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

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

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

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

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

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

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

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

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

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