On Sat, 05 Jun 2004 23:39:35 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
Then how can you explain what happens when you reverse the zip cord, plugging
one end in one way and the other with the prongs reversed ? Are you saying the
currents in this condition are seeing an open circuit?

I'm sorry, Walt, you lost me. What zip cord?

Cecil, you must not be reading my posts. Go back one or two to get the drift.

Walt

On Sat, 05 Jun 2004 23:36:07 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
I'm trying to locate my Hecht paper, but can't at the moment. Let's make sure
we're talking about the same set of of equations. The ones I'm saying are
invalid as stated in his article appear on Page 46, Col 2.

Walt, I'm sorry, all I have is the QEX CD and the pages are not the same as
they were in QEX magazine. Dr. Best's equations 13 and 15 are the classical
physics interference equations virtually identical to the irradiance equations
in _Optics_, by Hecht.

But Cecil, you CAN'T add V1 and V2 in any manner to obtain forward power,
because adding V1 and V2 does not yield forward voltage.

Yes, it does, Walt. Consider the following matched system similar to the
example in Dr. Best's article.

100w XMTR-----50 ohm line---x---1/2WL 150 ohm line---50 ohm load

P1 = 100W(1-rho^2) = 100(1-.25) = 75W

P2 = 33.33W(rho^2) = 33.33W*.25 = 8.33W

P1 = 75W, P2 = 8.33W, PFtotal = 133.33W

V1 = 106.07V, V2 = 35.35V, VFtotal = 141.4V

V1 + V2 = VFtotal, 106.07V + 35.35V = 141.4V

How many times do I have to explain that V1 + V2 does NOT equal VF total?

But it does, Walt. See above.


Well, Cecil, so far I haven't been able to follow the logic in the lines above.
Perhaps they are correct for that particular example, but I'm not so sure. For
example, if 100 w is available only 75 w will enter the 150-ohm line initially,
so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is
reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making
the re-reflected power 4 .69 w . I haven't had time to work through the
remaining steps to the steady state. So before I do I notice that you state that
Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by
using the example Steve used earlier in his Part 3, the one he took from
Reflections and then called it a 'fallacy'.

This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w
available from the source at 70.71 v. Matched at the line input, in the steady
state the total forward power Pfwd = 133.33 w and power reflected is Pref =
33.333 w. With these steady state powers the forward voltage is 81.65 v and
reflected voltage is 40.82 v. Due to the integration of the reflected waves the
source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady
state V1 = 81.65 v and V2 = 40.82 v.

Now using V1 and V2 in Eqs 7 and 8 we get P1 = 133.33 w and P2 = 33.333 w. So
far so good, these Eqs are correct and valid.

But now let's plug V1 and V2 into Eq 9:

(V1 + V2) = 122.47 v. Then 122.47^2/50 = 300 w. Something's amiss here. So let's
go a step further and plug P1 and P2 into Eq 13:

P1 = 133.33 w and P2 = 33.333 w.
Then using Eq 13, PF total = (sqrt 133.33 + sqrt 33.333)^2 = 300 w.
We get the same incorrect value as with Eq 9. Now why can this be?

For starters, P1 derived from Eq 7 is already the total forward power. So why is
the forward power value P1 plugged into Eq 13 do derive the total foward power
PFtotal?

Do you see where this is going, Cecil?

So now let's go one step backward to use Eq 11:

Plugging V1 and V2 into Eq 11 we get VFtotal^2 = 14 ,999.6983 (call it 15,000).
Therefore, VFtotal = sqrt VFtotal^2 = 122.4733 v.

This is the same incorrect value obtained using Eq 9.

So, Cecil, do you still believe these equations are valid for every matched
situation?

Walt

On Sun, 06 Jun 2004 15:33:31 GMT, Walter Maxwell wrote:

On Sat, 05 Jun 2004 23:36:07 -0500, Cecil Moore wrote:

Walter Maxwell wrote:

Well, Cecil, so far I haven't been able to follow the logic in the lines above.
Perhaps they are correct for that particular example, but I'm not so sure. For
example, if 100 w is available only 75 w will enter the 150-ohm line initially,
so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is
reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making
the re-reflected power 4 .69 w . I haven't had time to work through the
remaining steps to the steady state. So before I do I notice that you state that
Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by
using the example Steve used earlier in his Part 3, the one he took from
Reflections and then called it a 'fallacy'.

This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w
available from the source at 70.71 v. Matched at the line input, in the steady
state the total forward power Pfwd = 133.33 w and power reflected is Pref =
33.333 w. With these steady state powers the forward voltage is 81.65 v and
reflected voltage is 40.82 v. Due to the integration of the reflected waves the
source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady
state V1 = 81.65 v and V2 = 40.82 v.

Now using V1 and V2 in Eqs 7 and 8 we get P1 = 133.33 w and P2 = 33.333 w. So
far so good, these Eqs are correct and valid.

But now let's plug V1 and V2 into Eq 9:

(V1 + V2) = 122.47 v. Then 122.47^2/50 = 300 w. Something's amiss here. So let's
go a step further and plug P1 and P2 into Eq 13:

P1 = 133.33 w and P2 = 33.333 w.
Then using Eq 13, PF total = (sqrt 133.33 + sqrt 33.333)^2 = 300 w.
We get the same incorrect value as with Eq 9. Now why can this be?

For starters, P1 derived from Eq 7 is already the total forward power. So why is
the forward power value P1 plugged into Eq 13 do derive the total foward power
PFtotal?

Do you see where this is going, Cecil?

So now let's go one step backward to use Eq 11:

Plugging V1 and V2 into Eq 11 we get VFtotal^2 = 14 ,999.6983 (call it 15,000).
Therefore, VFtotal = sqrt VFtotal^2 = 122.4733 v.

This is the same incorrect value obtained using Eq 9.

So, Cecil, do you still believe these equations are valid for every matched
situation?

Walt

Cecil, one point I forgot in the msg above--the standing wave.

Remember I said earlier that V1 + V2 yields the standing wave, not forward
voltage Vfwd? We know that V1 + V2 = 122.47 v. This is the max value of the
standing wave.

Now V1 - V2 = 40.82 v. Notice that (V1+ V2)/(V1 - V2) = 122.47/40.82 = 3. 0.
Doesn't this look something like the SWR?

Walt

Walter Maxwell wrote:
Cecil Moore wrote:

Walter Maxwell wrote:
Then how can you explain what happens when you reverse the zip cord, plugging
one end in one way and the other with the prongs reversed ? Are you saying the
currents in this condition are seeing an open circuit?

I'm sorry, Walt, you lost me. What zip cord?

Cecil, you must not be reading my posts. Go back one or two to get the drift.

Are you talking about the two generators hooked up back to back.
If so, where are the two EM waves traveling in the same direction
which is a prerequisite for complete destructive/constructive
interference? The generated EM waves, in your example, are moving
in opposite directions which is simply not relevant to this discussion.

Please conjure up an experiment where two coherent waves are
moving in the same direction without affecting the generators.
--
73, Cecil http://www.qsl.net/w5dxp



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"Walter Maxwell" wrote in message
...
snip

Now V1 - V2 = 40.82 v. Notice that (V1+ V2)/(V1 - V2) = 122.47/40.82 = 3.
0.
Doesn't this look something like the SWR?

Walt


Walt
You really should write a book.
(rim shot) =]8*O

73
H.

Walter Maxwell wrote:
Well, Cecil, so far I haven't been able to follow the logic in the lines above.
Perhaps they are correct for that particular example, but I'm not so sure.

Please pick any example of your choice of matched systems. Dr. Best's equation
13 will be valid for any matched system. VFtotal = V1 + V2 will be valid for
any system, matched or not.

For
example, if 100 w is available only 75 w will enter the 150-ohm line initially,
so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is
reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making
the re-reflected power 4 .69 w . I haven't had time to work through the
remaining steps to the steady state.

When you do, you will get the same values as I.

So before I do I notice that you state that
Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by
using the example Steve used earlier in his Part 3, the one he took from
Reflections and then called it a 'fallacy'.

As I have said before, Steve's equations are correct but he simply drew the
wrong conclusions from them. His "fallacy" is a fallacy but has nothing to
do with his equations. He simply drew the wrong conclusions from valid
equations.

This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w
available from the source at 70.71 v. Matched at the line input, in the steady
state the total forward power Pfwd = 133.33 w and power reflected is Pref =
33.333 w. With these steady state powers the forward voltage is 81.65 v and
reflected voltage is 40.82 v. Due to the integration of the reflected waves the
source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady
state V1 = 81.65 v and V2 = 40.82 v.

Nope, you simply misunderstood what Steve said. V1 and V2 are *NOT* on the 50 ohm
line. In fact, the 1 WL 50 ohm line is irrelevant and just serves to confuse. V1
and V2 are voltages existing *at the match point* at the INPUT of the tuner. Steve
chose an example that is virtually impossible to explain or understand.

Let's take it step by step starting with Steve's example:

100W XMTR---50 ohm line---tuner---1WL 50 ohm line---150 ohm load

The 1WL 50 ohm lossless line is irrelevant except for power measurements so
eliminate it.

100W XMTR---50 ohm line---tuner---150 ohm load

The tuner can now be replaced by 1/4WL of 86.6 ohm feedline.

100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---150 ohm load
100W-- 107.76W--
--0W --7.76W
V1--
V2--

Now we have an example that is understandable and we haven't changed
any of the conditions.

The magnitude of the reflection coefficient is 0.268. That makes the magnitude
of the transmission coefficient equal to 1.268 (Rule of thumb for matched
systems with single step-function impedance discontinuities)

So V1 = 70.7 * 1.268 = 89.6V V2 = VF2 * 0.268 = 6.95V

VFtotal = V1 + V2 = 89.6V + 6.95V = 96.6V

PFtotal = 96.6V^2/86.6 = 107.76W

So, Cecil, do you still believe these equations are valid for every matched
situation?

Yes, they are, Walt, once one understands them. I don't blame you for being
confused about Steve's example. He chose the worst example possible and
didn't explain it very well at all.

I still maintain that you two are two inches apart and neither one of you
will budge an inch.
--
73, Cecil http://www.qsl.net/w5dxp






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Walter Maxwell wrote:
Remember I said earlier that V1 + V2 yields the standing wave, not forward
voltage Vfwd?

Dr. Best's V1 + V2 are moving in the same direction toward the load
and do NOT yield the standing wave. You have misunderstood what he
was trying to say (which is super easy to do).

Dr. Best's V1 is proportional to the S-parameter term, s21(a1)
The only difference is that all S-parameter voltages are normalized
to the square root of Z0.

Dr. Best's V2 is proportional to the S-parameter term, s22(a2)

Consider this easy-to-understand generalized example.

XMTR---Z01---x---1/2WL Z02---load=Z01
VF1-- VF2--
--VR1 --VR2

VF1 is the forward voltage in the Z01 section. VF2 is the forward
voltage in the Z02 section. VR1 is the reflected voltage in the
Z01 section. VR2 is the reflected voltage in the Z02 section.

RHO is the physical reflection coefficient and TAU is the physical
transmission coefficient which is equal to (1+RHO) at an impedance
discontinuity.

Dr. Best's V1 = VF1(TAU) His V2 = VR2(RHO)

Dr. Best's V1 is the part of VF1 that makes it through the impedance
discontinuity (coming from the source side). It is traveling toward
the load.

Dr. Best's V2 is the reflected voltage re-reflected from the match
point. It is also traveling toward the load.

V1 and V2 superpose into VF2 which Dr. Best calls VFtotal. And that
is indeed what happens. That Dr. Best was somewhat incapable of
explaining the equation is his fault, not yours. His explanations
were so obtuse, I'm not sure he understood them himself. He certainly
didn't recognize the classical interference equations as such.
--
73, Cecil http://www.qsl.net/w5dxp





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Walter Maxwell wrote:
Now V1 - V2 = 40.82 v. Notice that (V1+ V2)/(V1 - V2) = 122.47/40.82 = 3. 0.
Doesn't this look something like the SWR?

Well, let's see.

V1 = VF1(TAU)

V2 = VR2(RHO)

[VF1(TAU) + VR2(RHO)]/[VF1(TAU) - VR2(RHO)]

Does that look something like SWR?
--
73, Cecil http://www.qsl.net/w5dxp



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On Sun, 06 Jun 2004 12:44:38 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
Well, Cecil, so far I haven't been able to follow the logic in the lines above.
Perhaps they are correct for that particular example, but I'm not so sure.

Please pick any example of your choice of matched systems. Dr. Best's equation
13 will be valid for any matched system. VFtotal = V1 + V2 will be valid for
any system, matched or not.

For
example, if 100 w is available only 75 w will enter the 150-ohm line initially,
so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is
reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making
the re-reflected power 4 .69 w . I haven't had time to work through the
remaining steps to the steady state.

When you do, you will get the same values as I.

So before I do I notice that you state that
Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by
using the example Steve used earlier in his Part 3, the one he took from
Reflections and then called it a 'fallacy'.

As I have said before, Steve's equations are correct but he simply drew the
wrong conclusions from them. His "fallacy" is a fallacy but has nothing to
do with his equations. He simply drew the wrong conclusions from valid
equations.

This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w
available from the source at 70.71 v. Matched at the line input, in the steady
state the total forward power Pfwd = 133.33 w and power reflected is Pref =
33.333 w. With these steady state powers the forward voltage is 81.65 v and
reflected voltage is 40.82 v. Due to the integration of the reflected waves the
source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady
state V1 = 81.65 v and V2 = 40.82 v.

Nope, you simply misunderstood what Steve said. V1 and V2 are *NOT* on the 50 ohm
line. In fact, the 1 WL 50 ohm line is irrelevant and just serves to confuse. V1
and V2 are voltages existing *at the match point* at the INPUT of the tuner. Steve
chose an example that is virtually impossible to explain or understand.

Cecil, Steve's equations 4 thru 8 are correct, valid, and completely GENERAL. In
the text preceding the Eqs he even specifies Zo as 50 ohm, not that it matters.
He correctly defines V1 in Eq 7 and he correctly defines V2 in Eq 8. However, he
makes a vital error in the paragraph preceding these two equations.

He says incorrectly, "When two forward traveling waves add, general
superposition theory and Kirchoff's voltage law require that the total
forward-traveling voltage be the vector sum of the individual forward-traveling
voltages such that VFtotal = V1 + V2."

This statement is TOTALLY FALSE, and this error is the basis for the remainder
of his equations to be invalid. Kirchoff's voltage law does not apply in this
case of transmission line practice. There are times when circuit theory doesn't
hold and transmission line theory is required.

You are still ignoring his Eq 6 in Part 1, which is false and invalid because he
mistook the expression for determining the standing wave for the total forward
wave. Look back at my previous msg where I've shown that the addition of the
forward and reflected waves yields the standing wave, not the forward wave. The
standing wave is NOT the forward wave.

Refer to the text on your CD at the paragraph that begins: "A transmission line
system analysis must be performed with voltage and current, from which the power
is then derived." Read on from this point to where it reads, "If total
re-reflection of power occurs at the T-network, the re-reflected voltage must
have the same magnitude as the reflected voltage."

Please note the values of voltages and currents that appear in that entire
section, because this section is a direct copy of my example he took from
Reflections in an attempt to disprove it. Which he does in the next sentence:

"Therefore, based on the assumption that total power re-reflection and in-phase
forward-wave addition, the total forward-traveling wave of 81.65 v must be the
result of a voltage having a magnitude of 70.711 v adding in phase with a
voltage having a magnitude of 40.825 v. Two in-phase complex voltages having
magnitudes of 70.711 v and 40.825 v cannot add together such that the resulting
voltage has a magnitude 81.65 v."

OF COURSE IT CAN'T, because these two voltages CANNOT BE ADDED TOGETHER TO
DERIVE THE FORWARD VOLTAGE.. This is further evidence that he didn't realize
that Eq 6 in Part 1 was wrong, because it derives the standing wave voltage,
NOT THE FORWARD VOLTAGE.

He then goes on in an unsuccessful attempt to prove my method wrong, in which he
gets himself into trouble with those equations that don't work in general. If
you still don't see the problem, Cecil, please go back and review my analysis in
the earlier post and explain to me why you disagree with my results that show
conclusively that the equations don't work.

To conclude, I have shown you why I have not used his values of V1 and V2
incorrectly, as you say. If you can show that I'm wrong I'll take the time to
study the step-by-step in your example below.

Walt


Let's take it step by step starting with Steve's example:

100W XMTR---50 ohm line---tuner---1WL 50 ohm line---150 ohm load

The 1WL 50 ohm lossless line is irrelevant except for power measurements so
eliminate it.

100W XMTR---50 ohm line---tuner---150 ohm load

The tuner can now be replaced by 1/4WL of 86.6 ohm feedline.

100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---150 ohm load
100W-- 107.76W--
--0W --7.76W
V1--
V2--

Now we have an example that is understandable and we haven't changed
any of the conditions.

The magnitude of the reflection coefficient is 0.268. That makes the magnitude
of the transmission coefficient equal to 1.268 (Rule of thumb for matched
systems with single step-function impedance discontinuities)

So V1 = 70.7 * 1.268 = 89.6V V2 = VF2 * 0.268 = 6.95V

VFtotal = V1 + V2 = 89.6V + 6.95V = 96.6V

PFtotal = 96.6V^2/86.6 = 107.76W

So, Cecil, do you still believe these equations are valid for every matched
situation?

Yes, they are, Walt, once one understands them. I don't blame you for being
confused about Steve's example. He chose the worst example possible and
didn't explain it very well at all.

I still maintain that you two are two inches apart and neither one of you
will budge an inch.

Cecil, I've shown where we're apart. It's a lot more than 2 inches.

Walter Maxwell wrote:
To conclude, I have shown you why I have not used his values of V1 and V2
incorrectly, as you say. If you can show that I'm wrong I'll take the time to
study the step-by-step in your example below.

Steve is essentially doing an S-parameter analysis without the (square
root of Z0) normalization. Since we know that an S-parameter analysis of a
match point is indeed valid, a lot of Steve's equations are valid by
association.

Assuming the S-parameter equation is valid, we can use it to prove that

VFtotal = V1 + V2

V1/SQRT(Z0) = s21(a1) in the S-parameter analysis.

V2/SQRT(Z0) = s22(a2) in the S-parameter analysis.

VFtotal/SQRT(Z0) = b2 in the S-parameter analysis.

Given: b2 = s21(a1) + s22(a2) is a valid S-parameter equation.

Therefore, VFtotal/SQRT(Z0) = V1/SQRT(Z0) + V2/SQRT(Z0) is a valid equation
because there is an EXACT one-to-one correspondence to the S-parameter equation.

Therefore, if we multiply both sides by SQRT(Z0) we get VFtotal = V1 + V2

The only way for the above equation to be wrong is if the S-parameter equation
is wrong. The only difference in the S-parameter equation and Dr. Best's equation
is the normalization by [SQRT(Z0)].

Given a generalized matched system:

XMTR---Z01---x---1/4WL Z02---load
VF1-- VF2--
--VR1 --VR2

VF2 = VF1(TAU) + VR2(RHO) = V1 + V2 Dr. Best's equation

b2 = s21(a1) + s22(a2) S-parameter equation

Dr. Best is essentially quoting an S-parameter analysis
--
73, Cecil http://www.qsl.net/w5dxp





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