Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Jim Kelley wrote:

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

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)

Joules per second doesn't have phase, Cecil. Your claim was the value
was 1 joule/sec. And the equation is stated pretty clearly in AN-95-1
as the sum of the terms. It doesn't have a minus sign, and it there
isn't a cosine term in it. a1 and a2 are complex voltages, and their
relative phase take care of itself - unless of course you square the
equation. That's how you end up with nonsense like 0=4.

I'm done with this.

AC6XG

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
Once again you completely avoided the straightforward question.

No, I answered in a straightforward manner. You just didn't
like the answer.

RF energy in a transmission line is photonic. Photons
always move at the speed of light. Therefore, photonic
energy cannot stand still. QED in more ways than one.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Joules per second doesn't have phase, Cecil. Your claim was the value
was 1 joule/sec. And the equation is stated pretty clearly in AN-95-1
as the sum of the terms. It doesn't have a minus sign, and it there
isn't a cosine term in it. a1 and a2 are complex voltages, and their
relative phase take care of itself - unless of course you square the
equation. That's how you end up with nonsense like 0=4.

As you know, I have repeated all this to you before so
you cannot possibly still be confused. Why you continue
your obfuscation of the truth is really strange.

b1 = s11(a1@0deg) + s12(a2@180deg) = 0

1. When we square both sides of the equation, the
product of 2*s11(a1@0deg)*s12(a2@180deg) is at 180
degrees so the sign of the term is negative. (but you
already knew that)

2. Joules per second doesn't have phase. I have said that
over and over. The phase term in the power density
equation is the relative phase between the two associated
voltages. (but you already knew that)

3. The equation stated in AN-95-1 is a normalized voltage
equation which is the sum of phasors - no negative sign
required for phasor addition. (but you already knew that)

4. Phasor symbols don't have minus signs and the real value
of a phasor is understood to include a cosine term even
though it is omitted by convention. (but you already knew that)

5. The general case of the squared equation is the same
as the intensity (power density) equations from Hecht
and Born & Wolf. It doesn't contain a minus sign.

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)

A is the angle between the two associated normalized
phasor voltages, a1 and a2, and its cosine is sometimes
negative. (but you already knew that)

I did not end up with nonsense like 0=4. You ended up
with nonsense like 0=4 by engaging in deliberate
irrational obfuscation. Why you feel the need to
act that way is really strange.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Gene Fuller wrote:
Once again you completely avoided the straightforward question.

No, I answered in a straightforward manner. You just didn't
like the answer.

RF energy in a transmission line is photonic. Photons
always move at the speed of light. Therefore, photonic
energy cannot stand still. QED in more ways than one.

Cecil,

The question was about your claim that there are different forms of
electromagnetic energy. Now you are trying to switch the topic to some
babble about photons and the speed of light.

I guess you cannot answer the energy question. This is not surprising,
since there is no valid answer that brings in some additional form of
electromagnetic energy.

73,
Gene
W4SZ

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
b1 = s11(a1@0deg) + s12(a2@180deg) = 0

Actually, looking at it again, for b1 to be 0, the
following would have to be mathematically true.

b1 = |s11(a1)| - |s12(a2)| = 0

Now square both sides of the equation and see what
you get.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
The question was about your claim that there are different forms of
electromagnetic energy.

The electromagnetic force is one of the four fundamental
forces of nature. It can certainly exist in forms not
associated with photons. Excess electrons stored in a
capacitor is an example of electromagnetic energy devoid
of photons.

It is generally accepted that RF TEM waves are photonic in
nature being the result of accelerating and decelerating
electrons in a transmission line. If you agree with that
notion, then you must also admit that RF TEM wave energy
cannot stand still and the appearance of such in a
standing wave is just an illusion. Otherwise, you need
to disprove the notion that RF TEM waves are photonic.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

"Cecil Moore" wrote in message
et...
Jim Kelley wrote:
Joules per second doesn't have phase, Cecil. Your claim was the value was
1 joule/sec. And the equation is stated pretty clearly in AN-95-1 as the
sum of the terms. It doesn't have a minus sign, and it there isn't a
cosine term in it. a1 and a2 are complex voltages, and their relative
phase take care of itself - unless of course you square the equation.
That's how you end up with nonsense like 0=4.

As you know, I have repeated all this to you before so
you cannot possibly still be confused. Why you continue
your obfuscation of the truth is really strange.

b1 = s11(a1@0deg) + s12(a2@180deg) = 0

1. When we square both sides of the equation, the
product of 2*s11(a1@0deg)*s12(a2@180deg) is at 180
degrees so the sign of the term is negative. (but you
already knew that)

2. Joules per second doesn't have phase. I have said that
over and over. The phase term in the power density
equation is the relative phase between the two associated
voltages. (but you already knew that)

3. The equation stated in AN-95-1 is a normalized voltage
equation which is the sum of phasors - no negative sign
required for phasor addition. (but you already knew that)

4. Phasor symbols don't have minus signs and the real value
of a phasor is understood to include a cosine term even
though it is omitted by convention. (but you already knew that)

5. The general case of the squared equation is the same
as the intensity (power density) equations from Hecht
and Born & Wolf. It doesn't contain a minus sign.

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)

A is the angle between the two associated normalized
phasor voltages, a1 and a2, and its cosine is sometimes
negative. (but you already knew that)

I did not end up with nonsense like 0=4. You ended up
with nonsense like 0=4 by engaging in deliberate
irrational obfuscation. Why you feel the need to
act that way is really strange.
--
73, Cecil http://www.w5dxp.com

Isnt it true the Cos of the phase term will always be equal to -1 or plus
1, -1 for any value of SWR except 1:1 and 1 for a 1:1 SWR value.
I hope this dosnt seem stupid tommorow as I am somewhat sedated and it is
late.

Jimmie

Analyzing Stub Matching with Reflection Coefficients
 

Jimmie D wrote:
Isnt it true the Cos of the phase term will always be equal to -1 or plus
1, -1 for any value of SWR except 1:1 and 1 for a 1:1 SWR value.

The context of the present discussion is a Z0-match point
at which reflections toward the source are eliminated.
For the case of a Z0-match, the cosine of the phase term
will indeed always be +1 or -1 as the fields are either
in phase or 180 degrees out of phase.

If the SWR is 1:1, there is no reflected voltage and
therefore a2 = 0 and there is no assignable phase angle
between the voltages.

In the general case, where reflections are allowed to
reach the source, the cosine value can be anything
between +1 and -1 as the angle between a1 and a2 can
have any value between zero degrees and 360 degrees.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Gene Fuller wrote:
The question was about your claim that there are different forms of
electromagnetic energy.

The electromagnetic force is one of the four fundamental
forces of nature. It can certainly exist in forms not
associated with photons. Excess electrons stored in a
capacitor is an example of electromagnetic energy devoid
of photons.

It is generally accepted that RF TEM waves are photonic in
nature being the result of accelerating and decelerating
electrons in a transmission line. If you agree with that
notion, then you must also admit that RF TEM wave energy
cannot stand still and the appearance of such in a
standing wave is just an illusion. Otherwise, you need
to disprove the notion that RF TEM waves are photonic.


Cecil,

Keep going. You are getting really close to winning the RRAA KONS award.

(KONS = King Of Non-Sequiturs)

All you need to add to make the top ranking is a quote from Terman.

8-)

73,
Gene
W4SZ

Analyzing Stub Matching with Reflection Coefficients
 

Gene Fuller wrote:
Keep going. You are getting really close to winning the RRAA KONS award.

Apparently, you cannot afford to disagree with my posting.
Now consider the boundary conditions for a photonic wave
and you will probably agree with me that the energy in
a photonic wave cannot be stored in a capacitor without
an energy transformation from photonic energy to something
else. In short, photonic waves that appear to be standing
still are an illusion. Photonic waves cannot stand still.
--
73, Cecil http://www.w5dxp.com