Roy Lewallen wrote:
Cecil Moore wrote:
Why is it so hard to get a straight answer to this simple question?

The answer is simple. 25 watts.

That was a given, Roy. The question is: what is the feedline
loss in dB?

What do you think about Jim's definition of the power ratio
being (feedline losses)/(feedline losses + load power)?
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 01 Dec 2004 20:33:14 -0600, Cecil Moore
wrote:
what is the feedline loss in dB?
The rest of the world computes it at "per 100 feet." Why change?

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

Richard Clark wrote:

Cecil Moore wrote:
what is the feedline loss in dB?

The rest of the world computes it at "per 100 feet." Why change?

Is that the reason you make all your feedlines exactly 100
feet long? :-)
--
73, Cecil http://www.qsl.net/w5dxp

50 watts is entering the line, and 25 watts is exiting. The line loss is
3 dB. Surely you're able to make this calculation yourself.

I don't feel a need for a definition of a "power ratio". What you've
defined is indeed the ratio of two powers, but it escapes me of what use
it is except perhaps to cause confusion.

Roy Lewallen, W7EL

Cecil Moore wrote:

Roy Lewallen wrote:

Cecil Moore wrote:

Why is it so hard to get a straight answer to this simple question?


The answer is simple. 25 watts.


That was a given, Roy. The question is: what is the feedline
loss in dB?

What do you think about Jim's definition of the power ratio
being (feedline losses)/(feedline losses + load 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

The concepts of "forward" and "reflected" power are sometimes (but not
often) useful, but have to be carefully confined to a very specific set
of conditions and applications. When you start thinking of them as real
packets of power bouncing around inside and outside a cable, you can
easily be led into a number of traps which you can get out of only by
distorting reality and ultimately reaching conclusions which are more
and more wrong.

I strongly suggest forgetting completely about "forward" and "reverse"
power. If you must deal with directional waves, look at forward and
reverse voltage and current waves. Superimpose them as necessary, and
when you're done, calculate power from the result. Trying to separate
moving packets of average power will eventually force you into reaching
wrong conclusions, or at least to serious confusion.

The whole issue of power loss is extremely simple, and it provides a
good example of how trying to invent these moving packets of average
power can lead to unnecessary complication and confusion.

Roy Lewallen, W7EL

Jim Kelley wrote:


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

Roy Lewallen wrote:
50 watts is entering the line, and 25 watts is exiting. The line loss is
3 dB. Surely you're able to make this calculation yourself.

Surely you can appreciate the potential for one person to use the
signal generator output power as the reference for the dB loss
in the transmission line while another person may, as you do,
reference the line losses to the difference between the signal
generator output power and the power being dissipated in the
circulator load.

I suspect Bob and the ARRL are using different reference powers
in their calculations. My technician at Intel would have reported
the 25 watts of feedline loss above referenced to the signal
generator output power of 100 watts as was customary.

I don't feel a need for a definition of a "power ratio". What you've
defined is indeed the ratio of two powers, but it escapes me of what use
it is except perhaps to cause confusion.

Huh???? A "power ratio" is the *DEFINITION* of dB as applied to power.
Talking about dB is meaningless unless the two powers are defined.

A direct quote from _Reference_Data_For_Radio_Engineers_

"By *definition*, number of dB = 10*log(P1/P2)"

4th edition, page 40
--
73, Cecil http://www.qsl.net/w5dxp