Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:

Z0 didn't change. The load didn't change. Therefore, the
SWR didn't change. What exactly do you think changed?


I think the circuit changed. Don't you?


The circuit changed without changing the forward power,
reflected power, and SWR so nothing of interest to the
present topic (V/I ratio) changed.

Do you know what dictates the SWR in a distributed network?
Certainly not the length of the feedline or the removal
of a tuner (assuming lossless conditions).

Well, it's certainly true that both circuits are missing the 50 ohm
impedance discontinuity in the middle, which is at least one of the
"present topics" (and my point).

I'll just go ahead and say what you really want me to say:

You're good enough, you're smart enough, and doggone it, people like you!

:-)

ac6xg

Dave wrote:
i'm not agreeing with how you assume the 50 ohm impedance of the meter out
of the circuit.

I don't think you understand what I was saying. It takes a
certain length of feedline to establish a Z0 environment
for the forward and reverse traveling TEM waves, i.e. to
force Vfor/Ifor = Vref/Iref = Z0.

and i am not agreeing that there should be reflected power
measured by the meter in the case of the 50 ohm load on the end of a 1/2
wave 75 ohm line.

I never said it "should" measure that reflected power. What I
said is if it doesn't measure the reflected power that exists,
it is not giving a valid measurement of reflected power. A properly
calibrated meter will indeed measure the reflected power that exists.
--
73, Cecil http://www.qsl.net/w5dxp

Richard Clark wrote:
Any error of misunderstanding Owen's post is entirely your own.

Thanks, Richard, that really helps a lot.
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
.. .

The actual SWR on a lossless line doesn't change. Yet, in another
posting, I showed that moving the Bird 1/4WL closer to the load
caused a reported SWR change by the Bird from 1:1 to 2.25:1. How
could both results possibly be right?
--
73, Cecil http://www.qsl.net/w5dxp

The Bird does not know squat about transmission lines, foreward, or
reflected. It only cares about impedance. If you connect a 50 Ohm load to it
through 1/4 wave of 75 Ohm coax, the impedance the Bird sees will be
transformed to 112.5 Ohms; hence the 2.25 SWR. (Actually, a 2.25:1 impedance
ratio)

I have a computer simulation of that type of meter, which lets me do things
you can not do in the real world, for instance having a 50 KW source on each
side of the meter. It can separate out the signals in the two directions of
transmission, but only when it is terminated in 50 Ohms, when there are no
reflections.

Tam/WB2TT

Jim Kelley wrote:
Well, it's certainly true that both circuits are missing the 50 ohm
impedance discontinuity in the middle, which is at least one of the
"present topics" (and my point).

Jim, if the ratio of net voltage to net current is already 50 ohms,
then your Bird "impedance discontinuity" should cause that to change
but there is no evidence whatsoever that it changes anything. Conclusion:
The insertion of the Bird does not appreciably change the V/I ratio.
--
73, Cecil http://www.qsl.net/w5dxp

Owen Duffy wrote:
I fully expected someone to object, not only to object, but to do so
without any original experimental evidence ...

Owen, there was absolutely no reason for anyone else to do any
additional experiments. Your own experiment proved that the Bird
wattmeter was ignoring the 4.1667 watts of reflected energy flowing
through it when installed in the 75 ohm environment. You proved
exactly the opposite of what you were trying to prove.
--
73, Cecil http://www.qsl.net/w5dxp

Tam/WB2TT wrote:

"Cecil Moore" wrote:
The actual SWR on a lossless line doesn't change. Yet, in another
posting, I showed that moving the Bird 1/4WL closer to the load
caused a reported SWR change by the Bird from 1:1 to 2.25:1. How
could both results possibly be right?

The Bird does not know squat about transmission lines, foreward, or
reflected. It only cares about impedance. If you connect a 50 Ohm load to it
through 1/4 wave of 75 Ohm coax, the impedance the Bird sees will be
transformed to 112.5 Ohms; hence the 2.25 SWR. (Actually, a 2.25:1 impedance
ratio)

Yes, that's exactly what I said in the other posting. But some
people seem to believe that inserting a Bird into a transmission
line with a Z0 other than 50 ohms magically changes it to a 50
ohm environment. The 40mm of transmission line inside the Bird
is supposed to accomplish that miracle.
--
73, Cecil http://www.qsl.net/w5dxp

On Fri, 14 Oct 2005 00:24:09 GMT, Cecil Moore wrote:
Your own experiment proved that the Bird wattmeter was ignoring
the 4.1667 watts of reflected energy flowing through it
This fiction is entirely your own invention and trying to put Owen's
frame around your cartoon does not elevate this nonsense to art.

You have done nothing to offer any substance to your claim except the
time worn tradition of repetition, deflection, and ignoring the
observables by insisting error exists where it is not.

What reflections exist, exist at the input port of the Bird which is
50 Ohms by design, and which is attached to a 50 Ohm load by
deliberate selection. The Bird's response confirms the load, and the
rest stands on 50 years of Bird quality construction (he's not using
the poorer model with PL-259s).

Owen's work stands on its own, despite your chatter of a nabob of
negativity (apologies to our ex-VP-ex-convict of the GOP).

Richard Clark wrote:

On Fri, 14 Oct 2005 00:24:09 GMT, Cecil Moore wrote:

Your own experiment proved that the Bird wattmeter was ignoring
the 4.1667 watts of reflected energy flowing through it

This fiction is entirely your own invention ...

Sorry, I am just quoting Owen's results. There was 4.1667 watts
of reflected energy flowing back through the Bird. The Bird
indicated zero watts. That's an infinite percentage error on
reflected power for the Bird and proves my point that a 50 ohm
wattmeter used in a 75 ohm environment is like using a hammer
on a screw. To paraphrase: when the only tool one has is a 50
ohm Bird, every Z0 environment looks like 50 ohms.
--
73, Cecil http://www.qsl.net/w5dxp

Richard Clark wrote:
What reflections exist, exist at the input port of the Bird which is
50 Ohms by design, and which is attached to a 50 Ohm load by
deliberate selection.

But the 50 ohm load is attached through a piece of 75 ohm coax
supporting an SWR of 1.5:1. If you don't realize that fact, I
feel sorry for you.
--
73, Cecil http://www.qsl.net/w5dxp