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