Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:

[snip]

The "absence of energy flow" is an illusion. There is 100
joules/sec in the forward wave and 100 joules/sec in the
reflected wave. Since the waves are flowing in opposite
directions, you can argue that there is no *net* energy
flow, but the component wave energy flow is alive and well.


[snip]

Cecil,

You have made this claim about "component wave energy flow" or a similar
claim on numerous occasions. However, only a few minutes after the
message above, you wrote,

"Power and energy are scalars, Roy. Of course, scalars follow
different rules. Maybe the problem is that you are trying
to use phase math on power."

Help me out. How can we have scalars flowing in opposite directions? If
the waves can interact, as you claim, why does the associated energy
fail to interact and merely pass like ships in the night?

Conservation of energy does not really help the explanation in this
case, as your recently discovered new reference, Principles of Optics by
Born and Wolf, points out.

73,
Gene
W4SZ

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Tom Ring wrote:
Roy Lewallen wrote:
rules which constantly change to suit the needs of the moment.
Watterson fans will recognize the rules for propagating power waves
as closely resembling those of Calvinball.

Hopefully some of those here will get that. I would be surprised if
it's more than one in four.

The rules for propagating waves of EM energy have been
nailed down for generations. Optical physicists don't
have the luxury of measuring voltage and current. They
must necessarily measure average power density. They
are quite good at it and their average power density
equations are quite accurate and mature. They obviously
know a lot more about EM waves than most of the posters
here.

Cecil,

I guess you did not believe me when I said a day or two ago that optical
physicists have advanced well beyond the limited capabilities you allow
them. It is true that they don't measure "voltage and current". I don't
even know what voltage and current mean in the standard context of
optics. However, they do measure fields in many ways, and they do use
extensive calculations based on fields.

The measurement tools used at optical frequencies are a bit different
than the measurement tools used for RF, but they are no less capable.

Do you have any connection with "optical physicists" beyond your reading
of Hecht and the Melles Griot website?

73,
Gene
W4SZ

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
Do you see the common factor in your response about "wave interaction"?
In all of your examples there is an interface or some sort of
discontinuity. Nobody argues that waves are forever unchanging. However,
those changes take place only through interaction with interfaces or
other discontinuities.

I don't disagree and I have gone on record as saying that
reflections only occur at physical impedance discontinuities.
The interaction of s11(a1) and s12(a2) is caused by the
interaction of a1 and a2 with the impedance discontinuity.
There's no doubt about that.

a1 interacts with the impedance discontinuity to cause
s11(a1) and s21(a1). a2 interacts with the impedance
discontinuity to cause s12(a2) and s22(a2). s11(a1),
s12(a2), s21(a1), and s22(a2) are created as a result
(an effect) of the interaction of a1 and a2 with the
impedance discontinuity. I have *NEVER* said that
waves interact with each other in the absence of
an impedance discontinuity. Assertions to that effect
are obfuscations of what I have said.

But even you must realize that the wave component, s11(a1),
*originates* traveling *away from* the impedance discontinuity
as an *EFFECT* of the forward wave, a1, being incident upon
the impedance discontinuity. It is a1 that is incident upon
the impedance discontinuity, not s11(a1). s11(a1) *originates*
at the impedance discontinuity traveling *away from* it and is
*never incident* upon the impedance discontinuity. All s11(a1)
ever encounters is s12(a2) and is canceled on the spot if
s12(a2) is of equal magnitude and opposite phase.

The wave component, s12(a2), *originates* at the impedance
discontinuity traveling *away from* the impedance discontinuity
as an *EFFECT* of the reflected wave, a2, being incident upon
the impedance discontinuity. It is a2 that is incident upon
the impedance discontinuity, not s12(a2). s12(a2) *originates*
at the impedance discontinuity traveling *away from* it and
is *never* incident upon the impedance discontinuity. All it
ever encounters is s11(a1) and is canceled on the spot if
s11(a1) is of equal magnitude and opposite phase.

Arguing that s11(a1) and s12(a2) are incident upon the
impedance discontinuity is obviously false since they
originate traveling *away from* the impedance discontinuity.
It is impossible for signals that originate traveling away
from the impedance discontinuity, to ever be incident upon
the impedance discontinuity. The confusing of cause and effect
is obvious.

So how do you think that signals that originate traveling
*away from* the impedance discontinuity ever can be incident
upon the impedance discontinuity?
--
73, Cecil, w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil, W5DXP wrote:
"The number of joules in the stub, in both cases, is exactly the
magnitude needed to support the 100W forward wave and the 100W reflected
wave."

An energy source of frequency f feeds a 1/4-wave short-circuited stub.
The feedpoint of the stub receives a reflection from the short an
instant later which is practically equal in phase and magnitude to the
energy source.

How much energy will re-enter the source? Practically none because there
is no difference of potential between the reflection and the source.

How much energy will continue to enter the stub? Practically none
because there is no difference of potential between the source and the
reflection.

Best regards, Richard Harrison, KB5WZI

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
Help me out. How can we have scalars flowing in opposite directions? If
the waves can interact, as you claim, why does the associated energy
fail to interact and merely pass like ships in the night?

That is the nature of EM waves, Gene. EM waves flowing
in opposite directions do NOT interact. However, their
reflected and transmitted components traveling in the
same direction can and do interact at an impedance
discontinuity.

Let's take the following example using a signal generator
and 50 ohm circulator load. rho1 = s11 = 0.707

100W
SGCL---50 ohm line---+---1/2WL 291.4 ohm line---291.4 ohm load
Vfor1=70.7V-- Vfor2=120.7V--
--Vref1=50V --Vref2=0V

Vref1 is the reflected wave that *originates* traveling toward
the source. Its power gets dissipated in the circulator load
resistor. It *NEVER* travels toward the impedance discontinuity.

Vref1 = 0.707(70.7V) = 50V This is the wave that we have
to cancel when we switch the load to 50 ohms. This wave
gets canceled without ever encountering an impedance
discontinuity.

Would you please explain how a wave that originates traveling
away from the impedance discontinuity ever encounters the
impedance discontinuity?
--
73, Cecil, w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
I guess you did not believe me when I said a day or two ago that optical
physicists have advanced well beyond the limited capabilities you allow
them.

I know that, Gene. What I have been describing is a history
of the physics of optics. When the theories were first being
developed, all optical physicists could so was measure the
average power density of the light waves. The theories
based on those average power density measurements are
still good today so they must have known what they were
doing. However, if you would like to discredit those great
optical physicists, based on the better measurement
techniques available today, be my guest.
--
73, Cecil, w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Jim Kelley wrote:

Redistribution is an interaction....interesting. But, you were
telling us about how waves interact with other waves. I'm interested
to know what effect x has on y, and vice versa? We have x + y making
z. So after that, tell us how have x and y changed as a result of
their "interaction"?


In a transmision line, when z=0, x and y are permanently
changed. Their energy components combine into one re-reflected
wave. The separate identities of x and y disappear at the
instant that z becomes zero.

In order to measure s11 and s12, a2 is turned off. The result is:

a1----|
|----s21(a2)
s11(a1)----|

Note that s11(a1) has already reflected from the impedance
discontinuity and there are no other impedance discontinuities
between it and the source. Should be smooth sailing.

In order to measure s21 and s22, a1 is turned off. The result is:

|----s22(a2)
s12(a2)----|
|----a2

Note that s12(a2) has already passed through the impedance
discontinuity and there are no other impedance discontinuities
between it and the source. Should be smooth sailing.

s11(a1) and s12(a2) are your two waves. They exist and are
so measurable that their measurements results in knowing
the value of s11 and s12.

For b1 = s11(a1) + s12(a2) = 0, s11(a1) and s12(a2) must
be of equal magnitude and opposite phase. That's exactly
what happens at a Z0-match.

s11(a1) and s12(a2) *never* encounter an impedance discontinuity.
They are effects of a1 and a2 encountering an impedance discontinuity.
The only thing s11(a1) and s12(a2) encounter are each other and that
interaction completely changes those two waves. The two waves cancel
and their energy components are re-distributed in the opposite
direction. s11(a1) and s12(a2) never encounter an impedance
discontinuity.

It certainly is an interesting way of looking at things, Cecil. It's
certainly true that equal and opposite fields cancel. When that's the
case though it becomes problematic arguing that there are waves there.
Did you ever see the movie "A Brilliant Mind" by any chance?

73, ac6xg

Analyzing Stub Matching with Reflection Coefficients
 

Richard Harrison wrote:
How much energy will re-enter the source? Practically none because there
is no difference of potential between the reflection and the source.

Yes, the reflected wave from 50 ohm coax trying to enter Keith's
50 ohm transmitter will be 100% re-reflected in a lossless stub
in complete violation of his assertions which proves that the
reflected wave doesn't see the source impedance and instead sees
the source load-line impedance.

How much energy will continue to enter the stub? Practically none
because there is no difference of potential between the source and the
reflection.

In a lossless stub, zero energy will continue to enter the stub.
In fact, in a lossless stub, the source can theoretically be
completely disconnected and everything remains the same
including 100% re-reflection of reflected waves by the new
open circuit.

Of course, if the stub is real, the losses will take everything
to zero if the stub is disconnected because the source is
supplying the small amount of power lost to dissipation in
a real-world stub which doesn't have an infinite impedance.
--
73, Cecil, w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Correction: "A Beautiful Mind". Mine is neither beautiful nor
brilliant this morning I'm afraid.

jk

Jim Kelley wrote:

It certainly is an interesting way of looking at things, Cecil. It's
certainly true that equal and opposite fields cancel. When that's the
case though it becomes problematic arguing that there are waves there.
Did you ever see the movie "A Brilliant Mind" by any chance?

73, ac6xg

Analyzing Stub Matching with Reflection Coefficients
 

On Thu, 19 Apr 2007 14:56:33 GMT, Gene Fuller
wrote:

Do you have any connection with "optical physicists" beyond your reading
of Hecht and the Melles Griot website?

Hi Gene,

Aside from the number of us that have experience in the practice?

73's
Richard Clark, KB7QHC