You really don't know?

The power into the input end of the transmission line is 50 watts.
(Surely you can subtract the circulator resistor power from the source
power to find the power entering the transmission line. Can't you?)

The power exiting the load end of the feedline is 25 watts.

Therefore the transmission line loss is 25 watts.

It does seem that you've gotten yourself confused by your bouncing waves
of average power.

Roy Lewallen, W7EL

Cecil Moore wrote:
Roy Lewallen wrote:

Line loss itself is a comparison of two powers, those entering and
leaving the line. The line loss in watts is the power entering the
line at the input end minus the power leaving the line at the output end.


Here's an example: The source is a signal generator equipped with a
circulator-resistor that dissipates all reflected power.

100w SGCR--------------feedline------------------mismatched load

The signal generator is sourcing 100 watts. The load is dissipating
25 watts. The circulator resistor is dissipating 50 watts. The
feedline is dissipating 25 watts. What is the feedline loss in dB?

Is that 25 watts lost from the signal generator output power of
100 watts or lost from the NET power available which is 50 watts?
--
73, Cecil http://www.qsl.net/w5dxp

Roy Lewallen wrote:

You really don't know?

You really don't know how to calculate the dB loss in the feedline?
If you do, why haven't you done so?

The power into the input end of the transmission line is 50 watts.
(Surely you can subtract the circulator resistor power from the source
power to find the power entering the transmission line. Can't you?)

That's *NET* power. The power into the transmission line is the 100w of
measured source power. The power dissipated in the circulator resistor is
the measured power out of the transmission line reflected from the mismatched
load. If you say to calculate the dB loss in the feedline based on the NET
power, that's what I will do.

I am trying to understand the difference in Bob's results and the
results using the ARRL equations. The magnitude of NET power to which
the feedline losses are ratio'ed may be the key to understanding that
difference.

The power exiting the load end of the feedline is 25 watts.

That was given.

Therefore the transmission line loss is 25 watts.

That was given but didn't answer the question about dB.

It does seem that you've gotten yourself confused by your bouncing waves
of average power.

Nope, you seem to be confused about what the question was. You keep
answering with what was given in the original example. The question
is: What is the *dB* loss in the feedline? Using NET power input, the
dB loss in the feedline is 3 dB, i.e. half of (100w-50w) = 50 watts.
I'm satisfied with that definition but an wondering why nobody else
has said the feedline losses equal 3 dB. What was so difficult about
that?

Incidentally, that figure agrees with Jim's equation which doesn't
even mention source power.
--
73, Cecil http://www.qsl.net/w5dxp

Only you could take such a simple concept as power loss in a two port
device and muddle it with bouncing waves of average power.

I didn't explicitly give the loss in dB because I thought that you would
be able to do it yourself. You seem to be able to, but for some reason
regard it as some kind of major operation. I don't know how you do it,
but here's how I do.

From my previous posting, the power into the line is 50 watts and the
power out is 25 watts.

To find the loss in dB, take the ratio of input to output power, that
is, 50 divided by 25, to get 2. Now take the base ten logarithm of that.
(The Log key on a calculator is what I use for this complex operation. I
get about 0.301. Finally, multiply that by 10, to get 3.01 dB. 3 is
close enough for most of us.

Roy Lewallen, W7EL

Cecil Moore wrote:

Roy Lewallen wrote:

You really don't know?


You really don't know how to calculate the dB loss in the feedline?
If you do, why haven't you done so?

The power into the input end of the transmission line is 50 watts.
(Surely you can subtract the circulator resistor power from the source
power to find the power entering the transmission line. Can't you?)


That's *NET* power. The power into the transmission line is the 100w of
measured source power. The power dissipated in the circulator resistor is
the measured power out of the transmission line reflected from the
mismatched
load. If you say to calculate the dB loss in the feedline based on the NET
power, that's what I will do.

I am trying to understand the difference in Bob's results and the
results using the ARRL equations. The magnitude of NET power to which
the feedline losses are ratio'ed may be the key to understanding that
difference.

The power exiting the load end of the feedline is 25 watts.


That was given.

Therefore the transmission line loss is 25 watts.


That was given but didn't answer the question about dB.

It does seem that you've gotten yourself confused by your bouncing
waves of average power.


Nope, you seem to be confused about what the question was. You keep
answering with what was given in the original example. The question
is: What is the *dB* loss in the feedline? Using NET power input, the
dB loss in the feedline is 3 dB, i.e. half of (100w-50w) = 50 watts.
I'm satisfied with that definition but an wondering why nobody else
has said the feedline losses equal 3 dB. What was so difficult about
that?

Incidentally, that figure agrees with Jim's equation which doesn't
even mention source power.
--
73, Cecil http://www.qsl.net/w5dxp

Roy Lewallen wrote:

Only you could take such a simple concept as power loss in a two port
device and muddle it with bouncing waves of average power.

I didn't explicitly give the loss in dB because I thought that you would
be able to do it yourself. You seem to be able to, but for some reason
regard it as some kind of major operation. I don't know how you do it,
but here's how I do.

From my previous posting, the power into the line is 50 watts and the
power out is 25 watts.

I think the source of part of the confusion here is that some people
apparently interpret the 'forward power' reading on their meter to mean
the power into their transmission line.

The confusion I think stems from the contention that any 'reflected
power' (unfortunate nomenclature IMO) is first sourced and then after
reflection returned back into the source, or to a circulator load as the
case may be. The latter case is certainly correct. The former is
phenomenologically problematic.

73, Jim AC6XG

On Wed, 01 Dec 2004 21:27:25 -0800, Jim Kelley
wrote:

The confusion I think stems from the contention that any 'reflected
power' (unfortunate nomenclature IMO) is first sourced and then after
reflection returned back into the source, or to a circulator load as the
case may be. The latter case is certainly correct. The former is
phenomenologically problematic.

Hi Jim,

By that same logic it follows that the power "into" the transmission
line was in fact never "into" the line at all but into the circulator
input, and any power finding its way into the circulator load also
never found its way into the line, but was merely reflected at the
circulator/line interface.

Hence, the power loss of the line (in dB) is
the Same.

Hence any discussion of line loss and circulators, by omitting the
circulator, is a flawed argument of line loss.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
By that same logic it follows that the power "into" the transmission
line was in fact never "into" the line at all but into the circulator
input, ...

Very flawed logic as can be proven by observing the modulation
associated with the circulator. That modulation has obviously
made a round trip to the load and back. Sorry about that.
--
73, Cecil http://www.qsl.net/w5dxp

On Thu, 02 Dec 2004 08:47:23 -0600, Cecil Moore
wrote:
Very flawed logic as can be proven by observing the modulation
associated with the circulator.
And the initial condition violations just keep stacking up
When do we get to the unstated EMP analysis?

Richard Clark wrote:

On Wed, 01 Dec 2004 21:27:25 -0800, Jim Kelley
wrote:


The confusion I think stems from the contention that any 'reflected
power' (unfortunate nomenclature IMO) is first sourced and then after
reflection returned back into the source, or to a circulator load as the
case may be. The latter case is certainly correct. The former is
phenomenologically problematic.


Hi Jim,

By that same logic it follows that the power "into" the transmission
line was in fact never "into" the line at all but into the circulator
input, and any power finding its way into the circulator load also
never found its way into the line, but was merely reflected at the
circulator/line interface.

A circulator, being in general a three (or four) port directional
device, might have some trouble buying into that logic. ;-) The crux of
the phenomenological problem is that power does not flow or move, nor is
it something that is reflected. Hence Roy's (and Reg's) suggestion that
the voltages and currents resulting from the fields which propagate must
be analyzed. From that analysis (which involves the fields, or V and I,
propagating, reflecting, and interfering in both directions) one can
determine the quantities of energy being absorbed by the effected
dissipative loads in the circuit.

A transmission line circuit which includes a circulator w/load does
indeed provide a mechanism by which a portion of the energy produced by
a source can effectively be reflected from a mismatched load back toward
the generator. On encountering the circulator in the reverse direction,
it is then directed to the circulator load where it can be dissipated.
In a lossey transmission line, that reflected signal will be attenuated
and would in fact increase the total amount of energy the transmission
line dissipates. The amount of energy produced by the generator
increases by the amount lost to the circulator load and the transmission
line. **Absent the circulator, those energy losses would not be
realized - nor sourced.**

The argument that fields "have" or "contain" energy is misdirected and
misapplied. Obviously one can measure a field at each of the electrical
outlets in his house even when nothing is drawing energy from those
outlets. The potential to create a transfer of energy does not
necessarily equate with a transfer of energy. A mechanism must exist
which provides the conduit for a transfer of energy. It is that
mechanism, and the nature of the source and the load which determine the
amount of power being generated and transferred to the dissipating load.

73, Jim AC6XG

Jim Kelley wrote:
The crux of
the phenomenological problem is that power does not flow or move, nor is
it something that is reflected.

But energy does flow and move and is something that can be reflected.
You can easily see the energy packets using a TDR. Without energy,
those pulses wouldn't exist. The energy is obviously in the pulse,
where the voltage and current are.

And joules of energy flowing past a point is joules/sec, i.e. power,
by IEEE definition.

Incidentally, how do you explain the Poynting Vector and the Power
Flow Vector?

The argument that fields "have" or "contain" energy is misdirected and
misapplied.

A lot of people will be surprised to discover that their electromagnetic
ExH and ExB values are "misdirected and misapplied".
--
73, Cecil http://www.qsl.net/w5dxp

On Thu, 02 Dec 2004 11:25:32 -0800, Jim Kelley
wrote:

A circulator, being in general a three (or four) port directional
device, might have some trouble buying into that logic. ;-) The crux of
the phenomenological problem is that power does not flow or move, nor is
it something that is reflected.

Hi Jim,

I merely responded in like metaphors.

To this point, the meditation of the difference between Bob's results
and the ARRL table speculates that the ARRL used an unknown Intel 100W
circulating Signal Generator driving 1 second's worth of transmission
line where Poynters Theorem proves that the dB loss is - well, we
never get to the end, do we?

73's
Richard Clark, KB7QHC