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