On Dec 11, 11:24 pm, "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.- Hide quoted text -

To find the length of the 600 ohm line, do I not just plot -j567/600
and work from there? It seems to me that it yields the same
answer regardless of the content of the black box.

....Keith