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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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.

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

I gave you a break. The whole lot of it is nothing but Scattering
parameters. Might be fine for SHF and above. Useless for mobile
matching at HF. As predicted I couldn't understand one line of it. I
doubt if anybody except the several authors have ever worked right
through it. Have you? (2 smileys).

It's mid-day here. There's not a drop of wine in the house.
----
Your old pal, Reg.

Reg Edwards wrote:

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.

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

I gave you a break. The whole lot of it is nothing but Scattering
parameters. Might be fine for SHF and above. Useless for mobile
matching at HF.

"Useless" only in the sense of being unnecessarily complicated and
inconvenient to use. But not incorrect!

ALL valid methods will give the same correct result, if they are applied
correctly. That's how we confirm that a method is valid - by checking
its results for a series of test problems that can be solved by other
methods too.

Roy has posed a test problem that is very easy to understand, and can be
solved unambiguously by simple arithmetic. Solving it using S-parameters
will take time and some depth of understanding, but we can be confident
that they WILL give exactly the same result in the end.

The challenge for Cecil is to make his own theory do the same.


--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Ian White GM3SEK wrote:
Roy has posed a test problem that is very easy to understand, and can be
solved unambiguously by simple arithmetic. Solving it using S-parameters
will take time and some depth of understanding, but we can be confident
that they WILL give exactly the same result in the end.

It's not Roy's results that are flawed. It's his premises. If
one has a 100v source with a 50 ohm series impedance feeding
a 200 ohm resistor, Roy's results are perfect. But when we add
that 1/2WL of 200 ohm line, it changes things from a circuit
analysis to a distributed network analysis. Much more energy
is stored in the system, using the transmission line, than has
reached the load during steady-state. Roy tries to completely
ignore the stored energy and alleges that there is no energy in
the reflected waves. But there is *exactly* the same amount of
energy stored in the feedline as is required for the forward
waves and reflected waves to posssess the energy predicted by
the classical wave reflection model or an S-parameter analysis
or an analysis by Walter Maxwell of "Reflections" fame.

The challenge for Cecil is to make his own theory do the same.

"My theory" gives the exact same results as an S-parameter analysis
or a classical wave reflection model analysis. That's why I know it's
correct. Roy's (and Dr. Best's) models give the correct results for
voltage and current but not for power/energy. Remember Dr. Best's
assertion that 75w + 8.33w = 133.33w? It's been four years since
I told him here on this newsgroup that was ridiculous and that
75w + 8.33w + 50w of constructive interference = 133.33w
He responded that no interference existed or was necessary. That
can be verified by accessing Google, summer 2001.

Interference is built into the S-parameter model and the classical
wave reflection model but a lot of RF people don't recognize it. Dr.
Best's term, 2*SQRT(P1)*SQRT(P2), is known in the field of optics as
the "interference term" but he didn't know that at the time of
publication of his QEX article.
--
73, Cecil http://www.qsl.net/w5dxp

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Reg Edwards wrote:
I gave you a break. The whole lot of it is nothing but Scattering
parameters.

Exactly. There's nothing better for analyzing a transmission
line discontinuity. And there's nothing better for understanding
reflections.

It's mid-day here. There's not a drop of wine in the house.

Have some wine and you'll be able to understand the HP Ap Note.
That's how I first understood it. :-)
--
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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