W5DXP wrote:

Jim Kelley wrote:
I've never disputed the solutions produce the same number. I disputed
the validity of the approach you took to arrive at the equation.

Here's a question for you, Jim. How much power is transferred from the
mismatched load rearward by Pref2?

First I have to know where it is being transferred. I don't see two
ends of a path. Is there a load resistor hiding in a circulator
somewhere? Is the 33.33 watts extra power that I could drive another
antenna with, but is otherwise going unused?

Or is it instead, simply a field that is excited by the boundary
conditions at the load that is being measured.

source---50 ohm feedline---+---150 ohm feedline---load
Pfwd1=200W -- Pfwd2=133.33W --
-- Pref1=100W -- Pref2=33.33W

If, as you have asserted, Pref2 can transfer zero watts when it measures
33.33W, can it also transfer 100W when it measures 33.33W?

The meter measures voltage, assumes a transfer of energy, and displays
power based on that assumption.

73, ac6xg

Jim Kelley wrote:
First I have to know where it is being transferred.

This is the basic flaw in your argument. The power goes where it goes.
That you need to know something indicates that you believe what you
know dictates reality. Sorry, Jim, you are NOT that powerful. Mother
Nature doesn't care what you know or don't know. Reality is exactly
the same either way and will be exactly the same way after you are
six feet under. Why do you think you are so important that what you
know affects reality?
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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W5DXP wrote:

Jim Kelley wrote:
First I have to know where it is being transferred.

This is the basic flaw in your argument. The power goes where it goes.

So, where does the 33.33 watts in your question go?

73, ac6xg

Jim Kelley wrote:
But where does that 33.33 watts actually go? If you have 100 watts
coming from the source, and 100 watts traversing the first and second
boundaries, where does the 33.33 watts go?

Through a two step process, it merges in phase with the forward wave at
the Z0-match point. A directional wattmeter reads 133.33W forward power.
100W + 33.33W = 133.33W. 100W/|1-rho|^2 = 133.33W
Forward power - reflected power = power delivered to the load.
The standing waves confirm that there is indeed 133.33W forward and
33.33W reflected.

Let's look at the following system with two sources. Each source is a
signal generator equipped with a circulator and load. SGCL1 is equipped
with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm
circulator and load. The two signal generators are phase locked but can
be turned on and off independently.

rho=0.5
100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2
Pfwd1-- Pfwd2--
--Pref1 --Pref2

Using the principle of superposition: With SGCL1 on and SGCL2 off,
Pfwd1=100W, Pref1=25W, Pfwd2=75W, and Pref2=0W. SGCL2 dissipates 75W
(Pfwd1)(|rho|^2) and SGCL1 dissipates 25W (Pfwd1)(1-|rho|^2). This
is how the s11 and s21 s-parameters are measured.

With SGCL2 on and SGCL1 off, Pfwd1=0W, Pref1=25W, Pfwd2=8.33W,
Pref2=33.33W. SGCL2 dissipates 8.33W (Pref2)(|rho|^2) and SGCL1 dissipates
25W (Pref2)( 1-|rho|^2). This is how the s12 and s22 s-parameters are
measured.

Note that SGCL1 dissipates 25W in both of the above cases.

With SGCL1 and SGCL2 both on, Pfwd1=100W, Pref1=0W, Pfwd2=133.33W, and
Pref2=33.33W. This is similar to the earlier single source example. Note
that the two 25W waves obviously engage in wave cancellation and their
combined 50W of destructive interference joins the forward wave as constructive
interference. Assume there is a set of feedline lengths that will accomplish
the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the
impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm
feedline will do that if the voltages from the two sources are max+ at the
same time.

b1 = s11*a1 + s12*a2 = 0V s11*a1 and s12*a2 engage in destructive interference.

Pref1 = Pfwd1(|rho|^2) + Pref2(1-|rho|^2) - destructive interference
0W = 25W + 25W - 50W

Pfwd2 = Pfwd1(1-|rho|^2) + Pref2(|rho|^2) + constructive interference
133.33W = 75W + 8.33W = 50W
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Cecil, W5DXP wrote:
"Let`s look at the following system with two sources,"

Is this tempest about creation of an oppositely phased wave to cancel a
reflection?

A way to avoid reflected waves within a waveguide is mentioned by Terman
on page 148 of his 1955 edition:

"Create a reflected wave near the load that is equal in magnitude but
opposite in phase from the wave reflected by the load; in this way the
two reflected waves cancel each other."

"Some of these (matching arrangements) are analogous to the impedance
matching arrangements employed in transmission lines (described in Sec.
4-11) while others are unique to waveguides."

Best regards, Richard Harrison, KB5WZI

On Wed, 16 Jul 2003 12:04:17 -0500 (CDT),
(Richard Harrison) wrote:

Is this tempest about creation of an oppositely phased wave to cancel a
reflection?

Richard, you are spitting into the wind. Only Cecil would call that
reflection a fresh ocean mist. :-)

73's
Richard Clark, KB7QHC

W5DXP wrote:
A directional wattmeter reads 133.33W forward power.
100W + 33.33W = 133.33W. 100W/|1-rho|^2 = 133.33W
Forward power - reflected power = power delivered to the load.
The standing waves confirm that there is indeed 133.33W forward and
33.33W reflected.

So where does the 33.33 watts actually go? The 100 watt number seems to
account for all of the transfer of energy.

Let's look at the following system with two sources. Each source is a
signal generator equipped with a circulator and load. SGCL1 is equipped
with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm
circulator and load. The two signal generators are phase locked but can
be turned on and off independently.

rho=0.5
100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2
Pfwd1-- Pfwd2--
--Pref1 --Pref2

Using the principle of superposition: With SGCL1 on and SGCL2 off,
Pfwd1=100W, Pref1=25W, Pfwd2=75W, and Pref2=0W. SGCL2 dissipates 75W
(Pfwd1)(|rho|^2) and SGCL1 dissipates 25W (Pfwd1)(1-|rho|^2).


With SGCL2 on and SGCL1 off, Pfwd1=0W, Pref1=25W, Pfwd2=8.33W,
Pref2=33.33W. SGCL2 dissipates 8.33W (Pref2)(|rho|^2) and SGCL1 dissipates
25W (Pref2)( 1-|rho|^2). This is how the s12 and s22 s-parameters are
measured.

Note that SGCL1 dissipates 25W in both of the above cases.

With SGCL1 and SGCL2 both on, Pfwd1=100W, Pref1=0W, Pfwd2=133.33W, and
Pref2=33.33W. This is similar to the earlier single source example.

Except for the amount of power input. Although you didn't mention it
explicitly, I assume SGCL2 inputs 33.33 watts. Obviously there's that
much more power being input to the network in this example that the
others we've worked on.

Note
that the two 25W waves obviously engage in wave cancellation and their
combined 50W of destructive interference joins the forward wave as constructive
interference.

I note that interference takes place.

Assume there is a set of feedline lengths that will accomplish
the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the
impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm
feedline will do that if the voltages from the two sources are max+ at the
same time.

Well, you won't get a phase reversal from reflection at the 150/50 ohm
boundary as you would with the 150/450 ohm boundary in our other
problem, so I think the 150 ohm feedline would need to be an odd number
of 1/4WL's in order to put Vref1 out of phase with Vfwd1 at the first
boundary. But since we're really interested in the relative phase
between Vref2 and Vref1, and Vref1 becomes inverted at the 50/150 ohm
boundary, you would want to maintain the phase of Vref2 by using an even
multiple of 1/4 wave lengths of the 150 ohm line.

b1 = s11*a1 + s12*a2 = 0V s11*a1 and s12*a2 engage in destructive interference.

Pref1 = Pfwd1(|rho|^2) + Pref2(1-|rho|^2) - destructive interference
0W = 25W + 25W - 50W

Pfwd2 = Pfwd1(1-|rho|^2) + Pref2(|rho|^2) + constructive interference
133.33W = 75W + 8.33W = 50W

I'll trade you a plus sign for your equal sign. ;-)

We also know that 133.33W = 100W + 33.33W.
So 100W = 100W + 33.33W - 33.33W. Which is equally revealing. ;-)

73, Jim AC6XG

Jim Kelley wrote:
So where does the 33.33 watts actually go? The 100 watt number seems to
account for all of the transfer of energy.

I told you, the 33.33W joins the 100W forward wave at the impedance
discontinuity and heads toward the load where a new 33.33W is rejected
by the mismatched load. Maybe that is the point of your confusion. The
33.33W that makes a round trip to the impedance discontinuity and back
to the load is not the same 33.33W that is reflected by the load. TV
ghosting proves that to be true.


Let's look at the following system with two sources. Each source is a
signal generator equipped with a circulator and load. SGCL1 is equipped
with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm
circulator and load. The two signal generators are phase locked but can
be turned on and off independently.


rho=0.5
100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2
Pfwd1-- Pfwd2--
--Pref1 --Pref2

Except for the amount of power input. Although you didn't mention it
explicitly, I assume SGCL2 inputs 33.33 watts.

"33.33W SGCL2" seems pretty explicit to me.

Obviously there's that
much more power being input to the network in this example that the
others we've worked on.

We are also taking out more power than we previously were.

I note that interference takes place.

Please note that the 25W reflected when SGCL1 only is on is 35.36V at
zero deg, and 0.707A at 180 deg. When SGCL2 only is on, the 25W not
re-reflected is 35.36V at 180 deg, and 0.707A at zero deg. That's why
they cancel when both are on.


Assume there is a set of feedline lengths that will accomplish
the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the
impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm
feedline will do that if the voltages from the two sources are max+ at the
same time.

Well, you won't get a phase reversal from reflection at the 150/50 ohm
boundary as you would with the 150/450 ohm boundary in our other
problem, so I think the 150 ohm feedline would need to be an odd number
of 1/4WL's in order to put Vref1 out of phase with Vfwd1 at the first
boundary. But since we're really interested in the relative phase
between Vref2 and Vref1, and Vref1 becomes inverted at the 50/150 ohm
boundary, you would want to maintain the phase of Vref2 by using an even
multiple of 1/4 wave lengths of the 150 ohm line.

Nope, you are wrong about that but I'll let you figure our your own error.
The only voltage phase reversal in the given example is in the re-reflected
wave associated with the 8.33W. Vfwd1 travels 1WL while Vref2 travels 1/2WL.
That is enough to put them 180 degrees out of phase. Vfwd1(|rho|^2) undergoes
zero phase shift. Vref2(1-|rho|^2) undergoes zero phase shift.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Richard Harrison wrote:
Terman said about the same thing about cancellation of the reflected
wave in a waveguide. Terman offered several methods to generate the
offsetting wave in a waveguide, and there are several ways to generate a
wave which produces cancellation on a transmission line, too.

Trouble is, Terman didn't say anything about what happens to the
intrinsic energy that existed in the waves before they were
canceled. I say the energy in rearward-traveling canceled reflected
waves is reflected back toward the load. Jim (and others) disagree
even though we know that all the energy winds up incident upon the
load.
--
73, Cecil, W5DXP

Jim Kelley wrote:


W5DXP wrote:
The
33.33W that makes a round trip to the impedance discontinuity and back
to the load is not the same 33.33W that is reflected by the load. TV
ghosting proves that to be true.

TV ghosting is primarily the result of multipathing, which is external
to the antenna system.

:-) I'm talking about TV ghosting due to reflections on the transmission
line between a TV test generator and a TV receiver. The only "multipathing"
involved is the different paths taken by the reflected waves inside the
transmission line.

The only voltage phase reversal in the given example is in the re-reflected
wave associated with the 8.33W.

How are you determining your phase reversals? Don't forget that phase
reversal on reflection occurs when a wave encounters a more dense medium
(or a higher impedance).

Nope, a voltage phase reversal occurs when a wave encounters a *lower*
impedance. A voltage wave encountering a higher impedance does not
reverse phase. In the earlier example, the only lower impedance encountered
is by the rearward-traveling wave flowing back into the impedance discontinuity.
I repeat, Vref2(-rho) is the only voltage phase reversal.
--
73, Cecil, W5DXP