Keith Dysart wrote:
Cecil Moore wrote:
Here's a request to anyone capable of performing the
necessary math. Can anyone verify or disprove my
two voltage values in the following example?

Shortened Stub Example:

rho = -0.7143
--43.4 deg 600 ohm line--+--10 deg 100 ohm line--open
Vfor1--|--Vfor2 Vi--|
Vref1--|--Vref2

Assume the RMS voltage, Vi, incident upon the open
end of the stub is: Vi = 100 volts at 0 degrees

At point '+' 10 degrees back from the open,
Vfor2 = 100 volts at -10 degrees
Vref2 = 100 volts at +10 degrees

It appears to me that you have the signs wrong above.

I believe you are correct but since |Vfor2|=|Vref2| the
math turns out the same if consistency is maintained,
which it was.

Except for that, the computation seems correct,
though you have not explained your work.

I'm wondering how many people on this newsgroup can do
that computation that you say "seems correct".

Vfor1 = 143.33 volts at -46.6 degrees
Vref1 = 143.33 volts at +46.6 degrees

This step has no explanation.

Yes, because I wanted independent verification. Do you
know how to do the math? If so, see if you get the same
values that I did. Everyone else is also invited to perform
the math. It's pretty simple reflection superposition stuff.

Can you provide a justification from first principles?

I did that in another posting but I will repeat it here.
Here's the voltage superposition diagram. In fixed font:

Component Voltage Diagram

Vfor1--|
rho1--|--tau1

|--Vref2
tau2--|--rho2

Z01=600 ohms | Z02=100 ohms
|
Vfor1-------------|
|----Vfor1(tau1)
Vfor1(rho1)----| \
/ | \
/ (+)----Vfor2
Vref1----(+) /
\ | /
\ |----Vref2(rho2)
Vref2(tau2)----|
|-------------Vref2
|

Vref1 = Vfor1(rho1) + Vref2(tau2)

Vfor2 = Vfor1(tau1) + Vref2(rho2)

Two equations and two unknowns. Solve for Vfor1 and Vfor2

Or if you prefer s-parameters:

s-parameter diagram

Z01=600 ohms | Z02=100 ohms
|
a1-------------|
|----s21(a1)
s11(a1)----| \
/ | \
/ (+)----b2
b1----(+) /
\ | /
\ |----s22(a2)
s21(a1)----|
|-------------a2
|

b1 = s11(a1) + s12(a2) b2 = s21(a1) + s22(a2)

Next, what are the useful properties of this computed
phase shift.

The immediate useful property is that it settles the
argument over whether there is a phase shift at an
impedance discontinuity or not. The follow on argument
is that this is what happens at the loading coil to
stinger impedance discontinuity in a base-loaded mobile
antenna, i.e. it's related to the phase-shift-delay
through the loading coil. It will eventually explain
how loaded mobile antennas really work and may (or may
not) point to some improvement that can be made.

The phase shift between Vfor1 and Vfor2 is 36.6 degrees.

Silly math error? 36.6 + 46.6 +10 - 93.2, which does not
align with your previous posts.

Where did the 46.6 degrees come from? The 600 ohm section
is 43.4 degrees long. 36.6 + 43.4 + 10 = 90 degrees.

I remember where the 46.6 degrees came from.
43.4 + 46.6 = 90 degrees. 46.6 degrees is the electrical
length of the 600 ohm line needed to complete the 600 ohm
1/4WL stub function. But that was a different example.

The total RMS voltage at all points up and down the line
is 200 volts at 0 deg.

I don't follow this statement. Which line? From the original
experiment, is not the total voltage at the driven end zero?

That was a pre-coffee brain fart from a canceled posting. It
is a false statement. I changed that sentence, in the subsequent
posting, to something like:

At point '+', Vfor1+Vref1 = Vfor2+Vref2 = 196.96v at 0 deg.

The phasor diagram of that equation speaks volumes and is
something you previously asked for. I will draw it up and
post it on my web page.
--
73, Cecil http://www.w5dxp.com