Richard Clark wrote:
It is more than evident that your "objections" are inventions of your
own construction.

Perhaps you missed Keith's assertion that he believes in voltage
waves but not in power waves, thus implying that voltage waves are
not associated with (E x H) power.

You did not respond to your refutation of your own source, ...

Chipman's assertion of voltage doesn't imply that he doesn't believe
in current. He just didn't mention current. Exclusive Vs inclusive -
please learn the difference.
--
73, Cecil, W5DXP

On Fri, 19 Sep 2003 09:58:25 -0700, Cecil Moore
wrote:

Richard Clark wrote:
It is more than evident that your "objections" are inventions of your
own construction.

Perhaps you missed Keith's assertion that he believes in voltage
waves but not in power waves, thus implying that voltage waves are
not associated with (E x H) power.

You did not respond to your refutation of your own source, ...

Chipman's assertion of voltage doesn't imply that he doesn't believe
in current. He just didn't mention current.
He does mention current as Postulate 2 - you simply suffer from speed
reading past that passage.
Exclusive Vs inclusive -
please learn the difference.
Please try reading your references rather than using them as
cut-and-paste crutches to wobbly arguments.


Hi Cecil,

The difference is found by interchanging the names Keith and Chipman
(forgive me both) from these originals:
Keith's assertion that he believes in voltage
waves but not in power waves
and
Chipman's assertion of voltage doesn't imply that he doesn't believe
in current.

finds by your own logic that:
Keith's assertion of voltage doesn't imply that he doesn't believe
in current.
which is consistent with what he has published (not what you have
projected).
and
Chipman's assertion that he believes in voltage
waves but not in power waves
is literally stated in his postulates (of which you are admittedly
wholly ignorant).

Thus Chipman and Keith are not in contradiction, that is a product of
your own invention and your violations of Chipman's postulates.

73's
Richard Clark, KB7QHC

On Fri, 19 Sep 2003 13:15:05 -0700, Cecil Moore
wrote:
Thus, it is an absurd argument that Chipman never made.

Well Cecil,

Given you are eminently untutored in Chipman's greater work, and
choose only those portions that suit your cut-and-paste philosophy; it
comes as no surprise what arguments of invention you might project.

73's
Richard Clark, KB7QHC

Keith wrote:
"---it will turn out that there is no value in forward and reverse power
on a line with non-real Zo."

The Bird wattmeter works well enough on coax lines used between the
usual transmitter and antenna. Loss produces reactance in coax. We don`t
seek lossy coax. If it becomes lossy, the line is likely to be replaced.

If a transmitter isn`t matched when the load is matched to the coax, an
improved match between the transmitter and the coax may come from
mismatching the load.

A true conjugate match exists everywhere in the system between the
transmitter and the antenna. That is, one could check the impedances in
both directions at any place in the system and find that the two
inpedances are conjugates of each other.

There is reactance in every non-resonant length of coax which is not
terminated in its characteristic impedance value.

The vector sum of the incident and reflected wave voltages is less than
the arithmetic sum at a point 1/8 wavelength back from the reflection
point. At 1/8 wavelength back from a reflection, the incident and
reflected wave voltage vectors are 90-degrees apart. At 1/8 wavelength
in coax, the reactance is numerically equal to the Zo of the line, as a
piece of trivia.

At 1/4 wavelebgth back from the reflection, the incident and reflected
wave voltage vectors are 180-degrees apart. The line voltage total is
thus the arithmetic difference of the incident and reflected wave
voltages.

At 1/2 wavelength back from the reflection, the incident and reflected
wave phases have continued their phase changes in opposite directions
with distance back from the reflection until the voltages have reached
the in-phase condition. In a lossless line, the conditions at the
reflection point would be repeated at a point 1/2-wavelength back.

When a resistance load matches the Zo of its feedline, the reflection
coefficient is zero, so there is no reflection from the load. The load
absorbs all the incident wave. The effect of a reactive load is merely
to displace the positions of the minima and maxima along the line but
not with respect to each other.

Best regards, Richard Harrison, KB5WZI

Keith wrote:
"Not to mention the difficulties that arise on lines with complex Zo."

Transmission lines are strict enforcers. They only allow a volts to amps
ratio traveling the line of Zo.

Transmission lines are almost always selected and used for low-loss. The
Zo`s which are real (non-reactive) are not complex. The series
resistance is small as compared with te series inductive reactance. The
shunt conductance and dielectric loss are small as compared with
susceptance.

The square root of L/C is a nearly pure resistance and equals Zo, and
equals the square root of Z/Y. Zo is not complex.

The incident and reflected waves are both forced to have a V/I = Zo. In
the case of the reflected wave, the ratio is often written -Zo to
indicate a travel in the direction of the transmitter instead of toward
the load.

Keith also wrote:
"It is a voltage wave which does the travelling."

Sure. That`s why it is called a TEM wave. The E-field is all alone.

Best regards, Richard Harrison, KB5WZI

Ok, now take the capacitor off, and measure the voltage at the end
of the inductor. What do you get?

Afraid to measure it, eh? Go ahead, we can ignore the 15 pF for now,
as
the load was 100pF. What do you get?

You asked for an open. 15pf is about 1000 ohms at 10 Mhz. This is a long
way
from an open. In any case, I suggest it might be your turn to produce
some
experimental results.

Go ahead and measure it anyways, i'd like to know what you get. If you
think about it, this would be a
clue as to the forward Vi voltage.




Classic Rho gives -1, which is a short, and conjugate Rho gives
+1j, which is ALSO a short.

As I recall, the purpose of rho was to compute the reflected voltage so
that net voltage could be computed using:
Vnet = Vfwd + Vref
Using rho = -1 produces Vref = -Vfwd yielding Vnet = 0 as expected for
a short.
Using rho = +1j produces Vref = +1j * Vfwd which does not produce
Vnet=0.
This is not the expected result for a short.



You don't understand what i wrote. Rho = +1j is a short if Zo=50+j50 and
Zl= - j50

For our example, rho= -1 , but this is
NOT a short!! As it shouldn't be!

Hint: What is the center of the Smith Chart when it is normalized
to Zo=50+j50?

If you can't answer this question, that's
ok, most people don't understand the Smith very well.



Riddles just do not cut it.


My questions are indeed riddles for those whose don't understand the
concepts.




Feel free to point out the flaws. If you can find none, question why
you hold so tenaciously to revised rho when it does not work.


i just did point out the flaws.

And i will let go of this argument as
soon as someone can give me evidence
that i should. You certainly have not.


I will try the experiment when i get the chance.

Excellent. There is nothing better than seeing it with your own
eyes.


I don't need to see it with my eyes,
when i know you can't get a larger RMS voltage reflected from a capacitor, than
the RMS that charges it.


Slick

I disagree with the demonstration.

Feel free to point out the flaws in the method or the evaluation.


I already did. See other post.


Maybe you are a bit biased?

The experiment and its evaluation has been published in sufficient
detail so that anyone may replicate it. If your lab obtains different
results, please publish them and we can work on why this occurred.


You never measured the incident voltage. And you refused to measure the
end of the inductor, with the capacitor removed (even with the 15 pF, it should
tell us something about Vi).




This may be true, but are you saying that a capacitor can reflect an
RMS voltage wave that is greater than the one that charges it?

Yes indeed. Resonant circuits achieve this with ease.

...Keith


Absolutely incorrect! If capacitance is defined as Coulombs/Volt, then
how are you getting more coulombs than you put in? Remember, i said Root Mean
Square voltage.

How does a capacitor reflect more power than you feed it?

It's almost time for me to cut out of this discussion, if you still don't
understand me.


Slick

TEM stands for transverse electro-magnetic. It means that both the
electric (E) and magnetic (H) fields are transverse (at a right angle)
to the direction of propagation. It refers only to the orientation of
the fields relative to the direction of propagation.

A time-varying E field is never "all alone" -- it's always accompanied
by an H field. Consequently, a voltage wave on a transmission line is
always accompanied by a current wave. I'm certain that Keith didn't mean
to imply otherwise.

Roy Lewallen, W7EL

Richard Harrison wrote:
. . .
Keith also wrote:
"It is a voltage wave which does the travelling."

Sure. That`s why it is called a TEM wave. The E-field is all alone.

Best regards, Richard Harrison, KB5WZI

On Fri, 19 Sep 2003 22:08:01 +0000 (UTC), "Reg Edwards"
wrote:
Richard, is there a conjugate match immediately at the junction between the
transmitter output socket and whatever follows it ?


Even kids know to look both ways before crossing. ;-)

73's
Richard Clark, KB7QHC

In message , Reg Edwards
writes
"Richard Harrison" wrote in message
...
Reg wrote:
"How does each repeater generate its unique pilot tone when a trawler or
earthquake breaks the inner conductor?"

Well, I`ve checked and find that some subsea multiplex units do generate
pilot tones for the purpose I speculated and for others. There is an
"SD" model equipment in which each group modulator generates a pilot
tone used for switching, level adjustment, equalization, etc. If these
tones are being recorded and start losing power, I expect they fade away
but die all at once when the path is interrupted. By examining the
record, it should be possible to see which tones suffered sudden death
and which faded away. There is capacitance in the system and the power
is high-voltage and low current. My information says there is 5500 volts
at each end of the transatlantic cable and that the system current drain
is 389 milliamperes. It has 182 repeaters spaced at about 23 miles. An
equalizer follows each 10th repeater.

I have experience doing similar things with terrestrial radio multiplex.
We used cheap pressure stamping Rust-Rack event recorders to record
breaks in pilot reception. I`ve pored over years of these records, 24
hours at a time.

Best regards, Richard Harrison, KB5WZI

=======================================

OK, Richard, so we've located by some automatic means the cable damage to be
somewhere between the n'th and (n+1)'th repeaters. This is nowhere good
enough for the repair ship captain. He wants to be back in port again for
Christmas. And it costs a million dollars in lost revenue each day the cable
is out of service plus the cost of keeping an 8000-ton ship and its crew at
sea.

The repair ship is just leaving Southampton docks. There's an Atlantic storm
brewing. There are spare lengths of cable and repeaters aboard. This time
they remember to collect the cable jointer. The ship's doctor retires to his
cabin as he is always sea-sick. The ship's Chief Engineer Officer, always a
Scotsman, and the Cable Testing Officer, neither of whom ever seem to have
anything to do, are outside the duty-free bar waiting for it to open.

The captain is pacing up and down the bridge waiting impatiently for a radio
message, cursing the poor fellow at the cable terminal station 500 miles
away on the Scottish rocky Atlantic coast, who is sweating blood twiddling
knobs trying to balance the wideband, 0.05 to 50 Hz, reflection-coefficient
bridge to within plus or minus 0.1 miles along the articial line and
adjusting the artificial fault (R and C in parallel) to within 10 ohms with
a biassing current of 50 milliamps.

After another couple of hours with a sliderule, correcting for seasonal
water temperature, etc., he changes his mind for the tenth time, takes
another swig of scotch, plucks up courage, swallows, and calls the operator.
He asks to send an urgent radio telegram to a cable ship heading westwards
down the English Channel. The message is brief, polite and states "Fault
estimated to be 623.7 miles from Oban". The captain issues a brief
instruction to the duty officer and retires to the loneliness of his cabin.

It is deep dusk. The lights of Plymouth from where Drake set sail 400 years
before can be seen on the starboard horizon. The radar ppi display in the
darkness of the bridge sweeps round monotonously every three seconds clearly
displaying a dozen merchant ships sharing the roughening waters.

The ships engines quietly, monotonously, emit a steady thud, thud, thud,
thud . . . . . . until they merge unnoticed into everybody's
unconsciousness and on deck only the sea and wind can be felt and heard.
----
Reg.





Reg. Please be careful. I nearly choked with laughter on my cornflakes
as I read this this morning. Whatever you are taking, do I need a
prescription, or can I get it over the counter? A great start to the day
(but I'd better get my tax return done before the deadline of the end of
the month).
Ian.
--