Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"Certainly, the total V and I are in quadrature if the line is
terminated by an open, short, or purely reactive load. But not in any
other case."

Something else is at work. The reflection reverses direction of the wave
producing a 180-degree phase shift in either voltage or current, but not
both, if there is a reflection.

Yes.

Because the waves are traveling at the
sane speed in approaching each other, they produce a phase reversal in a
distance of only 90-degrees instead of 180-degrees. This places the
waves in quadrature to stay.

?? Which waves? Forward voltage and reverse current? Forward and reverse
voltage?

Terman shows the vector diagrams of
incident and reflected waves combined to produce a voltage distribution
on an almost lossless transmission line (Zo=R) for an open circuit case
and for a resistive load case where the load is Zo in Fig. 4-3 on
page 91 of his 1955 opus. Indeed, the angle between the incident and
reflected voltages is 90-degrees in either case.

Please look carefully at those diagrams. The horizontal axis is the
distance along the line. The diagrams are showing the relationship
between voltage and current envelopes as a function of position. The
graphs aren't showing the time phase of V and I, which is the matter
under discussion.

In Fig. 4-4 on page 92, Terman shows voltage and current distributions
produced on low-loss transmission lines by different load impedances and
in every case volts and amps are in quadrature.

I don't have that diagram in my 1947 Third Edition, but I'm sure that if
you'll study the diagram and accompanying text you'll find that it's
also a graph of peak amplitude vs position, not V or I as a function of
time.

Let me pose a very simple problem. Suppose you have a quarter wavelength
of 50 ohm transmission line terminated in 25 ohms and driven with a 100
volt RMS sine wave source. Consider the phase of the voltage source to
be the reference of zero phase angle.

1. What is the current at the line input? [Answer: 1 ampere, at 0 degree
phase]
2. What is the ratio of V to I at the line input? [Answer: 100 at an
angle of zero divided by 1 at an angle of zero = 100 + j0 ohms]
3. How do you resolve this with the graphs in Terman and your
explanation of the voltage and current being in quadrature everywhere
along the line?

Feel free to repeat this with any other line length. I'll wait very
patiently for the length which produces V and I in quadrature at the
input. A very interesting result of that would be that no power would be
consumed from the source, so if any reaches the load then we've created
power.

I'd be glad to post the equation relating Z (the ratio of V to I) at the
line input or any point along the line to the load and characteristic
impedances. But I'm afraid it would be wasted effort, since there's a
great reluctance here to actually work an equation or understand its
meaning. But good and accurate graphs of what Richard has claimed the
Terman graphs show (but don't) can be found in the _ARRL Antenna Book_.
In the 20th and 21st Editions, the graphs are Fig. 12 on p. 24-9. In
other editions, they're probably also Fig. 12 in the Transmission Lines
chapter. In the graphs, the angle between the I and E vectors is the
relative phase angle between the two, and also the angle of the
impedance of the point where the vectors are shown.

Roy Lewallen, W7EL

Roy Lewallen wrote:
3. How do you resolve this with the graphs in Terman and your
explanation of the voltage and current being in quadrature everywhere
along the line?

The standing wave voltage is *ALWAYS* in quadrature with
the standing wave current at all points and at all times.
If the total voltage is not in quadrature with the total
current, there is a traveling wave in the mixture.
--
73, Cecil http://www.w5dxp.com