Keith Dysart wrote:
So the next question is: What is the phase change
at the terminals of the black box?

It just occurred to me that you and I may be talking
about two different phases.

---Z01---+---Z02---

Vfor1--|--Vfor2

Vref1--|--Vref2

I am talking about the phase shift in the forward waves
across the impedance discontinuity, i.e. the phase shift
between Vfor1 and Vfor2. The list of phase shifts is
the phase shift in the forward voltages at the impedance
discontinuity. It is different for all the black boxes.

If you are talking about the phase between Vfor1 and Vref1,
then, yes, that phase is the same for all the black boxes.
It is impossible for it to be otherwise.
--
73, Cecil http://www.w5dxp.com

On Sun, 16 Dec 2007 08:06:18 -0600, Cecil Moore
wrote:

2) 36.6 degrees? (previous answer when it was 10 degrees of 100 ohm
line)
3) 0 degrees? (previous answer when it was 46.6 degrees of 600 ohm
line)

There's nothing wrong with those answers except maybe
a stupid math error.

Stupid math errors must be valid answers then?

On Sun, 16 Dec 2007 14:24:58 GMT, Cecil Moore
wrote:

It just occurred to me that you and I may be talking
about two different phases.

After weeks of this being explicitly stated by very many critics, it
just occurred to you?

Must be the onset of Netzheimers.

Cecil Moore wrote:
Keith Dysart wrote:
So the next question is: What is the phase change
at the terminals of the black box?

It just occurred to me that you and I may be talking
about two different phases.

---Z01---+---Z02---

Vfor1--|--Vfor2

Vref1--|--Vref2

I am talking about the phase shift in the forward waves
across the impedance discontinuity, i.e. the phase shift
between Vfor1 and Vfor2. The list of phase shifts is
the phase shift in the forward voltages at the impedance
discontinuity. It is different for all the black boxes.

If you are talking about the phase between Vfor1 and Vref1,
then, yes, that phase is the same for all the black boxes.
It is impossible for it to be otherwise.

I think that Vfor1 and Vref1 could also be understood to mean Vfor1 and
Vrefsum, with Vrefsum being the sum of all the reflected waves occurring
within the black box.

It strikes me that the concept of steady state AC is no different from
the concept of DC discussed earlier in this thread. Steady state AC has
no wave front to analyze. The impedance at the black box junction is a
fact, not something that can be analyzed with steady state waves. I
think you have said this a number of times.

The standing wave has no velocity, because we can not define a unit of
the wave that moves. I think you have also pointed this out.

Wave fronts must be used if we want to look into the black box, or at
least a MOVING sine wave so that we can look at the first reflection
separately. I think this is what you have said many times, but I used
different words.

73, Roger, W7WKB

Richard Clark wrote:
Cecil Moore wrote:
It just occurred to me that you and I may be talking
about two different phases.

After weeks of this being explicitly stated by very many critics, it
just occurred to you?

If it was ever stated, I missed it. I suspect it was never
stated and some people jumped to false conclusions. I don't
think anyone is stupid enough to assert that the phase shift
in a capacitor is the same as it is in the absence of any
physical impedance discontinuity.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Keith Dysart wrote:
So the next question is: What is the phase change
at the terminals of the black box?

It just occurred to me that you and I may be talking
about two different phases.

---Z01---+---Z02---

Vfor1--|--Vfor2

Vref1--|--Vref2

Continuing: What is the phase shift between Vfor1 and
Vfor2 for example:
(1) a capacitor with -j567 ohms impedance
(2) a 600 to 100 ohm dual-Z0 stub
(3) a single 600 ohm stub

I hope you are not going to tell us it's the same in
each case.
--
73, Cecil http://www.w5dxp.com

On Sun, 16 Dec 2007 19:12:50 GMT, Cecil Moore
wrote:

I don't
think anyone is stupid enough to assert that the phase shift
in a capacitor is the same as it is in the absence of any
physical impedance discontinuity.

Capacitance is not obtained in a physical impedance discontinuity?
or is it:
Physical impedance discontinuity is not obtained from a capacitor?
or is it:
Could be either is inside a box, supplying only the terminals to
either; specifically either of which is indeterminate at a single
frequency where the terminals might present 43.4 degrees? (or any
suitable angle)

There are any number of stupid choices available. The question is:
Has your netzheimers progressed so far as to add another one?

At 800 postings, the odds must be distinctly favoring stupid.

Make it the daily-double:
Does a stupid math error make the answer valid?

On Sun, 16 Dec 2007 19:17:48 GMT, Cecil Moore
wrote:

Cecil Moore wrote:
It just occurred to me that you and I may be talking
about two different phases.
Continuing: What is the phase shift

When you acknowledge there is some confusion as to which phase is
being talked about. Do you suppose you know enough to tell us which
phase you are talking about?

More to the matter, what TWO phases do you suppose there are to be
confused between?

At 800+ postings, you could continue to sail right on past these
questions in your cloud of netzheimer bliss and leave us with a 50%
risk -um- chance; answer your challenge; and watching you trying to
sort it out. Again.

Richard Clark wrote:
Cecil Moore wrote:
I don't
think anyone is stupid enough to assert that the phase shift
in a capacitor is the same as it is in the absence of any
physical impedance discontinuity.

Capacitance is not obtained in a physical impedance discontinuity?
or is it:

You missed the point. A terminating capacitor is a two
terminal network. The point where two pieces of feedline
are connected is a four-terminal network. A two-terminal
network is different from a four-terminal network.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
On Sun, 16 Dec 2007 19:17:48 GMT, Cecil Moore
wrote:

Cecil Moore wrote:
It just occurred to me that you and I may be talking
about two different phases.
Continuing: What is the phase shift

When you acknowledge there is some confusion as to which phase is
being talked about. Do you suppose you know enough to tell us which
phase you are talking about?

Funny. In the part you deleated, I said it was the phase
shift between Vfor1 and Vfor2. Your sneaky underhanded
deletion trick is noted.
--
73, Cecil http://www.w5dxp.com