The Rest of the Story
 

Keith Dysart wrote:
There was an 'if' there, wasn't there? Do you think
the 'if' is satisfied? Or not? The rest is useless
without knowing.

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

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

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

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

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

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

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

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

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

Nope, you are confused. I am saying absolutely nothing about
transients. Why do you think an instantaneous power analysis
during steady-state is not possible?

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

Are you saying that an analysis of instantaneous power
does not apply during steady-state? If that is true,
then all of your earlier analysis involving transmission
lines is bogus.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Sat, 5 Apr 2008 03:06:18 -0700 (PDT)
Keith Dysart wrote:

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

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

This is how I did it.

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

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

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

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

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

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

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

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

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

This returning power is all from the reflected wave. *

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

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

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

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

Pg(t) is the result of a standing wave, containing power from Pf(x) and Pr(x+90). Only the power from Pr(x+90) is available to at a later time Prs(t+90+delta). Power from Pf(x+delta) is found in the transmission line.


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

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

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

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

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

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

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

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

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

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

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

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


clip


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

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

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

...Keith

I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. Maybe if we pursue the trig identity
sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers?
--
73, Roger, W7WKB

The Rest of the Story
 

correction
On Sat, 5 Apr 2008 07:06:22 -0700
Roger Sparks wrote:

On Sat, 5 Apr 2008 03:06:18 -0700 (PDT)
Keith Dysart wrote:

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

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

This is how I did it.

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

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

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

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

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

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

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

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

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

This returning power is all from the reflected wave. *

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

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

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

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

Pg(t) is the result of a standing wave, containing power from Pf(x) and Pr(x+90). Only the power from Pr(x+90) is available to Rs at a later time Prs(t+90+delta). Power from Pf(x+delta) is found in the transmission line.
I omitted the Rs in "available to Rs at a later time Prs(t+90+delta)" Sorry!


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

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

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

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

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

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

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

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

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

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

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

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


clip


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

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

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

...Keith

I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. Maybe if we pursue the trig identity
sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers?
--
73, Roger, W7WKB


--
73, Roger, W7WKB

The Rest of the Story
 

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

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

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

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

Try the above example and maybe it will become clear.

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

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

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

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

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

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

Please note that all of these power components exist
when the two transmission lines are removed so this
analysis is probably the key to understanding what
is wrong with the earlier analysis.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Sat, 05 Apr 2008 10:01:13 -0500
Cecil Moore wrote:

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

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

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


I think this would be right on the average basis. Because any sine waves of identical frequency can ultimately be added to make one wave, we can describe a single sine wave on the source side and another wave on the load/reflection side. There is no need for constructive or destructive interference as part of the final sine wave description.

On the other hand, if we want to understand how the final wave is assembled for each side of the resistor, we need the idea of constructive and destructive interference. Your pictorial showed only the single reflection, but as you have explained previously, there are many more reflections between the resistor and +j50 points until the power on the +j50 side finally stabalizes to some power level.

It seems to me like my spreadsheet found at

http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf

captures a description of the forward and reflected waves as a single equation for each side of the resistor. The forward wave is y*sin(t) and reflected wave is y*sin(t+90).

If we want to learn how to find out how the source load begins at 100 ohms resistive and changes to 70.7 ohms reactive, we can either notice how the peak current has shifted from resistive to reactive (45 degree) from the combined waves described on my spreadsheet, or we can add all the reflections on each side of the resistor and come to the combined wave after many additions of ever smaller reflections.

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

Try the above example and maybe it will become clear.

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

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

Interference1 and interference2 would be the combined effects of successive ever smaller reflections. Frankly, I don't see that combining the smaller reflections in this way to be of much value. It is like saying that the first reflection is special and the subsequent reflections equally special so we will give subsequent refections a group name, "interference". To me, all the reflections are "interference" and must be either added or subtracted in the same manner.

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

We have to have a starting point, which must be the source power. How could we avoid having a starting point?

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

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

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

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

I think this equation is missing some terms, reflections 2,3,4...n.

--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
We have to have a starting point, which must be the source power. How could we avoid having a starting point?

We are analyzing a resistor isolated from the source and load
by 1WL of ideal feedline on each side. All we need are the
steady-state values of Pfor1, Pref1, Pfor2, and Pref2. It's
much like an s-parameter analysis of a 4-terminal box with
a 50 ohm series resistor. If we know a1 (Vfor1) and a2 (Vref2),
that is all that needs to be known to complete the analysis.
We don't actually need to know anything about the source and
the load if a1 and a2 are given.

I think this equation is missing some terms, reflections 2,3,4...n.

The equation already contains all of those "missing'
terms because it is steady-state and all of those
terms have already been added in to the total.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Sat, 05 Apr 2008 14:56:35 -0500
Cecil Moore wrote:

Roger Sparks wrote:
We have to have a starting point, which must be the source power. How could we avoid having a starting point?

We are analyzing a resistor isolated from the source and load
by 1WL of ideal feedline on each side. All we need are the
steady-state values of Pfor1, Pref1, Pfor2, and Pref2. It's
much like an s-parameter analysis of a 4-terminal box with
a 50 ohm series resistor. If we know a1 (Vfor1) and a2 (Vref2),
that is all that needs to be known to complete the analysis.
We don't actually need to know anything about the source and
the load if a1 and a2 are given.

I think this equation is missing some terms, reflections 2,3,4...n.

The equation already contains all of those "missing'
terms because it is steady-state and all of those
terms have already been added in to the total.
--
73, Cecil http://www.w5dxp.com

From your previous posting, you gave the equation

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

Where Pfor1 and Pref2 were toward the resistor, and Pfor2 and Pref1 were away from the resistor. This equation seems to assume that the power to the resistor is zero. This is because the same current must flow in Pfor1 and Pfor2, and the same current in Pref2 and Pref1. The voltage drop through the resistor only happens when the circuit is complete on both sides of the resistor.

If we try to separate them as you are suggesting, we would have to recognize that it is also correct to say that

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

This is true because power only flows when the circuit is complete, and we can not say all the power comes from the left, or from the right. The power only flows when both are connected so the power must come from both sides.

There will never be one answer for your equation, unless, as you suggest, a1 and a2 are given.

--
73, Roger, W7WKB

The Rest of the Story
 

On Sat, 5 Apr 2008 15:28:59 -0700, Roger Sparks
wrote:

BIG SNIPPAGE

Why don't you two take this to private email?

The Rest of the Story
 

Roger Sparks wrote:
Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 (all average)

Sorry for the obvious typo. The sign of P.Rs should
have been minus.

This equation seems to assume that the power to the resistor is zero.

No, these are really basic concepts. The energy flow away
from a junction is equal to the energy flow into the
junction minus the dissipation in the junction.

(Pfor1 + Pref2) is the energy flow into the junction

P.Rs is the dissipation in Rs

(Pfor2 + Pref1) is the energy flow away from the junction.

(Pfor1 + Pref2) - P.Rs = (Pfor2 + Pref1) or

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

There will never be one answer for your equation, unless, as you suggest, a1 and a2 are given.

a1 and a2 *are* given - at least indirectly so they can be
calculated. Since we know Vs, Rs, and ZL, and given the two
1WL feedlines, we can easily calculate a1 and a2.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Lets put the 'Rest of the Story' to bed before it becomes 'The Never Ending
Story' or are we too late already.
Mike. VK6MO
"Cecil Moore" wrote in message
. ..
Roger Sparks wrote:
Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 (all average)

Sorry for the obvious typo. The sign of P.Rs should
have been minus.

This equation seems to assume that the power to the resistor is zero.

No, these are really basic concepts. The energy flow away
from a junction is equal to the energy flow into the
junction minus the dissipation in the junction.

(Pfor1 + Pref2) is the energy flow into the junction

P.Rs is the dissipation in Rs

(Pfor2 + Pref1) is the energy flow away from the junction.

(Pfor1 + Pref2) - P.Rs = (Pfor2 + Pref1) or

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

There will never be one answer for your equation, unless, as you suggest,
a1 and a2 are given.

a1 and a2 *are* given - at least indirectly so they can be
calculated. Since we know Vs, Rs, and ZL, and given the two
1WL feedlines, we can easily calculate a1 and a2.
--
73, Cecil http://www.w5dxp.com