It is in the same class of parts with inductancless inductors,
and capacitanceless capacitors are in.
With lossless resistance they can form
a lossless, non-energy storing, extremely low/high Q, network.
No need for tuning either!

Heh...must be the principle behind this antenna...

http://cgi.ebay.com/ws/eBayISAPI.dll...tegory=48 692

Love that 10M beam!
A

Actually, not.
1. the line is mismatched at both ends, 450 balanced to 50 ohm singelended.
If you put in or assume transformers for the balanced to unbalanced 450 to
50 ohm conversion at both ends, then it is matched (still assuming that Tx
and Ant are 50 ohm resistive.) I have an equation that calculates the loss
from a text book. (if the section is mismatched it is not conjugally
matched)

2. If I have a 50 ohm load at each end, and a line of impedance Z, 1/2
wavelength long, what does Z have to be for max power transfer? Nothing
constrains Z to be a value. Z at 50 is great, Z at 500 is not bad (have to
do some matching)
This problem the way it is stated 50 1/2 line 50 is implying that any type
of line will do as long as it is 1/2 long, then Z can be 0.01 and Z=20,000

So as long as the matching transformers are in,(assuming the right ones) it
is matched max power too.
The Tx out 50 ohm unbalanced is matched to the 450 balanced line 1/2
wavelength long and another transformer back down to 50.

/////////////\\\\\\\\\\\\\//////////////\////////\\\\\\\\\\\\\\\\\\\\\\\\\\\
\\\\\\\\/////////////////////\/\/\/\\/\/\/\\/\\\/\\/\//\/\/\/\/\/\/\


"JDer8745" wrote in message
...
"Isn't a 50 ohm transmitter conjugately matched to a 50 ohm load
when fed through 1/2 wavelength of 450 ohm ladder-line?"

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

Heck yes, I'm assuming u mean 50 + j 0 Ohms.

73 de Jack, K9CUN

John Smith wrote:
"Actually not."

Actually so.

The input impedance of a 1/2-wave section of dissipationless line is a
repeat of its terminal impedance. The 50-ohm terminal impedance will be
repeated as the input impedance at the sending end of the 450-ohm line.
Likewise, the 50-ohm impedance connected to the sending end of the of
the 450-ohm line will be repeated at the output end of the 459-ohm line.
Cecil`s example is symmetrical. There are no lossless lines but
practical lines are close enough.

Best regards, Richard Harrison, KB5WZI

On Wed, 3 Mar 2004 22:08:54 -0600, "John Smith"
wrote:

Actually, not.
1. the line is mismatched at both ends, 450 balanced to 50 ohm singelended.
If you put in or assume transformers for the balanced to unbalanced 450 to
50 ohm conversion at both ends, then it is matched (still assuming that Tx
and Ant are 50 ohm resistive.) I have an equation that calculates the loss
from a text book. (if the section is mismatched it is not conjugally
matched)

I think we have some fertile ground here, ready for plowing.

Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

But using that "simplification", 1/2 wave repeats terminal impedance, allows
one to ignore line matching entirely.

It could be balanced lines of 2,000 ohms, or coax of 5 ohms, both of these
according to the simplification work the same as long as they are 1/2 wave.
Additionally, it could be zip cord, or wires just strung about, as the
impedance of the feed line is not used in the simplification, and can be
modeled by a transmission like of Z ohms.

Checkout "Reference Data for Radio Engineers", Transmission lines, Mismatch
and Transducer losses. There are formulas there for this case. The
simplification ignores the mismatches at the Tx and Antenna to the line,
which causes about 4.43 dB additional "mismatch" loss, in the case of
50,450, 50. This shows up as VSWR. For a 2000 ohm line, the mismatch loss
is 10.2 dB. Some of that loss is tuned out if you have an antenna tuner.

The simplification is valid only for 50 ohm lines, or near 50 ohms, perhaps
along the way that part got dropped.
(and also because of antenna tuners, Tx out is not 50 ohms, and Antennas are
not 50 ohms)


"Richard Harrison" wrote in message
...
John Smith wrote:
"Actually not."

Actually so.

The input impedance of a 1/2-wave section of dissipationless line is a
repeat of its terminal impedance. The 50-ohm terminal impedance will be
repeated as the input impedance at the sending end of the 450-ohm line.
Likewise, the 50-ohm impedance connected to the sending end of the of
the 450-ohm line will be repeated at the output end of the 459-ohm line.
Cecil`s example is symmetrical. There are no lossless lines but
practical lines are close enough.

Best regards, Richard Harrison, KB5WZI

John Smith wrote:
Checkout "Reference Data for Radio Engineers", Transmission lines, Mismatch
and Transducer losses. There are formulas there for this case. The
simplification ignores the mismatches at the Tx and Antenna to the line,
which causes about 4.43 dB additional "mismatch" loss, in the case of
50,450, 50. This shows up as VSWR.

There is little actual loss due to the "mismatch loss" since the
system is Z0-matched. Let's add a small length of 50 ohm line.

100W XMTR--50 ohm line---x---1/2WL 450 ohm line----50 ohm balanced load.

Since there are no reflections on the 50 ohm line, the system is Z0-matched
to 50 ohms at point x. (Assume a 9:1 balun at 'x') The 9:1 SWR on the 450 ohm
line causes very little loss and the "mismatch loss" is not really lost because
the forward power on the 450 ohm line is 278 watts, delivering virtually all of
the 100 watts of XMTR power to the load.

It can be illustrated that a steady-state destructive interference event toward
the source due to wave cancellation at point 'x' feeds energy to a constructive
interference event in the direction of the load. This has been understood in the
field of optics for many decades and is how non-glare glass works.
--
73, Cecil http://www.qsl.net/w5dxp



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Agreed that adding a 9:1 balun (impedance match) will have little mismatch
loss at x and all the Tx power will be "presented" to the antenna port. If
their is a matching section of 450 to 50 (another 9:1) balanced at the
antenna then there should be no reflections due to impedance mismatches at
the antenna, and all of the power goes out the antenna.

If the antenna is not 50 ohms, or the balun from 450 to 50 at the antenna is
missing, then part of the power is reflected back into the Tx output.

What the Tx source does with that depends upon the design of the final
Amplifier stage, some change output impedance, some go unstable, many cut
back Power out. If at a higher frequency one can use a circulator, and
dump the reflected power into a 50 ohm dummy load and keep the Tx safe (also
good in suppressing transmitter IM).

If the Tx source reflects the power back out, then the Tx source can "drive
that amount of power out into a higher VSWR", which means that Tx source can
accommodate loads around 50 ohms and keep Power output to spec. Which means
that there is no problem if all the mismatches from the antenna (not being
50 ohms), antenna/line mismatch, line/Tx mismatch etc. all add up to be less
than a 3:1 VSWR or so for mobile FM. Probably higher for Ham bands.

But one needs to keep an eye on the Tx output impedance as it changes, then
the load presented to the antenna (looking back into the line) will change,
not be close to 50 ohms, which can change antenna pattern (at higher
frequencies). One could put in a resistive pad to stabilize the impedance at
the Tx output, but why waste the power? Antenna tuners make sense at the
lower frequencies, they match up the strange loads, most of the mismatches
and Tx output,and maximize power output.

The optics example is good, as the anti reflective coatings are 1/4 wave
thick, and the coating constant (speed of light in the material) is the
square root of the product of Air and Glass, it is an optical matching
section. Need a coating on both sides of the glass.


"Cecil Moore" wrote in message
...
John Smith wrote:
Checkout "Reference Data for Radio Engineers", Transmission lines,
Mismatch
and Transducer losses. There are formulas there for this case. The
simplification ignores the mismatches at the Tx and Antenna to the line,
which causes about 4.43 dB additional "mismatch" loss, in the case of
50,450, 50. This shows up as VSWR.

There is little actual loss due to the "mismatch loss" since the
system is Z0-matched. Let's add a small length of 50 ohm line.

100W XMTR--50 ohm line---x---1/2WL 450 ohm line----50 ohm balanced load.

Since there are no reflections on the 50 ohm line, the system is
Z0-matched
to 50 ohms at point x. (Assume a 9:1 balun at 'x') The 9:1 SWR on the 450
ohm
line causes very little loss and the "mismatch loss" is not really lost
because
the forward power on the 450 ohm line is 278 watts, delivering virtually
all of
the 100 watts of XMTR power to the load.

It can be illustrated that a steady-state destructive interference event
toward
the source due to wave cancellation at point 'x' feeds energy to a
constructive
interference event in the direction of the load. This has been understood
in the
field of optics for many decades and is how non-glare glass works.
--
73, Cecil http://www.qsl.net/w5dxp



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John Smith wrote:
Agreed that adding a 9:1 balun (impedance match) will have little mismatch
loss at x and all the Tx power will be "presented" to the antenna port.

Oh Boy, I really screwed up. Replace that 9:1 balun with a 1:1 balun.
I was very stupid to have said 9:1. A 9:1 balun would reduce that 50
ohm Z0-match to 5.56 ohms thus creating a mismatch.
--
73, Cecil, W5DXP

Cecil Moore wrote:
100W XMTR--50 ohm line---x---1/2WL 450 ohm line----50 ohm balanced load.

Since there are no reflections on the 50 ohm line, the system is Z0-matched
to 50 ohms at point x. (Assume a 9:1 balun at 'x') ...

HEY EVERYONE, I DIDN'T MEAN A 9:1 BALUN AT 'X'. I MEANT A 1:1 BALUN
AT 'X'. THAT WAS A REALLY, REALLY STUPID BOO-BOO ON MY PART.
--
73, Cecil, W5DXP

John Smith wrote:
If the antenna is not 50 ohms, or the balun from 450 to 50 at the antenna is
missing, then part of the power is reflected back into the Tx output.

Just as I drew it, with a 1:1 balun, the system is Z0-matched. No
reflections reach the TX output. Saying a 9:1 balun was a huge mistake.
That's not what I meant. I certainly meant a 1:1 balun.

The optics example is good, as the anti reflective coatings are 1/4 wave
thick, and the coating constant (speed of light in the material) is the
square root of the product of Air and Glass, it is an optical matching
section. Need a coating on both sides of the glass.

Likewise, you can put 1/2WL of thin-film between air and air and still
have a match, i.e. no glare. That's what I did in the following diagram:

100w XMTR--50 ohm feedline--x--1/2WL 450 ohm line--50 ohm load

similar to:

Laser----air-----|---1/2WL thin-film---|---air---
--
73, Cecil, W5DXP