Cecil, Steve's equations are out there for all to see as general equations.
Nowhere has he made any caviats to the contrary, or that they only work for
reflection coefficients of 0.5. If they only work for one specific coefficient
it should be so noted that they are not for general use. And if equations are
not valid for general use they are then invalid.

Can you explain this?

Walt

Walter Maxwell wrote:
Cecil, Steve's equations are out there for all to see as general equations.
Nowhere has he made any caviats to the contrary, or that they only work for
reflection coefficients of 0.5. If they only work for one specific coefficient
it should be so noted that they are not for general use. And if equations are
not valid for general use they are then invalid.

Can you explain this?

They do work for any reflection coefficient, Walt. He chose 0.5 for
his examples because a 0.5 voltage reflection coefficient reflects
half of the voltage and that's easy to do in your head. The S-parameter
analysis will work for any reflection coefficient.

If you will take the time to perform an S-parameter analysis, you
will understand Steve's equations. His analysis is completely
different from yours.

Here's one with a voltage reflection coefficient of 0.707. You
can pick any Z0 for the 1/2WL line below and therefore choose
any reflection coefficient you want.

100W XMTR---50 ohm line---x---1/2WL 291.5 ohm line---50 ohm load
VF1=70.7V-- VF2=241.45V--
--VR1=0V --VR2=170.7V

Dr. Best's V1 = VF1(1+rho) = 70.7(1.707) = 120.7V

Dr. Best's V2 = VR2(rho) = 170.7(0.707) = 120.7V

VF2 = V1 + V2 = 120.7V + 120.7V = 241.4V (200W forward)

This is just an S-parameter analysis without normalizing to the
square root of Z0. In fact, we can *derive* Dr. Best's equations
from the S-parameter equations simply by multiplying by the
square root of Z0.

b2 = s21(a1) + s22(a2)

b2*Sqrt(Z0) = s21(a1)Sqrt(Z0) + s22(a2)Sqrt(Z0)

VF2 = b2*Sqrt(Z0), V1 = s21(a1)Sqrt(Z0), V2 = s22(a2)Sqrt(Z0)

VF2 = V1 + V2
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, how about doing me a favor and humor me a little.

How about solving my example, the one that Steve took from Reflections and then
trashed. To repeat, 100 w, 70.71 v source, 50-ohm lossline terminated in a 3:1
mismatch and perfectly matched at the input with a rho = 1.0.

Vfwd = 81.65 v and Vref = 40.83 V. Please solve this using Steve's method of
solution, Eqs 5 thru 13.

If this problem can't be solved with Steve's method, his method must be invalid,
not general.

Walt

Walter Maxwell wrote:
Cecil, how about doing me a favor and humor me a little.

How about solving my example, the one that Steve took from Reflections and then
trashed. To repeat, 100 w, 70.71 v source, 50-ohm lossline terminated in a 3:1
mismatch and perfectly matched at the input with a rho = 1.0.

Here's the configuration.

100W XMTR--50 ohm line--x-tuner-y--1wl 50 ohm line---150 ohm load

The tuner function is replaced by a 1/4WL series section which
performs *exactly* the same function so we can see what is really
happening inside the tuner. The match point is at 'x', not at 'y'.
The system is NOT matched at 'y'. Therefore, Dr. Best's system is
not matched at the output of the tuner. His choice to use a tuner
for that illustration, instead of a 1/4WL transformer, was a very
bad one and has obviously led to a lot of confusion and misunderstanding.

rho=0.268 rho=-0.268 rho=0.5
100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line--y--1wl 50 ohm line---150 ohm load
VF1=70.7V-- VF2=96.58V-- VF3=81.65V--
--VR1=0V --VR2=25.9V --VR3=40.82V
PF1=100W-- PF2=107.76W-- PF3=133.33W--
--PR1=0W --PR2=7.76W --PR3=33.33W

The mismatched impedance discontinuity at 'y' is interesting because
we have forward power and reflected power flowing to/from both directions.
That's what happens at a tuner output. The actual match point is at the
tuner input at 'x'.

At point 'y' on the 50 ohm line we have a V1y of 70.7V corresponding to
the 100W forward power that made it through the impedance discontinuity.
This can also be calculated by using the transmission coefficient. Since
the impedance is a step-down impedance, tau = 1-rho

V1y = VF2(1-rho) = 96.58(0.732) = 70.7V

V2y will be VR3(rho) = 40.82(0.268) = 10.94V

VF3 = 70.7V + 10.94V = 81.65V. (133.33W)

The modified S-parameter analysis works once again and this time it
is at an unmatched impedance discontinuity at point 'y'.

We could also do the modified S-parameter analysis at point 'x'.

V1x = VF1(1+rho) = 70.7V(1.268) = 89.6V

V2x = VR2(rho) = 25.9V(0.268) = 6.94V

VF2 = V1 + V2 = 89.64V + 6.94V = 96.58V (107.7W)
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Continuing and calculating interference energy at point 'y':

V1y = VF2(1-rho) = 96.58(0.732) = 70.7V (100W)

V2y will be VR3(rho) = 40.82(0.268) = 10.94V (2.39W)

VF3 = 70.7V + 10.94V = 81.65V. (133.33W)

The power equation at point 'y' looking toward the load is:

PF3 - P1y - P2y = constructive interference

Note that P2y + constructive interference = re-reflected power.

133.33W - 100W - 2.39W = +30.94W

A '+' sign on the interference term indicates constructive
interference.

There is 30.49 watts of constructive interference at 'y' looking
toward the load. That energy must come from somewhere.

Therefore, there is 30.49 watts of destructive interference at
'y' looking toward the source. The flow of energy from the
destructive interference event to the constructive interference
event is an interference-caused reflection.

The total re-reflected power at 'y' is 2.39W + 30.94W = 33.33W
In this case, everything except P1y is re-reflected. This is in
agreement with Walter Maxwell and in disagreement with Dr. Best.

V1x = VF1(1+rho) = 70.7V(1.268) = 89.6V (88.09W)

V2x = VR2(rho) = 25.9V(0.268) = 6.94V (0.556W)

VF2 = V1 + V2 = 89.64V + 6.94V = 96.58V (107.7W)

The power equation at point 'x' looking toward the load is:

PF2 - P1 - P2 = constructive interference

107.7W - 88.09W - 0.556W = +18.42W

The '+' sign tells us that there is 18.42 watts of constructive
interference at 'x' looking toward the load. That energy must
come from somewhere.

Therefore, there is 18.42 watts of destructive interference at
'x' looking toward the source. This is also an interference
reflection.

The total re-reflected power at 'x' is 0.556W + 18.42W = 18.98W
Everything except P1 is re-reflected at a match point.

Using the S-parameter equations:

b1 = s11(a1) + s12(a2)

b2 = s21(a1) + s22(a2)

s21(a1) is the only voltage term that doesn't get reflected
at least once.

s11(a1), s12(a2), and s22(a2) are all reflected terms. If the
energy in these terms winds up at the load, they have all been
re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 08 Jun 2004 23:09:32 -0500, Cecil Moore
wrote:

Walter Maxwell wrote:
Cecil, how about doing me a favor and humor me a little.

How about solving my example, the one that Steve took from Reflections and then
trashed. To repeat, 100 w, 70.71 v source, 50-ohm lossline terminated in a 3:1
mismatch and perfectly matched at the input with a rho = 1.0.

Thanks, Cecil, I'll be studying it carefully. Using a 1/4 wl transformer seems
like a great idea. I hadn't thought of that.

Walt

Hi Cecil,

Man, when I ask for you to humor me you certainly do it up brown!!

You've performed a magnificent job of detailing all the reflections and
re-reflections in the 1/4 wl transformer replacing the tuner. That performance
really took a lot of work, and I greatly appreciate it.

I haven't yet finished reviewing it to the point of verifying in my mind all the
various voltages, but I appreciate that the numbers came out correct.

However, I have another request for you. Can you find any forward voltages that
are appropriate to plug into Steve's Eq 9 and obtain 133.33 watts PFtotal ?

Walt

On Wed, 09 Jun 2004 19:16:39 GMT, Walter Maxwell wrote:

Hi Cecil,

Man, when I ask for you to humor me you certainly do it up brown!!

You've performed a magnificent job of detailing all the reflections and
re-reflections in the 1/4 wl transformer replacing the tuner. That performance
really took a lot of work, and I greatly appreciate it.

I haven't yet finished reviewing it to the point of verifying in my mind all the
various voltages, but I appreciate that the numbers came out correct.

However, I have another request for you. Can you find any forward voltages that
are appropriate to plug into Steve's Eq 9 and obtain 133.33 watts PFtotal ?

Walt

Cecil, I found the two Vfwd's we're looking for. They are derived from V1y and
V2y. 70.7 v and 10.94 v. The interesting part is that 10.94 v is the exact
increase in voltage resulting from adding 33.333 w to 100 w. This I knew many
moons ago. But one would never find this Vfwd without performing the complete
reflection analysis you performed. With this, plus the fact that a reflection
analysis on a stub matching configuration would not reveal this forward voltage
still makes Steve's Eq 9 invalid for general use. This is because the
re-reflected forward voltage there, 40.82 v added to the source voltage 70.71 v
does not equal the real total forward voltage of 81.65 v.

Therefore, my final comment on Eq 9 is that it works in specific cases but it
certainly is not valid in general.

Thanks again, Cecil, for the elegant reflection analysis, bravo!

Walt

Walter Maxwell wrote:
On Tue, 08 Jun 2004 23:09:32 -0500, Cecil Moore
wrote:


Walter Maxwell wrote:

Cecil, how about doing me a favor and humor me a little.

How about solving my example, the one that Steve took from Reflections and then
trashed. To repeat, 100 w, 70.71 v source, 50-ohm lossline terminated in a 3:1
mismatch and perfectly matched at the input with a rho = 1.0.

Thanks, Cecil, I'll be studying it carefully. Using a 1/4 wl transformer seems
like a great idea. I hadn't thought of that.

It's easier to do it in steps.

Step 1.
100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---150 ohm load

Calculate all the forward and reflected components.

Step 2.
100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---y---1WL 50 ohm line---150 ohm load

All other forward and reflected components remain the same.
Calculate the forward and reflected components on the 50 ohm line
going to the load.
--
73, Cecil http://www.qsl.net/w5dxp



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Walter Maxwell wrote:
However, I have another request for you. Can you find any forward voltages that
are appropriate to plug into Steve's Eq 9 and obtain 133.33 watts PFtotal ?

Here is a quote from an earlier posting:
----------------------
Continuing and calculating interference energy at point 'y':

V1y = VF2(1-rho) = 96.58(0.732) = 70.7V (100W)

V2y will be VR3(rho) = 40.82(0.268) = 10.94V (2.39W)

VF3 = 70.7V + 10.94V = 81.65V. (133.33W)

--------------- So ---

PFtotal = |VFtotal|^2/Z0 = 81.65V^2/50ohms = 133.33W

PFtotal = |V1+V2|^2/Z0 = (70.7V+10.94V)^2/Z0 = 133.33W

How's that?
--
73, Cecil http://www.qsl.net/w5dxp



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