Richard Clark wrote in message . ..

Thank you Bill. This is probably why they say that
an oscillator has negative resistance (postitive feedback),
and also why the stability circles have their centers located
outside the unit 1 Radius. Clearly, areas outside the
rho=1 circle on the Smith, are reserved for active networks.


Slick


Hi OM,

This does not follow logically. Active networks exhibit this behavior
and offer a library full of applications. Strained passive networks,
as evidenced by other correspondence, also reside in this "forbidden
region" but their number are significantly smaller and for good
reason.


Please give an example.


Slick

wrote in message ...
W5DXP wrote:

wrote:
But that does not address the issue with the generally
accepted formula; how can reflected power be greater
than incident?

The transmission coefficient equals (2*Z2)/(Z2-Z1). If
Z2 is large compared to Z1, the transmitted voltage can
be large compared to the incident voltage.

Yes indeed. And since this happens when the impedance
increases, energy can be conserved.


Agreed. Extremely important point.


But the discussion was about reflected voltage and how using
the generally accepted formula can result in the reflected
voltage exceeding the incident voltage. Converting this
voltage to power would result in a greater reflected power
than incident which would seem to violate conservation of
energy. Perhaps this would be an interesting place for
you to apply your energy analysis?

...Keith


Someone seems to understand the gist...


Slick

"William E. Sabin" sabinw@mwci-news wrote in message ...

For a moderately complex Z0 (moderately lossy
cable), the conventional formula for rho can have
magnitude values a little greater than 1.0. The
boundary of the Smith chart represents rho=1.0,
but a complex Z0 can push rho a little beyond the
circumference. Because SWR is extremely high at
the outer circle of the Smith chart, we should
avoid that region in our work with the Smith chart.


Incorrect, Bill. Rho greater than one is for active
networks only.


Slick

On 29 Aug 2003 20:00:30 -0700, (Dr. Slick) wrote:

Hi OM,

This does not follow logically. Active networks exhibit this behavior
and offer a library full of applications. Strained passive networks,
as evidenced by other correspondence, also reside in this "forbidden
region" but their number are significantly smaller and for good
reason.


Please give an example.


Slick

?

Seems you have been arguing over one for quite some time now. I see
no sense in flogging that dead mare. My money's on the bob-tailed
nag. Certainly no one's bet on the bay.

73's
Richard Clark, KB7QHC

On Fri, 29 Aug 2003 23:36:02 -0500, W5DXP
wrote:

Richard Clark wrote:

W5DXP wrote:
But: "Every time you make a measurement, you make an error." So said
my EE prof almost half a century ago. I assume it's still true. :-)

Yes.

What can I assume from the fact that you ignored the tough questions
in the rest of my posting?

Hi Cecil,

I have no idea what you can assume. You generally express it freely
anyway.

How do you measure the exact amount of
reflected power re-reflected from the source?

Dan's spreadsheet work seems like a good place to start. It is of no
particular interest to me and doesn't respond to the challenge in and
of itself. You might find an exact answer there. Simply going to the
bench would be adequate for amateurs, guessing would eclipse the
efforts of everyone else. ;-)

It should seem obvious that the power reflected at the source
interface is governed by the standard mechanics no one wishes to
impart to the source. In other words the power returning to the
source will be reflected through the same mechanics it met at the
mismatched load, with the "exact" value varying only by the reflection
coefficient of the source mismatch and the power incident upon it. As
the line presents a media of 50 Ohms, and the source presents a 100
Ohm discontinuity, the portion of power reflected is a rather trivial
computation - except for those who apparently employ "first
principles" in one direction only. Gad, I love that bit of irony!
And they do it for the sake of educating lurkers too!

Hi Ian, George, you will find this in chapter 9 from Chipman (got it
right that time George ;-). It also accounts for the increased loss
exhibited by the example of the Challenge that so baffles everyone. I
wonder what those so bench-shy would attribute that additional loss
to? Would their astrologers suggest the opposition of Mars? Hint:
the answer works at all aspects of any planets.

For those who closely follow their astrologer's advice, I once again
suggest this only matters to those interested in accuracy. With
equipment available across the counter in exchange for a credit card
charge, their purchase is not likely to suffer any issues brought to
the forum here and they can rest assured their SWR meters will work as
advertised given the likely source Z of being 50 Ohms or
insignificantly off from that value. (Many will gladly suffer 2:1
mismatch straight from the antenna connector so they can worry it at
the antenna.) Further, for inferior purchases at similar cost (how
would they know?), they will still be unaware barring some change in
the length of cabling that will have them muttering a moment or two
before they shrug it off anyway.

73's
Richard Clark, KB7QHC

"W5DXP" wrote in message
...
Tarmo Tammaru wrote:
...and all reflected power was re reflected.

Since the re-reflected waves are coherent with the source waves,
how do you know that? Would the amplitude of ghosting in a TV signal
support that assertion?
--
73, Cecil http://www.qsl.net/w5dxp

There was no actual power generated by the source. Some people seem to think
that *power* dissipated by the resistor due to current flowing in one
direction and *power* caused by current in the other direction cancel. Of
course, the experiment was moot, or 1/4 wave stubs wouldn't work. Ghosts
live in the realm of pulses, where things always work unambiguously. In
logic design, you often terminate the source, so there is only one
reflection. You can't series terminate the load, because the input impedance
is several K; you can't shunt terminate the load because the source can't
deliver enough current, although I have done things like shunt terminating
the load with a Zo resistor in series with 10PF or so.

Tam/WB2TT

Ian White, G3SEK wrote:
I believe you have specified more numbers than you can actually know.

That's the whole point, Ian. Exactly how are those things knowable
if they give identical readings? I could argue that either of those
conditions exist and you cannot prove otherwise. There are probably
an infinite number of configurations that will yield identical
measurements. What I don't understand is how any of them can be
knowable.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Tarmo Tammaru wrote:
There was no actual power generated by the source.

How do you know? Was it 100% efficient and consumed no
power?
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

W5DXP wrote:
Ian White, G3SEK wrote:
[Re-inserting my quote from Cecil]
1. The source is generating 100 watts and 20 watts of the 50 watts of
incident reflected power is being re-reflected in phase from the
source.
The forward power meter reads 120 watts. The reflected power meter
reads 50 watts.

The conventional rule-of-thumb has the source generating (100-30)=
70 watts since it is dissipating 30 watts of reflected power.

2. The source is generating 110 watts and 10 watts of the 50 watts of
incident reflected power is being re-reflected in phase from the
source.
The forward power meter reads 120 watts. The reflected power meter
reads 50 watts.

The conventional rule-of-thumb has the source generating (110-40)=
70 watts since it is dissipating 40 watts of reflected power.

How can you possibly distinguish between the above two identical
conditions caused by different source impedances?

I believe you have specified more numbers than you can actually know.

That's the whole point, Ian. Exactly how are those things knowable
if they give identical readings? I could argue that either of those
conditions exist and you cannot prove otherwise. There are probably
an infinite number of configurations that will yield identical
measurements. What I don't understand is how any of them can be
knowable.


Sorry, I didn't make myself clear. I believe you have specified more
numbers than you can know, in *each* example.

Therefore trying to distinguish between them is double-doomed :-))


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

On Sat, 30 Aug 2003 11:58:55 -0500, W5DXP
wrote:

Richard Clark wrote:
Simply going to the
bench would be adequate for amateurs, guessing would eclipse the
efforts of everyone else. ;-)

I'm ready to go to the bench but I don't know how to separate the
incident reflected power that is re-reflected from the source from
the generated power. That's what I am asking. How does one separate
those two coherent superposed forward waves?

It should seem obvious that the power reflected at the source
interface is governed by the standard mechanics no one wishes to
impart to the source.

I agree that the source obeys the wave reflection model rules. The
fly in the ointment is the unknowable source impedance encountered
by the incident reflected waves.

Worry about it if the example of the challenge presents more than 3+
dB loss. If not, there's hardly any point is there? We can work out
the differences then, but only if you care about accuracy and the lack
of confirming this exotic characteristic. I can appreciate that this
is not everyone's interest.


In other words the power returning to the
source will be reflected through the same mechanics it met at the
mismatched load, with the "exact" value varying only by the reflection
coefficient of the source mismatch and the power incident upon it. As
the line presents a media of 50 Ohms, and the source presents a 100
Ohm discontinuity, ...

I must have missed how you know the source presents a 100 ohm impedance
to incident reflected waves. Do you have a 100 ohm pad between the
transmitter output and the transmission line?

Hi Cecil,

I didn't say I did the example, it is drawn from a reference. We've
already been through the mechanics of how to do it employing a
variable transmission line. Consult our correspondence for specific
details, you already offered that your equipment could tolerate that
mismatch, however THAT discussion is separate and distinct from the
example of the challenge. The challenge merely offers another
approach. You can add a 50 Ohm Dummy Load in series to the output of
your rig (this, of course presumes it presents 50 Ohms characteristic,
but as many declaim that specification perhaps they could offer
another value - eh, unlikely).

Another issue of adding a series resistor is one of shielding and
common mode issues. You asked me in that earlier correspondence if I
considered this, and yes I did. That is why I used massive parallel
loads that insured an entirely shielded system. This is one of those
methods that one correspondent pondered:
There is no institutionalized ignorance, just a
lot of skepticism regarding the reliability of the
analysis methods and the measurement methods.
which is understandable from those not trained in the art of designing
measurement scenarios. I am trained but those still caught in the
quandary are immobilized by rejecting every method (institutionalized
ignorance). The measure of RF power is not simple by any means so it
is best left to those who are serious about accuracy - hardly an
amateur pursuit, and it hardly matters anyway as it is not a problem
with modern equipment.

I offered that you already had the tools to perform a simple first
pass approximation, you really need to consult that thread of
correspondence again. None of this with forced mismatches is all that
hard in the first place. The greater the mismatch at each end, the
more compelling the evidence. The simple fact of the matter is that
most rigs conform to 50 Ohm source Z and do not exhibit this issue.
If those who held to their cherished fantasy of source Z being other
than 50 Ohms, then they should be able to ace this test from the
beginning (the lack of their correspondence reveals the invalidity of
their claims).

I'm working on two repeater systems today (10M and GMRS), so enjoy.

73's
Richard Clark, KB7QHC