On Dec 9, 10:59 pm, "AI4QJ" wrote:
I do not understand why people will not accept that, when one connects a
length of transmission line of Zo1 to another length of transmission line
with Zo2, there can/will be a phase change at the discontinuity.

If there is a phase change on the Zo1 line by itself when the discontinuity
is 0 + j0 impedance (short) or 0 +j0 conductance (open), shouldn't there be
a phase shift between 0 and pi/2 when the impedance (or as you say,
"impedor") is added somewhere in between, i.e. when connecting the line of
Zo2, such as -j567 ohms at the electrical degree point of the termination?

This argument does seem to have a certain attractiveness.

It is partially understood in terms of a lumped component of -j567 but that
model describes the mismatch at the discontinuity only. It must be corrected
at the system level to also accouint for the known 10 degree additional
transmission line contribution of the 100 ohm line.

It seems to me that if connecting a -j567 impedance
produces a certain response, it should not matter how
that -j567 impedance is produced. Consider the following
cases:

- produce the -j567 with a lumped capacitor
- produce the -j567 with 10 degrees of 100 ohm line
- produce the -j567 with 46.6 degrees of 600 ohm line

The claim appears to be that despite the 43.4 degrees
of 600 ohm line being terminated in the exact same
-j567 impedance in all cases, the phase shift
experienced at the interface is different in each
case.

Does this really make sense?

....Keith

Keith Dysart wrote:
It seems to me that if connecting a -j567 impedance
produces a certain response, it should not matter how
that -j567 impedance is produced. Consider the following
cases:

(1) - produce the -j567 with a lumped capacitor
(2) - produce the -j567 with 10 degrees of 100 ohm line
(3) - produce the -j567 with 46.6 degrees of 600 ohm line

The claim appears to be that despite the 43.4 degrees
of 600 ohm line being terminated in the exact same
-j567 impedance in all cases, the phase shift
experienced at the interface is different in each
case.

They are terminated in the same value of impedance
but they are not terminated in identical impedances.
Please see the IEEE Dictionary for the three different
definitions of impedance. Each case involves a different
reflection coefficient at the -j567 point.
(I added a number to each case above.)

(1) involves an *impedor* with a reflection coefficient
of 1.0 at -93.2 degrees when connected to the 600
ohm line. The reflected wave at the feedpoint lags
the forward wave by 43.4 + 93.2 + 43.4 = 180 degrees.
The impedance is a real impedor, not a virtual impedance.

(2) involves an impedance discontinuity with a
reflection coefficient of -0.7143. With 10 degrees
of 100 ohm line, the reflected wave lags the forward
wave by 2(43.4 + 36.6 + 10) = 180 degrees. The phase
shift at the discontinuity is 36.6 degrees each way.
The impedance is caused by superposition of component
waves at the impedance discontinuity.

(3) with 46.6 degrees of 600 ohm line, the reflected
wave lags the forward wave by 2(43.4 + 0 + 46.6) = 180
degrees. The phase shift at the 600 to 600 ohm junction
is zero. The reflection coefficient at the 600 to 600
ohm junction is 0, different from (1) and (2) above.
The impedance is completely virtual and thus cannot
cause anything.
--
73, Cecil http://www.w5dxp.com

On Mon, 10 Dec 2007 23:43:58 -0500, "AI4QJ" wrote:

A lumped component is not enough to make the model correct.
Comments welcome.

Hi Dan,

It works both ways. However, lines can appear to be equivalent to
lumped components until you, say, double the frequency. The two
models clearly diverge. Where you could place a series resonant
lumped circuit in place of a resonant line (or versa vice) at a
fundamental frequency, the lumped series circuit would rarely
demonstrate parallel resonance at twice that frequency, but the line
would.

But it appears the model hasn't been constrained to less than an
octave bandwidth - until after Keith submitted the winning ticket to
the lottery commission. Cecil's lotteries are like that. To date
(more than 12 years now) you are the only winner, and I haven't seen
your prize awarded yet.

73's
Richard Clark, KB7QHC

AI4QJ wrote:
Once you get to -j567 at the discontinuity (travelling
10 degrees along the 100 ohm line), now you interface with the 600 ohm line.
At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945
on the smith chart (you normalize to Zo for it to calculate properly). This
abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43
degrees. I think the effect to look for is that the abrupt impedance change
when Zo changes.

Exactly. Plot -j567/100 and -j567/600 on a Smith Chart and
read the phase shift directly from the wavelength scale
around the outside of the chart.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Exactly. Plot -j567/100 and -j567/600 on a Smith Chart and
read the phase shift directly from the wavelength scale
around the outside of the chart.

Well, not exactly directly. One must multiply the
delta-wavelength by 360 degrees to get the phase
shift. If the angle of reflection coefficient
degree scale is used, it must be divided by two
to get the actual phase shift.
--
73, Cecil http://www.w5dxp.com

On Dec 10, 11:43 pm, "AI4QJ" wrote:
I may have misspoken. Once you get to -j567 at the discontinuity (travelling
10 degrees along the 100 ohm line), now you interface with the 600 ohm line.
At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945
on the smith chart (you normalize to Zo for it to calculate properly). This
abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43
degrees. I think the effect to look for is that the abrupt impedance change
when Zo changes. A lumped component is not enough to make the model correct.
Comments welcome

I follow the arithmetic, and it still has a certain attractiveness
but how can it make such a difference how the -j567 is produced.

What if you were offerred 3 black boxes, each labelled -j567?
Would it make much difference what was in them?
How does one compute the phase shift at the terminals?

....Keith

On Dec 10, 8:33 am, Cecil Moore wrote:
Keith Dysart wrote:
It seems to me that if connecting a -j567 impedance
produces a certain response, it should not matter how
that -j567 impedance is produced. Consider the following
cases:

(1) - produce the -j567 with a lumped capacitor
(2) - produce the -j567 with 10 degrees of 100 ohm line
(3) - produce the -j567 with 46.6 degrees of 600 ohm line

The claim appears to be that despite the 43.4 degrees
of 600 ohm line being terminated in the exact same
-j567 impedance in all cases, the phase shift
experienced at the interface is different in each
case.

They are terminated in the same value of impedance
but they are not terminated in identical impedances.
Please see the IEEE Dictionary for the three different
definitions of impedance. Each case involves a different
reflection coefficient at the -j567 point.
(I added a number to each case above.)

(1) involves an *impedor* with a reflection coefficient
of 1.0 at -93.2 degrees when connected to the 600
ohm line. The reflected wave at the feedpoint lags
the forward wave by 43.4 + 93.2 + 43.4 = 180 degrees.
The impedance is a real impedor, not a virtual impedance.

(2) involves an impedance discontinuity with a
reflection coefficient of -0.7143. With 10 degrees
of 100 ohm line, the reflected wave lags the forward
wave by 2(43.4 + 36.6 + 10) = 180 degrees. The phase
shift at the discontinuity is 36.6 degrees each way.
The impedance is caused by superposition of component
waves at the impedance discontinuity.

(3) with 46.6 degrees of 600 ohm line, the reflected
wave lags the forward wave by 2(43.4 + 0 + 46.6) = 180
degrees. The phase shift at the 600 to 600 ohm junction
is zero. The reflection coefficient at the 600 to 600
ohm junction is 0, different from (1) and (2) above.
The impedance is completely virtual and thus cannot
cause anything.

I see how you have done the arithmetic, and if I
understand correctly, you would claim these are
all systems with 90 "electical degrees".

What about the following...

Open circuit 46.6 degrees of 600 ohm line attached
to 80 degrees of 100 ohm line. The drive point impedance
will be 0 ohms. And the impedance at the junction will
be the same -j567. Using your arithmetic, there is
90-46.6-80 - -36.6 degrees of phase shift at the
junction. The physical length is 126.6 degrees
while the system is 90 "electrical degrees".

Is the definition of a system with 90 "electrical degrees"
becoming any system where the drive point impedance
has no reactive component?

....Keith

Keith Dysart wrote:
Open circuit 46.6 degrees of 600 ohm line attached
to 80 degrees of 100 ohm line. The drive point impedance
will be 0 ohms. And the impedance at the junction will
be the same -j567. Using your arithmetic, there is
90-46.6-80 - -36.6 degrees of phase shift at the
junction. The physical length is 126.6 degrees
while the system is 90 "electrical degrees".

Yes, you have just discovered that a 100 to 600 ohm
impedance discontinuity *loses degrees* when the 600
ohm end is open. That's why a center-loaded mobile
antenna takes more coil than a base-loaded mobile
antenna. Think about it.

Hint: if the stub is open, the low Z0 needs to be
placed at the open end to make the stub physically
shorter.

If the stub is shorted, the high Z0 needs to be
placed at the shorted end to make the stub
physically shorter.

You seem to be surprised at that. One wonders why?
--
73, Cecil http://www.w5dxp.com

AI4QJ wrote:
"Keith Dysart" wrote in message
...
On Dec 10, 11:43 pm, "AI4QJ" wrote:
I may have misspoken. Once you get to -j567 at the discontinuity
(travelling
10 degrees along the 100 ohm line), now you interface with the 600 ohm
line.
At that point you have to normalize the -j567 ohms to -j(567/600)
= -j0.945
on the smith chart (you normalize to Zo for it to calculate properly).
This
abrupt switch increases the angle from 10 degrees to arctan (0.945)) or
43
degrees. I think the effect to look for is that the abrupt impedance
change
when Zo changes. A lumped component is not enough to make the model
correct.
Comments welcome
I follow the arithmetic, and it still has a certain attractiveness
but how can it make such a difference how the -j567 is produced.

What if you were offerred 3 black boxes, each labelled -j567?
Would it make much difference what was in them?
How does one compute the phase shift at the terminals?

I use the smith chart in my response.

If you have 3 black boxes each labeled "input impedance = -j567", they could
contain a number of different things but since I am using the smith chart,
they will contain 3 different lengths of open transmission lines.

Box 1 contains 10 degrees length at 4MHz of 100 ohm transmission line. Based
on VF=1, this line would be 3/4*E10E2*10/360 = 2.08m long.

On the smith chart, plot from circle 10 degrees (transmission coefficient)
and read -j5.67. Normalize to 1 = 100. The impedance at the input of the
line is -j567. Label this box "input impedance = -j567".

Box 2 contains 19.2 degrees length at 4MHz of 200 ohm transmission line.
Based on VF=1, this line would be 3/4*10E2*19.2/360 = 4.0m long

On the smith chart, plot from infinte impedance circle 19.2 degrees
(transmission coefficient) and read -j2.84. Normalize to 1 = 200. The
impedance at the input of the line is -j(200*2.84) = -j567. Label this box
"input impedance = -j567".

Box 3 contains 27.2 degrees at 4MHz of 300 ohm transmission line. Based on
VF=1, this line would be 3/4*10E2*27.2/360 = 5.67m long.

On the smith chart, plot from infinte impedance circle 27.2 degrees
(transmission coefficient) and read -j1.89. Normalize to 1 = 300. The
impedance at the input of the line is -j(300*1.89) = -j567. Label this box
"input impedance = -j567".

So, do all boxres labeled "input impedance =-j567 ohm" transmission lines
behave the same when connected to the 600 ohm transmission line?

No.

For each of these, impedance at the discontinuity will be -j567. However,
each has different electrical lengths, thus the 600 ohm line connecting to
it will have to be cut to different electrical lengths, for all of the
degrees to add to 90 total.

You can only say that all -j567 of a given Zo will affect phase shifts in
connected 600 ohms lines in the same way.

So far, I find this very interesting. Not all -j567 impedors are equal when
it comes to transmission lines.

Wait a second.

You have three black boxes, all of which present an impedance of -j567
ohms at some frequency.

And you say that you can tell the difference between these boxes by
connecting a 600 ohm line to each one? If so, please detail exactly what
you would measure, and where, which would be different for the three
boxes. I maintain that there is no test you can devise at steady state
at one frequency (where they're all -j567 ohms) which would enable you
to tell them apart. I'll add a fourth box containing only a capacitor
and include that in the challenge.

If you agree with me that you can't tell them apart but that connecting
a 600 ohm line gives different numbers of "electrical degrees", then I'd
like to hear your definition of "electrical degrees" because it would
have to be something that's different for the different boxes yet you
can't tell the difference by any measurement or observation.

Roy Lewallen, W7EL

Roy Lewallen wrote:

...
You have three black boxes, ...
...
Roy Lewallen, W7EL

Hey!

Those ain't illegal citizens band (or, Chicken Band--for you
wanna-be-amateurs) linears (or leen-e-airs--as those properly schooled
in Chicken Banding would say/pronounce) are they? ;-)

And, three? Crud, I'd think a single 5kw would get ya by! Just turn
off the "foot warmer" and use the exciter for QRP! wink

Regards,
JS