Roger Sparks wrote:
So the question is "When does (x + y)^2 = x^2 + y^2 ?".

(x + y)^2 = X^2 + 2xy + y^2

X^2 + 2xy + y^2 = x^2 + y^2 only when either x or y = zero.

That's some "Food for Thought", Roger, but unfortunately
phasor math is more complex :-) than that. The "Rest of
the Story" is if x and y are phasors that are 90 degrees
out of phase with each other, is there another solution
besides the one you offered?

Given two phasors, 1v at 0 degrees and 1v at 90 degrees,
what is the sum of the square of the voltages vs the
square of the sum of the voltages. Hint: the phasor sum
of the voltages is 1.414.
--
73, Cecil http://www.w5dxp.com

K7ITM wrote:
Since it's a linear system, no matter where the "reflected" wave is
coming from (that is, whether it's actually a reflection, or from some
completely separate source which may or may not be on the same
frequency), I believe until someone shows me differently and proves
the multitude of analyses showing it to be so, that the reflection
coefficient going from the line back into the linear source really
does work, and it works at every instant in time.

What you seem to be missing, Tom, is that if the two signals
are not coherent, interference is not possible. Since we are
discussing interference effects between obviously coherent
forward and reflected waves, your observation seems to be a
moot point.

For instance, if the forward and reflected traveling waves
in Roy's "Food for Thought" page are replaced by two sources
that differ by 30% in frequency, there is no way for that
entry in his chart to reach the 400 watts dissipated in the
source resistor.

BTW, I apologize for my outburst last night. I just get friggin'
tired of the "Have you stopped beating your wife?" questions.
--
73, Cecil http://www.w5dxp.com

On Thu, 06 Mar 2008 14:29:25 -0600
Cecil Moore wrote:

Roger Sparks wrote:
So the question is "When does (x + y)^2 = x^2 + y^2 ?".

(x + y)^2 = X^2 + 2xy + y^2

X^2 + 2xy + y^2 = x^2 + y^2 only when either x or y = zero.

That's some "Food for Thought", Roger, but unfortunately
phasor math is more complex :-) than that. The "Rest of
the Story" is if x and y are phasors that are 90 degrees
out of phase with each other, is there another solution
besides the one you offered?

Given two phasors, 1v at 0 degrees and 1v at 90 degrees,
what is the sum of the square of the voltages vs the
square of the sum of the voltages. Hint: the phasor sum
of the voltages is 1.414.
--
73, Cecil http://www.w5dxp.com

OK, The vector sum is sqrt(1^2 + 1^2) = sqrt(2) = 1.414

I would think of the "square of the sum of the voltages" to be (1 + 1)^2 = 2^2 = 4

We must be very careful to not use scaler math when vectors are called for.

--
73, Roger, W7WKB

On Mar 6, 12:35 pm, Cecil Moore wrote:

What you seem to be missing, Tom, is that if the two signals
are not coherent, interference is not possible.

There is NO WAY I'm interested in moving to a one-dimensional world
that requires me to special-case a particular type of wave to get the
right answer and FORSAKE the multi-dimensional linear circuits world
I'm in, that quite accurately describes what happens regardless of the
content of forward and reverse. Your one-dimensional world apparently
limits you to thinking about interference in a way that mine does
not. I may post the results of a 'speriment I'm setting up in a day
or two that may spark some interesting discussion. Till then, I'm
outta here.

Cheers,
Tom

On Mar 6, 3:35*pm, Cecil Moore wrote:
What you seem to be missing, Tom, is that if the two signals
are not coherent, interference is not possible. Since we are
discussing interference effects between obviously coherent
forward and reflected waves, your observation seems to be a
moot point.

For instance, if the forward and reflected traveling waves
in Roy's "Food for Thought" page are replaced by two sources
that differ by 30% in frequency, there is no way for that
entry in his chart to reach the 400 watts dissipated in the
source resistor.

How coherent do the two signals have to be for interference to
occur?

You say that when the sources are coherent, interference occurs
but when the frequency differs by 30% it does not.

What happens if one of the sources has just a bit of phase noise,
or the frequency wanders just a bit, or is just offset a bit?

How much of a difference does there have to be for interference
to stop? What is the threshold? Phase noise? Wander? Offset?

And is it the mechanism that creates interference that stops working
once the threshold is crossed?
Or does the mechanism still work, but we just no longer call the
result interference? Why do we stop calling it interference once
the threshold is crossed?

What is the mechanism that creates the effect we call interference?

...Keith

Hi Keith,

I must still not be "getting" something because while I now follow your numbers and trig identity, it looks to me like you used Cecil's premise "PRs = 50w + Pref " to show that '50 W plus Pref' = 68 + 68cos(2wt-61.9degrees) watts, which would be correct for the 12.5 ohm case.

So, rather than disproving Cecil's premise, you successfully demonstrated that it was correct in the instantaneous case.

What am I missing?

On Thu, 6 Mar 2008 09:56:06 -0800 (PST)
Keith Dysart wrote:

On Mar 6, 11:41*am, Roger Sparks wrote:
On Thu, 6 Mar 2008 06:10:26 -0800 (PST)

Keith Dysart wrote:
On Mar 6, 1:17*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 21:37:06 -0800 (PST)
[snip]
Thanks Keith. *I see what you are doing now, although I still don't understand your logic in faulting Cecil on the instantaneous values. *I agree with you that the instantaneous values can be tracked, but don't see a fault in Cecil's presentation. *

For the special situation described in
http://www.w5dxp.com/nointfr.htm
Cecil is attempting to show that the reflected energy is dissipated in
the source resistor.

The logic he employs is:
- before the reflection arrives back at the generator, the source
resistor
* is dissipating X watts.
- the reflected wave has an energy flow of Y watts.
- after the reflection arrives back at the generator, the source
resistor
* is dissipating Z watts.
- since Z is equal to X + Y, the energy in the reflected wave is being
* dissipated in the source resistor.

In other words, since the dissipation in the source resistor increases
by the same amount as the power in the reflected wave, the energy in
the reflected wave must be being dissipated in the source resistor.

Your logic and conclusion seem correct to me.

Cecil analyzes the circuit for a number of load resistances and
suggests
that the equality holds for any load resistance.

For example, with a load resistance of 12.5 ohms, the original
dissipation
in the source resistor is 50 W which increases to 68 W when the 18 W
reflected wave arrives back at the generator. That is, X = 50, Y = 18
and Z = 68, so Z is equal to X + Y.

Cecil does all of this analysis using average powers.

But we know that the power dissipation varies as a function of time
and
that the power in the reflected wave is a function of time. It is my
contention that if it is the energy in the reflected wave that is
increasing the dissipation in the source resistor, the dissipation in
source resistor should occur at the same time that the reflected wave
delivers the energy.

In other words, not only should Z.average = X.average + Y.average,
but Z.instantaneous should equal X.instantaneous + Y.instantaneous
for if the dissipation in the source resistor is not tracking the
energy in the reflected wave, it can not be the energy in the
reflected wave that is heating the resistor.

Hard to follow this long sentence/paragraph. *The dissipation in the source resistor should be the sum of instantaneous energy flows from both source(forward) and reflected energy flows. *

Agreed. And that is what I was trying to say.

We would not expect that the instantaneous energy flow would be equal from both source and reflection because of the sine wave nature of the energy flow.

So using the same 12.5 ohm example,
X.inst = 50 + 50 cos(2wt)
Y.inst = 18 + 18 cos(2wt)
X.inst + Y.inst = 68 + 68 cos(2wt)
but
Z.inst = 68 + 68 cos(2wt - 61.9degrees)

So Z.inst is not equal to Y.inst + X.inst.

I think you are mixing average energy flow (the 50 and 18 watts) with instantaneous flow (50 cos(2wt) and 18 cos(2wt)).

I do not think so.

The instantaneous energy flow for the reflected wave is
Pr.g(t) = 18 - 18 cos(2wt) (see correction in previous post)
This means that at t=0, 0 energy is flowing.
When 2wt is 180 degrees, 36 joules per second are flowing.
The average over a full cycle is 18 W.

So I am summing the instantaneous flows for the two contributors of
energy.

The correct sum, however, is 68 + 32cos(2wt).

The actual dissipation in the source resistor is
Vrs(t) = 82.46 cos(wt -30.96 degrees)
Irs(t) = 1.649 cos(wt -30.96 degrees)
Prs(t) = Vrs(t) * Irs(t)
= 68 + 68 cos(2wt -61.92 degrees)

Whenever cos(2wt-61.92degrees) is equal to 1, 132 joules per second
are being dissipated and whenever it is equal to -1, 0 joules are
being dissipated. This follows from the periodic nature of the energy
flow.

I think Z.inst = x.inst + y.inst = 50 cos(2wt) + 18 cos(2wt). *The two energy components arrive 90 degrees out of phase (in this example) so we should write z.inst = 50 cos(2wt) + 18 cos(2wt + 90).

This means that the dissipation in the resistor is not happening
at the same time as the energy is being delivered by the reflected
wave, which must mean that it is not the energy from the reflected
wave that is heating the source resistor.

So while analyzing average power dissipations suggests that the
energy from the reflected wave is dissipated in the source resistor,
analysis of the instantaneous power shows that it is not.

...Keith

Either your math or mine is not correct. *Which is incorrect?

Both. But I think I have now corrected mine.

...Keith


--
73, Roger, W7WKB

"Keith Dysart" wrote in message
...

What is the mechanism that creates the effect we call interference?

superposition.

K7ITM wrote:
On Mar 6, 12:35 pm, Cecil Moore wrote:

What you seem to be missing, Tom, is that if the two signals
are not coherent, interference is not possible.

There is NO WAY I'm interested in moving to a one-dimensional world
that requires me to special-case a particular type of wave to get the
right answer and FORSAKE the multi-dimensional linear circuits world
I'm in, that quite accurately describes what happens regardless of the
content of forward and reverse.

Coherency, non-coherency, and interference is covered well
in "Optics" by Hecht and other textbooks. Optical physicists
have been tracking the EM energy flow for centuries. This
information may be new to you but it is old hat in physics.

What I have proved in Part 1 is that average reflected energy
is not always reflected back toward the load. Equally false
is the flip side old wives tale that says: "Reflected energy
is always dissipated in the source."

It's takes only one case to prove an old wives' tale to
be false. That's why I chose the special case of zero
interference. One needs to understand the special case of
zero interference before one tries to understand the general
case involving interference which will be Part 2 and Part 3
of my articles.

If you choose to remain ignorant, "NO WAY I'm interested",
then that's your choice and that's OK. But understanding
interference is the easiest way I know of to track the
energy flow.

Your one-dimensional world apparently
limits you to thinking about interference in a way that mine does
not. I may post the results of a 'speriment I'm setting up in a day
or two that may spark some interesting discussion. Till then, I'm
outta here.

If you come up with an easier analysis of average energy flow
and average power, that will be great.
--
73, Cecil http://www.w5dxp.com

On Mar 7, 8:30*am, Roger Sparks wrote:
Hi Keith,

I must still not be "getting" something because while I now follow your numbers and trig identity, it looks to me like you used Cecil's premise "PRs = 50w + Pref * " to show that '50 W plus Pref' = 68 + 68cos(2wt-61.9degrees) watts, which would be correct for the 12.5 ohm case.

So, rather than disproving Cecil's premise, you successfully demonstrated that it was correct in the instantaneous case.

What am I missing?

The actual dissipation in the source resistor was computed using
circuit theory
to derive the voltage and current through the resistor and then
multiplying them
together to get the power dissipation:
Vrs(t) = 82.46 cos(wt -30.96 degrees)
Irs(t) = 1.649 cos(wt -30.96 degrees)
Prs.circuit(t) = Vrs(t) * Irs(t)
= 68 + 68 cos(2wt -61.92 degrees)

This was then shown not to be equal to the results using Cecil's
hypothesis
because
Prs.before(t) = 50 + 50 cos(2wt)
Pref(t) = 18 - 18 cos(2wt)
which would give, using Cecil's hypothesis
Prs.cecil(t) = 68 + 32 cos(2wt)

So I accept the circuit theory result of
Prs.circuit(t) = 68 + 68 cos(2wt -61.92 degrees)
and conclude that, since the results using Cecil's hypothesis are
different, Cecil's hypothesis must be incorrect.

That is, the power dissipated in the source resistor after the
reflection
returns is not the sum of the power dissipated in the resistor before
the
reflection returns plus the power in the reflected wave.

Now it does turn out that the average power dissipated in the source
resistor is the sum of the average power before the reflection returns
plus the average power in the reflected wave since
Prs.circuit.average = average( 68 + 68 cos(2wt -61.92 degrees) )
= 68
This does agree with Cecil's analysis using average powers. But energy
flows must balance on a moment by moment basis if energy is to be
conserved so when we do the instantaneous analysis we find that
Cecil's
hypothesis does not hold.

...Keith

PS: To compute Vrs(t) and Irs(t) using circuit theory:
The generator output voltage
Vg(t) = Vf.g(t) + Vr.g(t)
where Vf.g(t) is the line forward voltage at the generator
and Vr.g(t) is the line reflected voltage at the generator.

The generator output current
Ig(t) = If.g(t) + Ir.g(t)
where If.g(t) is the line forward current at the generator
and Ir.g(t) is the line reflected current at the generator.

Where Vs(t) is the source voltage
Vrs(t) = Vs(t) - Vg(t)
Irs(t) = Ig(t)
and the power is
Prs.circuit(t) = Vrs(t) * Irs(t)

Keith Dysart wrote:
How coherent do the two signals have to be for interference to
occur?

Instead of spending hours typing the answer, I will point
you to "Optics", by Hecht, 4th edition, Chapter 12, Basics
of Coherence Theory.

In my example, the one source is an ideal single frequency
source. Thus all signals existing within the system are
completely coherent, by definition. Your questions are
irrelevant to the example provided.

You say that when the sources are coherent, interference occurs
but when the frequency differs by 30% it does not.

In between "completely coherent" signals and "completely incoherent"
signals is a very large gray area. Please read the reference.

What happens if one of the sources has just a bit of phase noise,
or the frequency wanders just a bit, or is just offset a bit?

Please read the reference. In my example, the source is ideal
so that problem doesn't exist.

How much of a difference does there have to be for interference
to stop? What is the threshold? Phase noise? Wander? Offset?

Please read the reference. In my example, the source is ideal
so those problems don't exist.

And is it the mechanism that creates interference that stops working
once the threshold is crossed?
Or does the mechanism still work, but we just no longer call the
result interference? Why do we stop calling it interference once
the threshold is crossed?

In "Optics", by Hecht, Chapter 7 is on superposition and
Chapter 9 is on interference.

Quoting "Optics", by Hecht, Chapter 12. "Thus far in our
discussion of phenomena involving the superposition of waves,
we've restricted the treatment to that of either completely
coherent or completely incoherent disturbances. ... There is
a middle ground between these antithetic poles, which is of
considerable contemporary concern - the domain of
*partial coherence*.

What is the mechanism that creates the effect we call interference?

"Interference", in this context, is not defined in the IEEE
Dictionary. The closest I can come to a definition is from
"Optics", by Hecht: "... optical [EM] interference corresponds
to the interaction of two or more lightwaves yielding a resultant
irradiance [average power density] that deviates from the sum of
the component irradiances [average power densities]."
--
73, Cecil http://www.w5dxp.com