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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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

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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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.

Roy Lewallen wrote:

We are not going to get anywhere until you admit there is 68
joules/sec in the feedline that haven't yet made it to the
load. Once you admit that fact, everything else will be moot.

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".

And it is easy to prove that the source has generated 50+18=68
watts that have not been delivered to the load. So I ask
you: Where are those 68 joules/sec located during steady-state
if not in the forward and reflected power waves? Why will
68 joules/sec be dissipated in the system *after* the source
power is turned off?

If those 68 joules/sec that have been generated by the source
but not delivered to the load are not in the forward and reflected
power waves, exactly where are they located? There's really no
sense in continuing this discussion until you answer that question.
Everything else is just a side argument.

The answer to that question will expose the errors in your premises.
You are apparently assuming there is not enough energy in the system
during steady-state to support the forward and reflected power waves.
But that exact amount of energy was supplied during the power-on
transient state and will be dissipated during the power-off transient
state. If it's not in the forward and reflected power waves, you
are going to have to store it somewhere else. Where is that
somewhere else?

The source has supplied 68 joules/sec that has not reached the
load. The forward and reflected power waves require 68 joules/sec.
That you don't see the logical connection between those two equal
energy values is amazing.

But I will get you started on an understanding of the component
powers using an S-parameter analysis.

How much of that 18 watts of reverse power is
going through the source resistor to reach the source to "engage in
destructive interference"?

reference the S-parameter equation: b1 = s11*a1 + s12*a2

I calculate 11.52 watts. (s12*a2)^2 = 11.52 watts

The other 6.48 watts are in (s22*a2)^2 where |a2|^2 = 18 watts

What does it interfere with?

From the S-parameter equation above, it obviously interferes
with s11*a1 .

Please reference HP App Note 95-1, available on the web. It should
answer most of your questions, in particular pages 16 & 17.
|a1|^2 = Power incident on the input of the network
|a2|^2 = Power incident on the output of the network
|b1|^2 = Power reflected from the input port of the network
|b2|^2 = Power reflected from the output port of the network
--
73, Cecil http://www.qsl.net/w5dxp


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Please reference HP App Note 95-1, available on the web.

There's only one in every half a million of radio amateurs who have a
copy of this document, or have ever heard of its existence, let alone
having any chance of finding a readable copy of it within the next
dozen years. By which time they will have changed their hobby to
keeping tropical fish or making Newtonian telescopes. Or just died.

In all likelihood they won't be able to make any sense out of it
anyway.

Cec, why do you bother to mention it? (smiley)
----
Reg.

Reg Edwards wrote:

W5DXP wrote:
Please reference HP App Note 95-1, available on the web.

There's only one in every half a million of radio amateurs who have a
copy of this document, or have ever heard of its existence, let alone
having any chance of finding a readable copy of it within the next
dozen years.

Give me a break, Reg. Download it for free and enjoy from:

http://www.sss-mag.com/pdf/hpan95-1.pdf

In all likelihood they won't be able to make any sense out of it
anyway.

My dog isn't able to make sense of it either - poor dog.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Roy Lewallen wrote:

[No, I didn't. Cecil wrote the following paragraph.]

We are not going to get anywhere until you admit there is 68
joules/sec in the feedline that haven't yet made it to the load. Once
you admit that fact, everything else will be moot.

I never said a word about how much energy is stored in the feedline;
it's irrelevant. It's stored there during the initial charging to the
steady state condition, and the same amount remains there until the
steady state condition no longer exists. It's exactly like the DC charge
on a capacitor or a DC current through an inductor (which, in fact, is
exactly what the feedline stored energy consists of) -- it doesn't have
any effect on an AC analysis.

[I did write this one.]

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".


And it is easy to prove that the source has generated 50+18=68 watts
that have not been delivered to the load.

Surely even you can do the basic circuit analysis which shows that the
source is continuously generating 40 watts, not 60. 32 of those are
delivered to the load and 8 to the source resistor. I guess you mean 68
joules -- but as I said, it's irrelevant.

So I ask you: Where are
those 68 joules/sec located during steady-state if not in the forward
and reflected power waves? Why will 68 joules/sec be dissipated in
the system *after* the source power is turned off?

The line's stored energy will be dissipated in either the source or load
resistor or both when the source power is turned off. But we're doing a
steady state analysis here.


If those 68 joules/sec that have been generated by the source but not
delivered to the load are not in the forward and reflected power
waves, exactly where are they located? There's really no sense in
continuing this discussion until you answer that question. Everything
else is just a side argument.

You tell me -- they can be anywhere you'd like. Just answer the simple
questions about the power "waves".


The answer to that question will expose the errors in your premises.

What exactly is my premise, please? All I've done is to give the
currents and powers at significant points in the circuit. Are any of the
values incorrect? It's you who has the premise, not me.

You are apparently assuming there is not enough energy in the system
during steady-state to support the forward and reflected power waves.

I'm making no such assumption. I'm questioning the existence of
traveling waves of average power, and so far you've failed to give any
evidence to convince me otherwise.

But that exact amount of energy was supplied during the power-on
transient state and will be dissipated during the power-off transient
state. If it's not in the forward and reflected power waves, you are
going to have to store it somewhere else. Where is that somewhere
else?

You tell me. My analysis doesn't need to consider the stored energy at
all. Apparently yours does, so have at it.

The source has supplied 68 joules/sec that has not reached the load.

I think you mean 68 joules.

The forward and reflected power waves require 68 joules/sec.

The "forward power" is 50 watts. The "reverse power" is 18 watts. It
requires 32 watts to sustain this. That's the amount of power flowing
through the transmission line, from source to load. That's 32
joules/second, not 68. If the line were open circuited, the forward and
reverse powers would be equal, and it would take no power to sustain them.

That you
don't see the logical connection between those two equal energy
values is amazing.

But I will get you started on an understanding of the component
powers using an S-parameter analysis.


Where will you find to put those all-important 68 joules in an
s-parameter analysis? That's a steady state analysis.

How much of that 18 watts of reverse power is going through the
source resistor to reach the source to "engage in destructive
interference"?


reference the S-parameter equation: b1 = s11*a1 + s12*a2

I calculate 11.52 watts. (s12*a2)^2 = 11.52 watts


Finally, an actual answer. So of the 18 watts of "reverse power", 11.52
watts is making it to the source to "engage in constructive
interference". Does any of it get dissipated in the source resistor, or
does it just slide through unscathed?

The other 6.48 watts are in (s22*a2)^2 where |a2|^2 = 18 watts

What's (s22*a2)^2? The forward power wave? The reverse power wave? If
it's something else, does it have a name? Where does it go? Is it
getting dissipated in the source resistor, reflected at the transmission
line/resistor interface, reflected at the source/resistor interface, get
radiated, or what?

What does it interfere with?


From the S-parameter equation above, it obviously interferes with
s11*a1 .

What's s11*a1? It must be something inside the source. The source is
just that, a source. It has (AC) voltage and current, 100 volts of
voltage and 0.4 amps of current. Does s11*a1 reside inside every source,
or only some special ones? Apparently 11.52 watts of this s11*a1 gets
cancelled by the reverse power wave. How much of it is left over?


Please reference HP App Note 95-1, available on the web. It should
answer most of your questions, in particular pages 16 & 17. |a1|^2 =
Power incident on the input of the network |a2|^2 = Power incident on
the output of the network |b1|^2 = Power reflected from the input
port of the network |b2|^2 = Power reflected from the output port of
the network

Nope.

Enough hand waving and evasion, we've been here before. [Of all the
questions, the sole quantitative answer was "(s12*a2)^2 = 11.52 watts".]
I'll leave you to the folks who regard this kind of gobbledegook as
convincing evidence. Have fun -- I've got actual work to do while you
take care of the visionary leadership part.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Cecil Moore wrote:
We are not going to get anywhere until you admit there is 68
joules/sec in the feedline that haven't yet made it to the load. Once
you admit that fact, everything else will be moot.

I never said a word about how much energy is stored in the feedline;
it's irrelevant.

Do you think it is random coincidence that the amount of energy
stored in the feedline is *EXACTLY* the amount of energy required
by the forward and reflected waves???? The fact that you think
it is irrelevant is simply a flight into a wet dream fantasy.

It's stored there during the initial charging to the
steady state condition, and the same amount remains there until the
steady state condition no longer exists. It's exactly like the DC charge
on a capacitor ...

What a coincidence! It's *EXACTLY* the amount of energy required
by the forward and reflected waves that you say don't exist - and
it's RF photons, not DC, so it must travel at the speed of light!
You cannot store photons in a capacitor.

Surely even you can do the basic circuit analysis which shows that the
source is continuously generating 40 watts, ...

Of course, but during the power-on transient period, 68 watts is NOT
delivered to the load. Please don't make me waste my time calculating
the forward and reflected power during the power-on transient period.
When you actually do those calculations, you will agree with me that,
during the power-on transient phase, 68 watts of power has been stored
in the feedline and remains there until the power-off transient phase.
It is *EXACTLY* the amount of power required by the forward and
reflected waves.

The line's stored energy will be dissipated in either the source or load
resistor or both when the source power is turned off. But we're doing a
steady state analysis here.

Which must necessarily include the energy stored in the feedline during
the power-on transient condition because it is *still there during
steady-state*. Ignoring the energy stored in the feedline during the
power-on transient phase is both irrational and illogical. It could
even be bad for your mental health.

What exactly is my premise, please?

from my earlier posting:

You are apparently assuming there is not enough energy in the system
during steady-state to support the forward and reflected power waves.

I'm making no such assumption.

I'm sorry, Roy, but that is just BS! Either you admit there is enough
energy to support the steady-state forward and reflected waves or you
don't.

My analysis doesn't need to consider the stored energy at all.

And that is exactly why your analysis is wrong. You have, once again,
been seduced by the steady-state model and are spreading old wives'
tales as a result. Hopefully, you don't really want to do that.

The source has supplied 68 joules/sec that has not reached the load.

I think you mean 68 joules.

68 joules/sec in your original example. 68 joules in my one-second-
long feedline example.

Where will you find to put those all-important 68 joules in an
s-parameter analysis? That's a steady state analysis.

The 68 joules were stored in the feedline during the power-on
transient phase. They are still there during steady-state. An
S-parameter analysis yields the correct results because it
includes the reflected power, |a2|^2, as an energy source,
something you deny. Wonder what the S-parameter analysis folk
know that you don't choose to admit?

Finally, an actual answer. So of the 18 watts of "reverse power", 11.52
watts is making it to the source to "engage in constructive
interference". Does any of it get dissipated in the source resistor, or
does it just slide through unscathed?

It never encounters the source resistor as it is re-reflected by
wave cancellation, not by an impedance discontinuity. I have a QEX
article coming soon that will explain the details. Stand by. With
an unprejudiced open mind, you might actually learn something.

What's (s22*a2)^2?

Same as (1-rho^2)*Pref2 in ham terms. Reflected power from the
load that is re-reflected back toward the load by an impedance
discontinuity.

What's s11*a1?

Same as Vfor1*rho in ham terms. Forward voltage that is reflected
back toward the source by an impedance discontinuity.

I'll leave you to the folks who regard this kind of gobbledegook as
convincing evidence.

Do you realize what intellectual shape you would be in if you had
adopted that attitude when you were one year old? :-)
--
73, Cecil http://www.qsl.net/w5dxp


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