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

Cecil Moore wrote:
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

You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of
the discussion that you've been pretty closed minded about.

Without a load on the circulator, you must know that no energy can be
flowing to it or through it. No power is being dissipated by it.
What effect should that have on the total energy in the transmission
line, and why?

I hope it's not true about Reg. That would be a huge loss. Perhaps
somebody will archive his website.

73, ac6xg

On Wed, 30 Aug 2006 10:17:44 -0700, Jim Kelley
wrote:

I hope it's not true about Reg. That would be a huge loss. Perhaps
somebody will archive his website.

Hi Jim,

The announcement came from his account, and his computer.

I've archived his site, but for others who wish to do the same, the
best tool for that purpose can be found at:
http://www.httrack.com

This is a website harvesting robot that will replicate an entire
website into the directory of your choice (changing links so that you
can browse it on your system).

73's
Richard Clark, KB7QHC

Jim Kelley wrote:

Cecil Moore wrote:
I honestly don't know what happens when the third port
of the circulator is unterminated

You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of the
discussion that you've been pretty closed minded about.

I know what happens if the circulator is not present. I
just don't don't know what happens with an unterminated
circulator.

Maybe you can help me out here. If the load resistors
are removed from the circulators in the previous experiment,
what is the forward power and reflected power readings on
the one second long lossless transmission line during
steady-state?

Without a load on the circulator, you must know that no energy can be
flowing to it or through it. No power is being dissipated by it. What
effect should that have on the total energy in the transmission line,
and why?

Let's remove the circulators and try to figure out what
happens. Let's assume the signal generator is a Thevenin
equivalent with an internal resistance of 50 ohms and not
equipped with a circulator.

200W SG---one second long 50 ohm lossless feedline---100W SG

Would you agree with me that during the first second, the
feedline is loaded with 300 joules because there have been
no reflections? What happens after that depends upon the
reflection coefficients at the signal generators. But the
$64k question is what happens to the original 300 joules?

I hope it's not true about Reg. That would be a huge loss. Perhaps
somebody will archive his website.

One of the posters to the news
of Reg's death was whom I suspect is
Reg's son (or nephew). Humans are mortal and can die at any time.
My son died when he was 8 months old. I wish he had lived as long
as Reg who certainly had a full, useful, and colorful life.
--
73, Cecil http://www.qsl.net/w5dxp

Jim Kelley wrote:
You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of the
discussion that you've been pretty closed minded about.

Here's a brainteaser for you, Jim. Assume the following example
under steady-state conditions:

XMTR--X--tuner--Y--one second long lossless feedline---291.42 ohms
Ps=100W Pfor=200W-- --100W=Pref PL=100W

The lossless tuner is tuned for a Z0-match so there is zero
reflected power at point 'X'. The Z0 of the feedline is 50
ohms. The voltage reflection coefficient at the load is
0.707 making the power reflection coefficient 0.5

We suddenly disconnect the source at point 'Y'. After the
source is disconnected, how many joules of energy are delivered
to the load? _____

How many seconds does it take to dissipate all the energy in
the feedline? _____

On a second by second basis, how much power is delivered to the
load after the disconnect? _____ _____ _____ _____ _____
--
73, Cecil, http://www.qsl.net/w5dxp