Several people (perhaps best not named) have said various things about a
"phase shift" occurring at the junction of two (lossless) transmission lines
of different characteristic impedances, if one of them is terminated in an
open circuit. In particular there has been discussion of a 100 ohm line with
an open circuit for its load, attached to a 600 ohm line, and with the
"electrical lengths" of the two lines chosen in ways making the other end of
the 600 ohm line appear to be a short circuit.

In one case, the 100 ohm line was 10 degrees long, the 600 ohm line was 43.4
degrees long, and the discontinuity at the junction between the two lines
was blamed for the missing 36.6 degrees. In another case, the 100 ohm line
was 5 degrees long, the 600 ohm line was 62.3 degrees long, and the
discontinuity at the junction between the two lines was blamed for the
missing 22.7 degrees.

I prefer to think of it this way. The open-ended 100 ohm line section of
length alpha degrees presents an impedance of -j*100*cot(alpha) to the 600
ohm line. (Call this 600 ohm line, which gets terminated by the 100 ohm
line, the first 600 ohm line section.) This impedance presented to the
(first) 600 ohm line section by the 100 ohm line section is the same
impedance that would be presented to the (first) 600 ohm line section if
instead it were to be terminated with another (second) length of 600 ohm
line, not of length alpha but rather of length (alpha + beta), so we have

-j*600*cot(alpha + beta) = -j*100*cot(alpha)

This equation can be solved for beta, which is going to represent the
(fictitious) phase shift. Now if we want the other end of the (first) 600
ohm line to look like a short circuit, then we will want the (first) 600 ohm
line to have length 90 - (alpha + beta) degrees.

All this really has nothing to do with the special values 100 and 600 ohms.
If the open-circuited line has characteristic impedance Z_1 and the other
line has characteristic impedance Z_2, then the Z_1 impedance line presents
a load of -j*(Z_1)*cot(alpha) to the Z_2 impedance line section, the same
load that would have been presented to the Z_2 ohm line section had it
instead been terminated by another piece of Z_2 ohm line of length
(alpha + beta), or in other words

-j*(Z_2)*cot(alpha + beta) = -j*(Z_1)*cot(alpha).

Dredging up from memory the formula for the tangent of the sum of two angles
(a formula which I probably last used in high school nearly 60 years ago)
and solving for beta,

beta = arctan{[(Z_2) - (Z_1)]*tan(alpha)/[(Z_1) + (Z_2)*(tan(alpha))^2]}

This formula agrees with the values previously given for the 10, or 5,
degree 100 ohm line, and 600 ohm line combinations. If Z_1 = 100, Z_2 = 600,
and alpha = 10 degrees, the formula gives beta = 36.6132921 degrees. Using
the 5 degree length instead for the 100 ohm line, beta = 22.6964176 degrees.

Note that beta depends only on Z_1, Z_2, and alpha (the length of the
open-circuited Z_1 ohm line), and not on the length (call it gamma) of the
Z_2 ohm line. If you want some other impedance, say W, instead of zero ohms,
to be presented at the input to the Z_2 ohm line then just make sure the Z_2
ohm line has length gamma, where

-j*(Z_2)*cot(alpha + beta + gamma) = W.

Of course, for lossless lines, W had better be a pure imaginary number.
(The choices W = 0 and W = infinity are OK.)

I really don't think beta is of any importance, nor that it has any physical
meaning. What is important, and does have meaning, is (alpha + beta). That's
the length that a (second) section of open-circuited Z_2 ohm line would have
to be, to have the same effect on the (first) section of Z_2 ohm line that
the open-circuited Z_1 ohm line section has. Once you mentally replace the
Z_1 ohm line section with such a Z_2 ohm line section, everything becomes
much simpler.

If you want more than just two line sections of different impedances, this
approach is easily adapted to that situation. Just take things one stage at
a time.

I think these formulas don't care whether the angle alpha is acute, whether
Z_2 is larger or smaller than Z_1 (or even the same), or whether the desired
input impedance W is that of a pure capacitance (negative imaginary W) or of
a pure inductance (positive imaginary W) or represents a short circuit or an
open circuit. If you don't like negative coax cable lengths, negative angle
values for gamma can be avoided by adding 180 degrees. Negative values for
beta are no problem, as long as (alpha + beta) is positive -- they just mean
that the Z_2 ohm cable thinks the open-circuited Z_1 ohm cable is shorter
than it really is, instead of longer. Notice that that is exactly what
happens when Z_1 is larger than Z_2 (assuming alpha is acute).

David, ex-W8EZE and retired SFU Math professor

--
David Ryeburn

To send e-mail, use "ca" instead of "caz".

David Ryeburn wrote:

...
David, ex-W8EZE and retired SFU Math professor


David:

That is a mouthful!

I will read, and re-read, this a few times ... QUITE OBVIOUSLY, you have
given this some SERIOUS thought ... I will attempt to make your efforts
meaningful--at least in the fact of it being "explored" by another,
thanks for the input ...

And, with that, good night, it is way past my bedtime--work has kept up
late.

Best regards,
JS