On Sun, 30 Jul 2006 16:04:10 -0700, "Bob Agnew" wrote:

Just a nit: You said:

If there are reflections, the voltage
and current are in phase only every 1/4 wavelength.

Actually, if there are reflections, the voltage and current ar NEVER in
phase.

In fact, voltage and current in the reflected wave are ALWAYS 180° out of phase,
while in the forward wave they are always in phase. Thus, along the line they
alternately add and subtract, first reinforcing and then cancelling each other
at every quarter wave, to form the standing wave.

Walt, W2DU

Walter Maxwell wrote:
On Sun, 30 Jul 2006 16:04:10 -0700, "Bob Agnew" wrote:
Actually, if there are reflections, the voltage and current ar NEVER in
phase.

In fact, voltage and current in the reflected wave are ALWAYS 180° out of phase,
while in the forward wave they are always in phase. Thus, along the line they
alternately add and subtract, first reinforcing and then cancelling each other
at every quarter wave, to form the standing wave.

All true, Walt, but I think we are discussing the net voltage
and net current which are in phase only every 1/4 wavelength
where the SWR circle crosses the horizontal purely resistive
line on the Smith Chart.
--
73, Cecil http://www.qsl.net/w5dxp

Walter Maxwell wrote:
On Sun, 30 Jul 2006 16:04:10 -0700, "Bob Agnew" wrote:

Just a nit: You said:

If there are reflections, the voltage
and current are in phase only every 1/4 wavelength.
Actually, if there are reflections, the voltage and current ar NEVER in
phase.

In fact, voltage and current in the reflected wave are ALWAYS 180° out of phase,
while in the forward wave they are always in phase. Thus, along the line they
alternately add and subtract, first reinforcing and then cancelling each other
at every quarter wave, to form the standing wave.

Walt, W2DU

And, if you assume the line is lossless, the voltage and current are in
phase every 90 degrees along the line regardless of the amount of mismatch.

This is easily illustrated with a Smith chart -- choose any point you'd
like, representing an arbitrary load impedance. Then draw a circle
through that point, with the center of the circle at the chart's origin.
Moving clockwise along this circle represents moving along the
transmission line from the load toward the source. You'll cross the
chart's axis, where the impedance is purely real, in a half revolution
(90 degrees of movement along the line) or less, and cross it each half
revolution (90 degrees) from then on.

Roy Lewallen, W7EL

Roy Lewallen wrote:
And, if you assume the line is lossless, the voltage and current are in
phase every 90 degrees along the line regardless of the amount of mismatch.

If the line has losses, the SWR circle becomes an SWR spiral
but the spiral still crosses the purely resistive axis like
the circle does, just not at the same points.

Does your answer imply that the number of degrees between
purely resistive crossings is not equal to 90 degrees when
the line is lossy? (Not a trick question)
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Roy Lewallen wrote:
And, if you assume the line is lossless, the voltage and current are
in phase every 90 degrees along the line regardless of the amount of
mismatch.

If the line has losses, the SWR circle becomes an SWR spiral
but the spiral still crosses the purely resistive axis like
the circle does, just not at the same points.

Does your answer imply that the number of degrees between
purely resistive crossings is not equal to 90 degrees when
the line is lossy? (Not a trick question)

After further thought, I think Roy's point is that a lossless
transmission line has a purely resistive Z0 so the voltage
and current are in phase every 90 degrees. Z0 is not purely
resistive for ordinary transmission lines. But real-world
distortionless lines are indeed lossy while possessing a
purely resistive Z0.
--
73, Cecil http://www.qsl.net/w5dxp