On Fri, 29 Aug 2003 08:25:06 -0700, W5DXP
wrote:

Now, be advised that when I say "accurately" that this is of concern
only to those who care for accuracy.

That's the part I don't understand. You can assume a whole range of
impedances for the source while the forward power and reflected power
remain the same. Is "accuracy" somehow involved with efficiency?
--

Hi Cecil,

This is a real (the example of the challenge) issue of the Mismatch
Uncertainty. It is only uncertain insofar as most folks are unaware
of the contribution of error the source presents, and doubly unaware
of the phase distances between the interfaces.

Accuracy is unrelated to efficiency, both valuations exist distinct
from the other. You can be efficient but express it with poor
accuracy or good accuracy, efficiency cares not a whit about your poor
situation to resolve it. You can also be pristine accurate and
horribly inefficient, here again, accuracy is not a function of
efficiency, accuracy cares not a whit about your waste.

As for your conjecture
You can assume a whole range of
impedances for the source while the forward power and reflected power
remain the same.
I assume nothing, but I can portray the range through the same
challenge's example. It merely involves moving the SWR measurement.
I have already demonstrated that.

73's
Richard Clark, KB7QHC

On Fri, 29 Aug 2003 08:27:09 -0700, W5DXP
wrote:

Richard Clark wrote:
You have expended far more time climbing Everest in search of the
guru, than simply doing the challenge at the bench.

I don't understand the challenge. Until I do, I would be wasting
my time on the bench. I'm sure I'm not the only one who doesn't
understand what you are trying to say.

Hi Cecil,

Yes, it is terribly cryptic: how much heat is lost in the line?

It requires advanced techniques in measuring power lost over a known
stretch of transmission line. This is an art unknown to many here and
baffles them in how to proceed when written directions are provided.

Even if they reject those methods and directions, they struggle
through massive tomes to come to no conclusion. What's more, the
example resides in one of their cherished scriptures that many would
swear by and would scream outrage at the suggestion that they might
read it.

The irony is delicious.

73's
Richard Clark, KB7QHC

On Fri, 29 Aug 2003 12:00:10 -0700, W5DXP
wrote:

Richard Clark wrote:
Accuracy is unrelated to efficiency, ...

This is what I thought. I just wanted to be sure you were thinking
the same way.

I assume nothing, but I can portray the range through the same
challenge's example. It merely involves moving the SWR measurement.
I have already demonstrated that.

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. :-)


Hi Cecil,

Yes.

73's
Richard Clark, KB7QHC

"Tarmo Tammaru" wrote in message ...
Why didn't you do the gamma for a shorted line using your formula? I think
you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes to -Z0*/Z0,
which, for Zo having phase angle b equates to -1 at angle(-2b).


Agreed. This doesn't prove anything, does it?


This is in more detail of what ACF do:

1. Normalize the RC to (Zn -1)/(Zn +1)
2. Let's say we agree that for biggest magnitude, this has to be in the left
hand plane.
3. Draw a vector for Zn = a at -135 deg
4. Draw vectors for Zn - 1 and Zn +1. Note that these 3 vectors have the
same y coordinate.
5. You now can draw two triangles with the corners at 0, Zn, and Zn -1 for
one, and 0, Zn, and Zn+1 for the other.
6. Solve for Zn+/-1 in terms of "a"
7. By plane geometry the magnitude of, of Zn -1=
SQRT(1 + a^2 + aSQRT(2))
8. The magnitude of Zn +1 is
SQRT(1 + a^2 - aSQRT(2))
9. Square both sides of the equation, and gamma^2=
(1 + a^2 + aSQRT(2))/(1 + a^2 - aSQRT(2))
10 This equates to 1 + (2SQRT(2))/((a + 1/a) - SQRT(2))
11 |Gamma| max occurs when (a + 1/a) is a minimum, which is 2 at a=1.
12|Gamma| max is 1 + SQRT(2)

They anticipate people being concerned about |Gamma| 1 and later come up
with a formula for time average power. I don't know that looking at it is
going to give anybody any insight, but for this is what they end up with ( I
am typing CM for Gamma):

P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)]

Remember, z is distance from the load.

Tam/WB2TT



ok, you've done a nice job of copying the text you sent me.

They also mention that the normalized load impedance Zn=Zr/Zo does
NOT have the same angle as Zr because Zo is complex in the general
case.

"They anticipate people being concerned about |Gamma| 1 and later come up
with a formula for time average power."

Really? Could you tell us more about that? Could you email me
more pages, especially the one that has Eq. 5.2b?



Slick

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

Given a line's primary characteristics, R,L,C,G,
length, or it's secondary characteristics Zo, dB, phase
angle, plus the line's terminatiing impedance it is
possible to calculate, by classical methods, all other
quantities of engineering interest - WITHOUT ANY
REFERENCE TO REFLECTION COEFFICIENT OR SWR which are
mere man-made notions supposed to assist understanding
of what goes on in the real world but, as exchanges on
this newsgroup show, are just a pair of bloody useless
nuisances.

Nevertheless, the outer circle of the Smith chart
is *always*
the locus of zero positive resistance and infinite
SWR, and a rho vector cannot terminate on, or
cross over, this circle when a load R0 is
present, regardless of the rest of the circuit,
including any possible combination of resistances
and reactances and complex Z0.



"gasp!" ...saved by a lifesaver of logic while drowning
in the Sea of Ignorance!

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



One can argue "ignore rho=1 and just jump over
it". This cannot be done in good mathematics.

Dismissing rho and SWR as "contrived nuisances" is
a convenient way to get rid of this problem, but
it does not "wash". Rho and SWR are fundamental
properties of transmission lines that do not go
away, and a non-zero R precludes rho=1.0.

Any attempt to circumvent (bypass) these small
inconveniences is doomed to failure, regardless of
the analytic geometry considerations.

Bill W0IYH

On 29 Aug 2003 13:41:35 -0700, (Dr. Slick) wrote:


"gasp!" ...saved by a lifesaver of logic while drowning
in the Sea of Ignorance!

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.

Those examples were "strained" for the purpose of your education and
rarely offer engineering coupes of design. More often they represent
the "gotcha's" of technical oversight and unplanned failure modes. As
such these side-bars offer insight, but hardly have the horsepower to
pull the load.

Both sides of the debate can whip that horse, but it ain't moving
another inch for anyone.

As for a horse of a different color, how well tutored are you to
handle specifying the loss in a system of two resistors and a hank of
line? (Line loss is the characteristic being sought.) No prizes are
offered, but you have and excellent prospect of beating the experts to
the crossing line.

It's as simple as giving even an incorrect answer! :-)

As the rich and powerful offer, the better part of success is just in
showing up.

73's
Richard Clark, KB7QHC

Bill:

[snip]
In the case of rather lossy cable, it is best to
abandon rho and SWR before they become "bloody
useless nuisances", and use the computer for
everything.

Bill W0IYH
[snip]

Amen Brother!

--
Peter K1PO
Indialatic By-the-Sea, FL.

"Dr. Slick" wrote in message
om...
"Tarmo Tammaru" wrote in message
...
Why didn't you do the gamma for a shorted line using your formula? I
think
you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes to -Z0*/Z0,
which, for Zo having phase angle b equates to -1 at angle(-2b).


Agreed. This doesn't prove anything, does it?

How does this allow for the sum of V+ and V- to be 0? That is what you have
across a short.

..........................................
P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)]

Remember, z is distance from the load.

Tam/WB2TT



ok, you've done a nice job of copying the text you sent me.

As I recall, you said you were not familiar with these diagrams and did not
understand them. What do you want me to do? derive it in a different way?


They also mention that the normalized load impedance Zn=Zr/Zo does
NOT have the same angle as Zr because Zo is complex in the general
case.

"They anticipate people being concerned about |Gamma| 1 and later
come up
with a formula for time average power."

Really? Could you tell us more about that? Could you email me
more pages, especially the one that has Eq. 5.2b?



Slick

I can't scan the whole chapter, and part of the next. You should be able to
get your library to borrow the book for you. Or, look at some other college
textbook under lossy lines. 5.2B is straightforward enough:

Zo=(1/Yo)=SQRT[(R+jwL)/(G+jwC)]=Ro+jXo

Tam/WB2TT

"W5DXP" wrote in message
...
Consider the following two examples:

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?
--
73, Cecil, W5DXP

Cecil,

I actually did a test along these lines. Believe we discussed that some
weeks ago before I did it.

1/4 wave 50 Ohm line shorted at the end. Freq ~ 2 MHz

50 Ohm generator in series with a 75 ohm resistor at the other end.

Looked at both ends of the 75 Ohms, both to ground, and differential.

At resonance, all voltages were in phase.

Unfortunately I used very low loss line, but to a good approximatio, the
only power supplied by the source could be explained by cable loss, and all
reflected power was re reflected. It would be interesting to do this with a
total source impedance of 50.

BTW, I also did the thing with measuring VSWR and then changing the source
impedance. Power changed, VSWR did not, for either 1:1 or 1.7:1.

Tam/WB2TT

In the case of rather lossy cable, it is best to
abandon rho and SWR before they become "bloody
useless nuisances", and use the computer for
everything.

Bill W0IYH
[snip]

Amen Brother!

--
Peter K1PO
Indialatic By-the-Sea, FL.

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

Amen ?

Peter, does "Amen" imply worship of Terman and the ARRL
Handbook has now been heretically switched to a mere
box of electronics ? Heaven help us!
----
Reg, G4FGQ