Richard Harrison wrote:
I don`t have a dog in this fight.

Neither do I, Richard. The ARRL Antenna Book seems to assume that
the loss in a line is proportional to forward power plus reflected
power. If reflected power equals zero, then the losses are matched
line losses. I posted an example through Google to which nobody
responded so I will re-post it he
************************************************** ********************
Here's an example. Assume 100w is delivered to the load for both the
matched and unmatched conditions. Assume 3dB matched line loss in the
transmission line. Assume an SWR of 5.83:1 (rho=0.707) at the load for
the mismatched condition.

Forward 200w------------3dB loss-------------Matched Load 100w


Forward 400w------------3dB loss-------------Mismatched Load 100w
Reflected 50w both Forward 200w
directions Reflected 100w

The ARRL equations give an additional loss of 5.44dB.
************************************************** ********************
It is obvious (to me) that if there are reflections at the load, then
more power must be sourced in order to deliver the same amount of power
to the load. But I am pretty simple-minded about obvious things. :-)
--
73, Cecil http://www.qsl.net/w5dxp

Cecil, W5DXP wrote:
"Assume 3 dB matched line loss in the transmission line. Assume an SWR
of 5.83:1 (rho=0.707) at the load for the mismatched condition."

Using 100 watts forward and 50 watts reflected into the "Antenna Book"
graph gives an SWR of 5 or 6. If I had my Bird SWR sliderule at hand, I
could be more precise without lifting a calculator.

According to the "additional line loss graph", this SWR adds about 2 dB
to a matched loss of 3 dB for a total loss of about 5 dB with such a
mismatch.

Best regards, Richard Harrison, KB5WZI

Reg Edwards wrote:
For example, in a large number of cases the single output quantity is
related to line loss, such as insertion loss in dB, or load power in watts,
or transmission efficiency in percent, or percent of input power lost in the
line itself.

But before this can be calculated it is essential to calculate input
impedance Rin+jXin for given attenuation in dB or nepers, given phase shift
in radians and given terminating impedance Rt+jXt. Then include generator
impedance Rg+jXg and internal generator volts. Having done this you are
half-way through.

Well Reg, I have a question about transmission line losses expressed in dB.
As you know, furnishing line losses in dB implies a power ratio. For a matched
line, transmission line losses in dB will be based on the ratio of power
not delivered to the load Vs the power being sourced. For instance, if
100 watts is being sourced and 95 watts are accepted by the load, then
5 watts of the 100 watts are line losses, or ~0.21 dB.

But to what power are the transmission losses referenced when reflected
power is allowed to be incident upon the source. For instance, assume
that at the source, we are reading 100 watts forward power and 85 watts
reflected power. That means the source is supplying a NET power of
15 watts to the system. If 10 watts of the total NET power of 15 watts
is being accepted by the load (implying 5 watts of line losses) does
that mean the power ratio for line losses is 33%, or 1.25 dB of losses?

Note from the matched line example above where 5 watts of line losses are
0.21 dB to the mismatched line example where 5 watts of line losses are
1.25 dB (for the same forward power of 100w), the same magnitude of loss
appears much higher as a dB value referenced to NET source power.
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 01 Dec 2004 10:17:05 -0600, Cecil Moore
wrote:
As you know, furnishing line losses in dB implies a power ratio.
dB is dimensionless. Such generalizations forced into logic become
naive paradoxes:
Note from the matched line example above where 5 watts of line losses are
0.21 dB to the mismatched line example where 5 watts of line losses are
1.25 dB (for the same forward power of 100w), the same magnitude of loss
appears much higher as a dB value referenced to NET source power.
with boundary conditions being violated with the substitution of
Mismatch Loss for Dissipative Loss - and done poorly too.

When I had you thank Bart for his teachings, you should have examined
the consequences of his lesson not simply his correction of your
error.

Richard Clark wrote:
Cecil Moore wrote:
As you know, furnishing line losses in dB implies a power ratio.

dB is dimensionless. Such generalizations forced into logic become
naive paradoxes:

Uhhhhhhh Richard, a power ratio *IS* dimensionless. Any ratio of like
quantities is dimensionless. dB is proportional to the log of the ratio
of two quantities. That ratio, e.g. of two powers, is dimensionless.

Note from the matched line example above where 5 watts of line losses are
0.21 dB to the mismatched line example where 5 watts of line losses are
1.25 dB (for the same forward power of 100w), the same magnitude of loss
appears much higher as a dB value referenced to NET source power.

with boundary conditions being violated with the substitution of
Mismatch Loss for Dissipative Loss - and done poorly too.

I certainly didn't intend to do that. The dissipative loss in each
of the examples was 5 watts. The NET source power was 100 watts in
the matched case and 15 watts in the mismatched case.

When I had you thank Bart for his teachings, you should have examined
the consequences of his lesson not simply his correction of your
error.

I'm asking an honest question, Richard. Why can't you provide an
honest answer?
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 01 Dec 2004 12:11:24 -0600, Cecil Moore
wrote:
I certainly didn't intend to do that.
The dog ate your homework? Shoot the dog, that excuse is too common.

Richard Clark wrote:
When I had you thank Bart for his teachings, you should have examined
the consequences of his lesson not simply his correction of your
error.

dB was not mentioned at all in Bart's posting so your comment
appears to be just an ad hominem shot in the dark. My news-
server indicates that Bart has not posted to this thread.

Let me ask my question again.

Given the dB equation for power, 10*log(P1/P2), when we say
the transmission line losses are 1 dB, what is P2 in the
denominator if it is not NET source power? There has to be
a standard. I just am not sure what that standard is. My
question is: What is the standard if not NET source power?

The magnitude of power available for being lost in transmission
line losses certainly cannot exceed the forward power minus the
reflected power, i.e. cannot exceed the NET source power. It seems
to follow that P2 in the equation must necessarily be NET power so
that transmission line losses are a percentage of the total power
available to be lost in transmission line. What am I missing?

In the previous examples, the 5 watt loss in the matched case was
5% of the NET source power of 100w. In the mismatched case, the 5
watt loss was 33% of the NET source power of 15w. It seems to follow
that the line loss in the matched case is about 0.2 dB of the 100w
of power available to be lost while the line loss in the mismatched
case is about 1.2 dB of the 15w of power available to be lost.

I'm still trying to explain the difference between Bob's dB loss
and the ARRL's dB loss. So an obvious question arose: Are they
using the same P2 in the denominator of the dB equation?
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 01 Dec 2004 13:12:39 -0600, Cecil Moore
wrote:
What am I missing?
... It seems to follow
Violation of initial conditions do not allow anything to follow but
distorted logic. And the example is poorly done (incorrect).

You are missing coming to terms with Bart's lesson - still.

Cecil asked,

But to what power are the transmission losses referenced when reflected
power is allowed to be incident upon the source.

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

Cecil, when there are several different power levels at different places in
a circuit, it is entirely up to you how you reference one to another in dB.
It's just a matter of convenience with the particular problem in hand.

But things are certain to become confused and indeterminate when your
imagination allows two or more different power levels at the same place at
the same time. For example: the so called forward and reflected power
levels.

Imagination is assisted by the existence of meters which claim to measure
these imaginary forward and reflected powers. Actually they do nothing of
the sort. They even purport to make measurements on non-existent
transmission lines.

Reflected volts - yes!
Reflected current - yes!
Reflected power - NO!

(PS : You can't deal with reflections without involving Distance, Velocity,
Place and Time.)
----
Reg.

Richard Clark wrote:
You are missing coming to terms with Bart's lesson - still.

It's a pretty simple question that, so far, no one has
answered, not even Bart. dB is always a comparison of
one thing to another. To what is the line loss in watts
being compared?
--
73, Cecil http://www.qsl.net/w5dxp

Reg Edwards wrote:
Cecil, when there are several different power levels at different places in
a circuit, it is entirely up to you how you reference one to another in dB.

Now Reg, that just cannot be true. Otherwise, there would exist no
conventions. It's a simple question: When you tell me that the losses
in a transmission line with reflections are 2 dB, exactly what power
are you referencing those losses against?

Reflected volts - yes!
Reflected current - yes!
Reflected power - NO!

Reflected volts and reflected current existing without any associated
joules/sec???? I've heard of waves without any trace of energy before,
Reg, but I certainly didn't expect to hear miracle metaphysics from you.

Whatever happened to V*I*cos(theta) being power? The power companies
would be surprised to learn that they are not transferring any joules/sec
to their customers. Do you think I can use your argument to get out of
paying my electric bill?
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 01 Dec 2004 14:47:55 -0600, Cecil Moore
wrote:
To what is the line loss in watts being compared?
That was established long ago in this thread, Bob expressed no
interest in Mismatch Loss. You can, of course, compare it to a light
bulb if you wish - life is like a box of chocolates isn't it?

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.
Line loss in dB is 10 times the base 10 logarithm of the power entering
the line at the input end divided by the power leaving the line at the
output end. These definitions can be and are used for any two-port device.

If you truly didn't know this, it's surely because you've befuddled
yourself with your forward and reflected waves of average power to the
point that you've lost complete track of the fundamentals.

Roy Lewallen, W7EL

Cecil Moore wrote:

Richard Clark wrote:

You are missing coming to terms with Bart's lesson - still.


It's a pretty simple question that, so far, no one has
answered, not even Bart. dB is always a comparison of
one thing to another. To what is the line loss in watts
being compared?
--
73, Cecil http://www.qsl.net/w5dxp

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

On Wed, 01 Dec 2004 16:50:00 -0600, Cecil Moore
wrote:
What is the feedline loss in dB?
The same. You haven't substituted it with a lightbulb have you?

Richard Clark wrote:
On Wed, 01 Dec 2004 10:17:05 -0600, Cecil Moore
wrote:

As you know, furnishing line losses in dB implies a power ratio.

dB is dimensionless. Such generalizations forced into logic become
naive paradoxes:

Note from the matched line example above where 5 watts of line losses are
0.21 dB to the mismatched line example where 5 watts of line losses are
1.25 dB (for the same forward power of 100w), the same magnitude of loss
appears much higher as a dB value referenced to NET source power.

with boundary conditions being violated with the substitution of
Mismatch Loss for Dissipative Loss - and done poorly too.

Well, I think he's assuming dissipative loss in both cases. But the 5
watt numbers were assumed, and there's nothing to indicate that the same
piece of transmission line would produce these losses under the
described conditions. Seems likely that the line losing 33% would have
to be quite a bit lossier. As for the second case, we don't know what
the source power was, so it presumably could also be a matched case at
lower power. The example, unfortunately, compared apples to oranges.

My take on dB attenuation figures in wire and cable table data:
dB loss = 10*log(power dissipated in the line / (power dissipated in the
line + power dissipated at load))

By way of reference, according to an old Standard Wire and Cable data
book of mine, the attenuation factor in dB per hundred feet is:

A=4.35*(Rsubt/Zo) + 2.78*sqrt(E)*p*F

Rsubt = .1*(1/d + 1/D)*sqrt(F) (total line resistance in ohms per 1000
ft) [*Note: there might be a misprint here. The .1 would seem to
indicate a conversion to ohms per hundred feet.]

E = dielectric constant

p = power factor of the dielectric

F = frequency in MHz

d = outside dia. center conductor [no units are indicated]

D = inside dia. outer conductor [no units are indicated]

The manual also discusses attenuation as a funtion of VSWR. But they do
not imply that the additional attenuation is due to anything but
reflection. In other words, there is no indication that reflected
'power' causes additional dissipative losses.


And.....I think Reg makes a good point about reflected power.
_____
$0.02

AC6XG

Richard Clark wrote:

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

The same. You haven't substituted it with a lightbulb have you?

The same as what? The feedline loss is 25 watts. What is the feedline
loss in dB? Please give me a number instead of a hard time.

A signal generator is pouring 100 watts into the line because its
source is seeing 50 ohms. The circulator load resistor is dissipating
50 watts of that source power. The load is dissipating 25 watts and
the feedline is dissipating 25 watts. What is the feedline loss in dB?

Is the dB loss in the feedline calculated using the signal generator
output power of 100 watts or is it calculated using the 50 watts of
NET power available as it would be (by definition) if the source were
a ham transmitter?

Why is it so hard to get a straight answer to this simple question?
--
73, Cecil http://www.qsl.net/w5dxp

Jim Kelley wrote:
Well, I think he's assuming dissipative loss in both cases. But the 5
watt numbers were assumed, and there's nothing to indicate that the same
piece of transmission line would produce these losses under the
described conditions.

The two transmission lines are completely separate and have absolutely
nothing to do with each other. The conditions are as stated.
--
73, Cecil http://www.qsl.net/w5dxp

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

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

The answer is simple. 25 watts.

Roy Lewallen, W7EL

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

Roy Lewallen wrote:
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.

The RF technician I worked with at Intel was directed by his manager to
report all powers in dB referenced to the signal generator output which,
in the above example, was 100 watts. His figure for dB loss in the transmission
line would be different from yours and you would be the one getting a bad
review for not following directions.

Like I said earlier, your refusal to define the powers being ratio'ed
may spread confusion instead of information.
--
73, Cecil http://www.qsl.net/w5dxp

Jim Kelley wrote:
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 comes about when the signal generator output power
is used as the reference, as is customary in a number of companies,
including the Intel cable group when I was there. Forward power is
*NOT* measured. Signal Generator Output Power is measured and all
dB measurements are referenced to that power.

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.

Another example of a physicist rejecting engineering principles.
I suggest you review HP's Ap Note 95-1 on S-Paramater analysis to
refresh your knowledge of the engineering definition of power.

|a1|^2 is forward power on the source side.
|b1|^2 is reflected power on the source side.
|a2|^2 is reflected power on the load side.
|b2|^2 is forward power on the load side.
--
73, Cecil http://www.qsl.net/w5dxp

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

"Cecil Moore" wrote
Reg Edwards wrote:

Cecil, when there are several different power levels at different places
in
a circuit, it is entirely up to you how you reference one to another in
dB.


Now Reg, that just cannot be true.

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

That's funny - _I_ don't have any difficulty in believing it! ;o)
---
Reg

Roy Lewallen wrote:
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.

Consider an earlier example made up of lossless lines:

100W XMTR---50 ohm---+---one second long 291.5 ohm---50 ohm load

The voltage reflection coefficient at the load is 0.707. The power
reflection at the load is 0.5, i.e. half the power is reflected.

After steady-state has been reached, the XMTR has output 300 more
joules than the load has accepted. A smaller real-world experiment
will easily verify that it is a fact that all energy sourced that
has not reached the load must necessarily be confined to circulating
energy or losses in the transmission line.

Question: In the above example, where are those 300 joules of energy
located and what is happening to them?

We know that 300 joules is wave energy and RF waves always move
at the speed of light, i.e. they cannot stand still. So please
determine how much energy is moving and in which of only two
possible directions.

The knife cuts both ways. Ignoring the energy gives one a very
warped view of reality where TV ghosting cannot exist, RADAR
can't possibly work, there's no such thing as non-glare glass,
and RF energy just sorta slowly mushes side-to-side inside a
transmission line. Can you spell M-A-G-I-C?

I strongly suggest forgetting completely about "forward" and "reverse"
power.

Good way to condemn yourself forever to a non-understanding of
energy movement in a transmission line. If that's your choice,
be my guest but please don't try to pass yourself off as an
energy expert by ignoring energy movement.

Many authors of transmission line textbooks disagree with your
stance.
--
73, Cecil http://www.qsl.net/w5dxp

On Thu, 02 Dec 2004 08:20:03 -0600, Cecil Moore
wrote:
My technician at Intel would have reported
the 11001 watts of feedline loss above referenced to the signal
generator output power of 1100100 watts as was customary.
Sure he would and this would have rendered 110dB
Common stuff that :-)

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?

On Thu, 02 Dec 2004 09:47:54 -0600, Cecil Moore
wrote:
100W XMTR---50 ohm---+---one second long 291.5 ohm---50 ohm load
And yet more violations of Initial Condition. What was wrong with the
original stub, didn't reach into a hyper dimension of virtual reality?
Do you suppose that the ARRL used such a long line for their table?
But if you are trying to prove they were in another dimension - it
follows. ;-)

Cecil Moore wrote:

The RF technician I worked with at Intel was directed by his manager to
report all powers in dB referenced to the signal generator output
which,
in the above example, was 100 watts. His figure for dB loss in the transmission
line would be different from yours and you would be the one getting a bad
review for not following directions.


Since his manager was Cecil, there may be a good case for worker's comp.
;-)


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Richard Clark wrote:

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

I know you consider it unfair to your position to bring up
the subject of TV ghosting but perhaps considering such would
enlighten you about EM RF wave reflection energy.

Or maybe you can just explain where the modulation energy was
hiding while it only apparently made a round trip to the load?
--
73, Cecil http://www.qsl.net/w5dxp