Keith Dysart wrote:
If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?

-jcot(5) = -j11.43 normalized to Z0=100 ohms

-j100(11.43) = -j1143 ohms at the junction

-j1143/600 = -j1.905 normalized to Z0=600 ohms

arctan(1.905) = 62.3 degrees of Z0=600 ohm line

Would the phase shift at the junction still be
36.6 degrees?

The new phase shift would be 90-5-62.3 = 22.7 deg.

62.3 + 5 + 22.7 = 90 degrees

This example would correspond to a larger coil and
a shorter stinger in a loaded mobile antenna.
--
73, Cecil http://www.w5dxp.com

On Dec 7, 1:00 pm, Cecil Moore wrote:
Keith Dysart wrote:
If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?

-jcot(5) = -j11.43 normalized to Z0=100 ohms

-j100(11.43) = -j1143 ohms at the junction

-j1143/600 = -j1.905 normalized to Z0=600 ohms

arctan(1.905) = 62.3 degrees of Z0=600 ohm line

Would the phase shift at the junction still be
36.6 degrees?

The new phase shift would be 90-5-62.3 = 22.7 deg.

62.3 + 5 + 22.7 = 90 degrees

This example would correspond to a larger coil and
a shorter stinger in a loaded mobile antenna.
--
73, Cecil http://www.w5dxp.com

The smith chart shows this well. If I go up only 5 degrees on the
outer circle of open-end (infinite impedance on the 100 ohm line),
lower values of electrical angle corresponds to higher reactance, in
this case -j1143 for the 5 degree line instead of -j567 for the 10
degree line at the junction. The new phase shift at the junction
should be, and is, now lower since the 100 ohm line has a higher
capacitive reactance at the junction. As the 100 ohm line is shortened
to 0 degrees, we have a 600 ohm transmission line that is open and now
the 600 ohm line must be lengthened to the full 90 degrees for 1/4W.
This would correspond to a coil with no stinger.

On Dec 7, 3:02 pm, wrote:
On Dec 7, 1:00 pm, Cecil Moore wrote:





Keith Dysart wrote:
If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?

-jcot(5) = -j11.43 normalized to Z0=100 ohms

-j100(11.43) = -j1143 ohms at the junction

-j1143/600 = -j1.905 normalized to Z0=600 ohms

arctan(1.905) = 62.3 degrees of Z0=600 ohm line

Would the phase shift at the junction still be
36.6 degrees?

The new phase shift would be 90-5-62.3 = 22.7 deg.

62.3 + 5 + 22.7 = 90 degrees

This example would correspond to a larger coil and
a shorter stinger in a loaded mobile antenna.
--
73, Cecil http://www.w5dxp.com

The smith chart shows this well. If I go up only 5 degrees on the
outer circle of open-end (infinite impedance on the 100 ohm line),
lower values of electrical angle corresponds to higher reactance, in
this case -j1143 for the 5 degree line instead of -j567 for the 10
degree line at the junction. The new phase shift at the junction
should be, and is, now lower since the 100 ohm line has a higher
capacitive reactance at the junction. As the 100 ohm line is shortened
to 0 degrees, we have a 600 ohm transmission line that is open and now
the 600 ohm line must be lengthened to the full 90 degrees for 1/4W.
This would correspond to a coil with no stinger.- Hide quoted text -

- Show quoted text -

Ooops, I am posting from the web. This is AI4QJ.

wrote:
The smith chart shows this well.

Dan, I'm glad you understand. Now it is your turn
to get nibbled down to nothing by the guru gaggle
of geese.
--
73, Cecil http://www.w5dxp.com

On Dec 7, 1:00 pm, Cecil Moore wrote:
Keith Dysart wrote:
If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?

-jcot(5) = -j11.43 normalized to Z0=100 ohms

-j100(11.43) = -j1143 ohms at the junction

-j1143/600 = -j1.905 normalized to Z0=600 ohms

arctan(1.905) = 62.3 degrees of Z0=600 ohm line

Would the phase shift at the junction still be
36.6 degrees?

The new phase shift would be 90-5-62.3 = 22.7 deg.

62.3 + 5 + 22.7 = 90 degrees

So sometimes a 600 to 100 ohm discontinuity
produces a 36.6 degree phase shift and sometimes
it produces a 22.7 degree phase shift (and probably
any value in between).

I suggest that "work[ing] up the phasor diagrams of
the component voltages (or currents) at the junction
where rho = (600-100)/(600+100) = 0.7143" will
not be useful for predicting the phase shift.

....Keith

On Dec 7, 3:05 pm, Keith Dysart wrote:
On Dec 7, 1:00 pm, Cecil Moore wrote:





Keith Dysart wrote:
If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?

-jcot(5) = -j11.43 normalized to Z0=100 ohms

-j100(11.43) = -j1143 ohms at the junction

-j1143/600 = -j1.905 normalized to Z0=600 ohms

arctan(1.905) = 62.3 degrees of Z0=600 ohm line

Would the phase shift at the junction still be
36.6 degrees?

The new phase shift would be 90-5-62.3 = 22.7 deg.

62.3 + 5 + 22.7 = 90 degrees

So sometimes a 600 to 100 ohm discontinuity
produces a 36.6 degree phase shift and sometimes
it produces a 22.7 degree phase shift (and probably
any value in between).

That's right, depending on the electrical (and physical) length of the
100 ohm line, it will have different values of reactance as seen by
the 600 ohm line, therefore different phase shifts, all the way to
zero when the length of the 100 ohm line is 0 degrees and the
reactance is infinite (the 600 ohm line sees an open circuit). Say at
0+ degrees, -jX(C) = 1000000000; this is where you see it headed when
you look at the smith chart.

At zero degrees of 100 ohm line, you have 90-0-90 = 0 degrees at the
discontinuity.

AI4QJ

Keith Dysart wrote:
So sometimes a 600 to 100 ohm discontinuity
produces a 36.6 degree phase shift and sometimes
it produces a 22.7 degree phase shift (and probably
any value in between).

Yes, of course - nobody said the phase shift wasn't
a variable. Why would you expect it to be a constant?
It is a variable that depends upon the phase of the
component forward and reflected waves.

I suggest that "work[ing] up the phasor diagrams of
the component voltages (or currents) at the junction
where rho = (600-100)/(600+100) = 0.7143" will
not be useful for predicting the phase shift.

It will be useful for reporting that particular phase
shift. If other conditions change, that phase shift
will change. What is unexpected about that?
--
73, Cecil http://www.w5dxp.com

On Dec 7, 4:09 pm, Cecil Moore wrote:
Keith Dysart wrote:
So sometimes a 600 to 100 ohm discontinuity
produces a 36.6 degree phase shift and sometimes
it produces a 22.7 degree phase shift (and probably
any value in between).

Yes, of course - nobody said the phase shift wasn't
a variable. Why would you expect it to be a constant?
It is a variable that depends upon the phase of the
component forward and reflected waves.

I suggest that "work[ing] up the phasor diagrams of
the component voltages (or currents) at the junction
where rho = (600-100)/(600+100) = 0.7143" will
not be useful for predicting the phase shift.

It will be useful for reporting that particular phase
shift. If other conditions change, that phase shift
will change. What is unexpected about that?

You implied that you were going to compute it
using just rho, which would mean it would be
constant for any pair of impedances.

With more inputs, it might be possible to
compute a number that, when added to the
actual electrical lengths of the lines, will
result in 90 degrees. I expect the algorithm
to be fairly complicated.

Of course, one can always just say it is equal
to 90 minus the sum of the electrical lengths
of the lines, though if there were two or more
impedance discontinuities, it might be difficult
to apportion the difference between them.

I await the algorithm.

....Keith

Keith Dysart wrote:
You implied that you were going to compute it
using just rho, which would mean it would be
constant for any pair of impedances.

No, I did not. Rho can be constant but the phase
angle of the incident voltage changes with
position. Therefore, the phase angle of the
reflected voltage changes with position.

This subject is already covered in my energy
analysis article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com