The Rest of the Story
 

On Apr 6, 7:02*am, "Mike" wrote:
Lets put the 'Rest of the Story' to bed before it becomes 'The Never Ending
Story' or are we too late already.

Too late. The first reference I could find in which Cecil is
chasing the imputed energy in reflected waves is "Is Maxwell
wrong?", posted December 31, 1994.

Soon to be a decade and a half.

...Keith

The Rest of the Story
 

Mike wrote:
Lets put the 'Rest of the Story' to bed before it becomes 'The Never Ending
Story' or are we too late already.

Do you have anything technical to contribute to
resolving the on-topic thread issues? If not,
it is very likely to continue until it is
resolved.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Soon to be a decade and a half.

One would think that by now someone would have
figured it out.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 6, 9:42*am, Cecil Moore wrote:
Keith Dysart wrote:
Soon to be a decade and a half.

One would think that by now someone would have
figured it out.

Perhaps, like the search for Sasquatch*, doomed to failure
because of the object's non-existance.

...Keith

* Substitute any of: Yeti, Nessie, Champ, Bigfoot,
Memphremagog Monster, Ogopogo, Reflected Wave Power.

The Rest of the Story
 

Keith Dysart wrote:
Perhaps, like the search for Sasquatch*, doomed to failure
because of the object's non-existance.

I agree that the conjugate of Sasquatch is
probably non-existent. :-)
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 5, 9:05*am, Cecil Moore wrote:
Keith Dysart wrote:
Please expand on what it tells us "about what is wrong
with the analysis so far".

You have not been able to tell where the instantaneous
reflected energy goes. This new example should help
you solve your problem.

It is not really my problem. It is only a problem for
those who expect the imputed power of a partial contributor
to superposition to represent a real energy flow.

If your steady-state equations
for the new example are not identical to the steady-
state equations for the previous example, then something
is wrong with your previous analysis.

I would fully expect the same results.

IMO, something
is obviously wrong with your previous analysis since
it requires reflected waves to contain something other
than ExH joules/sec, a violation of the laws of EM
wave physics.

Not at all a violation. Just as one does not expect the
partial values of volts and currents during superposition
to produce a power value that represents a real energy
flow, one should not expect it from Es and Hs which are
the partial values being superposed.

* * * * * feedline1 * * * * feedline2
source---1WL 50 ohm---Rs---1WL 50 ohm---+j50

The beauty of this example is that conditions at
the source resistor, Rs, are isolated from any
source of energy other than the ExH forward wave
energy in feedline1 and the ExH reflected wave
energy in feedline2. That's all the energy there
is available at Rs and that energy cannot be denied.

I suppose we shall see when you complete the arithmetic.

...Keith

The Rest of the Story
 

On Apr 5, 10:02*am, Cecil Moore wrote:
Keith Dysart wrote:
There was an 'if' there, wasn't there? Do you think
the 'if' is satisfied? Or not? The rest is useless
without knowing.

Under the laws of physics governing transmission lines
inserting an ideal 1WL line does not change the steady-
state conditions. If you think it does, you have invented
some new laws of physics.

You still have to explain where this destructive energy is stored
for those 90 degrees. Please identify the element and its energy
flow as a function of time.

Your request is beyond the scope of my Part 1 article.
If interference exists at the source resistor, the energy
associated with the interference flows to/from the source
and/or to/from the load. That condition is NOT covered in
my Part 1 article. Please stand by for Part 2 which will
explain destructive interference and Part 3 which will
explain constructive interference.

One advantage of moving the source voltage one wavelength
away from the source resistor is that it is impossible for
the source to respond instantaneously

You have previously claimed that the steady-state conditions
are the same (which I agree),

Glad you agree so there is nothing stopping you from an
analysis of the following example:

True, but as you say, the results will be the same.

source---1WL 50 ohm---Rs---1WL 50 ohm---+j50
* * * * * *Pfor1-- * * * * *Pfor2--
* * * * * *--Pref1 * * * * *--Pref2

Make Rs a 4-terminal network and a standard s-parameter
analysis is possible.

Yes, but that would be an average analysis and we have already
seen how averages mislead.

but now you have moved to discussing
transients, for which the behaviour is quite different.

Nope, you are confused. I am saying absolutely nothing about
transients.

You did say: "One advantage of moving the source voltage one
wavelength away from the source resistor is that it is impossible
for the source to respond instantaneously."

The words "respond instantaneously" suggested transient, rather
than waiting for the system to settle.

Why do you think an instantaneous power analysis
during steady-state is not possible?

Haven't said that. In fact, I think that is what I have been
doing.

If you want to claim similarity, then you need to allow the
circuit to settle to steady state after any change. Instantaneous
response is not required if the analysis is only steady-state.

...Keith

The Rest of the Story
 

On Apr 5, 10:06*am, Roger Sparks wrote:
On Sat, 5 Apr 2008 03:06:18 -0700 (PDT)

Keith Dysart wrote:
On Apr 4, 11:41*am, Roger Sparks wrote:
On Fri, 4 Apr 2008 06:30:08 -0700 (PDT)

Keith Dysart wrote:
On Apr 4, 1:29*am, Roger Sparks wrote:
[snip]
Another way to figure the power to the source would be by using the voltage and current through the source. *

This is how I did it.

Taking Esource.50[90..91] = 0.03046 J as an example ...

Psource.50[90] = V * I
* * * * * * * *= 0.000000 * 0.000000
* * * * * * * *= 0 W
Psource.50[91] = -2.468143 * -0.024681
* * * * * * * *= 0.060917 W

Using the trapezoid rule for numerical integration,
Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval
* * * * * * * * * *= ((0+0.060917)/2)*1
* * * * * * * * * *= 0.030459 J

The other powers and energies in the spreadsheet are computed
similarly.

There was an error in the computation of the 'delta's; the sign was
wrong.
The spreadsheet available athttp://keith.dysart.googlepages.com/radio6
has
now been corrected. (And, for Cecil, this spreadsheet no longer has
macros
so it may be downloadable.)

To use this spreadsheet to compute my numbers above, set the formulae
for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero.

Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J.

I could not match to the above data to any rows, so I can't comment.
But
perhaps the explanation above will correct the discrepancy.

I took a look at your revised spreadsheet entitled "Reflected45degrees-1.xls". Using the numbers from row 94 (91 degrees), the voltage developed at the source would be Vs = -2.468143v. *The current folowing through the source would be found from Ig which is 1.389317a in todays version of the spreadsheet (it was 1.439a previously). *The power flowing INTO the source is 2.468143* 1.389317 = 3.429032w. *This is the power Ps found in Column 11. *

This returning power is all from the reflected wave. *

I would not say this. The power *is* from the line, but this is Pg,
and it satisfies
the equation
* Ps(t) = Prs(t) + Pg(t)

The imputed power in the reflected wave is Pr.g(t) and is equal to
-99.969541 W, at
91 degrees. This can not be accounted for in any combination of Ps(91)
(-3.429023 W)
and Prs(91) (96.510050 W). And recall that expressing Cecil's claim
using instantaneous
powers requires that the imputed reflected power be accounted for in
the source
resistor, and not the source. This is column 26 and would require that
Prs(91) equal
100 W (which it does not).

Taking these numbers and adding the Psource.50[91] = -2.468143 * -0.024681*= 0.060917 W found previously above,
we have Total = 0.060917 + 3.429023 + 96.510050 = 99.99999. *Very close to 100w, but I am not sure of exactly what I am adding here.

Using my notation where Ptot = Pfor + Pref we have
Pg(t) = Pf.g(t) + Pr.g(t)
= 50 + 50 cos(2wt) + (-50 + 50 cow(2wt))
= 100 cos(2wt)
If you subtract Pr.g from Pf.g(t) you get 100. If you follow through
the math, you will
find that this is what you have done.
Ps.50[91] + (- Ps.0[91]) + Prs[91]
which is
2 * Pf.g[91] + (-Prs[91] - Pg[91]) + Prs[91]
which is
Pf.g[91] + Pf.g[91] - prs[91] - Pf.g[91] - Pr.g[91] + Prs[91]
which is
Pf.g[91] - Pr.g[91]
which is
100

I think you are right, that we should always be able to add the power on the line to 100w if we truely account for the power at any instant. *This results from the trig identity that sin(x) + cos(x) = 1. *Sin(x+90)= cos(x) so sin(x)+sin(x+90)= 1. *This explains why the sum of the forward power and reflected power should always equal 100w.

If I recall correctly, it is sin**2 + cos**2 = 1.

Pg(t) is the result of a standing wave, containing *power from Pf(x) and Pr(x+90). *

This is one way of thinking of it, but it is less misleading to
consider
that Pg(t) describes the actual energy flow, just as Vg(t) describes
the
actual voltage and Ig(t) describes the actual current. Using
superposition
Vf, If, Vr and Ir can be derived and from these Pf and Pr.

Only the power from Pr(x+90) is available to at a later time Prs(t+90+delta). *Power from Pf(x+delta) is found in the transmission line.

The source is acting like a resistor with an impedance of 2.468143/1.389317 = 1.776 ohms. *

This is not a good way to describe the source. The ratio of the
voltage to the current
is 1.776 but this is not a resistor since if circuit conditions were
to change, the
voltage would stay the same while the current could take on any value;
this being
the definition of a voltage source. Since the voltage does not change
when the current
does, deltaV/deltaI is always 0 so the voltage source is more properly
described as
having an impedance of 0.

As a result, the returning reflection does not truely see 50 ohms but sees 50 + 1.776 = 51.776 ohms. *

The returning reflection is affectively a change in the circuit
conditions. Using
the source impedance of 0 plus the 50 ohm resistor means the
reflection sees 50
ohms, so there is no reflection.

Using your approach of computing a resistance from the instantaneous
voltage and
current yeilds a constantly changing resistance. The reflection would
alter this
computed resistance. This change in resistance would then alter the
reflection
which would change the resistance. Would the answer converge?

The only approach that works is to use the conventional approach of
considering
that a voltage source has an impedance of 0.

The overriding issue is to account for all the power, which we are having a hard time doing. *The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. *That negates the idea that the source has an impedance of zero when we also assign the source a voltage.

When a current flows out of a voltage source, the source is providing
energy. When
the current flows into it, the voltage source is absorbing energy. A
charging
battery is reasonable real world example.

Of course the result is a another reflection. *Is this the idea you were trying to communicate Cecil?

To me, this is destructive interference at work, so all the power in the reflected wave does not simply disappear into the resistor Rs on the instant basis. *

I agree with latter, but not for the reason expressed. Rather, because
the imputed
power of the reflected wave is a dubious concept. This being because
it is impossible
to account for this power.

Does it help to notice that we are applying the power through two wires/paths. *We then divide the power in half and say that the power to the resistor from the close half flows seamlessly from source to resistor. *On the other hand, the second half of the source power flows through the transmission line and we say that it may never get to the resistor. *Something is wrong with this logic. *

If we can't account for the power, it is because we are doing the accounting incorrectly.

And the error in the accounting may be the expectation that the
particular set of powers chosen should balance. Attempting to
account for Pr fails when Pr is the imputed power from a partial
voltage and current because such computations do not yield powers
which exist.

clip

Yes. *I don't want to force my conventions onto you, so I am trying to understand yours while insisting that the answers from each convention must be the same. *

I think we both agree that the reflections are carrying power now.

Not I. Not until the imputed power can be accounted for.

...Keith

I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. *Maybe if we pursue the trig identity
sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers?

Except that it is sin squared + cos squared that equals one.

...Keith

The Rest of the Story
 

On Apr 5, 11:01*am, Cecil Moore wrote:
Roger Sparks wrote:
The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so

the source is absorbing power. *That negates the idea that the source
has an impedance of zero when we also assign the source a voltage.

Consider that what you are seeing is the flip side of the
interference at the source resistor. When a local source is
present, it can certainly absorb destructive interference
energy and supply constructive interference energy. The
following example has identical steady-state conditions
but brings Pfor1 and Pref1 into play for the instantaneous
values. I suspect that Pref1 is being completely ignored
in the present analysis.

Vs(t)---1WL 50 ohm---Rs---1WL 50 ohm---+j50
* * * * * Pfor1-- * * * * * Pfor2--
* * * * * --Pref1 * * * * * --Pref2

If we can't account for the power, it is because we are doing the accounting incorrectly.

Try the above example and maybe it will become clear.

Pref1 = Pfor1(rho1^2) + Pref2(1-rho2^2) + interference1

Pfor2 = Pfor1(1-rho1^2) + Pref2(rho2^2) + interference2

The source power doesn't appear directly in the equations
and need not be considered at all.

Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 * (all average)

I suspect the above equation will account for all the
energy components even at the instantaneous level such
that:

Pfor1(t) + Pref2(t) + P.Rs(t) = Pfor2(t) + Pref1(t)

Please note that all of these power components exist
when the two transmission lines are removed so this
analysis is probably the key to understanding what
is wrong with the earlier analysis.

When the two transmission lines are removed, how can
these imputed powers exist? What impedance did you
assign to the non-existant section of line to permit
you to compute these powers?

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Just as one does not expect the
partial values of volts and currents during superposition
to produce a power value that represents a real energy
flow, one should not expect it from Es and Hs which are
the partial values being superposed.

Given two coherent EM waves, W1 and W2, we know
that the power in W1 is E1xH1 and the power
in W2 is E2xH2. That is simple physics.

No matter what the results of superposition
of those two waves, the total energy in both
waves must be conserved. That is simple physics.

If you would think about what happens to two
coherent light waves in free space, you wouldn't
be able to justify your assertion above.
--
73, Cecil http://www.w5dxp.com