The Rest of the Story
 

Keith Dysart wrote:
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)

Where do you take into account that the forward wave
is 90 degrees out of phase with the reflected wave?

Z.inst = 68 + 68 cos(2wt - 61.9degrees)

Where does the 61.9 degrees come from?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

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. 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 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?

--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
The dissipation in the source resistor should be the
sum of instantaneous energy flows from both source
(forward) and reflected energy flows. 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.

The key question: Is the square of the sum of the
two voltages equal to the sum of the squares of
the two voltages? If yes, there is no interference
and it is valid to add the powers directly as
Keith has done.

If no, interference exists and it is *INVALID* to
add the powers directly as Keith has done. Every
EE was warned about superposing powers at the
sophomore level if not before. This is why.

So what we need to know is:

Is [Vfor(t) + Vref(t)]^2 equal to
Vfor(t)^2 + Vref(t)^2 ???

Does [70.7v*cos(wt) + 42.4v*cos(wt+90)]^2 equal
[70.7v*cos(wt)]^2 + [42.4v*cos(wt+90)]^2 ???

Is 2[70.7v*cos(wt)*42.4v*cos(wt+90)] always zero???

The answer is obviously 'NO' so Keith's direct addition
of powers, i.e. superposition of powers, is invalid as
it always is when interference is present.

When the interference term is properly taken into
account, the instantaneous dissipation in the source
resistor will no doubt equal the dissipation from
the forward wave plus the dissipation from the
reflected wave plus the interference term which is
minus for destructive interference and plus for
constructive interference. The interference will
average out to zero over each single complete cycle.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 6, 11:12*am, Cecil Moore wrote:
Keith Dysart wrote:
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)

Where do you take into account that the forward wave
is 90 degrees out of phase with the reflected wave?

Z.inst = 68 + 68 cos(2wt - 61.9degrees)

Where does the 61.9 degrees come from?

The computation of all of these can be seen in the spreadsheet at
http://keith.dysart.googlepages.com/...d%2Creflection

Recall that: cos(a)cos(b) = 0.5( cos(a+b) + cos(a-b) )

The power in the resistor is computed by multiplying the voltage
Vrs(t) = 82.46 cos(wt -30.96 degrees)
by the current
Irs(t) = 1.649 cos(wt -30.96 degrees)
yielding
Prs(t) = 68 + 68 cos(2wt -61.92 degrees)

For reflected power at the generator
Vr.g(t) = 42.42 cos(wt +90 degrees)
Ir.g(t) = 0.8485 cos(wt -90 degrees)
Pr.g(t) = Vr.g(t) * Ir.g(t)
= -18 + 18 cos(2wt)

Note that I previously made a transcription error in the sign
of one of the terms.

So X.inst + Y.inst is actually 68 + 32 cos(2wt).

...Keith

The Rest of the Story
 

Cecil Moore wrote:
Is 2[70.7v*cos(wt)*42.4v*cos(wt+90)] always zero???

This is a test to see if interference exists. It turns
out that constructive interference exists for the
first 90 degrees and third 90 degrees of the forward
wave cycle. Destructive interference exists for the
second and fourth 90 degrees of the cycle. The magnitude
of the destructive interference exactly equals the
magnitude of the constructive interference as expected
so the net interference is zero as expected.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

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

The Rest of the Story
 

Keith Dysart wrote:
The computation of all of these can be seen in the spreadsheet at
http://keith.dysart.googlepages.com/...d%2Creflection

The error in your calculations has been diagnosed.
Since interference exists at the instantaneous level,
it is invalid to add the instantaneous powers directly
as you have done.

Hint: if (V1^2 + V2^2) is not equal to (V1 + V2)^2
then it is invalid to add powers directly. Everyone
should have learned that fact in EE201 when the
professor said: "Thou shalt not superpose powers".
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
So I am summing the instantaneous flows for the two contributors of
energy.

An invalid thing to do when interference is present.
Reference EE-201 to find out when and why

[V1(t)^2 + V2(t)^2] is NOT equal to [V1(t) + V2(t)]^2.

The whole premise and assertion that the dissipation
in the source resistor is equal to 50w plus the
reflected power is based on the condition that

(V1^2 + V2^2) is equal to (V1 + V2)^2

Since you did NOT satisfy that condition, it
logically follows that the assertion would not
apply unless that necessary condition is met.

Please get back to us when you meet the above
condition. In attempting to do so, you will
realize why Eugene Hecht said that instantaneous
power is "of limited utility".
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Thu, 06 Mar 2008 11:21:46 -0600
Cecil Moore wrote:

Roger Sparks wrote:
The dissipation in the source resistor should be the
sum of instantaneous energy flows from both source
(forward) and reflected energy flows. 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.

The key question: Is the square of the sum of the
two voltages equal to the sum of the squares of
the two voltages? If yes, there is no interference
and it is valid to add the powers directly as
Keith has done.

If no, interference exists and it is *INVALID* to
add the powers directly as Keith has done. Every
EE was warned about superposing powers at the
sophomore level if not before. This is why.

So what we need to know is:

Is [Vfor(t) + Vref(t)]^2 equal to
Vfor(t)^2 + Vref(t)^2 ???

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.



Does [70.7v*cos(wt) + 42.4v*cos(wt+90)]^2 equal
[70.7v*cos(wt)]^2 + [42.4v*cos(wt+90)]^2 ???

Is 2[70.7v*cos(wt)*42.4v*cos(wt+90)] always zero???

The answer is obviously 'NO' so Keith's direct addition
of powers, i.e. superposition of powers, is invalid as
it always is when interference is present.

When the interference term is properly taken into
account, the instantaneous dissipation in the source
resistor will no doubt equal the dissipation from
the forward wave plus the dissipation from the
reflected wave plus the interference term which is
minus for destructive interference and plus for
constructive interference. The interference will
average out to zero over each single complete cycle.
--
73, Cecil http://www.w5dxp.com

Thanks for creating part 1 of this series, Cecil.
--
73, Roger, W7WKB

The Rest of the Story
 

On Mar 6, 6:10 am, Keith Dysart wrote:

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.

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.

The paragraph quoted above, then, begs the question: if not in the
resistor, where? The answer should be perfectly clear.

Cheers,
Tom

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

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

The Rest of the Story
 

"Keith Dysart" wrote in message
...

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

superposition.

The Rest of the Story
 

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

The Rest of the Story
 

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)

The Rest of the Story
 

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

The Rest of the Story
 

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

Roger, my premise has nothing to do with the instantaneous
case. I have made no assertions about instantaneous values.
My formula applies *only to average power*. Using that
formula on instantaneous values is an invalid thing to do,
a misuse of the tool.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Dave wrote:
"Keith Dysart" wrote:
What is the mechanism that creates the effect we call interference?

superposition.

Not disagreeing - just expanding:
Superposition is certainly necessary but superposition
alone is not sufficient. Superposition can occur with
or without interference.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
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.

Keith, please stop using innuendo to try to discredit
me. My hypothesis does NOT apply to instantaneous values,
never has applied to instantaneous values, and never will
apply to instantaneous values. Please cease and desist
with your unethical innuendos.

If you have to stoop to lying about what I have said,
you will only discredit yourself.

My hypothesis is correct for average values of powers
and *applies only to average values of powers* just as
the irradiance equation from optical physics applies
only to average power densities. To the best of my
knowledge, there is no such thing as instantaneous
irradiance.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
To the best of my
knowledge, there is no such thing as instantaneous
irradiance.

The definition of irradiance is the "average energy
per unit area per unit time". Any deviation away
from "average energy per unit area per unit time"
when discussing what I have said is a straw man
diversion, not in the spirit of a good will
discussion.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:


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.


Cecil,

You may or may not already know this, but a lot of detailed optical
analysis these days is done with full 3-D electromagnetic simulation,
starting from Maxwell equations and boundary conditions. Interference,
coherence, energy flow, and all of the other stuff you like to discuss
can be *output* from that analysis, but those items are not part of the
input. The "centuries old" optics simply does not get the job done. The
"centuries old" stuff may work in the (impossible) cases where
everything is completely lossless and ideal, but it doesn't give the
right answers in the real world.

73,
Gene
W4SZ

The Rest of the Story
 

Cecil Moore wrote:


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*.

"Contemporary" is an interesting word choice. Partial coherence has been
recognized for a long time. There is a very widely referenced paper by
H. H. Hopkins on partial coherence in optical imaging systems that was
published in 1950. He did not invent the concept, but he did popularize
the standard formulation still used today. The math is a bit messy, with
4 dimensional integrals and other complications, but numerical solutions
are widely done.

I would be quite surprised if the radar and other RF experts don't use
the same type of analysis.

Oh, by the way, even a completely monochromatic wave can be partially
coherent if the source is extended. Indeed, that is one of the more
common configurations.

73,
Gene
W4SZ

The Rest of the Story
 

Cecil Moore wrote:
Cecil Moore wrote:
To the best of my
knowledge, there is no such thing as instantaneous
irradiance.

The definition of irradiance is the "average energy
per unit area per unit time". Any deviation away
from "average energy per unit area per unit time"
when discussing what I have said is a straw man
diversion, not in the spirit of a good will
discussion.

The definition of irradiance, according to NIST, is power per unit area.
The standard units are W/m2 or lumen/m2.

You can add average, peak, instantaneous, or whatever you like to
further define your quantity of interest. Such additions, however, are
not part of the standard definition of irradiance.

73,
Gene
W4SZ

The Rest of the Story
 

On Fri, 07 Mar 2008 15:15:08 GMT
Cecil Moore wrote:

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

Roger, my premise has nothing to do with the instantaneous
case. I have made no assertions about instantaneous values.
My formula applies *only to average power*. Using that
formula on instantaneous values is an invalid thing to do,
a misuse of the tool.
--
73, Cecil http://www.w5dxp.com

I find myself surprised at your insistance that the instantaneous case must be seperated from the average case in our example of waves on a transmission line. Our ability to measure within the confines of the sine wave is much greater than what is possible at optic frequencies so we are led to expect much more than averages.

It is informative to look at each problem from many angles. Keith's method is one way. Another way is to observe that when we begin considering what is happening with the source resistor, we are really bringing the source resistor into the circuit, i.e., bringing the source resistor outside of the 'black box'. Once we do that, we can see that the source prereflection load is not the same as the post reflection load. From this perspective, the equation "PRs = 50w + Pref" becomes a target to which we adjust the source voltage to acheive. We would accomplish that by using your equation but Keith's method.

It seems to me like Keith is using interference, both constructive and destructive, to calculate what the source load would be on the instantaneous basis. To me it seems like a validation of your premise.

Did I misunderstand your premise, and you were really trying to say that the inclusion of a 50 ohm source resistor would prevent the source from ever 'seeing' anything but a 100 ohm load? I don't think that was your intent.
--
73, Roger, W7WKB

The Rest of the Story
 

Gene Fuller wrote:
You may or may not already know this, but a lot of detailed optical
analysis these days is done with full 3-D electromagnetic simulation,
starting from Maxwell equations and boundary conditions. Interference,
coherence, energy flow, and all of the other stuff you like to discuss
can be *output* from that analysis, but those items are not part of the
input. The "centuries old" optics simply does not get the job done. The
"centuries old" stuff may work in the (impossible) cases where
everything is completely lossless and ideal, but it doesn't give the
right answers in the real world.

Ideal examples are time-honored ways of discussing concepts
and getting away from the vagaries of the real world. If one
understands the ideal examples, one is in a position to then
proceed to understanding the real world. If one fails to
understand the conceptual principles underlying the ideal
examples, one cannot possibly understand the real world.

Your posting seems to reflect your usual sour grapes attitude.
I will expect you to object to every example that uses lossless
transmission lines from now on including ones by Ramo & Whinnery,
Walter Johnson, Walter Maxwell, J. C. Slater and Robert Chipman.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Gene Fuller wrote:
The definition of irradiance, according to NIST, is power per unit area.
The standard units are W/m2 or lumen/m2.

Exactly how much power can exist in a zero unit of
time?

You previously objected to things that don't match
the real world. Instantaneous irradiance would rely
on an infinitesimally small amount of time, something
that doesn't match reality very well. One would think
you would therefore object to the concept of instantaneous
irradiance since it cannot be measured in reality and
exists only in the math model in the human mind.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Hi Keith,

Thanks for the additional explaination. I am wondering if I misunderstood Cecil's original premise.

On Fri, 7 Mar 2008 06:26:46 -0800 (PST)
Keith Dysart wrote:

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)


--
73, Roger, W7WKB

The Rest of the Story
 

Cecil Moore wrote:
"The definition of irradiance is "the average energy per unit time. Any
deviation away---."

As energy per unit time is power, Cecil`s definition agrees with what my
dictionary says:
"Irradiance-The incident radiated power per unit area of a surface; the
radiometric counterpart of illumination, usually expressed in watts/cm,
squared."

This is different from Poynting which uses instantaneous values.

Best regards, Richard Harrison. KB5WZI

The Rest of the Story
 

Roger Sparks wrote:
On Fri, 07 Mar 2008 15:15:08 GMT
Cecil Moore wrote:
My formula applies *only to average power*.

I find myself surprised at your insistance that the instantaneous
case must be seperated from the average case in our example of waves
on a transmission line.

You have misunderstood what I am objecting to - which is:
Please don't say or imply that I have said or implied
anything about instantaneous values. I have NOT done so!
The formulas and concepts being used in the discussion
of instantaneous values are NOT mine! My objection is
to the false statements being made about what I have
posted.

Please go ahead and have your instantaneous discussion with
Keith but please don't say or imply that anything associated
with that discussion is about anything I have posted. Saying
or implying such a thing is simply false.

It is informative to look at each problem from many angles.

I agree - just stop saying that it is based on my assertions.
You or Keith may be right or wrong but either way, it is not
associated with anything I have posted. Please leave my name
out of any discussion concerning instantaneous values.

Did I misunderstand your premise, and you were really trying to say that
the inclusion of a 50 ohm source resistor would prevent the source from
ever 'seeing' anything but a 100 ohm load? I don't think that was your
intent.

Please add a dimension to your thinking. No matter what the
value of the load resistor, the source delivers 100 watts.
There are an infinite number of loads that the source could
"see" besides 100+j0 ohms that will make that condition true.

When the load is 0 ohms, the source "sees" 50+j50 ohms.
When the load is 12.5 ohms, the source "sees" 73.5+j44.1 ohms.
When the load is 25 ohms, the source "sees" 90+j30 ohms.
Only when the load is 50 ohms, does the source "see" 100+j0 ohms.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Roger Sparks wrote:
Thanks for the additional explaination. I am wondering if I misunderstood Cecil's original premise.

Roger, if you thought it involved any instantaneous
values then, yes, you misunderstood my premise.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Richard Harrison wrote:
Cecil Moore wrote:
"The definition of irradiance is "the average energy per unit time. Any
deviation away---."

Richard, your newsreader dropped part of what I
wrote which was:

The definition of irradiance is the "average energy
per unit area per unit time".
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 7, 8:17 am, Gene Fuller wrote:
Cecil Moore wrote:

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.

Cecil,

You may or may not already know this, but a lot of detailed optical
analysis these days is done with full 3-D electromagnetic simulation,
starting from Maxwell equations and boundary conditions. Interference,
coherence, energy flow, and all of the other stuff you like to discuss
can be *output* from that analysis, but those items are not part of the
input. The "centuries old" optics simply does not get the job done. The
"centuries old" stuff may work in the (impossible) cases where
everything is completely lossless and ideal, but it doesn't give the
right answers in the real world.

73,
Gene
W4SZ

You can sure say that again...in fact, Maxwell doesn't really do it
either when you get to quantum mechanical effects. But that's a story
for another day.

Certainly, those who design and build FTIR spectrometers know
perfectly well that interference does not depend on a narrow-band
coherent source. Blackbody radiation works just fine, thank you. But
it doesn't take much beyond belief in linear systems to understand
that. I recall explaining to a company VP how it worked in terms of a
linear system, and it was very gratifying to see the virtual light
bulb lighting up in his head...he really got it.

Cheers,
Tom

The Rest of the Story
 

K7ITM wrote:
Certainly, those who design and build FTIR spectrometers know
perfectly well that interference does not depend on a narrow-band
coherent source.

How narrow-band? How coherent? In the irradiance (power
density) equation, Ptot = P1 + P2 + 2*sqrt(P1*P2)cos(A),
if the angle 'A' is varying rapidly, what value do you
use for cos(A)?

A constant average sustained level of destructive
interference cannot be maintained between two waves
unless they are coherent. If they are not coherent
the interference will average out to zero.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

K7ITM wrote:
... interference does not depend on a narrow-band coherent source.

OK Tom, here's a challenge for you. Given a system with
a constant steady-state destructive interference magnitude.
Exactly how can that constant steady-state destructive
interference magnitude be maintained if the two interfering
signals are not coherent?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

K7ITM wrote:
And exactly which part of "linear system" do you fail to understand?

I understand the meaning of your question now and
here is one for you:

Exactly which part of a constant, average, steady-state
condition of destructive interference do you fail to
understand?

Given two coherent signals interfering whose results are
10 watts of constant, average, steady-state destructive
interference, how do you propose to accomplish that
outcome when the signals are not coherent?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 7, 1:16*pm, Cecil Moore wrote:
Roger Sparks wrote:
Thanks for the additional explaination. I am wondering if I misunderstood Cecil's original premise.

Roger, if you thought it involved any instantaneous
values then, yes, you misunderstood my premise.

My understanding of your claim was that for the special case of
a 45 degree line supplied from a matched source, the energy
in the reflected wave is dissipated in the source resistor.

This sentence fragment from your document suggests this:
"reflected energy from the load is flowing through the source
resistor, RS, and is being dissipated there".

As "proof" of this, you computed average powers and showed that
the dissipation in the source resistor increased by the same
amount as the computed average power in the reflected wave.

But when an attempt it made to validate your claim using
instantaneous energy flows, the claim is proved false because
the dissipation in the source resistor does not occur at the
correct time to be absorbing the energy from the reflected
wave.

To prove your claim, I can see two paths:
- find some element in the circuit that stores the energy from
the reflected wave and releases it into the source resistor
at the correct time
- allow the violation of the principle of conservation of
energy

On the other hand, if you want to modify your claim to simply
be that the numerical value of the dissipation in the source
resistor has the same value as the pre-reflection dissipation
plus the numerical value of the energy in the reflected wave,
then the discrepancy is resolved. But then you can not claim
that the energy in the reflected wave is dissipated in the
source resistor.

...Keith