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