wrote:
So how are you taking into account the stored energy in the ideal
autotuner?

The ideal autotuner is lossless. The delay through the
tuner compared to one second is negligible. The energy
in the tuner components is negligible compared to 300
joules.

The principles are just as easy to comprehend if the
tuner is replaced by a circulator and load.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
wrote:
So how are you taking into account the stored energy in the ideal
autotuner?

The ideal autotuner is lossless. The delay through the
tuner compared to one second is negligible. The energy
in the tuner components is negligible compared to 300
joules.

Since the Z0-match is achieved at the ideal autotuner
input, just consider the lossless autotuner to be part
of the one second lossless transmission line system.
--
73, Cecil http://www.qsl.net/w5dxp

What's the complex Poynting vector in the tuner? Why should I believe
that the delay through the tuner is negligible? Why should I believe
that the stored fields are negligible? It's not obvious to me.

For that matter, why does the ideal autotuner not reciprocal? The
impedance in the presence of the standing waves at the tuner-line
junction is R+jX. That is transformed by the autotuner so the source
sees 50 ohms, but that also means that the output impedance of the
generator is transformed as viewed by a wave coming down the line.

It's not transformed to infinity. If it is a circulator and load, this
is a distinctly nonreciprocal system. If it's a tuner affecting a
match, it is reciprocal. If your definition of an ideal autotuner
includes the nonreciprocity, please send me a schematic of such a
device.

I think I calculated the Poynting vector for this situation, I think
the Poynting vector divided by the group velocity of the waves and
integrated over the area of the coax gives the energy density per unit
length in the steady state, and if you've tuned the line length so that
the imaginary part of the Poynting vector goes to zero then the
explanation you give falls apart.

The tuner matters here. This is the problem with English language
discussions, especially disagreements, about electrical systems.
"There exists an ideal autotuner at the input" is insufficient
specification of the mathematical problem. If you place a device that
does everything you say it does at the source-line junction, maybe you
do get 300J in the one-second line.

Why isn't it reciprocal? Why is its phase shift negligible? Is it
just because it's physically small? I seem to remember a certain hotly
contested paper that I would think would have certainly put that sort
of idea out of your head. I think the crux of the misunderstanding is
in the tuner. I think it has stored energy and nonnegligible phase
delay, else it is non-reciprocal and dissipative (circulator+load).

What *IS* an ideal autotuner? I think you've stuffed too many
conditions into its operation for it to be a physically realizable
device. I'm pretty sure that it breaks the time reversal symmetry of
the system. Send out a pulse into your one second misterminated line
via your ideal autotuner. Wait half a second. FREEZE! Now run the
film backward while obeying the rules of your autotuner. The pulse
doesn't go back into the source. It bounces. What's up?

73,
Dan

wrote:
What's the complex Poynting vector in the tuner? Why should I believe
that the delay through the tuner is negligible? Why should I believe
that the stored fields are negligible? It's not obvious to me.

The delay calculates out to be in nanoseconds compared to
the one second delay in the lossless line. The stored
fields calculate out to be in microjoules compared to
the 300 joules stored in the one second long feedline.

The real question is, allowing for all the things to which
you have objected, why is there very close to the number
of joules in the feedline necessary to support the forward
and reflected waves? What makes you think there are not
enough joules in the line to support the forward and
reflected waves?

For that matter, why does the ideal autotuner not reciprocal? The
impedance in the presence of the standing waves at the tuner-line
junction is R+jX. That is transformed by the autotuner so the source
sees 50 ohms, but that also means that the output impedance of the
generator is transformed as viewed by a wave coming down the line.

It can be proved that a 50 ohm Z0-match, because of destructive
interference on the 50 ohm side, reflects 100% of the reflected
energy back toward the load.

It's not transformed to infinity. If it is a circulator and load, this
is a distinctly nonreciprocal system. If it's a tuner affecting a
match, it is reciprocal. If your definition of an ideal autotuner
includes the nonreciprocity, please send me a schematic of such a
device.

Please read the Worldradio magazine article from my web page.
What you are calling nonreciprocity is merely destructive
interference on the source side of the Z0-match and constructive
interference on the load side of the Z0-match.

I think I calculated the Poynting vector for this situation, I think
the Poynting vector divided by the group velocity of the waves and
integrated over the area of the coax gives the energy density per unit
length in the steady state, and if you've tuned the line length so that
the imaginary part of the Poynting vector goes to zero then the
explanation you give falls apart.

Not if one deals with the forward Poynting vector and reflected
Poynting vector *separately* as does Ramo/Whinnery.

Pz+ = 200 watts of forward power in a matched line.

Pz- = 100 watts of reverse power in a matched line.

Superpose the two coherent waves and count the joules in the
superposed waves.

The tuner matters here. This is the problem with English language
discussions, especially disagreements, about electrical systems.
"There exists an ideal autotuner at the input" is insufficient
specification of the mathematical problem. If you place a device that
does everything you say it does at the source-line junction, maybe you
do get 300J in the one-second line.

Why isn't it reciprocal?

Because a Z0-match allows energy flow in
only the forward direction. That's how antenna tuners work.
Destructive interference keeps reflected energy from flowing
past the Z0-match toward the source. That's also how 1/4WL
thin-film anti-reflective glass works.

Why is its phase shift negligible? Is it
just because it's physically small?

All it has to do is shift the impedance from 300 ohms to 50 ohms.
A few microhenries and a few picofarads will do that. Its phase
shift would be in the nanoseconds.

What *IS* an ideal autotuner?

It's a concept to be thought about and discussed. If one accepts
a one second lossless feedline, why not an ideal autotuner for the
sake of conceptual discussion? Then a real autotuner can be
substituted and the losses, delays, and energy components estimated.

But like I said, the concept that I am presenting doesn't require
the ideal autotuner. It is actually easier to understand using
a circulator and load. There is exactly the amount of joules in
the transmission line needed to support the forward and reflected
energy waves.
--
73, Cecil http://www.qsl.net/w5dxp

wrote:
Send out a pulse into your one second misterminated line
via your ideal autotuner. Wait half a second. FREEZE! Now run the
film backward while obeying the rules of your autotuner. The pulse
doesn't go back into the source. It bounces. What's up?

I certainly didn't say that. The steady-state voltage and
current component values have to exist for 100% destructive
interference to occur at a Z0-match. On the source side of
the Z0-match, in order to eliminate the reflected energy
wave, two reflected wave components are required to be
traveling in the same direction, be equal in magnitudes
and opposite in phase. Short pulses make those conditions
impossible.

In your above pulse example, there is no interference at
the impedance discontinuity and therefore a Z0-match doesn't
even exist. A Z0-match depends upon a steady-state supply
of RF waves.

A Z0-match at the input of a tuner works exactly like the
1/4WL thin-film anti-reflected glass coating. If one sends
a 1/2WL pulse of light at that piece of glass, there will
be reflections. Anti-reflective glass depends upon a
steady-state supply of light waves. One of the cancellation
components is 180 degrees older than the other one.

Let's return to the Signal Generators equipped with Circulators
and Loads (SGCL). Would you agree that the feedline contains
300 joules in the following example?

200W SGCL----one second long lossless feedline----100W SGCL
200W-- --100W
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:

In your above pulse example, there is no interference at
the impedance discontinuity and therefore a Z0-match doesn't
even exist. A Z0-match depends upon a steady-state supply
of RF waves.

A Z0-match at the input of a tuner works exactly like the
1/4WL thin-film anti-reflected glass coating. If one sends
a 1/2WL pulse of light at that piece of glass, there will
be reflections. Anti-reflective glass depends upon a
steady-state supply of light waves. One of the cancellation
components is 180 degrees older than the other one.

You really do need to choose your words a bit more carefully, Cecil.

Let's return to the Signal Generators equipped with Circulators
and Loads (SGCL). Would you agree that the feedline contains
300 joules in the following example?

200W SGCL----one second long lossless feedline----100W SGCL
200W-- --100W

Yes. But how much is there with the circulator load removed? (Not
replaced by an autotuner.) Numbers please.

Why you suppose I keep asking this question, Cecil?

73, ac6xg

Jim Kelley wrote:
You really do need to choose your words a bit more carefully, Cecil.

Your statement is a personal opinion without technical content.

200W SGCL----one second long lossless feedline----100W SGCL
200W-- --100W

Yes. But how much is there with the circulator load removed? (Not
replaced by an autotuner.) Numbers please.

OK, let's remove the circulator loads and replace them
with unknown devices or objects inside black boxes.

Here's the same one second long lossless feedline with
unknown black boxes connected at each end (where BB stands
for black box).

BB#1------one second long lossless feedline------BB#2
200W-- --100W

Given: all conditions on the one second long lossless feedline
are identical in both of the above examples.

The laws of physics tell us that the joules in the two
lines with identical conditions are equal, i.e. 300
joules in these two cases. It doesn't matter what exists
or doesn't exist at the two ends of the line. 200 watts
forward and 100 watts reverse requires 300 joules/sec.

In other words, given a purely resistive Z0 and a properly
calibrated wattmeter, it is impossible to get 200 watts
of forward power and 100 watts of reverse power without
there being 300 joules in the one second long lossless line.
If there wasn't 300 joules in that line, we wouldn't
measure 200 watts forward and 100 watts reverse.
--
73, Cecil, http://www.qsl.net/w5dxp

Jim Kelley wrote:
Why you suppose I keep asking this question, Cecil?

Because you believe in intelligent waves that somehow
sense their ultimate fate and therefore know how much
energy to carry or not?

Wave#1: "I am going to be dissipated in a circulator
load and therefore, I must contain N joules of energy."

Wave#2: "I am not going to be dissipated in a circulator
load and therefore, I must contain zero energy."

Given the following two configurations where BB stands
for Black Box:

BB#1----one second long lossless feedline----BB#2
200W-- --100W

BB#3----one second long lossless feedline----BB#4
200W-- --100W

Given: All voltages, currents, fields, and powers
are identical in both systems.

Please present us with a set of Black Boxes that will
satisfy your assertions and result in different energy
magnitudes in the two lines.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Jim Kelley wrote:

Why do you suppose I keep asking this question, Cecil?


Because you believe in intelligent waves that somehow
sense their ultimate fate and therefore know how much
energy to carry or not?

No. That's something you keep insisting on talking about. Try to
keep your mind focussed on the fact that we're discussing the steady
state. (Or as you might proffer, after the waves have "decided" what
they're going to do.)

How about we just get back to the question. Energy in your 1 sec.
transmission line, only with no load on the circulator (and no
autotuner). How much? Explain. Try it!

73 de ac6xg

Jim Kelley wrote:
How about we just get back to the question. Energy in your 1 sec.
transmission line, only with no load on the circulator (and no
autotuner). How much? Explain. Try it!

I honestly don't know what happens when the third port
of the circulator is unterminated but it is irrelevant
because the steady-state forward and reflected powers
would not be the same. I concede to you the fact that
one can always make an example too complicated to
analyze presumably for the purpose of diverting the
issue. But why would you want to divert the issue?

The real question is: Do two identical feedlines with
identical signal characteristics contain the same
amount of energy as in the following two examples.
Again SGCL is a Signal Generator with a Circulator and
Load resistor. Z0 is 50 ohms and the load is 291.42 ohms.

200W SGCL-----one second long lossless feedline---load
200W-- --100W
lossless
100W---tuner--one second long lossless feedline---load
200W-- --100W

It is easy to prove that the first example contains 300
joules in the feedline after two seconds.

It is easy to specify that the conditions in the feedline
in the two examples are identical during steady-state.

Are you asserting that two identical feedlines with identical
conditions contain a different number of joules?

I have another example in another posting.
--
73, Cecil http://www.qsl.net/w5dxp