Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:

Cecil Moore wrote:
Yes, and HF also observes that, e.g.:

|s11(a1)|^2 = 1 joule/sec

|s12(a2)|^2 = 1 joule/sec

I don't think they do.

Take a look at pages 16 & 17. Quoting:

"Another advantage of s-parameters springs from
the simple relationship between the variables
a1, a2, b1, and b2, and VARIOUS POWER WAVES."

Definitions of |a1|^2, |a2|^2, |b1|^2, & |b2|^2
follow.

"The previous four equations show that s-parameters
are simply related to power gain and mismatch loss,
quantities which are often of more interest than
the corresponding voltage functions."

Definitions of |s11|^2, |s22|^2, |s21|^2, & |s12|^2
follow.

On page 10 it says: "The s-parameters s11 and s22
are the same as optical reflection coefficients;
s12 and s21 are the same as optical transmission
coefficients."

In fact, the intensity equation from optics can
be had from squaring both sides of:

b1 = s11(a1) + s12(a2)
--
73, Cecil, http://www.qsl.net/w5dxp

Analyzing Stub Matching with Reflection Coefficients
 

Dr. Honeydew wrote:
So you don't believe in any of the S-parameter analysis done on linear
small-signal amplifiers, then.

They certainly work for systems designed to be linear
as are *linear* small-signal amplifiers, sources with
circulators, and padded sources as I have said over
and over.

Amateur transmitters are not designed to be linear
and they are NOT linear. Adding a ten cent resistor
to them is not going to make them linear.
--
73, Cecil, http://www.qsl.net/w5dxp

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 23, 1:18 pm, Cecil Moore wrote:
Keith Dysart wrote:
Were one to solve the differential equation for power
on the open circuited line at a location of a voltage
null, the answer would be zero. And all computed without
the need for forward and backward waves.

The *NET* power at any point on a lossless stub is
zero so that is no big deal. The standing wave
current is always 90 degrees out of phase with
the standing wave current so cos(90) = 0. At a
voltage node, all of the energy has simply moved
into the magnetic field.

You really don't WANT to think outside of the box that
you have built for yourself, do you. Anyway, reread
what I wrote and work out why your response is a
non-sequitor.

Calculate the number of electromagnetic joules in
the line and get back to us on how they are reflected
from your above purely virtual impedance at a
voltage "null" and how they can even exist without
a velocity of c(VF).

You really are stuck in a box. Consider a capacitor.
It has no difficulty storing energy without velocity.
And lest you say, "but that is not a transmission
line", consider an open circuited transmission line
excited with a step function. After settling: constant
voltage, no current, energy stored in capacitance,
and the DC coupled directional wattmeter will
tell you there are the same number of 'Bird watts'
going east as there are 'Bird watts' going west

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Jim Kelley wrote:


Cecil Moore wrote:

Yes, and HF also observes that, e.g.:

|s11(a1)|^2 = 1 joule/sec

|s12(a2)|^2 = 1 joule/sec


I don't think they do.


Take a look at pages 16 & 17. Quoting:

"Another advantage of s-parameters springs from
the simple relationship between the variables
a1, a2, b1, and b2, and VARIOUS POWER WAVES."

Definitions of |a1|^2, |a2|^2, |b1|^2, & |b2|^2
follow.

"The previous four equations show that s-parameters
are simply related to power gain and mismatch loss,
quantities which are often of more interest than
the corresponding voltage functions."

Definitions of |s11|^2, |s22|^2, |s21|^2, & |s12|^2
follow.

On page 10 it says: "The s-parameters s11 and s22
are the same as optical reflection coefficients;
s12 and s21 are the same as optical transmission
coefficients."

In fact, the intensity equation from optics can
be had from squaring both sides of:

b1 = s11(a1) + s12(a2)

But I still don't see where HP said that either s11(a1) or s12(a2) is
equal to the square root of 1 joule/sec. Which pretty much confirms
what I said. I think you might have made that part up to try to sound
authoratative.

But I will concede that if you had been right, it may have lent
support to your contention that waves and energy are moving in the
direction of b1 while not moving in the direction of b1. :-)

73, Jim AC6XG

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
You really are stuck in a box. Consider a capacitor.
It has no difficulty storing energy without velocity.

Uhhhh Keith, a capacitor cannot store an EM wave which
is photonic energy. If a capacitor stores the energy in
an EM wave, it must first convert the EM wave energy to
some other form of energy, e.g. electronic energy. That
is certainly possible, but the energy ceases to be EM
energy and cannot be recovered as the same EM energy
coherent with that which was stored. If thinking outside
the box requires a complete denial of reality and the
laws of physics, then I think I will pass.

To summarize, converting EM energy to electronic energy
causes the EM energy to lose coherence. It would take
heroic measures to recover that coherency and those
measures simply do not exist inside a transmission line.

It certainly seems that you guys don't understand the
difference between photonic energy and electronic
energy and you treat them as if they were interchangeable
with no conversion process required. Hint: When energy
is converted from one form to another, the previous
form disappears and is generally not recoverable in
a form coherent with the original. Yet, all the energy
inside a transmission line remains coherent. You have
generated a logical and technical contradiction.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Cecil Moore wrote:
In fact, the intensity equation from optics can
be had from squaring both sides of:

b1 = s11(a1) + s12(a2)

But I still don't see where HP said that either s11(a1) or s12(a2) is
equal to the square root of 1 joule/sec.

Good grief, Jim, are you too lazy to perform the math???

Squaring both sides of the above equation to obtain
joules/sec:

b1^2 = |s11(a1)|^2 + 2*s11(a1)*s12(a2) + |s12(a2)|^2

Does that center term look familiar yet? It should since
it is the interference term in the intensity (power density)
equations from "Optics" by Hecht and from "Principles of
Optics", by Born and Wolf.

Let |s11(a2)|^2 = P1 Then s11(a1) = SQRT(P1)

Let |s12(a2)|^2 = P2 Then s12(a2) = SQRT(P2)

Is your mental light bulb coming on yet? Assuming the phase
angle between a1 and a2 is zero, we can directly write the
equation for total constructive interference.

Ptot = P1 + P2 + 2*SQRT(P1*P2)
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 23, 6:07 pm, Cecil Moore wrote:
Uhhhh Keith, a capacitor cannot store an EM wave which
is photonic energy.

Could you clarify; At what frequency does the signal in the
transmission line become "EM" energy? A step function has
high frequency components, and yet the end result is DC.
Did it stop being "EM" energy at some point? Which point?

Or how about a square wave at 0.00001 Hertz? "EM" energy?
Or not? Harmonics are only limited by the rise time. But my
meter will have trouble knowing its not DC.

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

On 23 Apr 2007 15:30:36 -0700, Keith Dysart wrote:

On Apr 23, 6:07 pm, Cecil Moore wrote:
Uhhhh Keith, a capacitor cannot store an EM wave which
is photonic energy.

Could you clarify; At what frequency does the signal in the
transmission line become "EM" energy?

A Question for the Ages!

(another, less profound question: "When did caps stop working in tank
circuits?")

73's
Richard Clark, KB7QHC

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
Could you clarify; At what frequency does the signal in the
transmission line become "EM" energy?

This is not the proper forum to try to teach you the difference
between a photonic wave and a DC electronic circuit. Please feel
free to return when you have educated yourself of the considerable
technical information available on the subject.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Richard Clark wrote:
(another, less profound question: "When did caps stop working in tank
circuits?")

When the tank gets hit by a Scud missile?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
Could you clarify; At what frequency does the signal in the
transmission line become "EM" energy?

Additional Hint: Try storing the energy in a light wave
in a capacitor.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

On Mon, 23 Apr 2007 18:03:20 -0500, Cecil Moore
wrote:
(another, less profound question: "When did caps stop working in tank
circuits?")
When the tank gets hit by a Scud missile?

Ah yes! The only thing we have here is entertainment value.
Unfortunately this is how vaudeville died.

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:

Good grief, Jim, are you too lazy to perform the math???

Squaring both sides of the above equation to obtain
joules/sec:

Voltage squared does not EQUAL joules/sec (still).

b1^2 = |s11(a1)|^2 + 2*s11(a1)*s12(a2) + |s12(a2)|^2

and from an earlier post:

HP would simply observe that what b1 = 0 means is that there is no
energy and there are no waves moving in the direction of b1.

Yes, and HF also observes that, e.g.:

|s11(a1)|^2 = 1 joule/sec

|s12(a2)|^2 = 1 joule/sec

From which it follows that 0 = 4.

You might want to consider performing the math yourself before you
post it.

73, Jim AC6XG

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:

Keith Dysart wrote:

Could you clarify; At what frequency does the signal in the
transmission line become "EM" energy?


Additional Hint: Try storing the energy in a light wave
in a capacitor.

Ever hear of a CCD camera?

73, ac6xg

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Squaring both sides of the above equation to obtain
joules/sec:

Voltage squared does not EQUAL joules/sec (still).

That shows exactly how much you understand the s-parameter
analysis. The s-parameter voltages have been normalized such
that the square of the normalized voltage is indeed equal to
joules/sec. If you had paid attention, or were half as smart
as you think you are, or even read HP Ap Note 95-1, you would
have realized that fact a long time ago and not be showing
your ignorance here and now.

From which it follows that 0 = 4.

Obviously either an ignorant response or an obfuscation.
It follows that the result of destructive interference is
zero while the result of constructive interference is 4,
just as Born and Wolf asserted in "Principles of Optics",
Chapter VII.

Imin = I1 + I2 +2*SQRT(I1*I2) = 0 equation (16b)

Imax = I1 + I2 + 2*SQRT(I1*I2) equation (16a)

If I1=I2, then Imax = 4*I1 equation (17)

There's your 0 and 4, Jim, straight from Born and Wolf.
Now you can assert that Born and Wolf have been discredited
just as you asserted that Hecht has been discredited.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:

Cecil Moore wrote:
Additional Hint: Try storing the energy in a light wave
in a capacitor.

Tsk, tsk! You unethically trimmed the context, Jim. Try
storing the energy in a light wave in a capacitor and then
recovering that same energy coherent with the original wave.

Ever hear of a CCD camera?

I've never heard of a CCD camera that stored EM waves.
Exactly how long does a CCD camera store EM waves?
Does it store them in a delay line or what?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:

Cecil Moore wrote:
Squaring both sides of the above equation to obtain
joules/sec:

Voltage squared does not EQUAL joules/sec (still).

What dimensions do you get when you perform the
following operation using the s-parameter dimensions
of a1, a2, b1, and b2?

[volts/SQRT(Z0)]^2 = _____________

Apparently, that is a really tough one for you since
you have failed to comprehend it for weeks now.

[volts/SQRT(Z0)]^2 = volts^2/Z0

Maybe if I simplified it to V^2/R, you would recognize
it as joules/sec?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Lux wrote:
Sources are linear.

Can we agree that class-B, class-C, class-D,
and class-E sources are not linear?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Imin = I1 + I2 +2*SQRT(I1*I2) = 0 equation (16b)

Typo there - should be:

Imin = I1 + I2 - 2*SQRT(I1*I2) = 0 equation (16b)
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 23, 6:23 pm, Cecil Moore wrote:
Jim Kelley wrote:

Cecil Moore wrote:
Additional Hint: Try storing the energy in a light wave
in a capacitor.

Tsk, tsk! You unethically trimmed the context, Jim.

Hi Cecil -

I included your entire post - Keith's remarks and yours. Please
refrain from making ad hominem remarks. Particularly baseless ones.

Ever hear of a CCD camera?

I've never heard of a CCD camera that stored EM waves.

The context of your remark, which you have now deleted, was "energy
from EM waves". The little capacitors in a charge coupled device store
the light energy until it is read-out and converted to digital.

73, Jim AC6XG

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 23, 6:40 pm, Cecil Moore wrote:

Maybe if I simplified it to V^2/R, you would recognize
it as joules/sec?
--
73, Cecil http://www.w5dxp.com

What I said was, V^2 does not have units of Joules per second.
There's nothing to argue about here, Cecil.

ac6xg

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
The context of your remark, which you have now deleted, was "energy
from EM waves". The little capacitors in a charge coupled device store
the light energy until it is read-out and converted to digital.

The context was "EM wave energy that remains unchanged
without being converted to some other form of energy".
If that context was not clear to you, I apologize.

When I say "EM wave energy", I am not talking about extracting
energy from EM waves. I am talking about energy continuing
to exist in the form of EM waves while obeying all the
boundary conditions for EM waves.

The energy in RF standing waves is recovered during the
power-down transient period as EM wave energy, not as
a DC charge on a capacitor. That means that the energy
in the standing waves is not standing still since EM
energy must necessarily move at the speed of light, c(VF).
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Cecil Moore wrote:
Maybe if I simplified it to V^2/R, you would recognize
it as joules/sec?

What I said was, V^2 does not have units of Joules per second.
There's nothing to argue about here, Cecil.

I said:
Squaring both sides of the above equation to obtain
joules/sec:

b1^2 = |s11(a1)|^2 + 2*s11(a1)*s12(a2) + |s12(a2)|^2

you responded:
Voltage squared does not EQUAL joules/sec (still).

Nice CYA, Jim, but the context was the normalized
voltages used in the s-parameter analysis. I posted
the s-parameter equation, b1 = s11(a1) + s12(a2) = 0,
and said that squaring it yields joules/sec. Your
statement was in direct response to that statement
of mine.

If you knew that squaring an s-parameter normalized
voltage yields joules/sec and still made that misleading
statement, you were deliberately engaging in false
implications. Why are you more interested in
obfuscation than in the truth?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Additional Hint: Try storing the energy in a light wave
in a capacitor.

On Tue, 24 Apr 2007 09:59:12 -0500, Cecil Moore
wrote:

When I say "EM wave energy", I am not talking about extracting
energy from EM waves.

Hmmm. energy in a light wave is separate and distinct from energy from
EM waves. Must be the subtle distinction between "in" and "from"
waves.

Well, now that Vaudeville is dead, the entertainment value has gone to
hell.

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Jim Kelley wrote:
The context of your remark, which you have now deleted, was "energy
from EM waves". The little capacitors in a charge coupled device store
the light energy until it is read-out and converted to digital.

The context was "EM wave energy that remains unchanged
without being converted to some other form of energy".
If that context was not clear to you, I apologize.

When I say "EM wave energy", I am not talking about extracting
energy from EM waves. I am talking about energy continuing
to exist in the form of EM waves while obeying all the
boundary conditions for EM waves.

The energy in RF standing waves is recovered during the
power-down transient period as EM wave energy, not as
a DC charge on a capacitor. That means that the energy
in the standing waves is not standing still since EM
energy must necessarily move at the speed of light, c(VF).


Cecil,

The generally accepted description for energy density in an
electromagnetic field is:

W = 1/2 (E dot D + B dot H)

All the variables on the right side are vectors. The total energy can be
determined by integrating over the volume of interest, of course. This
relationship holds for either static or time-dependent fields. You can
find this equation in virtually any E&M textbook, but I can provide
specific references if needed.

Do you have a definition for "EM wave energy" that differs from the
standard equation? Are there multiple forms of electromagnetic energy?
Is is necessary to convert "to some other form of energy" when the
entire problem is about electromagnetic fields?

Inquiring minds want to know.

73,
Gene
W4SZ

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Jim Kelley wrote:

Cecil Moore wrote:

Maybe if I simplified it to V^2/R, you would recognize
it as joules/sec?


What I said was, V^2 does not have units of Joules per second.
There's nothing to argue about here, Cecil.


I said:
Squaring both sides of the above equation to obtain
joules/sec:

b1^2 = |s11(a1)|^2 + 2*s11(a1)*s12(a2) + |s12(a2)|^2

you responded:
Voltage squared does not EQUAL joules/sec (still).

Nice CYA, Jim, but the context was the normalized
voltages used in the s-parameter analysis.

CYA?

I posted
the s-parameter equation, b1 = s11(a1) + s12(a2) = 0,
and said that squaring it yields joules/sec. Your
statement was in direct response to that statement
of mine.

Yes. The statement was intended to point out to you that squaring an
equation expressed in units of volts does not convert the expression
to power in units of Joules per second.

If you knew that squaring an s-parameter normalized
voltage yields joules/sec and still made that misleading
statement, you were deliberately engaging in false
implications. Why are you more interested in
obfuscation than in the truth?

You can't possibly believe that the impedances in the problem we're
discussing are all 50 ohms.

ac6xg

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:

Cecil Moore wrote:

Imin = I1 + I2 +2*SQRT(I1*I2) = 0 equation (16b)


Typo there - should be:

Imin = I1 + I2 - 2*SQRT(I1*I2) = 0 equation (16b)

So, was it also a typo when you squared s11(a1) + s12(a2) = 0 and
claimed that s11(a1)^2 = s12(a2)^2 = 1 Joule/sec? :-)

ac6xg

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:

If that context was not clear to you, I apologize.

The context, and your misunderstanding, were both perfectly clear to me.

The apology should be for the unfounded character assaults you
routinely administer to people here on the newsgroup who attempt to
engage you in reasoned discourse.

73, Jim AC6XG

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
Is is necessary to convert "to some other form of energy" when the
entire problem is about electromagnetic fields?

Yes, it is necessary to convert to some other form
of energy if the energy is not moving. EM wave energy
cannot stop moving.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
The statement was intended to point out to you that squaring an
equation expressed in units of volts does not convert the expression to
power in units of Joules per second.

But the equation that I squared was *NOT* expressed
in volts. So what was your agenda in implying a
falsehood?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
So, was it also a typo when you squared s11(a1) + s12(a2) = 0 and
claimed that s11(a1)^2 = s12(a2)^2 = 1 Joule/sec? :-)

Not at all. s11 and s12 are dimensionless. The units of
a1 and a2 are volts/SQRT(Z0). What do you get when you
square volts/SQRT(Z0)?

Hint: [volts/SQRT(Z0)]^2 = V^2/Z0 = joules/sec.

How many times do I have to tell you this before
you start to comprehend it?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
The apology should be for the unfounded character assaults ...

"Unfounded"? You've got to be kidding. You are still
harping and carping at me about me squaring s11(a1)
to get joules/sec. To continue to imply that it is
an invalid thing to do is unethical. When you stop
such behavior, I will stop pointing it out.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Gene Fuller wrote:
Is is necessary to convert "to some other form of energy" when the
entire problem is about electromagnetic fields?

Yes, it is necessary to convert to some other form
of energy if the energy is not moving. EM wave energy
cannot stop moving.


Cecil,

OK, I provided a standard equation that covers all sorts of
electromagnetic energy.

What is this "some other form" of electromagnetic energy? Do you have
any references other than your standard "cannot stop moving"? Do you
have a reference for THAT statement?

73,
Gene
W4SZ

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Jim Kelley wrote:

So, was it also a typo when you squared s11(a1) + s12(a2) = 0 and
claimed that s11(a1)^2 = s12(a2)^2 = 1 Joule/sec? :-)


Not at all. s11 and s12 are dimensionless. The units of
a1 and a2 are volts/SQRT(Z0). What do you get when you
square volts/SQRT(Z0)?

Hint: [volts/SQRT(Z0)]^2 = V^2/Z0 = joules/sec.

How many times do I have to tell you this before
you start to comprehend it?


Here's what you're asking us to "comprehend", Cecil:

0^2 = (s11(a1) + s12(a2))^2

= s11(a1)^2 + 2*s11(a1)*s12(a2) + s12(a2)^2

0 = 4

ac6xg

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
What is this "some other form" of electromagnetic energy?

Transverse electromagnetic *wave* energy is photonic
and moves at the speed of light. If it is not moving
at the speed of light, it is not TEM *wave* energy.
If the energy in RF standing waves is not moving, it
is not photonic TEM wave energy. If the energy in RF
standing waves is not moving, that energy is not photonic
and must have been converted to some other form of energy
besides photonic TEM waves. I'm asking you to explain
what other form that energy in RF standing waves takes
only to be converted back to photonic TEM wave energy
during the transient power-down period.

Do you have
any references other than your standard "cannot stop moving"? Do you
have a reference for THAT statement?

From "Optics", by Hecht: "Photons are stable, chargless,
massless elementary particles that exist only at the
speed c."

The validity of my statement seems obvious. Do you have
a reference that says photonic TEM energy waves can stand
still?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Here's what you're asking us to "comprehend", Cecil:

0^2 = (s11(a1) + s12(a2))^2

= s11(a1)^2 + 2*s11(a1)*s12(a2) + s12(a2)^2

0 = 4

Sorry Jim, you got it wrong (again). s11(a1) and
s12(a2) are 180 degrees out of phase so the |b1|^2
equation is actually:

|b1|^2 = [s11(a1) - s12(a2)]^2

Expanding:

0^2 = [s11(a1)]^2 - 2*s11(a1)*s12*(a2) + [s12(a2)]^2

0 = 1 - 2 + 1 = 0 (but you knew that already)

That's total destructive interference in action just
as Hecht and Born & Wolf explain. Your equation
above is actually for |b2|^2 in the direction of
the load where the constructive interference is
taking place.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:

Jim Kelley wrote:

Here's what you're asking us to "comprehend", Cecil:

0^2 = (s11(a1) + s12(a2))^2

= s11(a1)^2 + 2*s11(a1)*s12(a2) + s12(a2)^2

0 = 4


Sorry Jim, you got it wrong (again). s11(a1) and
s12(a2) are 180 degrees out of phase so the |b1|^2
equation is actually:

|b1|^2 = [s11(a1) - s12(a2)]^2

But everyone knows that's not what you wrote, Cecil. I copied the
equations right from your posts.

Expanding:

0^2 = [s11(a1)]^2 - 2*s11(a1)*s12*(a2) + [s12(a2)]^2

0 = 1 - 2 + 1 = 0 (but you knew that already)

I did already know that. And apparently now we both do.

Though, you can't just arbitrarily change the sign in an equation in
order to make the answer come out right.

73, Jim AC6XG

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Though, you can't just arbitrarily change the sign in an equation in
order to make the answer come out right.

Please stop your unethical practice of trying to imply
that something is wrong when you know it is not wrong.

For b1 = 0, s11(a1) and s12(a2) *must* be 180 degrees
out of phase.

cos(180) IS NOT AN ARBITRARY CHANGE!

cos(180) = -1 (but you knew that already)
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Gene Fuller wrote:
What is this "some other form" of electromagnetic energy?

Transverse electromagnetic *wave* energy is photonic
and moves at the speed of light. If it is not moving
at the speed of light, it is not TEM *wave* energy.
If the energy in RF standing waves is not moving, it
is not photonic TEM wave energy. If the energy in RF
standing waves is not moving, that energy is not photonic
and must have been converted to some other form of energy
besides photonic TEM waves. I'm asking you to explain
what other form that energy in RF standing waves takes
only to be converted back to photonic TEM wave energy
during the transient power-down period.

Do you have any references other than your standard "cannot stop
moving"? Do you have a reference for THAT statement?

From "Optics", by Hecht: "Photons are stable, chargless,
massless elementary particles that exist only at the
speed c."

The validity of my statement seems obvious. Do you have
a reference that says photonic TEM energy waves can stand
still?


Cecil,

Once again you completely avoided the straightforward question. I gave
you a universal equation for electromagnetic energy. You come back with
all sorts of waffling about "photonic TEM energy waves", but with zero
substance.

Forget about standing still or moving; do you have any reference that
mentions "photonic TEM energy waves"? If so, what behavior is attributed
to such beasts? What is the equation that describes the energy in
"photonic TEM energy waves". Is there a conversion factor to convert
into plain ol' ordinary electromagnetic energy?

Do you simply make this stuff up as you go along, or do you have notes
from years of fantasy?

73,
Gene
W4SZ

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
But everyone knows that's not what you wrote, Cecil. I copied the
equations right from your posts.

The equations were phasor equations, Jim - no negative
sign necessary at all. (But you already knew that)

Please stop the obfuscations. You know full well that
cos(180) can be replaced by -1 and that subtraction signs
are never necessary for phasor subtraction. There is no
problem using a plus sign in the following equation.

b1 = s11(a1) + s12(a2) = 0

It just means that s11(a1) and s12(a2) are 180 degrees
out of phase with each other. (But you already knew that)
--
73, Cecil http://www.w5dxp.com