On Mon, 8 Mar 2004 15:35:46 -0600, "Steve Nosko"
wrote:

I wonder what happened to my last post with questions to Richard Harrison.
I don't see it. Boy! I'm not doing well at all on this Usenet thing
today.....

Comments inserted below.

Hi Steve,

Newsgroup material moves across the internet in a bucket brigade style
of relay using NNTP as the control language. It passes from the
source back towards you (such that you can see you own posting) with
delay that is variable.

For me, that delay is on the order of 1 second to three or four
minutes. On bad days it can take as many hours. On Google, the delay
is more pronounced.

73's
Richard Clark, KB7QHC

Steve Nosko wrote:
In my nomenclature, all "resistance" dissipates power as heat.

Even SQRT(L/C)???? The Z0 of transmission line is a resistance,
e.g. 50 ohms, i.e. the "real part of impedance".

Seems the two "non-equivalent" IEEE definitions resolve the
contradictions in your posting.
--
73, Cecil http://www.qsl.net/w5dxp



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Steve Nosko wrote:
For some reason, the above post does not show on my reader. I had to go
to Google to see what I was missing.

You must be using a computer with Motorola chips. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Steve Nosko wrote:
I won't get
insulting (nudge, nudge, wink, wink), but I have this bone in my head which
makes me WANT to say "I know I'm right, I tried to help you understand so
you go prove it to yourself." without sounding insulting...

Well, Steve, maybe you can tell us exactly what happens at the Z0-match
point in the following system. What changes the direction and momentum
of the reflected power wave from the load?

100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load
--
73, Cecil http://www.qsl.net/w5dxp



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"Richard Clark" wrote in message
news
On Mon, 8 Mar 2004 15:35:46 -0600, "Steve Nosko"
wrote:

I wonder what happened to my last post

Newsgroup material moves across the internet ...
delay that is variable.

For me, that delay is on the order of 1 second to three or four

I posted on Friday and couldn'r see it on Monday. I went to Googls and at
least saw Richard H.s comment but not my own. I may have goofed. For the
last few weeks now, i ALWAYS get an error message when posting though they
have always shown up in a few seconds (if I look).
73, Steve

"Cecil Moore" wrote in message
...
Steve Nosko wrote:
I won't get
insulting (nudge, nudge, wink, wink), but I have this bone in my head
which
makes me WANT to say "I know I'm right, I tried to help you understand
so
you go prove it to yourself." without sounding insulting...

Well, Steve, maybe you can tell us exactly what happens at the Z0-match
point in the following system. What changes the direction and momentum
of the reflected power wave from the load?

100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load

Well, Cecil... First, I don't know what point you refer to as "Z0-match
point".

If it is the "---x---", I proceed...
For the way you pose the problem (with no other constraints), the Z @ "x"
seen toward the load is 50 ohms, so all is happy in the universe. The 100W
enters the ladder line and exits at the "50 ohm load". I'm obviously
assuming I don't have balanced-to-unbalanced-problems and I know the good
'ole G5RV does this.
So I'd have to ask you just what point are you examining? Is it the
use of balanced T-Line as a matching transformer in an unbalanced system, or
something else?
I think this setup is a sub-optimal way to do it...but Hey! if it
works, use it. Manu suboptimal things are used all the time.


I don't know from "momentum" of power. If you asking what is going on
within the section of ladder line... The easy explanation is that the waves
act such that the two ends look line 50 ohms (OH YES, this DOES assume the
TX has a 50 ohm Zout... I think). I'd have to study the detail for a while
to come up with my mental model and be able to put it into words. I know
that the subject of "which way is power flowing" and "what happens at the
source" and "how many reflections really occur" are big topics here. I
don't get into these discussions because 1) it's been a long time since I
studied it and 2) the path and nitty-gritty detail has faded into the far
recesses of my mind and I would surely stumble around for a proper answer.

I am currently teaching a class in communications which will get into
T-lines later. If I need to get this deeply into the subject, perhaps I'll
come back and post my take on it.
--
Steve N, K,9;d, c. i My email has no u's.

"Cecil Moore" wrote in message
...
Steve Nosko wrote:
In my nomenclature, all "resistance" dissipates power as heat.

You need to include the rest of my post to to it justice. I discuss
the terminology distinction. I go on the say that the term "real part (of
an impedance)" is better suited (as a name) to what some like to call
"loss-less resistance".

In the chopper case that Richard Harrison posed, there is no parallel to the
T-line situation. There, he was comparing a chopper with 50% duty cycle to
an equal valued resistor and calling the chopper a "loss-less resistance"



Even SQRT(L/C)???? The Z0 of transmission line is a resistance,

I don't consider that term (resistance) suitable for this situation.
"real part of Z" is better. I think to some this is the same thing, but
obviously it is not. I believe this is what is causing all the confusion.
Remember, the Z0 is properly called "Characteristic Impedance" or "Surge
Impedance". I think this distinction makes the subject easier to understand
since it eliminates the confusing term "loss-less resistance".


Seems the two "non-equivalent" IEEE definitions resolve the
contradictions in your posting.

I'd have to read the full this before having an opinion!


--
Steve N, K,9;d, c. i My email has no u's.

I see this paper as a variation on a line of reasoning that goes like
this:

1) A conjugate match results in maximum power delivered to a load, so
it is good.

2) A connection where the load has much higher resistance than the
Thevenin equivalent source resistance results in high efficiency, so
it is also good.

3) Since (1) and (2) are both good, they must be equivalent to each
other. Therefore a conjugate match is what it is not. This is an
apparent contradiction.

4) The contradiction is resolved by postulating a special kind of
resistance that adds to the source resistance. However, it has no
physical effect and exists only to resolve the contradiction in (3).

All good matches are now conjugate matches and everyone is happy!

73--Nick, WA5BDU






"Dave" wrote in message ...
have you guys read this one yet?

www.qsl.net/w9dmk/MPTT.pdf

"Nick Kennedy" wrote in message
om...
I see this paper as a variation on a line of reasoning that goes like
this:

1) A conjugate match results in maximum power delivered to a load, so
it is good.

2) A connection where the load has much higher resistance than the
Thevenin equivalent source resistance results in high efficiency, so
it is also good.

3) Since (1) and (2) are both good, they must be equivalent to each
other. Therefore a conjugate match is what it is not. This is an
apparent contradiction.

4) The contradiction is resolved by postulating a special kind of
resistance that adds to the source resistance. However, it has no
physical effect and exists only to resolve the contradiction in (3).

All good matches are now conjugate matches and everyone is happy!

73--Nick, WA5BDU

nick,
I like your summary. You have captured the essence of my original
efficiency puzzle...except for your #4. My puzzle was not intended to be an
equivlent to maximum power transfer. Only an "efficiency enhancement"
technique/concept.idea/proposal. I did not intend to imply that my solution
was transferring "maximum power" only a higher efficiency than the alternate
case that I described.
Your #3 is clearly only for the discussion. The "they are both good,
therefore, they are equivalent" concept clearly can't be a serious
conclusion, just an argument tool.


--
Steve N, K,9;d, c. i My email has no u's..

Steve Nosko wrote:
"Cecil Moore" wrote:
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load

I don't know from "momentum" of power. If you asking what is going on
within the section of ladder line... The easy explanation is that the waves
act such that the two ends look line 50 ohms

It's a pretty simple question. The energy wave reflected from the load
possesses energy and momentum, both of which must be conserved. There
is no reflected energy on the 50 ohm coax and therefore no momentum in
the reflected waves on the 50 ohm coax. So what changes the direction
and momentum of the energy wave reflected from the load?
--
73, Cecil http://www.qsl.net/w5dxp



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