Keith wrote:
"How does Terman first write of the power in these waves?"

I don`t remember chapter and verse anymore. So, I can`t say where Terman
first writes of power in these waves. On page 88 of Terman`s 1955
edition he says:
"At radio frequencies it is nearly always permissible to assume that
omega L R, and omega C G. To the extent that it is true, one can
rewrite Eqs. (4-5) and (4-7) as follows:
Zo = sq rt L/C

This means current is in-phase with voltage in the waves in both
directions.

In-phase volts and amps are real power, not apparent power.

Best regards, Richard Harrison, KB5WZI

Reg,

I was merely agreeing with your comment about reading imaginary SWR on a
nonexistent transmission line. I tried the same thing with the HF
transmitter, which is connected to a Kenwood meter. Same results. In looking
at the Kenwood schematic, it looks like in the absence of any current flow,
it reverts to being a voltmeter, with both diodes getting the same drive.
The thing reminds me of old HP analog voltmeters that always had a dbm
scale. Of course, the dbms were nonsense, unless you were measuring across a
600 Ohm load.

Interesting about you looking at rho max = 1 +SQRT2 10 years before Chipman.
That beats Adler, Fano, and Chu who had it in their 1960 book.

Tam/WB2TT
"Reg Edwards" wrote in message
...
"Tarmo Tammaru" wrote -
"Reg Edwards" wrote -
It is possible to imagine the so-called SWR meter is telling you the
imaginary value of the SWR on a non-existent transmission line. But
it's
hardly of educational value when novices, even experienced engineers,
are
trying to understand what the reading really means. It's as confusing
and
as untruthful as Blair.

Why don't we accept the simple fact that the meter tells us only
whether
the transmitter is loaded with a resistance of a particular value or
not.
Which is no more nor less than what the instrument on the front panel
of your transceiver is provided for.

Then we can forget all about SWR, fwd and reflected power, until
needed
on real ines. Change the name of the meter to TLI.

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

Amen to that. I turned down the power on my transmitter, and measured P
forward and P rev while feeding about 100 feet of unterminated 9913. I
then
REMOVED the coax; i.e. there was nothing connected to the output side of
the
meter. Still measured the same Pf and Pr. (Daiwa meter)

Tam/WB2TT

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

Tam, what did I tell you ?

Your meter twice indicated the correct valuable information : In neither
case was your transmitter loaded with 50 ohms!

To complete the experiment connect a good, + or - 5 percent, 50-ohm
dummy-load to the meter. At 1.9, 3.8 or 7 MHz you can check the quality
of
the dummy load with a DC ohmeter. The actual DC resistance will depend on
how hot is the dummy load. It can probably be adjusted over an appreciable
range just by varying its temperature.

If the meter indicates an SWR less than about 1.15 then again it provides
the correct valuable information : The transmitter load is in the right
ball park. (Over here, Euro-side, we say in the "right street".)

To calibrate the meter, to make it even more "right", there's a little
preset resistor or capacitor inside the meter box. Twiddle it until a
minimum value of SWR is indicated at 7 MHz when the dummy load is
connected.
Or if the user is accustomed to work mostly at higher frequencies then it
can be twiddled at 21 MHz.

A meticulous amateur, proud of his shack and workmanship, may remove the
redundant SWR scale graduations and paint in their place two coloured
bands,
green = good, red = bad, according to artistic preferences.

Incidentally, to accurately determine transmitter power output just use a
diode+capacitor+DC voltmeter across the dummy load and calculate P =
(V-squared) / 100 watts. At HF it will usually be found to be within +/-
10
percent of what a reputable transceiver manufacturer says it is. And it
can
then be forgotten about. A 10 percent difference in power corresponds to
1/15th of an S-unit.

Tam, I'm sure you're already familiar with the foregoing procedures. I
describe them on these walls for the benefit of novices, CB-ers, etc., and
anyone else who studiously reads through these interminable threads in the
vain hope of learning anything from (Z-Zo)/(Z+Zo).

I have a guilty conscience - it was I who began the rho = 1+Sqrt(2) and
rho
= (Z-Zo) / (Z+Zo) threads. My excuse is that I was measuring and
investigating values of rho greater than unity 10 years before Chipman
issued his book in October 1968. But by then I had migrated to an entirely
different field.
----
Reg, G4FGQ

Keith wrote:
"But the quantities being measured are line volts and line amps and
neither is proportional to incident or reflected watts."

Bird states:
"Where should the wattmeter be inserted? Again referring to the
formulas, we see that the elements extract a voltage proportional to
either Ef or Er, while the total E varies along an improperly terminated
50-ohm line, the component voltages do not. This is simply another way
of saying that the energy in the forward wave remains the same from the
source to the load, where some or all of it is reflected (unless the
load is 50 ohms) and that reflected energy remains constant from the
load back to the source. Our directional power meters can, therefore, be
placed anywhere between the source and the load."

I don`t guarantee a perfect copy of Bird`s Technical Series extract. You
should request your own copy from Bird.

Best regards, Richard Harrison, KB5WZI

wrote:
My interest is with highlighting the lack of generality of the
reflected power model.

Every model, including yours, has limitations and should not be
misapplied.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:

Keith wrote:
"How does Terman first write of the power in these waves?"

I don`t remember chapter and verse anymore. So, I can`t say where Terman
first writes of power in these waves. On page 88 of Terman`s 1955
edition he says:
"At radio frequencies it is nearly always permissible to assume that
omega L R, and omega C G. To the extent that it is true, one can
rewrite Eqs. (4-5) and (4-7) as follows:
Zo = sq rt L/C

This means current is in-phase with voltage in the waves in both
directions.

In-phase volts and amps are real power, not apparent power.

Maybe. Sort of. Not necessarily.

I agree completely when the volts and amps are the measurable
resultant volts and amps.

But when the volts and amps are contributary terms to a
superposition solution, then, in general (using my definition
of general), multiplying them to produce watts is an
invalid operation since the watts terms can not be superposed
to obtain resultant watts.

It is, I suspect, for this reason that Johnson is fuzzy about
the power in forward and reverse waves; superposition, does
not, in general, apply for power.

The question is, was Terman similarly circumspect?

....Keith

Richard Harrison wrote:

Keith wrote:
"But the quantities being measured are line volts and line amps and
neither is proportional to incident or reflected watts."

Bird states:
"Where should the wattmeter be inserted? Again referring to the
formulas, we see that the elements extract a voltage proportional to
either Ef or Er, while the total E varies along an improperly terminated
50-ohm line, the component voltages do not. This is simply another way
of saying that the energy in the forward wave remains the same from the
source to the load, where some or all of it is reflected (unless the
load is 50 ohms) and that reflected energy remains constant from the
load back to the source.

It is not surprising that they should write this, after all they do
mark their instruments in watts, so they likely believe it.

But is it correct?

Our directional power meters can, therefore, be
placed anywhere between the source and the load."

This part is true.

I don`t guarantee a perfect copy of Bird`s Technical Series extract. You
should request your own copy from Bird.

Since the whole question is about whether what Bird does is correct,
they hardly stand as someone who can resolve the question.

....Keith

Cecil Moore wrote:

wrote:
... the quantities being measured are line volts and line amps
and neither is proportional to incident or reflected watts.

line volts = Vfwd + Vref, line amps = Ifwd + Iref

load power = Vfwd*Ifwd - Vref*Iref at the load

Not with my math....

power = volts * amps
= (Vfwd + Vref) * (Ifwd + Iref)
= Vf*If + Vf*Ir + Vr*If + Vr*Ir

Seems you've lost a couple of terms in there.

This is why, in general (using my definition of general), superposition
does not hold for power. Those extra terms get in the way.

Much happens to the measured voltage and current before the result is
inaccurately interpreted as forward and reverse power.

If you wade through the math you will see why it works for low-loss
feedlines.

You have started to state this caveat.

Does this mean that forward and reverse waves only have power in
the special case of low-loss feedlines?

Where is that example that proves your assertion that reflected waves
don't exist? How do standing waves occur when there are only forward-
traveling waves?

With my model, incident and reflected VOLTAGE waves and CURRENT waves
do exist. This is in common, I think, with most authors on the subject.
And this all works fine since superposition holds for voltage
and current.

What does not exist in general (using my definition of general) is
POWER in these incident and reflected voltage and current waves.
While a power term can be computed, and correct results will be
obtained in some special cases, in general (using my definition
of general), this is an invalid operation because superposition
does not hold for power.

This, I suspect, is the reason that at least one author has
carefully chosen fuzzy words when describing the 'power' in
the waves.

To recap. Everything with the incident and reflected wave model
works as long as you stick to voltage and current waves. It is only
when extended to include power (as done by Bird and others), that
the model starts to deteriorate.

So to get the right answers in the general case (using my definition
of general), compute your forward and reverse voltages and currents.
Use superposition to derive the resultant voltages and currents at
any point on the line and then use p(t) = v(t) * i(t) to compute
the power, which you may then average if you desire.

In the special case of low loss RF, the shortcut of 'regarding'
the waves as having power will produce an answer that is good
enough. But remembering that you have taken a shortcut and
by only 'regarding' the waves as having power, you will indicate
an enhanced understanding of the subject.

....Keith

This is in common, I think, with most authors on the subject.

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

. . . . . . and 1 million housewives can't be wrong!

Reg Edwards wrote:

This is in common, I think, with most authors on the subject.

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

. . . . . . and 1 million housewives can't be wrong!

Touche. I generally (using Cecil's definition of general) dislike
appeals to authority (or the masses), but occasionally let them
slip in when dealing with someone who does.

But knowing that you are watching, I will be more cautious in
the future.

....Keith

Keith wrote:
"I agree completely when the volts and amps are the measurable resultant
voolts and amps."

There is a big problem with resultant volts and amps. It is the
resultant variation in amplitude which is position dependent.

The only average variation in forward and reflected powers is a decline
with distance caused by line loss. Power flows at a constant average
rate into, through, and out of a transmission line. Line loss causes
decline in power along a lossy line.

The convenient way to get useful numbers is to separate energy by its
direction of travel and to measure these. A directional coupler is
needed and the Bird among others does this.

Best regards, Richard Harrison, KB5WZI