Cecil Moore wrote:
Roy Lewallen wrote:

In my third example, where does the other 10 watts of reflected power
go? If it goes to the load and back, why does it reflect off the
source resistor?


I have read the third example, which is NOT steady-state,
and I don't understand the question. Give me some steady-
state values and I will discuss it.

It most certainly is steady state, as are the other two examples.
Perhaps I wasn't clear in saying I was referring to my recent posting.
Here it is again:

[The setup is a 100 volt zero impedance voltage source in series with a
50 ohm resistor driving a half wavelength of 50 ohm transmission line.]

---------

Well, shoot, maybe the source resistor dissipates all the reverse power
*plus* some more power that comes from somewhere else. So let's try a
200 ohm load. Now the current is 0.4 amp, the power in the 200 ohm load
resistor is 32 watts, and the power in the 50 ohm source resistor is 8
watts. The SWR is 4:1, the forward power is 50 watts, and the reverse
power is 18 watts. Oops, the source resistor is only dissipating 8 watts
but the reverse power is 18 watts. Not only isn't it dissipating all the
reverse power, but it isn't even dissipating that extra power that came
from somewhere else when we connected the 16.67 ohm resistor. Wonder
where the other 10 watts of reverse power went?

---------

Must have bounced off the forward wave, then? If so, did it bounce only
when it reached the source, and not all along the line? Why? How much of
it bounced back at the load and why? Can you show me how you calculate
the various source resistor dissipations for this and the other two
examples using your bouncing average power model?

I'd also welcome H.'s comments on this.

Roy Lewallen, W7EL

Richard

I've seen 3-500Z where the solder had melted out of the filament pins.

I have some 811 from my 30L-1 that have holes in the plates, they still work
quite well. It's been 15 years since I recall finding the holes in the
plate sides and replacing. In additon I can't recall the reason for the
holes but it had to be either
1. overdriving (which I don't think I did, 100 watt Icom)
2. high swr which I know has been over 10 to 1, but I can't see how that
could do it..
3. Light loading

But in any event for something to melt we need dissipation and only
resistance can do that!

--

73
Hank WD5JFR
Still learning, un-learning and re-learning

Roy Lewallen wrote:
Cecil Moore wrote:

Roy Lewallen wrote:
Does your analysis produce the result of 2.3 dB loss claimed by H.
for a 1.7:1 SWR?

Cheap friggin' damn shot, Roy, after my posting where I disagreed
with H. and agreed with your calculations.

Sorry, it was an honest question. I saw your posting re the SWR
calculations, but guess I missed the one you mention. H's calculations
remain a mystery to me, as apparently they do to you.

Roy Lewallen, W7EL

Really strange response since you already responded to the posting
where I agreed with you. Are you as senile as I? :-)
--
73, Cecil http://www.qsl.net/w5dxp

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Roy Lewallen wrote:
... and the power in the 50 ohm source resistor is 8 watts.

Sorry, Roy, the Thevenin equivalent source model expressly forbids
you from making that assumption. You are going to have to do better
than that. What is the DC power input to the source electronics?
--
73, Cecil http://www.qsl.net/w5dxp

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On Mon, 06 Jun 2005 20:52:52 GMT, "Henry Kolesnik"
wrote:

But in any event for something to melt we need dissipation and only
resistance can do that!


Hi Hank,

Unfortunately as much as you and I agree on that bedrock principle,
others with Simpson Ohmmeter in hand would glare goggle eyed at us and
say that plate has no resistance to speak of and that no amount of
current through its Ohmic resistance could ever bring about enough
heat to produce the effects so obviously witnessed. One of the most
enraging questions I've asked
"If it is not the value I've offered, what value is it?"

Well, I've never been given a quantitative answer, however I've seen
enough carefully crafted mathematical proofs in this group to replace
substantive results so easily seen. There is some irrefutable logic
in circulation that clearly reveals that what we've experienced just
couldn't be.

Glasses will be need to be readjusted for such extreme myopic
aberrations. There are two principles involved in what is called
Plate Resistance, and the first and foremost is not even related to
the plate at all. It is called the work function of the cathode
emissivity. So, in fact it is more proper to refer to this usual loss
as Cathode Resistance, not Plate Resistance. The cathode is the
fundamental limit on power generated.

What Plate Resistance is, is the ill termed substitution for Plate
dissipation. If folks want to work their Simpson, they would blow an
aneurysm trying to measure the resistance from cathode (filaments have
the same work function issue too) to plate. In fact, the hobby horse
argument of it is not resistance at all, but some figurative charting
artifice called a "load line" usually appears in the last gasp.

Plate Dissipation is resistance clear and simple in spite of the
failure of conventional tools to measure a common physical property.
Newton would have recognized it, it is called inertia.

Once the work function is overcome (the job of the grid), then Plate
voltage dominates through the acceleration of charge beyond the grid,
toward the plate. That stream of electrons (and there is no doubt
about actual current flow in easily counted, significant populations
of electrons) is elevated to 90% the speed of light. This current
flow is entirely different from what current flows in the remainder of
the Plate load. That is also known as displacement current and
electrons are shuffling along at a placid meter per second rate.
Plate current and displacement current are equal in amplitude and
phase, but not in motion nor kinetics.

NOW. When that same stream encounters the Plate - WHAM! If anyone
here has walked into the wall, and NOT encountered resistance, then we
will call you Casper.

Inertia reveals that to slow a mass in a distance results in
acceleration (negative in this instance) and that property is called
Force. Force over time expends calories and is expressed in any
number of systems and units - Watts is one, Degrees is another. We
could abstract to Horsepower and Candelas (the plate glows too).

We know the speed, not many here would give it much though, but none
would know the length interval of going at that speed to going zero
(0). It is roughly two atoms distance into the metal of the plate. I
will leave those calculations of Force to the student to compute or I
can provide it from notes of correspondence with Walt Maxwell and
Richard Harrison from a round robin discussion several years ago.

Hank, does this fulfill your earlier question as to "what" is
happening? I first gave you many examples, I hope this segue into
real physics fills in their actuality. Too many correspondents demand
that I open the source and point at a 50 Ohm carbon composition
resistor that is the "source resistance."

73's
Richard Clark, KB7QHC

Cecil Moore wrote:
Roy Lewallen wrote:

... and the power in the 50 ohm source resistor is 8 watts.


Sorry, Roy, the Thevenin equivalent source model expressly forbids
you from making that assumption. You are going to have to do better
than that. What is the DC power input to the source electronics?

Who said anything about a Thevenin equivalent? The circuit I proposed
isn't intended to be an equivalent of anything. It's a very simple
circuit made with components whose characteristics are well defined and
well known. The source is an AC steady state source (which I naively
assumed was obvious). There is no DC power input to it. There are no
"source electronics" -- there are no electronics at all. All elements in
my example are simple electrical circuit components, as described in any
elementary electrical circuits textbook.

There are 18 watts of "reverse power" in the transmission line, as
calculated by those who embrace this concept. The source resistor
matches the Z0 of the line. The source resistor dissipates 8 watts.
Nothing "forbids" calculation of the dissipation of the resistor -- it's
V^2/R, I^2*R, or V*I, take your choice. Any EE freshman, and I'd like to
think most amateurs, including you, should be able to calculate it. The
total power from the source equals the sum of the dissipation in the two
resistors. The power dissipated by the load is the difference between
the "forward power" and "reverse power", as you can easily see with a
small amount of simple arithmetic. There is nothing "forbidden" about
this simple circuit, except perhaps explanation by means of your theory.

Where's the "reverse power" going?

Any theory that doesn't work in a circuit composed of simple elemental
electrical circuit elements is suspect to say the least. Are you saying
yours doesn't?

Roy Lewallen, W7EL

Cecil,

Nice try.

You first.

Describe how you set up this coherent wave/anti-wave pair that happily
travel together for some indeterminate distance. Then I will describe
what happens when at some arbitrary point and time they decide to
annihilate.

I will repeat one more time, since you did not seem to understand
previously.

* Maxwell's equations are all that is needed.

* Interference is *derived* from the correct application of Maxwell's
equations. It is not an independent physical entity.

* Interference may be "intuitive" and it may help your understanding,
but it adds precisely nothing to the physics of the problem. Everything
you believe is buried in the magic of interference is already built into
Maxwell's equations.

73,
Gene
W4SZ



Cecil Moore wrote:
Gene Fuller wrote:

There is no need to invoke interference or wave cancellation to
explain anything, ...


But there is, Gene. It's the only way to correct the present
misinformation and old wives' tales being promoted on this
newsgroup. It is obvious that the r.r.a.a poster who
understands the role that interference plays in the
conservation of RF energy is very rare.

There is a conspiracy to keep this information from
surfacing - "Nothing new", "no need", "irrelevant",
"who cares?" Why are you guys afraid to discuss the
technical details?

This should be an easy question to answer. If two coherent
waves of 50 joules/sec each, are traveling in the same
path in the same direction and are 180 degrees out of
phase, they cancel in that direction of travel. What
happens to their 50+50 joules/sec? Hint: energy doesn't
cancel and there are only two possible directions. Can
you spell R-E-F-L-E-C-T-I-O-N?

Roy Lewallen wrote:
Where's the "reverse power" going?

Since we can calculate 23 watts of constructive interference
occuring toward the load, using the conservation of energy
principle as explained by Hecht in "Optics", we can deduce
that the reflected power is engaged in destructive interference
inside the black box source. Using Dr. Best's conventions:
V1 = 32v, V2 = 18v, Vfor = 50v, Vref = 30v

Any theory that doesn't work in a circuit composed of simple elemental
electrical circuit elements is suspect to say the least. Are you saying
yours doesn't?

It works just fine. Pfor = P1 + P2 + constructive interference.
But that still looks like a Thevenin equivalent to me, you know,
the one we cannot trust for internal power calculations.
--
73, Cecil http://www.qsl.net/w5dxp


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Gene Fuller wrote:

Cecil,

Nice try.

You first.

Describe how you set up this coherent wave/anti-wave pair that happily
travel together for some indeterminate distance. Then I will describe
what happens when at some arbitrary point and time they decide to
annihilate.

Sure, here's the two coherent reflected waves that cancel at a
Z0-matched impedance discontinuity in a transmission line.

b1 = s11*a1 + s12*a2 = 0

I'm sure you recognize the S-parameter equation for the reflected
voltage flowing toward the source which is the phasor sum of two
other reflected voltages.

They don't travel together for some indeterminate distance. They
are cancelled within the first dl and dt. And they don't annihilate.
They simply cancel in the rearward direction.

Incidentally, if you square both sides of the equation you get

b1^2 = s11^2*a1^2 + s12^2*a2^2 + 2*s11*a1*s12*a2

Pref1 = rho^2*Pfor1 + (1-rho^2)*Pref2 + interference

The forward voltage equation toward the load is b2 = s21*a1 + s22*a2
--
73, Cecil http://www.qsl.net/w5dxp


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Your use of "Dr. Best's conventions" only muddles the matter -- neither
I nor probably any of the other readers have any idea what this is. In
any event, the numbers you produced are volts. I and many others know
how to calculate forward and reflected voltages and currents, and their
sums. What's at issue here is where the imaginary waves of average power
are going, what they're bouncing off, and why. Correct me if I'm wrong,
but power is generally expressed in watts, BTUs per hour, or other more
arcane units, but not volts.

Let's try again. The source is providing 40 watts, 32 watts of which is
delivered to the transmission line. The transmission line is
transferring this 32 watts of power to the load. In the transmission
line, we can calculate that there's 50 watts of "forward power", and 18
watts of "reverse power". How much of that 18 watts of reverse power is
going through the source resistor to reach the source to "engage in
destructive interference"? What does it interfere with? How does
whatever it interferes with get there? Does any of the power going
either way, forward or reverse, get dissipated in the source resistor?
If so, how much and why? If not, why not?

In your "explanation", I don't see a single figure in watts, except that
"we have 23 watts of constructive interference occuring toward the
load". (Where is this interference occurring, that is, just where is the
point "toward the load" located? Where did the 23 watt figure come from?
How much of it is "forward power" and how much "reverse power"?) It's
not an explanation at all, but hand-waving.

And why do you insist that every combination of voltage source and
resistor be a "Thevenin equivalent"? I suggest you go back and read your
basic circuits texts, where you'll find that a Thevenin equivalent is a
circuit which is used to substitute for a more complex linear circuit to
simplify analysis. The electrical circuit components used here are not a
substitute for anything. And there's no rule, except something
apparently stuck firmly in your mind(*), which prohibits calculating the
power dissipated in a resistor. No matter what it's connected to. I make
no claim that the power dissipated by the resistor represents anything
but the power dissipated in a resistor, that the resistor represents
anything but a resistor and the voltage source anything but a voltage
source. It is not a Thevenin equivalent, it's a painfully simple
electrical circuit (alas, so simple it's difficult to obfuscate). It
doesn't matter if you can trust a Thevenin equivalent for internal power
calculations. There is no "internal power" here -- it's all out in the
open where we can easily measure and calculate it.

There's nothing in that "black box source" except an ideal voltage
source. You can find a description of this fundamental electrical
circuit component, including its complete terminal characteristics, in
the circuit analysis textbook of your choice. People skilled in the art
are able to calculate the power it produces by multiplying v across it
times i flowing from it, and the average power by applying the
mathematical definition of average to the calculated power. In this
case, it produces an average of 40 watts (100 volts times 0.4 amp).

So please tell us how many watts are going where, what they do when they
get there, and why. If you can't, you don't have a theory at all.

(*)Forgive me, I just can't shake the image of a certain memorable scene
from the movie "The Long Kiss Goodbye" as I write this.

Roy Lewallen, W7EL

Cecil Moore wrote:
Roy Lewallen wrote:

Where's the "reverse power" going?


Since we can calculate 23 watts of constructive interference
occuring toward the load, using the conservation of energy
principle as explained by Hecht in "Optics", we can deduce
that the reflected power is engaged in destructive interference
inside the black box source. Using Dr. Best's conventions:
V1 = 32v, V2 = 18v, Vfor = 50v, Vref = 30v

Any theory that doesn't work in a circuit composed of simple elemental
electrical circuit elements is suspect to say the least. Are you
saying yours doesn't?


It works just fine. Pfor = P1 + P2 + constructive interference.
But that still looks like a Thevenin equivalent to me, you know,
the one we cannot trust for internal power calculations.