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