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

Ummmm... sorry, Richard.

I am very familiar with choppers, having designed and built quite
a few in my time. But no engineer I have ever met would refer to
a chopper as a "lossless resistor."

An (ideal) resistor is in the class of passive, linear, time-invariant
devices. A "lossless resistor" would be a 0 Ohm resistor.

A chopper is a time-variant device, and depending on the details,
could be modelled as linear or non-linear on top of that.

The two devices (resistor and chopper) have very little in common,
besides being usable in electronic circuits. One of many differences
(a BIG one for us radio guys) is that the resistor would generate
no radio noise (other than thermal); but the chopper would generate
horrific radio hash, from the chopper frequency to daylight.

As for analogies to power amplifiers, an ideal Class A amplifier
would normally be modelled as an active, linear, time-invariant
device; so it would have some commonality with a resistor.
But the difference between passive and active is huge, of course.

A Class C amplifier would normally be modelled as an active,
non-linear, time-invariant device. There is very little similarity
left to a resistor, except for the time-invariant part. But a sidewalk
is normally time-invariant, too. So we could just as well say a
sidewalk is like a "lossless resistor." Or maybe that would be a
sidewalk with ice on it. ;-)

But to each his own...

Regards, Ed

"Richard Harrison" wrote in message
...
Old Ed wrote:
"Couild you explain this concept, and/or provide some references?"

Suppose we adjust a variable d-c supply to full-scale indication on an
external meter. Next, install a chopper (lossless on-off circuit) driven
at a high frequency to produce a square wave interruption of the d-c
with a 50% duty cycle, and insert the chopper contacts in series with
the external meter.

The chopper connects the external meter 50% of the time and disconnercts
it 50% of the time. The meter reads 50% of full scale.

Another way to reduce the scale reading to 50% is to insert a resistor
in series with the meter. If it is an 0-1 ma meter and if it has an
internal resistance of 1000 ohms, insertion of a 1000-ohm resistor in
series with the meter will reduce the meter scale indication to 50%.

The chopper as part of the meter source eliminates current to the meter
50% of the time.

The resistor which has the same effect and produces the same scale
indication as the chopper on the effective output current exacts its
loss of 0.0005 amp x 0.5 volt or 0.25 milliwatt during 100% of the time.

The chopper eliminates the power-losing resistor by substituting
off-time in the power source. The source only supplies the power used by
the load. With a resistor limiting power to the load, the source
supplies its loss and the load power.

The power in the load, a meter in our example, is the same using either
the resistor or the chopper. The resistor is analogous to a Class-A
amplifier. The chopper is analogous to a Class-C amplifier.

The off-time has the same effective opposition to current to a load as a
dissipative resistance. As the time-limited currented opposition to load
current consumes no power, it is called a dissipationless resistance.

Best regards, Richard Harrison, KB5WZI

OOPS!

Richard,

Now I understand the source of your "lossless resistance" remark in the
class C thread. I respectfully disagree. There is nothing of the sort. I
will admit, however that in a Switched case, the concept of resistance *may*
be thought of this way, but it is not a good way to look at it. I believe
you are taking two things which are not similar and calling them the same.
I will point out the error below. It can be verified by experiment rather
simply if you wish.



"Richard Harrison" wrote in message
...
Old Ed wrote:
"Couild you explain this concept, and/or provide some references?"

...d-c supply to full-scale indication on an external meter.

Therefore we have the situation whe
V out = Ifs x Rm
V out set to the IR drop of the meter at fullscale. All ok so far.


Next, install a chopper (lossless on-off circuit) ... with a 50% duty
cycle...
... The meter reads 50% of full scale.

Still ok so far.


Another way to reduce the scale reading to 50% is to insert a resistor

Yep, still ok.



The resistor which has the same effect ...

Start of the error. I disagree with "same effect". One is a steady
state current (resistor), the other a square wave of current (chopper),
right? The time average current IS the same. ...HOWEVER...



and produces the same scale
indication as the chopper on the effective output current exacts its
loss of 0.0005 amp x 0.5 volt or 0.25 milliwatt during 100% of the time.

The word "effective" here is not standard as it is the same word we
use to denote "producing the same power". Again, it certainly is that the
AVERAGE current is obviously the same for both cases.



The chopper eliminates the power-losing resistor by substituting
off-time in the power source. The source only supplies the power used by
the load. With a resistor limiting power to the load, the source
supplies its loss and the load power.

This is indeed correct. The chopper DOES change produce the same
AVERAGE current as the resistor. The added resistor looses some power, the
chopper does not. Now the zinger.


The power in the load, a meter in our example, is the same using either
the resistor or the chopper.

Here is where the error is. The POWER in the load (meter) in the
chopper case is not the same as it is in the resistor case! It is...Hold on
to your hat... TWICE as much as in the resistor case!

In the meter only case load power is 1 miliwatt.
In the resistor case the load power is 0.25 miliwatt (half voltage x half
current = 25%)
Or P=I^2 * R which is also 0.25 miliwatt.
In the chopper case the load power is 0.5 Miliwatt!

In the chopper case, the RMS current is = Imax x square-root(duty-cycle).
This is 1ma * 0.707 or 0.5 miliwatt.

See
http://www.irf.com/technical-info/an949/append.htm
For verification of the square root of the duty cycle formula.

You can easily verify this since a power difference should be easily
observed by finger-and -clock method. Set up the experiment with a resistor
and power supply which gives some power which can be felt as a temperature
rise within, say 10 seconds. I'm guessing somewhere around 0.2 to 1.0 watt.
This is with the two resistors in the circuit. Then subsitute the chopper
(a relay). Twice the power should be evident in a faster rise in
temperature. It should heat up in half the time. I maintain that if it
doesn't, then you did it incorrectly.



The off-time has the same effective opposition to current to a load as a
dissipative resistance. As the time-limited currented opposition to load
current consumes no power, it is called a dissipationless resistance.

This is what I believe would be called a self conflicting term.
Resistance removes power from the circuit in question in all cases, period.
And YES on the proverbial infinite T-line, the power is gone from the
source, never to return. It looks just like a resistor (Steve says with
great confidence, knowing full well that others may disagree, but
nonetheless firm in his beliefe) It *is* LOST.

Perhaps a concept called "Average resistance" might be a subject to discuss.
In matching a class C power amplifier, there has been myriad of words spent
on just what, if anything, is being matched by the output matching network.
Is there an "output" impedance there or not and if so, what is
it....(rhetorical, of course)

Sorry Richard, but I believe you have it wrong this time. Steady state and
pulsed must be understood with slightly different principles.
--
73, Steve N, K,9;d, c. i My email has no u's.

"John Smith" wrote in message
...
Actually, not.
1. the line is mismatched at both ends,

John, You missed something. The problem was stated;

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


The TRANSMITTER "sees" 50 ohms therefore it is. Some discussion can be made
for the exact phrasing which says, in effect; "...50 ohm transmitter ...to a
50 ohm load". By this I mean using the phrase "to a 50 ohm load" is a bit
misdirecting. In the stated case, the Tx is NOT actually connected *to* the
load. It is connected to a 1/2 wave of line. At the Tx output, there is a
conjugate match.
I believe the intended situation is that a 1/2 wave of ANY line repeats the
load Z at its input. From your comment #2, it appears that you do not know
this.


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?

Actually, it IS anything! Yea, Yea there are other *practical*
considerations...balanced...unbalanced....


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

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

and *I* got suckered in... Oh well...

-- 73
Steve N, K,9;d, c. i My email has no u's.

"Robert Lay W9DMK" wrote in message
...
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

OK, OK, OK. How come voltage dB and power dB are different, huh? Yhy is it
that power dB carry twice the weight of voltage dB?

dB= 10 LOG(power ratio) = 20 LOG(voltage ratio)

Knock yourself out.

--
Steve N, K,9;d, c. i My email has no u's.
he he he
"J. McLaughlin" wrote in message
...
When a candidate for an EE faculty position visits, someone, usually at
lunch, will bring the conversation around to the MPTT. If he or she
does anything other than giggle they do not get my vote.

Let us leave this tar-baby out in the field. 73 Mac N8TT

--
J. Mc Laughlin - Michigan USA

Cecil --

I think your right, 1:1 should work fine, with 1/2 wavelength, low loss
line.

I was matching to a long line with charestic impedance of 450. But, I think
the length of the line must be long (or have loss) before the charestic
impedance of the line shows up to really matter. I used to have a program
that showed this effect. I installed a 450MHz 500 foot line in Saudi, and
measurements had to be careful, because we had a lot of line loss. (we were
heating up the Heliax, more than radiating power out the antenna)

But on a short run, lower frequency, there is not enough of the line to make
a large effect. (I think it has to be more than about 2-3 dB line loss...),
so the Tx output is seeing primarily the load(antenna). And then the 1/2
wavelength will causing a "null" at the Tx output. When I run across the
program/formula I post it. It is probably in the same book, same section,
too.
'''''''''''''''''''''''''''''''''''''''''''''''''' ''''''''''''''''''''''''''
''''''''''''''''''

"Cecil Moore" wrote in message
...
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

Steve Nosko wrote:
OK, OK, OK. How come voltage dB and power dB are different, huh? Yhy is it
that power dB carry twice the weight of voltage dB?

dB= 10 LOG(power ratio) = 20 LOG(voltage ratio)

Because log(v^2) = 2*log(v)? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Seems to me this thread has gone off track, as many do after the first three or
four responses.

This thread was started with Bob's fine treatise on proof of the existence of
non-dissipative resistance, a resistANCE that cannot be duplicated with a
resisTOR. But the thread went astray.

Non-dissipative resistance is not well accepted or understood by many otherwise
well informed engineers, because it has had little or no (or even incorrect)
treatment in EE courses. This is because the profs by habit generally used only
the classical generator as the source in treating network theory. The classical
generator is always considered to have a dissipative internal resistive source.
Consequently, the profs never considered treatment of a source having other than
a 50 percent maximum efficiency. All the equations I'm familiar with show
clearly that 50 percent is all yer gonna get.generally

Unfortunately, the profs of my ken never tried to understand the reason the
efficiencies of Class B and C amps exceed 50 percent. Fortunately, Bob Lay comes
along with proof that there really is resistance established only by the RATIO
of voltage to current with no dissipation whatever to heat or radiation.

Bob's paper furnishes further proof of my own proof that dissipationless
resistance exists. I presented my proof in Chapter 19, and Appendices 9, 10, and
11 in Reflections 2. For those who are interested in reviewing my writings
there, but who don't have a copy of Reflections 2, those references are on my
web page at http://home.iag.net/~w2du.

For a short preview, take note that in Class B and C amps the DC source power
goes to only TWO places, 1) the power that is dissipated in the cathode/plate
resistance, which transitions to heat, and 2) the power dissipated in the load.
Note also, the source resistance at the output of the tank circuit is linear,
time invariant, and determined solely by the voltage/current ratio at the output
of the tank. The reason is that that the energy storage of the tank isolates the
non-linear input portion of the amp from the linear output portion. With
sufficient Q the output wave form is a near-perfect sine wave, thus verifying
linearity.

Also, data obtained from my measurements of the source resistance of Class B and
C power amps shows that when the plate tuning and loading conrtrols are adjusted
to deliver all the available power at a nominal drive level, the source and load
resistance are equal. Consequently, if the power dissipated in the cathode/plate
resistance plus the power dissipated in the load equals the DC input power,
there can be no power dissipated in the source resistance.

And finally, with the source resistance non-dissipative, it cannot absorb any
power reaching it from a mismatched termination down stream, which is why no
reflected power is absorbed in the amp.

Eric Nichols, KL7AJ, operates multi megawatt ionospheric sounding rigs in
Alaska. The water cooling the water-cooled amp tubes is carefully measured
calorimetrically. He has observed over a period of many years that the water
temperature remains the same, no matter what the SWR at the input of load line
is, proving there is no reflected power absorbed in the amplifier. Of course the
tank circuit elements are re-adjusted to deliver all the available power, what
ever the input impedance of the line may be.

Howabout if we give Bob's paper a somewhat more open-minded review. Crap? That
criticism of Bob's paper is where the crap is!

Walt, W2DU

Steve Nosko wrote:
"How come voltage dB and power dB are different?"

They aren`t.

10log(power ratio) = 20log(voltage ratio)

They are equal because power is proportional to voltage squared. To
square with a logarithm, you multiply by (2).

Best regards, Richard Harrison, KB5WZI

Steve Nosko wrote:
"The power in the load (meter) in the chopper case is not the same as it
is in the resistor case!"

A square wave has the same heating value as d-c because it has identical
amplitude and both alternations are equally effective in producing heat.
Power is indifferent to polarity.

The sine wave has an amplitude 1.414X its effective value. The heat
produced over a period is the average of the sum of the instantaneous
powers of increments within a cycle.

Someone argued that resistance is an agency which removes power from the
scene. Thus, Zo is a resistance.

I would rather define resistance as the ratio of the applied emf to the
resulting current in the circuit. The current must be in-phase with the
applied voltage..

With my definition, the Zo of the transmission line as defined by the
square root of L/C makes sense to me. It is a lossless resistance.

Best regards, Richard Harrison, KB5WZI

Richard Clark wrote:
"John Smith" wrote:
Check for reflected power in either direction.

You would be surprised how hard that is for some, especially when an
argument about the direction of current flow becomes part of the
turmoil.

If none, then it is conjugally matched exactly.

Still too simple, not enough gristle to chew on. ;-)

If some, it is not conjugately matched?????

If it is about -14 dB down, that is good and close enough.

In three sentences you've said enough to resolve the matter. But not
enough to end the debate....

Resolve what matter?
--
73, Cecil http://www.qsl.net/w5dxp



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Old Ed wrote:
"Lossless resistance?" Would that be zero resistance,
or perhaps a negative resistance, as in the active part of
a tunnel diode's V-I characteristic?

I am a career EE, with a couple of graduate EE degrees;
and this is something entirely new to me. Could you explain
this concept, and/or provide some references?

How about an example? If L and C are lossless, then SQRT(L/C)
will be lossless with a dimension of ohms, i.e. resistance.
--
73, Cecil http://www.qsl.net/w5dxp



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John Smith wrote:
But on a short run, lower frequency, there is not enough of the line to make
a large effect.

For a 1/2WL line, as was the example, the impedance is transformed
around one shallow spiral on the Smith Chart. For long, lossy
runs, the SWR "circle" becomes multiple spirals that spiral inward
toward Z0 which is in the center of the Smith Chart.

For a short relatively lossless run of feedline, if the SWR is not
1:1, the impedance seen by the transmitter can never be Z0.
--
73, Cecil, W5DXP

Very good Richard. If I had a prize, you'd get it. (:-(


"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"How come voltage dB and power dB are different?"

They aren`t.

10log(power ratio) = 20log(voltage ratio)

They are equal because power is proportional to voltage squared. To
square with a logarithm, you multiply by (2).

Best regards, Richard Harrison, KB5WZI

Working very hard to resist his long-winded gene, and apparently failing,
Steve responds:



Hi Richard,



"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"The power in the load (meter) in the chopper case is not the same as it
is in the resistor case!"

Sorry Richard, but I show how this is true below. Stay tuned...

However, first I'll address this digression. You say:
A square wave has the same heating value as d-c because it has identical
amplitude and both alternations are equally effective in producing heat.
Power is indifferent to polarity.

I believe you have changed your premise. The chopper example you
gave is not this type of square wave. Your example has what might be called
a "ONE sided alternation". Again, more below.

The sine wave has an amplitude 1.414X---
I want to stay on the chopper example you posed.

Someone argued that resistance is an agency which removes power from the
scene. Thus, Zo is a resistance.

I did. I maintain that *to the sourse* - that is, from the source's point
of view, an R or a Zo (all real) are indistinguishable. The current is in
phase with the voltage and power is removed from the source. Yes the power
isn't lost from the system, yet, so I do understand your resistance (sorry)
to calling it a loss. It is traveling down the line, but it IS gone from the
source.

I would rather define resistance as the ratio of the applied emf to the
resulting current in the circuit. The current must be in-phase with the
applied voltage.
With my definition, the Zo of the transmission line as defined by the
square root of L/C makes sense to me. It is a lossless resistance.
Best regards, Richard Harrison, KB5WZI
So, you hold true the two means to the definitions. That's ok, but I
maintain that the chopper example is correctly analyzed by my method. I see
that you are trying to build mental models of these situations, but I think
your model does not do justice to the physics... IT seems a reasonable
conclusion that - because the value of Zo is the ratio of two lossless
parameters that therefore the Zo itself is "lossless". In the sence that
the power is not lost from the system ans delivered to some type of load,
this is true. So I understand this analogy, though I believe it does not
allow a good understanding of electronics.

Now, I return to the Meter and resistor/chopper premise you posed earlier:



Rather than succumbing to jokes, I submit that while you
base your understanding on the facts, your approach is from the wrong
direction. I suggest that your "loss-less resistance" concept is the wrong
way to look at it. When we analyze a situation we must be careful to have
all the correct basics - and not violate any basic "laws". I believe there
is only one kind of resistance (as far as the circuit is concerned) and
therefore all resistance follows the same rules -- all the time. I don't
need one understanding for loss-full resistance and another for loss-less
resistance.

I actually have pondered trying to use a model called
"average resistance", but I think it is not the right way to go. Here's how
I came to this conclusion.... Short answer: it does not work.



Long answer:

I DO know that one not-to-be-violated principle is that ALL
resistance removes or looses power - that there is no such thing as
loss-less resistance. If we place ourselves 'inside' a circuit, any time
some of the I is in phase with V (across an element of the circuit), power
exits out that element. A real Zo does this, a resistor does this and a
motor does this. The power exits the source and becomes energy traveling
down a T-line, heat or mechanical energy, respectively. (I hope I am using
all the correct terminology here) This MUST be an overriding principle and
my solution must not violate it. If it does, then I went in the wrong
direction. I am not going to say that the switched situation is more
complex because it is actually quite simple. You just went the wrong way in
a desire to understand it.



You give three situations which can still be used to get to the correct
conclusion.



1) A simple 1V. source and a 1K load (a meter) [ IL=1 ma. PL= 1 mW ]

2) The above with an added 1k resistor [ IL=0.5 ma. VL=0.5 V. PL=0.25
mW. ]

3) #1 with a 50% chopper.[ ILavg.=0.5 ma. ... PL=0.5mW. ]



You tried to understand #3 (chopper) by comparing it to #2
(added resistor) both having the same AVERAGE current. You say they are the
same. This sounds reasonable, at first,, but switched or pulsed circuits
can be misleading.



I expected a response from someone to the effect that I
threw in a square root out of nowhere and was confusing the matter by
throwing math at it. Bear with me, and I will show that the added resistor
and the chopper situations are definitely NOT the "same". While I knew the
root is correct, I actually puzzled over this. Why isn't #2 (added
resistor) and #3 (chopper) the correct comparison? I do know that the
chopper situation does have a square root in the power equation. In fact,
when doing RMS calculations there will always be a root in the equation
which is not in the 'average' equation. Here's a way to look at it which
gives the correct conclusion.



To understand #3 (chopper), it is better to compare it to _#1_. Like
this:

When the switch is closed, Power in the load is PL= 1.0 mW. When open,
PL= 0.0. Therefore based on an intuitive understanding of how things work,
it surely seems that the average Power in the load must be half or 0.5 mW.
Well, this IS correct.

Now look closely at case #2 (added resistor) . The Power in the load is
0.5 V times 0.5 ma., or 0.25 mW. So while the average current is the same
as #3 (chopper), the average power is not (it's half that of the chopper
case). Therefore, the chopper case and the added resistor case are not
really the same. Something is definitely different. It is the power.

Applying the RMS formula to the chopper case gives the same result. The
chopper case is not your regular square wave with a positive and negative
"alteration". You are correct that the standard square wave is = to DC.
The chopper case IS NOT.

It turns out that when we move into the RMS realm, a square root always
appears in the formula. In case you can't see that Int' Rect paper, the
chopped case RMS is:

RMS = Imax * sqr-root(duty-cycle ratio).

So for a 50-50 chopper it is Irms = 0.707ma. This gets the 0.5 mw number.



Intl Rect paper URL:

http://www.irf.com/technical-info/an949/append.htm

Formulas roughly in mid document.



OK, so what about the "resistance" of the chopper? Can we assign it a
numerical value? I suggest that while this is a very tempting path , it
should be discarded and you only think in terms as I describe. That is, for
the steady state cases (#1 & 2) we use ohms law and for #3 (chopper) we
shift gears slightly and apply principles (still correct ones) which
properly describe that situation.



Another stumbling point is the idea of current being "In-Phase" for the
chopper case. Phase (when we start getting into power factor
considerations) is a property of sine signals and applying it to the chopper
case, I believe, is a confusing use. I can imagine a concept where - the
wave shape of the current being the same shape as the voltage wave indicates
a resistive circuit - as being valid.





Now, we may want to ask; "OK. Then what IS the "Equivalent resistance"
of this 50-50 chopper. Since it is limiting the current to this average
value of 50%, it MUST have some "resistance" associated with it--shouldn't
it? Well, let us look two ways. If there is a equivalent resistance, then
we should get the same value any way we figure it.



Looking at the AVERAGE current, we'd be led to think it is 1k since the
"current" is cut in half. However, looking at the power, it appears to be
414.43 ohms. (this is what it would take to get 0.5 mW. in the 1k load)

Now this looks strange. That is, a 50-50 duty cycle chopper and two
different ways to arrive at this "equivalent resistance". Pick other
chopper duty cycles and I think we may find that a square root crops up (I
haven't derived it, yet). There is no square-root in the average current
formula. It is simply the duty cycle ratio. This suggests that the concept
of an "equivalent resistance" has some problem. If there is a truly
equivalent resistance, then any method of analysis should arrive at the same
value. I didn't need any "loss-less resistance" to get the numbers.



I submit that it is a bad mental model and should be discarded.



We all must form what I call "mental models" of electronics. If our models
are good, then we can use them and come to the correct conclusion as we look
at new or more complex situations. The formulas always work when properly
applied. I guess the catch is: which formulas are the correct ones. I
think some may think that arrogance tells me which is which. If we pick the
wrong view, we can be led astray. This is the case here.



I believe the conclusion should be:

In the simpler steady state DC (or AC) situations, we use ohms law and think
in terms of our normal understanding.

In the switch-mode situation, we still use ohms law. However, we base our
understanding on different concepts which do not overlap completely with the
steady state case, but are nonetheless correct basics (and equivalent or
loss-less resistance are not among them). Also that using two applicable,
but different methods, we arrive at only one answer.



For a VERY light look at the chopper situation, the "equivalent resistance"
idea may be ok for some simplistic explanation to a non technical person,
but for someone trying to understand electronics, it falls apart.



Are you swayed, or am I whistling up a drain pipe? If we agree to disagree,
so be it, but how do you justify the difference in *power* between case 2
and 3 when you maintain they are the "same"? I would like to focus on that.
--
Steve N, K,9;d, c. i My email has no u's.

Steve Nosko wrotew:
"3) #1 with 50% chopper, [ILavg =0.5 ma...
PL = 0.5 mW]"

At 1/2-scale, the current is 0.0005 amp.
P = 0.0005 squared x 1000 = 0.00025 watt.

As the meter reads the same, with the chopper or resistor, the series
drop from a series resistor in place of the chopper is the same. The
load resistance inside the meter is the same. Therefore, the power in
both cases is identical.

Lossy resistance or lossless resistance produce the same effect in the
load.

Best regards, Richard Harrison, KB5WZI

*** MY inserted comments ***


"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"3) #1 with 50% chopper, [ILavg =0.5 ma... PL = 0.5 mW]"

At 1/2-scale, the *** AVERAGE *** current is 0.0005 amp.
P = 0.0005 squared x 1000 = 0.00025 watt.

***
Power is calculated from *RMS* values of voltage and/or current. Do you
disagree with this last statement? You previously quoted the RMS character
of the sine wave, so I know you understand the concept, yet you seem to be
denying that the pulsed wave need not follow the same laws. Or is it that
that you disagree that the 50-50 pulse has the .707 RMS value.


As the meter reads the same, with the chopper or resistor, the series
drop from a series resistor in place of the chopper is the same.

***
It is not. In the chopper case, the series drop is NEVER 0.5 volt.
It is either zero or one volt - changing with time. This mucks about with
your reasoning. Load resistance and *RMS* load current are used to
calculate the power, no? We do this for the sine wave, why not the pulse.

The load resistance inside the meter is the same. Therefore, the power in
both cases is identical.

**** Last sentence is incorrect.



Lossy resistance or lossless resistance produce the same effect in the
load.
Best regards, Richard Harrison, KB5WZI

********************
I am afraid this is incorrect. Yes, The AVERAGE current is, as you say,
the same, but the Effective or RMS current is as I said = 0.000707. Do you
wish to address your interpretation of the pulse RMS formula and the
apparent agreement with my power analysis copied again here?

----------------------------------------------------------------------------
--
When the switch is closed, Power in the load is PL= 1.0 mW. When open,
PL= 0.0. Therefore...the average Power in the load must be half or 0.5 mW.

----------------------------------------------------------------------------
--

If you stay with your chopper PL=0.25mw claim, then what do you say
about my argument, which uses the average power as a measure rather than the
average current? Do you claim that power does not average?
Why is the average current the correct parameter to arrive at the
correct answer and the average power not? And what do you say about the
commonly accepted square pulse formula P=Imax * SQUARE-ROOT(Duty-cycle)?

Imax - 1ma. ROOT(.5) = .707

I am quite puzzeled by your conclusion. Consult with any respected expert
of your choice.
73 & good DX (if desired) (:-) , Steve
--
Steve N, K,9;d, c. i My email has no u's.

I believe this thread originated with the paper on non-dissipative resistance by
Bob Lay. The original thread was entitled, 'Max power transfer theorem'. It
seems to have gone off track after a few postings.

I posted a msg on the original thread yesterday, 3-4-04, which at present is the
last posting in that thread. Since my posting there is pertinent to this
thread, I'd appreciate it if you'd all take a look at my thread there, rather
than have me repost it here, because it's rather long, but what I believe is an
important contribution to the thread.

TIA,

Walt Maxwell, W2DU

wrote:
Non-dissipative resistance is not well accepted or understood by many otherwise
well informed engineers, because it has had little or no (or even incorrect)
treatment in EE courses.

Yet the IEEE recognizes those two types of resistances with different definitions.
Definition (A) talks about "dissipation or other permanent loss". Definition (B)
simply says "The real part of impedance." Then a note: "Definitions (A) and (B)
are *NOT* equivalent ..." (emphasis mine)

The resistance in a resistor satisfies definition (A). The characteristic
impedance of a transmission line satisfies definition (B).
--
73, Cecil http://www.qsl.net/w5dxp



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Steve Nosko wrote:
"Power is calculated from "RMS" values of voltage and current."

Steve is correct! I apologize.

If you have a square wave of minimum peak to peak value of zero volts,
its maximum peak to peak value must be 1.414 x the d-c value. Then,
1.414 Vd-c x 1.414 Id-c = 2 Pd-c.

2 Pd-c x 1/2 t = Pd-c for an average.

Best regards, Richard Harrison, KB5WZI

Hi Walter,
I am sorry for carelessly screwing up on my lossless resistance
postings. The arithmetic wasn`t the center of my attention. I should
have been more careful. I`ve apologized to Steve Nisko.

Starving the load for power during a percentage of the cycle limits
power as surely as limiting current all of the time.

Time limitation can be nearly lossless. Resistor limitation is lossy and
reduces efficiency.

Best regards, Richard Harrison, KB5WZI

Hi Walt,

I was the one who called this paper "crap", and I will stick to that
characterization. (The wording is a bit strong, and I apologize for that.)

I have no argument with the notion of maximum power transfer or
non-dissipative resistance. My comment was based on the extensive use of
a goofy analogy to steam turbines. This sort of extended mechanical
analogy is pointless and unprofessional.

Unless we are attempting to retrain displaced marine propulsion
engineers, why should anyone assume that the reader will be more
knowledgeable about steam systems than electrical systems? It is likely
that the targeted reader is already more conversant with electrical systems.

This sort of analogy proves nothing. There is no "proof" that
non-dissipative resistance exists. This term is a "definition", not
something that can be proven. If one looks carefully, all of the "proof"
arguments are circular in nature. Indeed, this is the only possibility
when dealing with a definition. Self-consistency is all we can hope to
achieve.

The 29 pages of the subject paper could be reduced to less than 10
without any loss of important content. The other 19 pages are crap.
Majority rules.

73,
Gene
W4SZ


Walter Maxwell wrote:
I believe this thread originated with the paper on non-dissipative resistance by
Bob Lay. The original thread was entitled, 'Max power transfer theorem'. It
seems to have gone off track after a few postings.

I posted a msg on the original thread yesterday, 3-4-04, which at present is the
last posting in that thread. Since my posting there is pertinent to this
thread, I'd appreciate it if you'd all take a look at my thread there, rather
than have me repost it here, because it's rather long, but what I believe is an
important contribution to the thread.

TIA,

Walt Maxwell, W2DU

Gene Fuller wrote:
There is no "proof" that non-dissipative resistance exists. This term
is a "definition", not something that can be proven.

Gene, seems to me that the necessity of two "non-equivalent"
definitions of "resistance" in the IEEE dictionary is proof
of something that needs differentiating in the language.

Did trees exist before the word "tree" was invented?
--
73, Cecil http://www.qsl.net/w5dxp



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On Fri, 05 Mar 2004 23:11:00 -0600, Cecil Moore
wrote:
two "non-equivalent"
definitions of "resistance" in the IEEE dictionary is proof
of nothing

Huhhh???

Cecil Moore wrote:

Gene Fuller wrote:

There is no "proof" that non-dissipative resistance exists. This term

is a "definition", not something that can be proven.

Gene, seems to me that the necessity of two "non-equivalent"
definitions of "resistance" in the IEEE dictionary is proof
of something that needs differentiating in the language.

Did trees exist before the word "tree" was invented?

Richard Clark wrote:
wrote:
two "non-equivalent"
definitions of "resistance" in the IEEE dictionary is proof

of nothing

On the contrary, it is proof that the IEEE thinks it is
necessary to differentiate between the two definitions.
--
73, Cecil, W5DXP

On Sat, 06 Mar 2004 11:53:46 -0600, Cecil Moore
wrote:
On the contrary, it is proof
of nothing

On Fri, 05 Mar 2004 21:50:55 -0600, Cecil Moore
wrote:

wrote:
Non-dissipative resistance is not well accepted or understood by many otherwise
well informed engineers, because it has had little or no (or even incorrect)
treatment in EE courses.

On Sat. 06 Mar 2004 Cecil Moore wrote:

Yet the IEEE recognizes those two types of resistances with different definitions.
Definition (A) talks about "dissipation or other permanent loss". Definition (B)
simply says "The real part of impedance." Then a note: "Definitions (A) and (B)
are *NOT* equivalent ..." (emphasis mine)

The resistance in a resistor satisfies definition (A). The characteristic
impedance of a transmission line satisfies definition (B).

What you said above is true, Cecil, but one more statement applies to Definition
(B). Although Definitions (A) and (B) are not equivalent, Definition (B) does
include the real part of the impedance of a dissipative resistor. The only way
to tell which is which is to determine which develops heat.

I still maintain that many otherwise well qualified engineers not aware of
Definition (B), and therefore reject the concept of a resistance that doesn't
dissipate power. And this applies to much more than the Zo of transmission
lines.

And because there still remains many who believe the RF power amplifier absorbs
and dissipates reflected power, I chose to try again to dispel that notion in my
post in the 'max power theorem' thread.

Walt Maxwell, W2DU

Richard Clark wrote:
wrote:
On the contrary, it is proof

of nothing

Richard, how much do you pay for your blinders?
--
73, Cecil, W5DXP

On Sat, 06 Mar 2004 18:19:20 -0600, Cecil Moore
wrote:
Richard, how much do you pay for your blinders?
Trying to sell yours?