On May 16, 6:03*am, Wimpie wrote:
I am not ignoring a problem (as you suggested), I am just using the
right tool to solve a problem.

I'm sorry, Wim, that is just not true. When you convert a V/I ratio to
a lumped circuit impedance, you are switching models in mid-example
and it changes everything in one direction (while keeping conditions
the same in the other direction). Switching to a model that doesn't
recognize reflections at all when a question about reflections arises
is an obvious logical diversion. The lumped-circuit model does not
recognize reflected energy and therefore does not allow the tracking
of reflected energy. The distributed network/wave reflection model
does allow for the tracking of reflected energy and was developed
because of the limitations of the lumped circuit model.

I will buy your assertion that one can use voltage and current to
achieve the same thing if one is careful not to violate the known laws
of EM wave physics. In my example, the reflected power on the 100 ohm
line is 12.5 watts. The reflected voltage is 35.35 volts. The
reflected current is 0.3535. The reflected voltage and reflected
current are 180 degrees out of phase so the power is real, i.e.
cos(180)=1.0, and the reflected wave has a Poynting vector magnitude
of 12.5 watts (per coax cross-sectional area).

In my very simple example, there is no reflected power on the 50 ohm
feedline yet there is (35.35)(0.3535)=12.5 watts of reflected power
from the load incident upon the 100/50 ohm impedance discontinuity. I
ask you again: Exactly what phenomenon of EM wave physics causes the
reflected wave to *reverse* its momentum and direction of energy flow
when the magnitude of the reflection coefficient is 0.3333? As long as
you refuse to answer this simple question about such a simple example,
this discussion will go nowhere.

Cecil, I think you have sufficient knowledge to form an opinion
without hiding behind others. You also have the equipment to figure
out some things yourself, and I gave some hints to help you. *The only
question is, are you willing to do this?

If you cannot answer the simple question about what happens to the
reflected energy at a simple passive impedance discontinuity, I am not
about to trust your assertions about what happens inside an active
source. Trying to introduce a more complicated example while refusing
to deal with the very simple example that I provided is an obvious and
typical logical diversion. Again, I am not going to cooperate in your
attempts at diversions. If you don't know why reflected energy
reverses momentum and direction at a passive Z0-match, just say so.

Here's an easier example: Two EM waves superpose in a Z0=100 ohm
environment. Each wave is 100 volts at 1 amp = 100 watts. The phase
angle on wave1 is +60 degrees and the phase angle on wave2 is -60
degrees. The superposition results in a new wave of 100 volts at 1 amp
= 100 watts with a phase angle of zero degrees.

We superposed two 100 watt waves and the result was one 100 watt wave.
What happened to the other 100 watts?
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On 16 mayo, 16:15, Cecil Moore wrote:
On May 16, 6:03*am, Wimpie wrote:

I am not ignoring a problem (as you suggested), I am just using the
right tool to solve a problem.

I'm sorry, Wim, that is just not true. When you convert a V/I ratio to
a lumped circuit impedance, you are switching models in mid-example
and it changes everything in one direction (while keeping conditions
the same in the other direction). Switching to a model that doesn't
recognize reflections at all when a question about reflections arises
is an obvious logical diversion. The lumped-circuit model does not
recognize reflected energy and therefore does not allow the tracking
of reflected energy. The distributed network/wave reflection model
does allow for the tracking of reflected energy and was developed
because of the limitations of the lumped circuit model.

I will buy your assertion that one can use voltage and current to
achieve the same thing if one is careful not to violate the known laws
of EM wave physics. In my example, the reflected power on the 100 ohm
line is 12.5 watts. The reflected voltage is 35.35 volts. The
reflected current is 0.3535. The reflected voltage and reflected
current are 180 degrees out of phase so the power is real, i.e.
cos(180)=1.0, and the reflected wave has a Poynting vector magnitude
of 12.5 watts (per coax cross-sectional area).

In my very simple example, there is no reflected power on the 50 ohm
feedline yet there is (35.35)(0.3535)=12.5 watts of reflected power
from the load incident upon the 100/50 ohm impedance discontinuity. I
ask you again: Exactly what phenomenon of EM wave physics causes the
reflected wave to *reverse* its momentum and direction of energy flow
when the magnitude of the reflection coefficient is 0.3333? As long as
you refuse to answer this simple question about such a simple example,
this discussion will go nowhere.

Cecil, I think you have sufficient knowledge to form an opinion
without hiding behind others. You also have the equipment to figure
out some things yourself, and I gave some hints to help you. *The only
question is, are you willing to do this?

If you cannot answer the simple question about what happens to the
reflected energy at a simple passive impedance discontinuity, I am not
about to trust your assertions about what happens inside an active
source. Trying to introduce a more complicated example while refusing
to deal with the very simple example that I provided is an obvious and
typical logical diversion. Again, I am not going to cooperate in your
attempts at diversions. If you don't know why reflected energy
reverses momentum and direction at a passive Z0-match, just say so.

Hello Cecil,

I answered a simple question requested several times by Walt (I had to
reply to it!). It wasn't my statement but I did it. The solution I
gave involved both lumped circuit theory (to calculate the net power)
and transmission line theory (to calculate forward and reflected power
in a 50 Ohms environment).

Even other people had to help Walt to understand a voltage divider
(the 212.1*50 issue). Maybe you can comment whether my simple
solution (not involving momentum, Poynting vector or optics) is
correct or not. I am familiar with the use (and mis-use by others)
of the Poynting vector, but I don't discharge 10 kJ through a mosquito
when using a newspaper does the job also.

Regarding reflections:
Does a PA see difference between:

1. 100 Ohms lumped circuit load
2. RC = +0.33333 (for 50 Ohms reference)
3. VSWR = 3 (voltage minimum, for a 300 Ohms reference)

The answer is no, all can be converted to 100 Ohms lumped circuit.
What is in between the PA and the actual load is not relevant, what
matters (for the PA) is what it sees at its SO239 socket.



Wim
PA3DJS
www.tetech.nl

Hello Cecil,

Here's an easier example: Two EM waves superpose in a Z0=100 ohm
environment. Each wave is 100 volts at 1 amp = 100 watts. The phase
angle on wave1 is +60 degrees and the phase angle on wave2 is -60
degrees. The superposition results in a new wave of 100 volts at 1 amp
= 100 watts with a phase angle of zero degrees.

We superposed two 100 watt waves and the result was one 100 watt wave.
What happened to the other 100 watts?

If you want this question answered, please open a new thread as it is
not relevant to the original question.

Maybe people will ask you a circuit diagram showing the sources and
the combiner circuitry to enable calculation of the net power
delivered by each source. Otherwise people may consider your problem
as a single incident wave problem (as for these type of steady state
signals you first add complex amplitudes, then calculate powers).

I did respond to Walt's request because it is on topic and I stated
that such thing can happen (without given a numerical example).


Wim
PA3DJS
www.tetech.nl

On May 16, 11:19*am, Wimpie wrote:

The solution I gave involved both lumped circuit theory (to calculate
the net power) and transmission line theory (to calculate forward
and reflected power in a 50 Ohms environment).

Lumped circuit theory presupposes that waves do not exist and that RF
energy travels instantaneously, faster than the speed of light. You
seem to be ignoring the numerous laws of physics violated by the
lumped circuit theory. You also seem to be ignoring the fact that when
lumped circuit theory yields different results than the distributed
network theory, distributed network theory always wins because it is
closer to Maxwell's equations.

I gave an earlier CLC Pi-Network Tuner example that proved the lumped
circuit model fails when reflections are present. EZNEC results are
nothing alike when using the lumped inductance option vs the helical
wire option for an inductance in a standing wave antenna.

If you want this question answered, please open a new thread as it is
not relevant to the original question.

Asserting that it is not relevant for the purpose of diversion will
not make it go away. If one cannot understand, explain, and solve the
simplest passive interference problem, how is one ever going to
understand, explain, and solve the multiple levels of interference
possible within an active source being invaded by reflected energy? If
one cannot add one plus one, one is not likely to be able to add two
plus two and something akin to that is what I am seeing here. Until
one understands exactly how a Z0-match reverses the direction and
momentum of a reflected wave, one will not understand what is
happening inside a source with incident reflected energy. What is
actually happening in reality is revealed when one sticks with the
distributed network/wave reflection theory throughout the analysis.
People who whine that such is too difficult have to be satisfied with
a certain level of ignorance and inaccuracy.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Cecil,

It seems that you on purpose remove/ignore things that you don't like,
but (you know) are true.

A CLC pi filter doesn't know the difference between:

1. 100 Ohms lumped circuit load
2. RC = +0.33333 (for 50 Ohms reference)
3. VSWR = 3 (voltage minimum, for a 300 Ohms reference)

It seems you don't want to notice that.

That it is convenient to use transmission line theory to calculate the
load as seen by a PA when transmission line sections are involved, is
OK, I didn't deny that.

That lumped circuit theory has limitations is fully understood.
Frequently transmission line effects are modelled using parasitic L
and C additions yielding accurate models valid up to GHz frequencies
(depending on the size of the component). We are below 30 MHz (for
this topic).

Here the experience of the Engineer comes into play: when you can use
a lumped circuit model and when you need to use transmission line
models (the particle/wave issue is similar)?

A helical inductor of an antenna no longer small w.r.t. wavelength may
be better modelled with transmission line theory, but that is OT.
Even the L of the CLC filter, you can model with a lumped circuit
equivalent with more than sufficient accuracy. This is daily business
for manufacturers of inductive components.

Generally, converting results from transmission line models to
impedance in combination with lumped circuit theory to calculate the
load as seen by the active device, is daily practice. Especially here,
as we are dealing with narrow band signals and don't have to model the
behavior for harmonics. But for some reason you don't want to see
that, and you elevate transmission line theory to a goal.

So again, once you did the conversion to Z, you no longer have to
worry about transmission line issues in the load or cabling (including
reflection coefficient) when treating your PA's CLC pi filter.

Now speed of light becomes important in a CLC pi filter for a HF PA,
when becomes "Gaussian" of importance (and may lose all the readers of
this topic)?

With kind regards,



Wim
PA3DJS
www.tetech.nl

On May 16, 2:58*pm, Wimpie wrote:
A CLC pi filter doesn't know the difference between:

1. 100 Ohms lumped circuit load
2. RC = +0.33333 (for 50 Ohms reference)
3. VSWR = 3 (voltage minimum, for a 300 Ohms reference)

It seems you don't want to notice that.

It is not worth wasting my time to notice since *everyone* already
knows that a CLC pi filter is not alive and doesn't have a brain so it
must necessarily be dumb as a dead stump. You, OTOH, hopefully being
smarter than the average CLC pi filter, should know that the
conditions existing within a resistor are different from the
conditions existing within an antenna with the same feedpoint
impedance. Hint: If you don't know what is in the box, alleviate your
ignorance by looking inside the box. If you put on the blinders and
refuse to look, then you will make errors like you did earlier while
measuring an s11 of zero when it was actually 0.3333.

Even the L of the CLC filter, you can model with a lumped circuit
equivalent with more than sufficient accuracy.

When the task is to determine the exact delay through the inductor,
how the heck can the lumped circuit model tell you that?
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On May 16, 3:29*pm, Cecil Moore wrote:
When the task is to determine the exact delay through the inductor,
how the heck can the lumped circuit model tell you that?

Wim, I forgot to note that using your stated methods, W8JI "measured"
a 3ns delay through a 10" long, 2" diameter, 100 turn, 100uh, 80m
mobile loading coil. Doesn't a 4 MHz RF wave traveling the length of a
large 100uH air-core 80m loading coil in 3 ns give you some pause for
reconsidering your methods? Every wonder why computer manufacturers
don't install 100uh coils in series with their computer bus lines to
speed up their computers? :-)
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On 16 mayo, 22:29, Cecil Moore wrote:
On May 16, 2:58*pm, Wimpie wrote:

A CLC pi filter doesn't know the difference between:

1. 100 Ohms lumped circuit load
2. RC = +0.33333 (for 50 Ohms reference)
3. VSWR = 3 (voltage minimum, for a 300 Ohms reference)

It seems you don't want to notice that.

It is not worth wasting my time to notice since *everyone* already
knows that a CLC pi filter is not alive and doesn't have a brain so it
must necessarily be dumb as a dead stump. You, OTOH, hopefully being
smarter than the average CLC pi filter, should know that the
conditions existing within a resistor are different from the
conditions existing within an antenna with the same feedpoint
impedance. Hint: If you don't know what is in the box, alleviate your
ignorance by looking inside the box. If you put on the blinders and
refuse to look, then you will make errors like you did earlier while
measuring an s11 of zero when it was actually 0.3333.

Even the L of the CLC filter, you can model with a lumped circuit
equivalent with more than sufficient accuracy.

When the task is to determine the exact delay through the inductor,
how the heck can the lumped circuit model tell you that?

Just via the capacitance to ground (for example a CLC model of an
inductor well below the first self resonance frequency). But when
looking to a PA, there is often an additional capacitance left and
right of the inductor that causes the most of the phase shift. I did
some tesla coiling, so I am aware of the various models for single
layer inductors. You are further drifting away from the main subject
(PA output impedance and what mismatch will do).

--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK


Wim
PA3DJS
www.tetech.nl