Gene Fuller wrote:
I explained this to Walt in some private e-mails about 18 months ago,
and I wrote similar explanations in this group. I was vilified by both
Walt and you.

I still believe that neither model is perfectly consistent. I do
believe that an S-parameter analysis is perfectly consistent.

Both the Walter Maxwell model and the Steve Best model for this
transmission line problem work correctly. They are internally self
consistent, and they give the same physical answers.

But how can Steve deny the 100% re-reflected energy premise when
all the energy winds up at the load? How did it get to the load
if it was not re-reflected at the match point? TV ghosting proves
that it is indeed re-reflected at the match point.
--
73, Cecil http://www.qsl.net/w5dxp



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On Mon, 07 Jun 2004 23:40:03 GMT, Gene Fuller wrote:

Cecil,

I am happy to see that we are in agreement on this long-standing argument.

I explained this to Walt in some private e-mails about 18 months ago,
and I wrote similar explanations in this group. I was vilified by both
Walt and you.

I will summarize once again.

Both the Walter Maxwell model and the Steve Best model for this
transmission line problem work correctly. They are internally self
consistent, and they give the same physical answers. They obey the
standard laws of physics and mathematics. However it is not possible to
mix-and-match the models, using equations and definitions from one model
in the other.

Walt continues to mis-read Steve Best's QEX articles. He needs to throw
away all of his pre-conceived definitions about the meaning of specific
Vx components and read exactly what Steve wrote.

73,
Gene
W4SZ

Hello again Gene,

What Steve wrote exactly, Gene, and I have read it exactly, over and over again,
is based entirely on an invalid premise that you and all the rest who, like you,
have expressed the notion that Steve's paper is the 'most brilliant and
definitive paper ever written on the subject. Unfortunately, it's both ironic
and tragic that a person so brilliant a mathematician, and knowledgeable in
transmission line technique, should make such a grievous error in formulating
the basis for his entire paper. It's also tragic that all of you who think his
article is so great also made the same error in not recognizing the problem in
the First Part of the article, his Eqs 6, 7, and 8, because they are invalid.

Steve used an equation in Johnson's "Transmission Lines and Networks" as the
foundation for the entire paper. But unfortunately he misunderstood the meaning
of the equation. He misunderstood it as early as 1998 when I pointed out why he
misunderstands it, but he denied that he misunderstands it , stating that it is
I who misunderstands it. Not a chance.

The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.

The difference between 'E' and Vfwd is so significantly different as to be
unbelievable that he, of all people, used it for this purpose, and it is also
unbelievable and shameful that you experienced engineers failed to recognize,
not only the error in Steve's equation, but the ramifications it had on the
domino effect on the remainder of his paper, especially making his equations 9
thru 15 in Part 3 invalid for general use.

I'd like for you and Cecil, and any other reader who believes Steve's paper is
correct, to rethink the definitions of the standing wave and the forward voltage
with the view toward another review of the paper. Then make that review and see
if you still believe the material appearing there is correct.

If you then still believe the paper is correct I'd like for you to let me know
the technical reason why you believe it's correct so that I can discuss it with
you. It's really worth that effort--don't just blow it off as another rant and
rave from Walt Maxwell.

Thanks for listening to what I think is an important issue.

Walt

Good luck. There are a few "engineers" here I would let touch my toaster.

tom
K0TAR

Walter Maxwell wrote:
On Mon, 07 Jun 2004 23:40:03 GMT, Gene Fuller wrote:


Cecil,

I am happy to see that we are in agreement on this long-standing argument.

I explained this to Walt in some private e-mails about 18 months ago,
and I wrote similar explanations in this group. I was vilified by both
Walt and you.

I will summarize once again.

Both the Walter Maxwell model and the Steve Best model for this
transmission line problem work correctly. They are internally self
consistent, and they give the same physical answers. They obey the
standard laws of physics and mathematics. However it is not possible to
mix-and-match the models, using equations and definitions from one model
in the other.

Walt continues to mis-read Steve Best's QEX articles. He needs to throw
away all of his pre-conceived definitions about the meaning of specific
Vx components and read exactly what Steve wrote.

73,
Gene
W4SZ


Hello again Gene,

What Steve wrote exactly, Gene, and I have read it exactly, over and over again,
is based entirely on an invalid premise that you and all the rest who, like you,
have expressed the notion that Steve's paper is the 'most brilliant and
definitive paper ever written on the subject. Unfortunately, it's both ironic
and tragic that a person so brilliant a mathematician, and knowledgeable in
transmission line technique, should make such a grievous error in formulating
the basis for his entire paper. It's also tragic that all of you who think his
article is so great also made the same error in not recognizing the problem in
the First Part of the article, his Eqs 6, 7, and 8, because they are invalid.

Steve used an equation in Johnson's "Transmission Lines and Networks" as the
foundation for the entire paper. But unfortunately he misunderstood the meaning
of the equation. He misunderstood it as early as 1998 when I pointed out why he
misunderstands it, but he denied that he misunderstands it , stating that it is
I who misunderstands it. Not a chance.

The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.

The difference between 'E' and Vfwd is so significantly different as to be
unbelievable that he, of all people, used it for this purpose, and it is also
unbelievable and shameful that you experienced engineers failed to recognize,
not only the error in Steve's equation, but the ramifications it had on the
domino effect on the remainder of his paper, especially making his equations 9
thru 15 in Part 3 invalid for general use.

I'd like for you and Cecil, and any other reader who believes Steve's paper is
correct, to rethink the definitions of the standing wave and the forward voltage
with the view toward another review of the paper. Then make that review and see
if you still believe the material appearing there is correct.

If you then still believe the paper is correct I'd like for you to let me know
the technical reason why you believe it's correct so that I can discuss it with
you. It's really worth that effort--don't just blow it off as another rant and
rave from Walt Maxwell.

Thanks for listening to what I think is an important issue.

Walt

Walter Maxwell wrote:
The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.

Johnson's equation contains all possible components - both forward and
reflected, so Johnson's equation predicts the total net standing-wave
voltage. (Note that Johnson's equation contains both positive and negative
exponents.)

Dr. Best's equation contains only the forward components of Johnson's
equation. Therefore, Dr. Best's equation predicts *only* the forward-
traveling voltage. (Note that Dr. Best's equation contains only negative
exponents and represents the E+ half of Johnson's equation while omitting
the E- half of Johnson's equation.)

In other words, the two equations do *not* predict the same quantity
and they are *not* supposed to be the same equations. As I said before,
Dr. Best made a lot of conceptual errors but his equations seem to be
valid. If you break Johnson's equation into two parts representing E+
and E-, the E+ part will be Dr. Best's Eq 6.

Please consider the following matched system with a 1/2 second long lossless
transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V.

100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load
VF1=70.7V-- VF2--
--VR1 --VR2

V1 is equal to 70.7(1.5) = 106.06V and is a constant value
V2 starts out at zero and builds up to 35.35V
VF2=V1+V2 assuming they are in phase

Here is (conceptually simplified) how VF2 builds up to its steady-state value.
VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t)

Time in
Seconds V1 VR2 V2 VF2
0 106.06 0 0 106.06
1 106.06 53.03 26.5 132.56
2 106.06 66.26 33.13 139.19
3 106.06 69.58 34.79 140.85
4 106.06 70.4 35.2 141.26
5 106.06 70.63 35.32 141.38
6 106.06 70.69 35.34 141.39
7 106.06 70.7 35.35 141.4 very close to steady-state

My rounding is a little off but above is a close approximation to what happens.
Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V.

Steady-state VF2 = 106.06 + 26.5 + 6.63 + 1.66 + 0.41 + ...
This is indeed of the form 106.06(1 + 1/4 + 1/16 + 1/64 + 1/256 + ...)
or VF2 = V1(1 + A + A^2 + A^3 + A^4 + ...)
--
73, Cecil http://www.qsl.net/w5dxp



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Hash: SHA1

"Tom" == Tom Ring writes:

Tom Good luck. There are a few "engineers" here I would let touch my
Tom toaster. tom K0TAR

I had a similar thought, but it involved a bathtub.

Jack.
- --
Jack Twilley
jmt at twilley dot org
http colon slash slash www dot twilley dot org slash tilde jmt slash
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On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.

Johnson's equation contains all possible components - both forward and
reflected, so Johnson's equation predicts the total net standing-wave
voltage. (Note that Johnson's equation contains both positive and negative
exponents.)

Dr. Best's equation contains only the forward components of Johnson's
equation. Therefore, Dr. Best's equation predicts *only* the forward-
traveling voltage. (Note that Dr. Best's equation contains only negative
exponents and represents the E+ half of Johnson's equation while omitting
the E- half of Johnson's equation.)

In other words, the two equations do *not* predict the same quantity
and they are *not* supposed to be the same equations. As I said before,
Dr. Best made a lot of conceptual errors but his equations seem to be
valid. If you break Johnson's equation into two parts representing E+
and E-, the E+ part will be Dr. Best's Eq 6.

Please consider the following matched system with a 1/2 second long lossless
transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V.

100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load
VF1=70.7V-- VF2--
--VR1 --VR2

V1 is equal to 70.7(1.5) = 106.06V and is a constant value
V2 starts out at zero and builds up to 35.35V
VF2=V1+V2 assuming they are in phase

Here is (conceptually simplified) how VF2 builds up to its steady-state value.
VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t)

Time in
Seconds V1 VR2 V2 VF2
0 106.06 0 0 106.06
1 106.06 53.03 26.5 132.56
2 106.06 66.26 33.13 139.19
3 106.06 69.58 34.79 140.85
4 106.06 70.4 35.2 141.26
5 106.06 70.63 35.32 141.38
6 106.06 70.69 35.34 141.39
7 106.06 70.7 35.35 141.4 very close to steady-state

My rounding is a little off but above is a close approximation to what happens.
Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V.

Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the
max voltage of the standing wave. This is the same mistake that Steve made. I
asked you to rethink what the forward voltage and you have come up wrong.

You are using circuit theory for superposition, as Steve did, when circuit
theory fails to apply in certain transmission line cases. This is one of those
cases. In circuit theory you can superpose the voltages from two sources and V1
+ V2 equals Vtotal. But the re-reflected voltage CANNOT be added to the source
voltage in the transmission line case to obtain Vfwd, because there is only ONE
source.

I'll say it again, Cecil, V1 + V2 = maxV of the standing wave--V1 + V2 does NOT
equal Vfwd because V2 is not a second source, it came fr om ONE source, the
transceiver, and therefore superposition of V1 and V2 does not apply to
establish Vfwd.

Therefore, it is true that maxV may be incident on the load if the relative
phase between the reflected and forward waves permits, The only time the forward
wave is incident on the load is when the load = Zo.

Please, Cecil, go back to the drawing board and come up with the correct
Vfwd--it is not equal in any way to the right-hand side of Johnson's Eq 4.23;.
Remember, I said earlier that when rho = 0.5 and the circuit is matched, Vfwd =
Vsource x 1.1547. When you discover where the 1.1547 came from when rho = 0.5
you will have discovered the source of Vfwd.

Steady-state VF2 = 106.06 + 26.5 + 6.63 + 1.66 + 0.41 + ...
This is indeed of the form 106.06(1 + 1/4 + 1/16 + 1/64 + 1/256 + ...)
or VF2 = V1(1 + A + A^2 + A^3 + A^4 + ...)

Wrong, Cecil. Change VF2 to 'E', the max value of the standing wave, and the
values obtained from the sum of the terms in the geometric series will be
correct.

Walt

On Tue, 08 Jun 2004 15:54:50 GMT, Walter Maxwell wrote:

On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.

snip
Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the
max voltage of the standing wave. This is the same mistake that Steve made. I
asked you to rethink what the forward voltage and you have come up wrong.

You are using circuit theory for superposition, as Steve did, when circuit
theory fails to apply in certain transmission line cases. This is one of those
cases. In circuit theory you can superpose the voltages from two sources and V1
+ V2 equals Vtotal. But the re-reflected voltage CANNOT be added to the source
voltage in the transmission line case to obtain Vfwd, because there is only ONE
source.

I'll say it again, Cecil, V1 + V2 = maxV of the standing wave--V1 + V2 does NOT
equal Vfwd because V2 is not a second source, it came fr om ONE source, the
transceiver, and therefore superposition of V1 and V2 does not apply to
establish Vfwd.

I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

Way 1. Vfwd = sqrt(Power fwd x Zo)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]

Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

Walt

Therefore, it is true that maxV may be incident on the load if the relative
phase between the reflected and forward waves permits, The only time the forward
wave is incident on the load is when the load = Zo.

Please, Cecil, go back to the drawing board and come up with the correct
Vfwd--it is not equal in any way to the right-hand side of Johnson's Eq 4.23;.
Remember, I said earlier that when rho = 0.5 and the circuit is matched, Vfwd =
Vsource x 1.1547. When you discover where the 1.1547 came from when rho = 0.5
you will have discovered the source of Vfwd.

Steady-state VF2 = 106.06 + 26.5 + 6.63 + 1.66 + 0.41 + ...
This is indeed of the form 106.06(1 + 1/4 + 1/16 + 1/64 + 1/256 + ...)
or VF2 = V1(1 + A + A^2 + A^3 + A^4 + ...)

Wrong, Cecil. Change VF2 to 'E', the max value of the standing wave, and the
values obtained from the sum of the terms in the geometric series will be
correct.

Walt

I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

Way 1. Vfwd = sqrt(Power fwd x Zo)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]

Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

Walt

Cecil, let's take it one step further and include current, forward current, that
is, Ifwd.

Ifwd = Isouce x sqrt[1/(1 - rho^2)]

Now let's see what happens with a souce voltage of 70.71 v at 100 w on a 50-ohm
line. The source current is 1.414 a.

Now find Ifwd x Vfwd and see what comes up.

Walt

Walter Maxwell wrote:
On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:


Walter Maxwell wrote:

The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.

Johnson's equation contains all possible components - both forward and
reflected, so Johnson's equation predicts the total net standing-wave
voltage. (Note that Johnson's equation contains both positive and negative
exponents.)

Dr. Best's equation contains only the forward components of Johnson's
equation. Therefore, Dr. Best's equation predicts *only* the forward-
traveling voltage. (Note that Dr. Best's equation contains only negative
exponents and represents the E+ half of Johnson's equation while omitting
the E- half of Johnson's equation.)

In other words, the two equations do *not* predict the same quantity
and they are *not* supposed to be the same equations. As I said before,
Dr. Best made a lot of conceptual errors but his equations seem to be
valid. If you break Johnson's equation into two parts representing E+
and E-, the E+ part will be Dr. Best's Eq 6.

Please consider the following matched system with a 1/2 second long lossless
transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V.

100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load
VF1=70.7V-- VF2--
--VR1 --VR2

V1 is equal to 70.7(1.5) = 106.06V and is a constant value
V2 starts out at zero and builds up to 35.35V
VF2=V1+V2 assuming they are in phase

Here is (conceptually simplified) how VF2 builds up to its steady-state value.
VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t)

Time in
Seconds V1 VR2 V2 VF2
0 106.06 0 0 106.06
1 106.06 53.03 26.5 132.56
2 106.06 66.26 33.13 139.19
3 106.06 69.58 34.79 140.85
4 106.06 70.4 35.2 141.26
5 106.06 70.63 35.32 141.38
6 106.06 70.69 35.34 141.39
7 106.06 70.7 35.35 141.4 very close to steady-state

My rounding is a little off but above is a close approximation to what happens.
Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V.

Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the
max voltage of the standing wave. This is the same mistake that Steve made. I
asked you to rethink what the forward voltage and you have come up wrong.

Nope, I haven't, Walt. The maximum voltage of the standing wave equals the
maximum forward voltage plus the maximum reflected voltage. The forward
voltage is 141.4V. The reflected voltage is 70.7V. The maximum standing-wave
voltage is 141.4V + 70.7V = 212.1V. The minimum standing-wave voltage is
141.4V - 70.7V = 70.7V. The VSWR = 212.1V/70.7V = 3:1. Rho = 0.5
SWR = (1+Rho)/(1-Rho) = 1.5/0.5 = 3:1 Everything is perfectly consistent.

So the maximum standing-wave voltage is 212.1V, not 141.4V.

Please reconsider - here's the steady-state powers for the above
matched example:

100W XMTR-----50 ohm line---x---150 ohm line---50 ohm load
PF1=100W-- PF2=133.33W--
--PR1=0 --PR2=33.33W

The forward power is 133.33W and Z0=150 ohms. Last time I looked,
the square root of [133.33W(150 ohms)] was 141.4V.

141.4V^2/150 ohms equals 133.33W. The forward voltage is indeed 141.4V.

Until you discover your mental block, whatever it is, this discussion is
not going to progress. The above example is exceptionally simple so your
mistake must also be simple. You, yourself, have used the above example
and always agreed that 100W + 33.33W = 133.33W, i.e. forward power equals
generated power plus reflected power.
--
73, Cecil http://www.qsl.net/w5dxp



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Walter Maxwell wrote:
I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

OK, here's the example:

100W XMTR---50 ohm line---x----1/2WL 150 ohm line---50 ohm load
PF1=100W-- PF2=133.33W--
--PR1=0W --PR2=33.33W

Way 1. Vfwd = sqrt(Power fwd x Zo)

So Vfwd = sqrt(133.33W x 150ohms) = 141.4 volts (that's what I said)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]
Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

OK, Vfwd = 70.7V x sqrt[1/0.75] = 81.6 volts (something wrong there)
We know Vfwd has to be 141.4V as calculated above using Ohm's law.

Walt, your two ways don't yield the same value. Perhaps that is the
source of your confusion about what Dr. Best said. Your Way 2 appears
to be a calculation of the voltage delivered to a mismatched load.
--
73, Cecil http://www.qsl.net/w5dxp




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