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

Keith Dysart wrote:
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)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged? If not, why do you
require such for the example under discussion?

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.
--
73, Cecil http://www.w5dxp.com

On Sat, 22 Mar 2008 14:14:35 GMT
Cecil Moore wrote:

Keith Dysart wrote:
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)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged? If not, why do you
require such for the example under discussion?

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

= [SQRT(pf.g(t) + SQRT(pr.g(t)]^2

and/or= [SQRT(pf.g(t) - SQRT(pr.g(t)]^2

How can we justify calling the +/- 2*SQRT[Pf.g(t)*Pr.g(t) term the "interference term"?

Am I correct in assuming that this equation describes the instantaneous power delivered to Rs?

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.
--
73, Cecil http://www.w5dxp.com


--
73, Roger, W7WKB

Roger Sparks wrote:
How can we justify calling the +/- 2*SQRT[Pf.g(t)*Pr.g(t) term the "interference term"?

Page 388, "Optics", by Eugene Hecht, 4th edition:
"The interference term becomes I12 = 2*SQRT(I1*I2)cos(A)"
where 'I' is the Irradiance (power density)[NOT Current]
Later Hecht says +2*SQRT(I1*I2) is the total constructive
interference term and -2*SQRT(I1*I2) is the total
destructive interference term. Chapter 9 is titled
"Interference" - recommended reading.

Am I correct in assuming that this equation describes the instantaneous power delivered to Rs?

Yes, if Tom, K7ITM, is correct about the equation working
for instantaneous power densities, not just for average
power densities as I had first assumed.

Let's say the instantaneous forward voltage dropped across
the source resistor is +50 volts and the instantaneous
reflected voltage across the source resistor is -30 volts.
The source resistor is 50 ohms.
Pf.rs(t) = (+50v)^2/50 = 50w
Pr.rs(t) = (-30v)^2/50 = 18w
Prs(t) = Pf.rs(t) + Pr.rs(t) - interference
Prs(t) = 50w + 18w - 2*SQRT(50*18) = 8 watts

If Tom is correct, that should be the actual dissipation
in the source resistor at that time which includes 60 watts
of destructive interference that will be dissipated 90 degrees
later when 2*SQRT(50*18) = +60 watts.
--
73, Cecil http://www.w5dxp.com

On Sun, 23 Mar 2008 09:03:31 -0500
Cecil Moore wrote:

Roger Sparks wrote:
How can we justify calling the +/- 2*SQRT[Pf.g(t)*Pr.g(t) term the "interference term"?

Page 388, "Optics", by Eugene Hecht, 4th edition:
"The interference term becomes I12 = 2*SQRT(I1*I2)cos(A)"
where 'I' is the Irradiance (power density)[NOT Current]
Later Hecht says +2*SQRT(I1*I2) is the total constructive
interference term and -2*SQRT(I1*I2) is the total
destructive interference term. Chapter 9 is titled
"Interference" - recommended reading.

Am I correct in assuming that this equation describes the instantaneous power delivered to Rs?

Yes, if Tom, K7ITM, is correct about the equation working
for instantaneous power densities, not just for average
power densities as I had first assumed.

Let's say the instantaneous forward voltage dropped across
the source resistor is +50 volts and the instantaneous
reflected voltage across the source resistor is -30 volts.
The source resistor is 50 ohms.
Pf.rs(t) = (+50v)^2/50 = 50w
Pr.rs(t) = (-30v)^2/50 = 18w
Prs(t) = Pf.rs(t) + Pr.rs(t) - interference
Prs(t) = 50w + 18w - 2*SQRT(50*18) = 8 watts

If Tom is correct, that should be the actual dissipation
in the source resistor at that time which includes 60 watts
of destructive interference that will be dissipated 90 degrees
later when 2*SQRT(50*18) = +60 watts.
--
73, Cecil http://www.w5dxp.com

Thanks for your thoughtful reply.

TanH(30/50) = 30.96 degrees. This takes us back to the 12.5 ohm load example.

Is it possible that in your example here, the reflected voltage acts in series with Rs but arrives 90 degrees out of phase with the forward voltage? If so, then

Vrs = sqrt(50^2 + 30^2) (the reflected voltage should ADD to the source voltage)

= sqrt(3400) = 58.31v

The power to Rs would be

Prs = (V^2)/50 = 3400/50 = 68w

We previously found that 32w was used at the 12.5 ohm load, so 32 + 68 = 100w. The entire output from the source is accounted for.

If this is the case, we have here an example of constructive interference, and complete accounting for the power.

You might wonder why I would consider this alternative. If the destructive interference included a 90 degree delay, how would I know whether the 30v was the delayed voltage or exactly in phase with the source? It must be delayed by 90 degrees because the forward voltage is always 90 degrees ahead of the reflected wave (in 45 degree line length example).

Your example certainly works as written, but it also introduces a dilemma. Where is the power stored for 90 degrees?

To answer that question, I see two possiblities: The source voltage causes a reflection so the 60w is stored as an additional reflected wave on the transmission line.

Or second, the 60w is stored in the source.

--
73, Roger, W7WKB

Roger Sparks wrote:
(the reflected voltage should ADD to the source voltage)

If you graph the two voltages you will find that half the time
the reflected voltage adds to the source voltage and half the
time the reflected voltage subtracts from the source voltage.
Both are true half the time. You can point out either case on
the graph. That's why the average interference term is zero
for this special case and therefore why 100% of the average
reflected power is dissipated in the source resistor for
this special case.

You might wonder why I would consider this alternative.
If the destructive interference included a 90 degree delay,
how would I know whether the 30v was the delayed voltage or
exactly in phase with the source?

By looking at the graphs?

Where is the power stored for 90 degrees?

In the equivalent reactance of the transmission line.

That's what reactances do in AC circuits. They store
energy and deliver it back to the system some time
later.
--
73, Cecil http://www.w5dxp.com

On Sun, 23 Mar 2008 19:57:24 GMT
Cecil Moore wrote:

Roger Sparks wrote:
(the reflected voltage should ADD to the source voltage)

If you graph the two voltages you will find that half the time
the reflected voltage adds to the source voltage and half the
time the reflected voltage subtracts from the source voltage.
Both are true half the time. You can point out either case on
the graph. That's why the average interference term is zero
for this special case and therefore why 100% of the average
reflected power is dissipated in the source resistor for
this special case.

You might wonder why I would consider this alternative.
If the destructive interference included a 90 degree delay,
how would I know whether the 30v was the delayed voltage or
exactly in phase with the source?

By looking at the graphs?

Where is the power stored for 90 degrees?

In the equivalent reactance of the transmission line.

That's what reactances do in AC circuits. They store
energy and deliver it back to the system some time
later.
--
73, Cecil http://www.w5dxp.com

I did not use a graph, but created a spreadsheet that calculated Vrs for the short circuit, 45 degree long line. It shows the 90 degree transfer of power that you described. I posted the spreadsheet in PDF format at http://www.fairpoint.net/~rsparks/Reflect_short.pdf.

To me, this shows that my traveling wave analysis on an instant basis is not correct because the energy can not be located precisely on a degree-by-degree scale. Yes, it is correct on the average over 360 degrees, but not instantaneously. We are missing something.

Central to traveling waves is the assumption that the wave is not compressable. The energy is assumed to flow in a consistantly predictable mannor that is linear and described by a sine wave. That assumption is violated when energy is delayed for reasons other than distance of travel, which is demonstrated in this example.

I am not ready to suggest a cure for my traveling wave analysis. I only see that it does not work to my expectations.

Thanks for providing the examples and comments.
--
73, Roger, W7WKB

On Mar 22, 10:14*am, Cecil Moore wrote:
Keith Dysart wrote:
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)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged?

Are you making this a trick question by using the
word "dissipated"?

If not, then yes. Consider my laptop which has an
external power supply connected by a cord to the
laptop.

Conservation of energy means that the instantaneous
energy flow (i.e. power) along this cord into the
laptop is always exactly equal to the sum of
- the sum of energy being dissipated as heat
in each individual component
- the increase in energy being stored in
components such as capacitors, inductors
and batteries, minus any stored energy
being returned from components such as
capacitors, inductors and batteries
- energy being emitted such as
- light (e.g. display)
- sound (e.g. fans, speakers)
- RF (e.g. Wifi antennas, RFI)
- etc.

So yes, the phrase "conservation of power" is
appropriately descriptive and follows from
conservation of energy.

If not, why do you
require such for the example under discussion?

I require it for both.

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.

Where is "cos(theta)" in this?
And what "theta" is to be used?
Is it the same theta that was used to conclude that
this was a "no interference" example?
See "A Simple Voltage Source - No Interference" at
http://www.w5dxp.com/intfr.htm

So the plus/minus in the equation aboves means that
for part of the cycle we should add and part of the
cycle we should subtract.
This equation is ambiguous and is therefore incomplete
since it does not tell us when we should add and when
we should subtract.
Could you provide a mathematically precise description
of when we should do each?

Where is this destructive interference energy stored
while waiting to be dissipated constructively later?
If it is in a capacitance, which one? Does the
voltage on the capacitance increase appropriately
to account for the energy being stored?
Similarly if it is stored in an inductance.

...Keith

Keith Dysart wrote:
So yes, the phrase "conservation of power" is
appropriately descriptive and follows from
conservation of energy.

You have a contradiction built into your concepts.
You have argued that the instantaneous power dissipated
in the source resistor is not equal to the instantaneous
forward power component plus the instantaneous reflected
power because power must be conserved at each instant of
time. That is simply not true.

I'm telling that energy must be conserved at each instant
of time but power does not have to be conserved at each
instant of time. Energy can obviously be stored in a
battery or network reactance for dissipation later in
time.

Do you require that the power used to charge a battery
be instantaneously dissipated in the battery? Of course
not! That's true for a dummy load but NOT for a battery.
There is no reason to require dissipation of power at
each instant of time to balance. Since energy can be
stored, there is no such thing as conservation of
instantaneous power, only of instantaneous energy.

Where is "cos(theta)" in this?
And what "theta" is to be used?

How many times do I have to explain this? For instantaneous
values of voltage, if the sign of the two interfering voltages
are the same, theta is zero degrees and the cosine of theta is
+1.0. If the sign of the two interfering voltages are opposite,
theta is 180 degrees and the cosine of theta is -1.0.

Is it the same theta that was used to conclude that
this was a "no interference" example?
See "A Simple Voltage Source - No Interference" at
http://www.w5dxp.com/intfr.htm

That theta is the phase angle of the phasor. If you
would switch over to phasor notation for your values
of instantaneous voltages, it might make things clearer
for you. You have been dealing only with real part of
the phasors which limits theta to either zero or 180
degrees.

So the plus/minus in the equation aboves means that
for part of the cycle we should add and part of the
cycle we should subtract.
This equation is ambiguous and is therefore incomplete
since it does not tell us when we should add and when
we should subtract.

I have told you about five times now. If the sign of the
two interfering voltages are the same, the interference
term has a positive sign (constructive). If the sign of
the two interfering voltages are opposite, the interference
term has a negative sign (destructive). Instead of me
having to keep posting this over and over how about you
keep reading it over and over until you understand it?

Could you provide a mathematically precise description
of when we should do each?

See above. If we deal with phasors, theta is the phase
angle between the two interfering voltages. If we deal
only with real values, theta is limited to either zero
(cosine = +1.0) or 180 degrees (cosine = -1.0).

Where is this destructive interference energy stored
while waiting to be dissipated constructively later?

In our 45 degree shorted stub example, it is stored in
the equivalent inductive reactance of the stub. For each
instant in which destructive interference energy is stored,
there is an instant in time 90 degrees later when that same
energy is dissipated as constructive interference power
in the source resistor. That's why 100% of the average
reflected power is dissipated in the source resistor for
these special cases of zero average interference.
--
73, Cecil http://www.w5dxp.com

On Mar 23, 10:31 am, Cecil Moore wrote:
Keith Dysart wrote:
So yes, the phrase "conservation of power" is
appropriately descriptive and follows from
conservation of energy.

You have a contradiction built into your concepts.
You have argued that the instantaneous power dissipated
in the source resistor is not equal to the instantaneous
forward power component plus the instantaneous reflected
power because power must be conserved at each instant of
time.
That is simply not true.

Bzzzzzzzzzzztt.

I'm telling that energy must be conserved at each instant
of time but power does not have to be conserved at each
instant of time.

Bzzzzzzzzzzztt.

Energy can obviously be stored in a
battery or network reactance for dissipation later in
time.

Indeed. And forgetting to include such flows in the summation
would be a serious error.

For example, from the example,
Ps(t) = Prs(t) + Pg(t)
includes Pg(t) which accounts for the energy stored in the
line and later returned. There are no missing flows in
this equation. And Ps(t) also accounts for energy absorbed
in the voltage source. Only the source resistor has a
unidirectional energy flow.

Do you require that the power used to charge a battery
be instantaneously dissipated in the battery?

The energy flow into the battery is exactly and always
accounted for by the energy flow that heats the battery
and the energy flow consumed in the reversable chemical
reaction.

The instantaneous flows always sum appropriately to
satisfy conservation of energy. Of course, if one forgets
a flow, then the sum will not balance.

Of course
not! That's true for a dummy load but NOT for a battery.
There is no reason to require dissipation of power at
each instant of time to balance. Since energy can be
stored, there is no such thing as conservation of
instantaneous power,

Of course there is, but you must include the flows into
the elements that store energy as I have done.

only of instantaneous energy.

Where is "cos(theta)" in this?
And what "theta" is to be used?

How many times do I have to explain this? For instantaneous
values of voltage, if the sign of the two interfering voltages
are the same, theta is zero degrees and the cosine of theta is
+1.0. If the sign of the two interfering voltages are opposite,
theta is 180 degrees and the cosine of theta is -1.0.

A strange of way of looking at it. It seems easier just
to say that there is no theta. And add the voltages.

But no matter, I have figured out where your extra term
comes from.

Let us a consider a simple circuit with two voltage sources
(V.s1 and V.s2) in series, connected to a resistor R.

Using superposition we have
V.s1 = R * Ir.s1
and
V.s2 = R * Ir.s2

So
Vr.tot = V.s1 + V.s2
and
Ir.tot = I.s1 + I.s2
This is superposition, and all is well.

The power dissipated in the resistor is
Pr = (Vr.tot)**2 / R
but we could also derive Pr in terms of V.s1 and V.s2
Pr = (V.s1 + V.s2) (V.s1 + V.s2) / R
= ((V.s1)**2 + (V.s2)**2 + (2 * V.s1 * V.s2) ) / R
= (V.s1)**2 / R + (V.s2)**2 / R + (2 * V.s1 * V.s2) / R

Now some people attempt to compute a power for each of
the contributing voltages across the resistor and obtain
Pr.s1 = (V.s1)**2 / R
Pr.s2 = (V.s1)**2 / R
When these are added one obtains
Pr.false = (V.s1)**2 / R + (V.s2)**2 / R
which, by comparison with Pr above can easily be seen
not to be the power dissipated in the resistor. Pr.false
is missing the term ((2 * V.s1 * V.s2) / R) from Pr. It
is for this reason that it is said that one can not
superpose powers. Simply stated, when powers are dervived
from the constituent voltages that are superposed, it is
not valid to add the powers together to derive the total
power.

Of course for the most part, powers being added are not
powers derived from the constituent voltages of a total
voltage, so in most cases it is quite valid to add powers
and expect them to sum to the total power.

But what do you do if a circuit is superposing two
voltages and you are presented with information about
the circuit in terms of powers. Well then you can add
the powers and include a correction term.

Assume
Pr = Pr.false + Pr.correction
= Pr.s1 + Pr.s2 + Pr.correction
But can we find a Pr.correction? It has to correct for
the term missing from Pr.false but present in Pr,
i.e. ((2 * V.s1 * V.s2) / R).

Restated in terms of power, ((2 * V.s1 * V.s2) / R)
becomes
2 * sqrt(P.s1 * P.s2)
But sqrt has two solutions so we have to write
Pr = Pr.s1 + Pr.s2 +/- 2 * sqrt(P.s1 * P.s2)
which should look very familiar.

As you have correctly pointed out, the sign of
Pr.correction is negative when the signs of the
constituent voltages are different and positive when
they are the same. The reason for this can easily
be seen from the derivation of Pr.correction.

This Pr.correction term has nothing to do with
interference, it is the correction required when it is
desired to add two powers computed from the superposing
constituent voltages of an actual total voltage across
an element and derive the energy flow into the
element.

Note that there is no hint that Pr.correction needs to be
stored when it is negative nor come from somewhere when
it is positive. It is, after all, just a correction that
needs to be applied when one wants to compute the total
power given two powers derived from the constituent
voltages of superposition.

Since one needs to know the constituent voltages to
determine the sign of Pr.correction, why not just use
superposition to compute the total voltage and then
derive the power? It would be much simpler. With no
need for Pr.correction, interference, storage and
release of interference 'energy', ...
(Of course, over in optics land it is difficult to
measure the voltage, so suffering the pain of using
powers is probably appropriate).

This analysis also makes clear the nature of powers
computed from the constituent voltages of superposition.
These powers do not represent real energy flows. As
discussed far above, real energy flows can be summed
to test for conservation of energy. When energy flows
do not sum appropriately, then either an energy flow
is missing, or one is attempting to sum powers which
are not real, for example, having been computed from
the constituent voltages of superposition (e.g. Pfor
and Pref in a transmission line).

....Keith