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 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.

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

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?

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

On Sun, 16 Dec 2007 16:11:05 -0600, Cecil Moore
wrote:

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.

More signs of your confusion. How would you know how many terminals
if its inside a box?

Richard Clark wrote:
Cecil Moore
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.

More signs of your confusion. How would you know how many terminals
if its inside a box?

If you were locked in a black box, you would be ignorant
of night and day. Does that mean that night and day would
not be happening? Hint: No, it would just mean you are
ignorant. Your ignorance changes absolutely nothing
outside of the box in which you are locked.

Why does my ignorance of what's in the black box change
the reality of what's in the black box? Hint: it doesn't.

The fact that I am ignorant of the four terminal network
in the box doesn't change the fact that it is a four
terminal network. The fact that you won't allow me to
open the box and measure the phase shift at the impedance
discontinuity doesn't change the fact that it is 36.6
degrees.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
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.

Cecil,

This appears to be an unusual definition. How does the "point where two
pieces of feedline are connected" become a four-terminal network? One
typically thinks of a four-terminal network as having inputs and
outputs, with something between. What is that "something between" in the
case of two connected feedlines? In your models this "something" seems
to have no dimensions and no characteristics other than a phase shift.

Are you suggesting that every simple connection is now a four-terminal
network? Do all of the textbooks need to be re-written?

73,
Gene
W4SZ

Gene Fuller wrote:
Cecil Moore wrote:
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.

This appears to be an unusual definition.

Not unusual at all, Gene. The two input terminals to the
black box are on one side. The two output terminals from
the black box are on the other side. The impedance
discontinuity is inside the box. The black box is extremely
small.

Give me the four s-parameters, s11, s12, s21, and s22
and I can tell you virtually everything about what is
inside the black box without even applying a signal.
--
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

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