On Sun, 06 Jun 2004 17:32:24 -0500, Cecil Moore wrote:

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.

snip

Dr. Best is essentially quoting an S-parameter analysis

Cecil, if the S-parameteri analysis is applied correctly the results of the
S-parameter analysis should agree with the results of mine that appears in my
earlier posts. You have not responded to the results of my analysis that proves
Steve's use of the equations 9 thru 15 is incorrect. I've proved that these
equations do not work in general. Referring to my analysis, please show me
where I went wrong, if that's your position.

Walt

Walter Maxwell wrote:
Cecil, if the S-parameteri analysis is applied correctly the results of the
S-parameter analysis should agree with the results of mine that appears in my
earlier posts. You have not responded to the results of my analysis that proves
Steve's use of the equations 9 thru 15 is incorrect. I've proved that these
equations do not work in general. Referring to my analysis, please show me
where I went wrong, if that's your position.

I thought I did that, Walt. Your V1 and Dr. Best's V1 are NOT the same quantity.
Your V2 and Dr. Best's V2 are NOT the same quantity. It is no wonder that you
didn't get the same results. The 1WL 50 ohm line in Dr. Best's example is
absolutely irrelevant. Calculating anything on that line is a waste of effort.

Please center your calculations around the match point.

Plug any values into the following generalized matched system:

Z0-match
XMTR-----Z01-----x-----1/4WL Z02-----load
VF1-- VF2--
--0V --VR2

VF2 = VFtotal in Dr. Best's article traveling toward the load

VF1(TAU) = V1 in Dr. Best's article traveling toward the load

VR2(RHO) = V2 in Dr. Best's article traveling toward the load

VR2 will always equal VF1(TAU) + VR2(RHO) = V1 + V2

just like b2 will always equal s21(a1) + s22(a2)
--
73, Cecil http://www.qsl.net/w5dxp



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On Sun, 06 Jun 2004 18:03:31 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
Cecil, if the S-parameteri analysis is applied correctly the results of the
S-parameter analysis should agree with the results of mine that appears in my
earlier posts. You have not responded to the results of my analysis that proves
Steve's use of the equations 9 thru 15 is incorrect. I've proved that these
equations do not work in general. Referring to my analysis, please show me
where I went wrong, if that's your position.

I thought I did that, Walt. Your V1 and Dr. Best's V1 are NOT the same quantity.
Your V2 and Dr. Best's V2 are NOT the same quantity. It is no wonder that you
didn't get the same results. The 1WL 50 ohm line in Dr. Best's example is
absolutely irrelevant. Calculating anything on that line is a waste of effort.

Cecil, it seems like we're going around in cirles.

If Steve's equations are valid they should work in general. It doesn't matter
whether we use the values from his T network section that comes later, or the
values in my example that he attempts to prove incorrect.

What does matter is that the equations must deliver the correct answers
regardless of the values used in the equations. I have proved that valid values
plugged into his equations don't yield the correct answers.

Cec;il, why are you avoiding trying to understand the basis for his erroneous
concept of adding forward and reflected voltages to obtain total forward
voltage? You don't even respond to my discussion on this point.

Walt

Walter Maxwell wrote:
Cecil, why are you avoiding trying to understand the basis for his erroneous
concept of adding forward and reflected voltages to obtain total forward
voltage? You don't even respond to my discussion on this point.

I'm not trying to avoid it, Walt. Dr. Best simply doesn't do that. V1 is
a *forward-traveling voltage*. V2 is a *forward-traveling voltage*. Their
sum is VFtotal, the *total forward-traveling voltage*. He does NOT add a
forward voltage to a reflected voltage. V2 is the *forward-traveling* re-
reflected voltage equal to VR2(RHO).

When the reflected voltage is acted upon by the reflection coefficient, it
becomes a forward-traveling voltage. That you think Dr. Best is adding forward
and reflected voltages, is the source the present misunderstanding. The
individual Poynting Vector for V1 points toward the *load*. The individual
Poynting Vector for V2 points toward the *load*. V1 and V2 are coherent
component waves, both flowing toward the load so, of course, they superpose.

Again, consider the following *matched* configuration where RHO is
the reflection coefficient and TAU is the transmission coefficient.

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

There are four superposition components that occur. Two of them are
traveling toward the load and two of them are traveling toward the
source.

V1 = VF1(TAU) traveling toward the load
V2 = VR2(RHO) traveling toward the load

Adding these two forward-traveling voltages yields VF2 = V1 + V2

V3 = VF1(RHO) traveling toward the source
V4 = VR2(TAU) traveling toward the source

Adding these two rearward-traveling voltages yields VR1 = V3 + V4
which, in a matched case is zero because V3 = -V4.

VF1 breaks up into two components, V1 toward the load and V3
toward the source.

VR2 breaks up into two components, V2 toward the load and V4
toward the source.

Collect and superpose the two forward-traveling terms and you get
the total forward-traveling voltage.

Collect and superpose the two rearward-traveling terms and you get
the total rearward-traveling voltage.
--
73, Cecil http://www.qsl.net/w5dxp



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On Sun, 06 Jun 2004 20:44:46 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
Cecil, why are you avoiding trying to understand the basis for his erroneous
concept of adding forward and reflected voltages to obtain total forward
voltage? You don't even respond to my discussion on this point.

I'm not trying to avoid it, Walt. Dr. Best simply doesn't do that. V1 is
a *forward-traveling voltage*. V2 is a *forward-traveling voltage*. Their
sum is VFtotal, the *total forward-traveling voltage*. He does NOT add a
forward voltage to a reflected voltage. V2 is the *forward-traveling* re-
reflected voltage equal to VR2(RHO).

When the reflected voltage is acted upon by the reflection coefficient, it
becomes a forward-traveling voltage. That you think Dr. Best is adding forward
and reflected voltages, is the source the present misunderstanding. The
individual Poynting Vector for V1 points toward the *load*. The individual
Poynting Vector for V2 points toward the *load*. V1 and V2 are coherent
component waves, both flowing toward the load so, of course, they superpose.

Cecil, I know V2 is the re-reflected voltage, but what I'm trying to persuade
you is that they do NOT superpose to form the forward voltage--they superpose
only to form the standing wave. You've go to accept that the standing wave
voltage is NOT the forward voltage. If you can't come to realize this is the key
to the problem I'm going to have to give up.

Incidentally, you say tau is 1+ rho as the transmission coefficient, which when
muliplied by input voltage yields forward voltage. I thought (1 - rho^2) is the
transmission coefficient. These two terms are not equal. Can you explain the
difference?

Walt

Walter Maxwell wrote:
Cecil, I know V2 is the re-reflected voltage, but what I'm trying to persuade
you is that they do NOT superpose to form the forward voltage--they superpose
only to form the standing wave. You've go to accept that the standing wave
voltage is NOT the forward voltage. If you can't come to realize this is the key
to the problem I'm going to have to give up.

I'm sorry, Walt,
Your belief that V2 is a reflected wave is the root of the misunderstanding.
V2 is a re-reflected wave and is therefore forward-traveling toward the load.
V2 is equal to the reflected wave voltage multiplied by the reflection
coefficient. V1 and V2 are traveling in the same direction, toward the load.

Incidentally, you say tau is 1+ rho as the transmission coefficient, which when
muliplied by input voltage yields forward voltage. I thought (1 - rho^2) is the
transmission coefficient. These two terms are not equal. Can you explain the
difference?

(1-rho^2) is the POWER transmission coefficient. For a single impedance
discontinuity situation, 1+rho is the VOLTAGE transmission coefficient.
From the IEEE Dictionary: "transmission coefficient, ... Note 2, An interface
is a special case of a network where the reference planes associated with
the incident and transmitted waves become coincident; in this case the
voltage transmission coefficient is equal to one plus the voltage
reflection coefficient."
--
73, Cecil http://www.qsl.net/w5dxp



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On Sun, 06 Jun 2004 23:18:37 -0500, Cecil Moore
wrote:
Walter Maxwell wrote:
voltage is NOT the forward voltage. If you can't come to realize this is the key
to the problem I'm going to have to give up.

I'm sorry, Walt,
Your belief that V2 is a reflected wave is the root of the misunderstanding.
V2 is a re-reflected wave
Another way of saying "You are right, Walt, you are wrong."

Let's see, you two have passed this SAME thing back and forth 39
times, cannot agree about what each thinks about ONE particular, and
each of you insist you know what a third party meant.

Well, I'm off to Foggy Bottom (D.C.) again to where they do this kind
of thing for a living and call it law. ;-)

Let's see if a week improves this chowder.

73's
Richard Clark, KB7QHC

Richard Clark wrote,

Let's see, you two have passed this SAME thing back and forth 39
times, cannot agree about what each thinks about ONE particular, and
each of you insist you know what a third party meant.

Well, I'm off to Foggy Bottom (D.C.) again to where they do this kind
of thing for a living and call it law. ;-)

Let's see if a week improves this chowder.

73's
Richard Clark, KB7QHC


The triumph of hope over experience.
73,
Tom Donaly, KA6RUH

On Sun, 06 Jun 2004 23:18:37 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
Cecil, I know V2 is the re-reflected voltage, but what I'm trying to persuade
you is that they do NOT superpose to form the forward voltage--they superpose
only to form the standing wave. You've go to accept that the standing wave
voltage is NOT the forward voltage. If you can't come to realize this is the key
to the problem I'm going to have to give up.

I'm sorry, Walt,
Your belief that V2 is a reflected wave is the root of the misunderstanding.
V2 is a re-reflected wave and is therefore forward-traveling toward the load.
V2 is equal to the reflected wave voltage multiplied by the reflection
coefficient. V1 and V2 are traveling in the same direction, toward the load.

Cecil, please read me in the first paragraph. By Steve's own words he says the
re-reflected wave must equal the reflected wave. This means the system is
matched in his account. Therefore, V2, is not the root of any misunderstanding
in my part.

You are still not getting the picture concerning that V1 and V2 cannot be added
to establish the forward wave, as Steve incorrectly believes, they add only to
form the standing wave.

I'm sorry that I didn't think of this earlier that Steve copied his Eq 6 in Part
1 from Johnson, except he placed Vfwd, the forward voltage, instead of E for the
standing wave voltage. Look it up in your Johnson on Pages 99 and 100. He
derives Eq 4.23 (the Eq Steve misunderstands) on Page 99, and expresses it on
Page 100. However, note on Page 98, the beginning of Sec 4.2:

"The equations for E and I along the line can be expressed...."

So Cecil, please understand that this equation does NOT yield the forward
voltage, as Steve believes, which is the root of his misunderstanding throughout
his entire paper.

Concerning tau, I've seen it described in an HP App Note, which I didn't bring
to Michigan, but I've never used it. However, if the power transmission
coefficient is (1 - rho^2) the coefficient is 0.75 for rho = 0.5. Therefore, for
100 w forward only 75 w are delivered. This condition is shown valid
experimentally.

Now let's use tau = 1 + rho as the voltage transmission coefficient. I interpret
this to mean tau x input voltage = forward voltage arriving at a mismatched
load. For a 100 w source at 50 ohms with the same rho as above, we have 70.71 x
1.5 = 106.07 v. But we know that the forward voltage on a matched 50-ohm line
with rho = 0.5 is 81.65 v. Why the difference?

I should have been more aware of the explanation in the HP App Note--there must
be a reason shown there to explain the difference.

Walt

Richard Clark wrote:
On Sun, 06 Jun 2004 23:18:37 -0500, Cecil Moore
wrote:
I'm sorry, Walt,
Your belief that V2 is a reflected wave is the root of the misunderstanding.
V2 is a re-reflected wave

Another way of saying "You are right, Walt, you are wrong."

It's a very minor mistake, Richard, and one easily made. If you have
a copy of Dr. Best's QEX article, please feel free to express your
take on this discussion.

Let's see if I can present superposition in ASCII graphics. rho
is the voltage reflection coefficient and tau is the voltage
transmission coefficient. Assume VF1 has a phase angle of zero
degrees. Phase angles are important in the following but since
the system is matched, all phase angles are either at zero degrees
or at 180 degrees at the match point. So a sign change is equivalent
to a 180 degree phase shift.

100W XMTR---50 ohm line---x---1/2WL 150 ohm line---50 ohm load
VF1=70.7V-- VF2=141.4V--
--VR1=0V --VR2=70.7V

According to the rules of superposition, the two voltages incident
upon 'x', VF1 and VR2, can be considered separately and then added.

----------------------------------------------------------------

Breaking VF1 down into its two superposition components yields:

x
|
VF1=70.7V (100W)--|
|-- V1=106.06V (75W)
V3=35.35V (25W)--|
|

VF1 = 70.7V at zero degrees (100W)

V1 = VF1(forward-tau) = 70.7(1.5) = 106.06V at zero degrees (75W)

V3 = VF1(forward-rho) = 70.7(0.5) = 35.35V at zero degrees (25W)

Note that PF1 = 100W = P1 + P3 = 75W + 25W

-----------------------------------------------------------------

Breaking VR2 down into its two superposition components yields.


x
|
|-- VR2=70.7V (33.33W)
V4=35.35V (25W)--|
|-- V2=35.35V (8.33W)
|

VR2 = 70.7V at 180 degrees (33.33W)

V2 = VR2(reverse-rho) = 70.7(-0.5) = 35.35V at zero degrees (8.33W)

V4 = VR2(reverse-tau) = 70.7(0.5) = 35.35V at 180 degrees (25W)

Note that PR2 = 33.33W = P2 + P4 = 25W + 8.33W

-----------------------------------------------------------------

Now, following the rules of superposition:

To get the total forward voltage, add V1 + V2

VF2 = V1 + V2 = 106.06V + 35.35V = 141.4V (133.33W)

To get the total reflected voltage, add V3 and V4

VR1 = V3 + V4 = 35.35V - 35.35V = zero volts (0W)

Note: Dr. Best neglected to mention P3 and P4 in his QEX article.
P3+P4 is the interference joules/sec. They are scalar values.

All voltages are consistent and all powers are consistent.

So, in this matched system, all reflected power is re-reflected:

PF2 = 133.33W = P1 + P2 + P3 + P4 = 75W + 8.33W + 25W + 25W

PF2 = P1 + P2 + (complete constructive interference) = 133.33W

PR1 = P3 + P4 - (complete destructive interference) = zero watts

Note that the voltage forward-rho = (150-50)/(150+50) = +0.5

reverse-rho = (50-150)/(50+150) = -0.5 (180 deg phase shift)
--
73, Cecil http://www.qsl.net/w5dxp



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