On Sun, 16 Dec 2007 22:14:49 GMT, Cecil Moore
wrote:

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.

You still show signs of confusion. Phase shift in what?

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:
You still show signs of confusion. Phase shift in what?

I have explained it twice already. Given the following
4-terminal network impedance discontinuity at '+':

---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open
Vfor1--|--Vfor2

What is the phase shift in the forward voltage at the
impedance discontinuity? The forward voltage on each
side of the impedance discontinuity , '+', is not
equal in magnitude or phase.

We know the stub causes a 90 degree phase shift end to
end. Since there is 43.4 deg phase shift in the 600 ohm
line and 10 deg phase shift in the 10 ohm line, guess
what the phase shift in the forward voltage has to be
at the impedance discontinuity?

Also previously explained, if you prefer - in s-parameter
terms: What is the phase shift between a1 and b2 in the
s-parameter equation:

b2 = s21*a1 + s22*a2
--
73, Cecil http://www.w5dxp.com

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

On Mon, 17 Dec 2007 15:22:44 GMT, Cecil Moore
wrote:

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.

OK, So you are ignorant of what is inside the box. Rather a long
ramble to such a simple conclusion.

Are you still ignorant of which of two phases? Your rambling
confusion has yet to come to terms even to the point of not being able
to name both of them in one posting!

Like trying to wake a sleep-walker, I hesitate to offer a dangerous
suggestion with you in your condition: Could they be "phase" shift
and "phase" length? In your somnambulant state of foggy recall
It just occurred to me that you and I may be talking
about two different phases.
can you tell the group what the two different phases are? This could
be an important moment in recovery (it may take only 11 steps more).

Take your time, we appreciate that your sudden catharsis can jog your
mind into curious responses...even though they would be
indifferentiable from the several hundred that preceded. ;-)

Richard Clark wrote:
Are you still ignorant of which of two phases?

I have spelled out the desired phases three times now.
This will be the forth time.

--43.4 deg 600 ohm line--+--10 deg 100 ohm line--open
Vfor1--|--Vfor2

The phase shift we are looking for is between Vfor1
and Vfor2. I might as well perform the calculations
for you since you seen to be incapable of doing so.

If we assume a reference of 100 volts at zero degrees
incident upon the open end of the stub, then back at
the impedance discontinuity:

Vfor2 = 100 volts at -10 deg.
Vfor1 = 143.33 volts at -46.6 deg

The phase shift between Vfor1 and Vfor2 is 36.6 degrees
just as it has to be for a 90 degree phase shift to
occur end to end in the above 1/4WL stub.

Vref2 = 100 volts at 10 deg
Vref1 = 143.33 volts at 46.6 deg
--
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

On Mon, 17 Dec 2007 15:11:36 GMT, Cecil Moore
wrote:

Richard Clark wrote:
You still show signs of confusion. Phase shift in what?

I have explained it twice already.

You have many explanations that don't actually answer questions. Your
confusion, as evidenced in the sudden realization:
It just occurred to me that you and I may be talking
about two different phases.
doesn't really tell us what phase shift. Perhaps if you could state
what the two are, and which you are using, the rest of us would be
satisified you are no longer confused.

Or maybe everyone is actually on the same page with only one phase
being mentioned, and some alternate phase expression (yet to be
revealed by you) is tormenting your imagination and corrupting your
answers with their math errors.

Cecil Moore wrote:

I(x,t) = Imax sin(kx) cos(wt)

For any point location 'x', it can be seen that the standing
wave current is not "flowing" in the ordinary sense of the word
but rather, is just oscillating in place at that fixed point.

According to the equation you provide above, for any point location
'x', the phase of the current varies continuously with t. Presumably
that is what it means to just oscillate in place.

73, ac6xg

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