the line, and wishing to derive Vf and Vr we have the
following four equations:

V = Vf + Vr
I = If - Ir
Zo = Vf / If
Zo = Vr / Ir

rearranging and substituting
Vf = V - Vr
= V - Zo * Ir
= V - Zo * (If - I)
= V - Zo * (Vf/Zo - I)
= V - Vf + Zo * I
= (V + Zo * I)/2
similarly
Vr = (V - Zo * I)/2

Pf = Vf * If
= Vf**2 / Zo
= ((V + Zo * I)(V + Zo * I)/4)/Zo
= (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)

Pr = Vr * Ir
= (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)

So, comtemplating that
P = Pf - Pr
and substituting
P = (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)
- (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)
= 4(V * Zo * I) / (4 * Zo)
= V * I
as required.

So when Zo is real, i.e. can be represented by R, it is
clear that P always equals Pf - Pr. And it does not even
matter which value of R is used for R. It does not have
to be the characteristic impedance of the transmission
line, the subtraction of powers still produces the correct
answer.

But when Zo has a reactive component, it still cancels
out of the equations. So why is this not a proof that
also holds for complex Zo.

I suspect it has to do with complex Zo being a concept
that only works for single frequency sinusoids, but am
having difficulty discovering exactly where it fails.

And if it is related to Zo and single frequency sinusoids,
does that mean that P = Pf - Pr also always works for
single frequency sinusoids?

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Roger Sparks wrote:

Ooooppps. That should have been Vg(t) = Vf.g(t) + Vr.g(t)

but doesn't this describe the standing wave?

Not by itself. It is simply the function describing the voltage at
point g.

The net total voltage at point g is the standing wave
voltage that is present at that point.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Actually the flip-flopping I was referring to was the
constant changes in your view of the limitations that
apply to your claim.

As I said, I corrected an error in my thinking. You
are free to consider that to be a flip-flop, I consider
it to be a step forward.

Are you now agreeing that the
energy in the reflected wave is only dissipated in the
source resistor for those instances when Vs is 0?

Yes, for instantaneous reflected energy exactly as I
previously stated, but not true for average reflected
energy. 100% of the average reflected energy is
dissipated in the source resistor when the transmission
line is 45 degrees long. That intra-cycle interference
exists, thus delaying the dissipation by 90 degrees,
is irrelevant to where the net energy winds up going.

Your conservation of power principle would have you
demanding that the power sourced by a battery charger
must be instantaneously dissipated. Everyone except
you seems to realize that is an invalid concept. The
dissipation of energy can be delayed by a battery
or a reactance. In the present example, the dissipation
of the instantaneous reflected energy is delayed by
90 degrees by the reactance.

I could not find them in the archive. Could you kindly provide
them again, showing where the 'cos(A)' term fits in the
equations:

Go back and read the part where I said cos(0)=+1 and
cos(180)=-1. There is no such thing as conservation of
power. Your equations assume a conservation of power
principle that doesn't exist in reality. The forward
power is positive power. The reflected power is positive
power. The only negative power is destructive interference
which must be offset by an equal magnitude of constructive
interference. When the instantaneous interference power is
negative, the two voltages are 180 degrees out of phase and
that is your cos(180)=-1. When the instantaneous interference
power is positive, the two voltages are in phase and that
is your cos(0)=+1. Why don't you already know all this
elementary stuff?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 22, 9:36*am, Cecil Moore wrote:
Keith Dysart wrote:
Roger Sparks wrote:
Ooooppps. That should have been Vg(t) = Vf.g(t) + Vr.g(t)

but doesn't this describe the standing wave? *

Not by itself. It is simply the function describing the voltage at
point g.

The net total voltage at point g is the standing wave
voltage that is present at that point.

I suppose that is a name you could use, but do you still
call it "standing wave voltage" when there is no
reflected wave, i.e. Vr is 0 for all time. There is
no standing wave when Vr is always 0.

...Keith