Keith Dysart wrote:
But the rules for black boxes do not allow measurements
on the inside. This is how they help clarify the thinking.

So instead of sweeping technical facts under the rug,
you hide them in a black box. In both cases, the only
apparent purpose is to maintain ignorance.

It seems that whatever part of the system you don't
understand, you draw a black box around it so you
don't have to understand it.

So I ask you once again, given the following two stubs:

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

--43.4 deg 600 ohm line--+--46.6 deg 600 ohm line--open
Vfor1--|--Vfor2

What are the phase shifts between Vfor1 and Vfor2 for
the two cases?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:


It seems that whatever part of the system you don't
understand, you draw a black box around it so you
don't have to understand it.


Cecil,

Interesting comment, especially since you frequently reference
s-parameter analysis.

A direct quote from AN-95-1, the slide version, is:

Two-port, three-port, and n-port models simplify the input / output
response of active and passive devices and circuits into "black boxes"
described by a set of four linear parameters.

If you deny the legitimacy of "black boxes" do you need to give up the
use of s-parameters?

73,
Gene
W4SZ

Gene Fuller wrote:
A direct quote from AN-95-1, the slide version, is:

Two-port, three-port, and n-port models simplify the input / output
response of active and passive devices and circuits into "black boxes"
described by a set of four linear parameters.

Thank you, Gene. That contradicts what you said before
about the black box not being allowed to have two of
the four terminals on the other side. Play silly games
with the facts and you tend to get caught.

If you deny the legitimacy of "black boxes" do you need to give up the
use of s-parameters?

No, you need to give up your assertion that a four-
terminal black box doesn't have two terminals on
the other side. Your black box and HP's are two
entirely different concepts.

HP puts a black box around a 4-terminal network to
enhance understanding of the contents of the black
box. You put a black box around a stub to promote
ignorance of the contents of the black box.

I have said before. Specify that the black boxes be
supplied with the four measured s-parameters stamped
on them and I can probably tell you which box is which
without even applying a signal.

Or, more logically, forget the black box entirely since
it is totally irrelevant to the subject being discussed.

Exactly what is it that you think you have proved by
using black boxes. Please be specific.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
A direct quote from AN-95-1, the slide version, is:

Two-port, three-port, and n-port models simplify the input / output
response of active and passive devices and circuits into "black boxes"
described by a set of four linear parameters.

Thank you, Gene. That contradicts what you said before
about the black box not being allowed to have two of
the four terminals on the other side. Play silly games
with the facts and you tend to get caught.

If you deny the legitimacy of "black boxes" do you need to give up the
use of s-parameters?

No, you need to give up your assertion that a four-
terminal black box doesn't have two terminals on
the other side. Your black box and HP's are two
entirely different concepts.

HP puts a black box around a 4-terminal network to
enhance understanding of the contents of the black
box. You put a black box around a stub to promote
ignorance of the contents of the black box.

I have said before. Specify that the black boxes be
supplied with the four measured s-parameters stamped
on them and I can probably tell you which box is which
without even applying a signal.

Or, more logically, forget the black box entirely since
it is totally irrelevant to the subject being discussed.

Exactly what is it that you think you have proved by
using black boxes. Please be specific.

I said no such thing about "black boxes" being unable to have more than
two terminals. What I said is that the "black boxes" defined by Keith
and Roy have only two terminals.

If you want to drift off into some other irrelevant chatter, go right
ahead. It might make you feel good, but it won't change the real world.

Gene Fuller wrote:
I said no such thing about "black boxes" being unable to have more than
two terminals. What I said is that the "black boxes" defined by Keith
and Roy have only two terminals.

But the "black boxes" defined by me have either two
terminals or four terminals. Here they are again:

--43.4 deg 600 ohm line--+--10 deg 100 ohm line--open

--43.4 deg 600 ohm line--+--46.6 deg 600 ohm line--open

--43.4 deg 600 ohm line--+--j567 impedor

The black boxes are drawn around the '+' point just as
I said before. That makes the first two examples four-
terminal networks and the last example a two-terminal
network.

Now tell us again how s11, s12, s21, and s22 are identical
for those three black boxes.
--
73, Cecil http://www.w5dxp.com

Gene Fuller wrote:
If you want to drift off into some other irrelevant chatter, go right
ahead. It might make you feel good, but it won't change the real world.

As if imaginary black boxes exist in the real world. :-)
Get real, Gene. Any black box you provide, I can open.
--
73, Cecil http://www.w5dxp.com

Gene Fuller wrote:
Cecil Moore wrote:


It seems that whatever part of the system you don't
understand, you draw a black box around it so you
don't have to understand it.


Cecil,

Interesting comment, especially since you frequently reference
s-parameter analysis.

A direct quote from AN-95-1, the slide version, is:

Two-port, three-port, and n-port models simplify the input / output
response of active and passive devices and circuits into "black boxes"
described by a set of four linear parameters.

If you deny the legitimacy of "black boxes" do you need to give up the
use of s-parameters?

This is simply a diversion to deflect the discussion away from the
sticky questions about "electrical degrees" which his theory is unable
to resolve. Phase reference is another, and we can expect more.

Roy Lewallen, W7EL

Roy Lewallen wrote:
This is simply a diversion to deflect the discussion away from the
sticky questions about "electrical degrees" which his theory is unable
to resolve. Phase reference is another, and we can expect more.

There were no black boxes in the original example so the
black box was the original diversion. Coming back from that
diversion, can you calculate the current amplitude and phases
in the original example? I would be very surprised if you
could do it. I would be even more surprised if you did it
and published the results. Roy, here's your chance to nail
me to the wall. Simply prove that the phase shift between
Vfor1 and Vfor2 below is something other than 36.6 degrees.
(All of Roy's worshipers hold their breath for a response. :-)

This is not "my" theory - this is standard distributed
network reflection theory that I learned at Texas A&M
in the 50's. And the theory is certainly capable of
resolving the electrical degree problems.

Here's the original example again - no black box necessary.

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

Assuming 100v at 0 deg incident upon the open at the
end of the stub, what is the phase shift between
Vfor1 and Vfor2?

Vfor2 = 100v at -10 deg
Vfor1 = 143.33v at -46.4 deg

The phase shift between Vfor1 and Vfor2 is 36.6 degrees just
as predicted originally. Roy, you are always advising me to
use voltages so I did. The results are easy to verify if you
know how. But I don't think you know how.

Everyone is invited to use any valid model you want to and
prove me either right or wrong.

I predict that Roy will be silent on this subject and rely
on his political power to try to suppress those results.
The emperor has no clothes. The emperor's worshipers have
no clothes.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
But the rules for black boxes do not allow measurements
on the inside. This is how they help clarify the thinking.

So instead of sweeping technical facts under the rug,
you hide them in a black box. In both cases, the only
apparent purpose is to maintain ignorance.

It seems that whatever part of the system you don't
understand, you draw a black box around it so you
don't have to understand it.


No, it is a perfectly normal technique to test a theory or model. The
black box reveals just enough information to solve the problem, and
nothing more.

In this particular case, the impedance at the terminals of the black box
is the only *necessary* information to solve the transmission-line
problem (in the steady state, at one frequency). It is not necessary to
know how that impedance was created.

Conventional transmission-line theory handles this situation
effortlessly, thus proving that no more information is needed. Any
theory that claims to need more information has failed the test - for
somewhere it has a soft centre that means it cannot be trusted.

Professional scientists and engineers are quite ruthless about this.
They don't wait for other people to propose such tests - they do it
themselves, beating hardest on their own ideas, to find out what they're
good for and where the limits are. Any ideas that don't stand up to this
treatment are ruthlessly discarded.

That isn't always easy, but a professional scientist or engineer has to
have the clarity and integrity to know when it has to be done. That is
why the professionals are very careful not to keep ideas as pets. As in
farming, it's only the amateurs who can afford that self-indulgence.


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Ian White GM3SEK wrote:
In this particular case, the impedance at the terminals of the black box
is the only *necessary* information to solve the transmission-line
problem (in the steady state, at one frequency). It is not necessary to
know how that impedance was created.

Black boxes have their function but I doubt that any proponent
of black boxes will admit that the function proposed here is
to obscure technical facts because those technical facts are
distasteful to some people.

Ian, the entire problem (as stated previously by me) is to
ascertain the phase shift at the impedance discontinuity
between Vfor1 in the 600 ohm line and Vfor2 in the 100 ohm
line at point '+' in the following example. That is the
problem as stated. It's a straight forward problem - no
black box necessary.

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

Assuming the voltage incident upon the open end of the stub
is 100 volts at 0 degrees, I calculate the following voltages
at point '+'.

Vfor2 = 100 volts at -10 degrees

Vfor1 = 143.33 volts at -46.6 degrees

The phase shift between Vfor1 and Vfor2 is 36.6 degrees.

You should be able to prove or disprove those values. In
fact, you seem to be frothing at the mouth wanting to
disprove them. Well, go ahead and prove me wrong (if you
can).

Instead of performing the calculations to disprove my figures,
you attempt to sweep part of the problem under the rug by
putting everything from point '+' to the end of the stub in
a black box thus making the stated problem impossible to solve.
I'm sorry, but that is an unethical diversion away from the
stated problem.

I have already stated that no matter what is in the black box,
if the impedance or impedor is -j567 then the conditions external
to the black box are identical. But that diversion has nothing
to do with solving the original problem.

Why are you afraid to solve the problem as stated? I am going
to keep repeating this posting until someone provides a solution
to the original problem.

My voltage calculations above are either right or wrong. If
they are wrong, as you suggest, please prove it. If they are
right, I don't blame you for trying your best to suppress the
technical facts by hiding things in a black box but now the
whole world is aware of your attempted suppression of
technical facts, not a good reputation to have for a technical
editor.
--
73, Cecil http://www.w5dxp.com