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

So how does a 1/2WL piece of transmission line driving a
50 ohm load wind up with the voltage and current in phase
no matter what the SWR?
--

If the characteristic impedance of the transmission line is 50 ohms, then
there are no reflections; furthermore the current and voltage are in phase
at every point along the line, There are no standing waves in this case.

In fact it doesnt matter how long the line is as long as it is terminated in
its charcteristic impedance. This case corresponds to the circle in the
middle of the Smith Chart on which the impedance is constant.


"Cecil Moore" wrote in message
. com...
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.
c 73, Cecil http://www.qsl.net/w5dxp

Bob Agnew wrote:

So how does a 1/2WL piece of transmission line driving a
50 ohm load wind up with the voltage and current in phase
no matter what the SWR?
--

If the characteristic impedance of the transmission line is 50 ohms, then
there are no reflections; furthermore the current and voltage are in phase
at every point along the line, There are no standing waves in this case.

In fact it doesnt matter how long the line is as long as it is terminated in
its charcteristic impedance. This case corresponds to the circle in the
middle of the Smith Chart on which the impedance is constant.

It doesn't matter what the characteristic impedance of the
transmission line is as long as it is an electrical 1/2WL. As such,
you can have reflections, feeding a 50 ohm resistive load, and the
voltage and current will be in phase 1/2WL back fron the load. Antenna
matching with transmission line transformers use different impedance
transmission lines to wind up with a perfect match.
73 Gary N4AST

I haven't taught this subject at the University level for a few years now,
so my terminology is a little rusty. VSWR = (1 + abs(rho))/(1 - abs(rho))

When the reflection coefficient rho is +/- 1 the VSWR is infinite. When rho
= 0 the VSWR is 1:1. A VSWR of 1:1 corresponds to the unit circle at the
center of the Smith Chart.

wrote in message
oups.com...

Bob Agnew wrote:

So how does a 1/2WL piece of transmission line driving a
50 ohm load wind up with the voltage and current in phase
no matter what the SWR?
--

If the characteristic impedance of the transmission line is 50 ohms, then
there are no reflections; furthermore the current and voltage are in
phase
at every point along the line, There are no standing waves in this case.

In fact it doesnt matter how long the line is as long as it is terminated
in
its charcteristic impedance. This case corresponds to the circle in the
middle of the Smith Chart on which the impedance is constant.

It doesn't matter what the characteristic impedance of the
transmission line is as long as it is an electrical 1/2WL. As such,
you can have reflections, feeding a 50 ohm resistive load, and the
voltage and current will be in phase 1/2WL back fron the load. Antenna
matching with transmission line transformers use different impedance
transmission lines to wind up with a perfect match.
73 Gary N4AST

Bob Agnew wrote:
I haven't taught this subject at the University level for a few years now,
so my terminology is a little rusty. VSWR = (1 + abs(rho))/(1 - abs(rho))

When the reflection coefficient rho is +/- 1 the VSWR is infinite. When rho
= 0 the VSWR is 1:1. A VSWR of 1:1 corresponds to the unit circle at the
center of the Smith Chart.

Hi Bob, I have never taught at the University level myself, this is an
"Amateur" newsgroup. If you look at the Smith Chart, 1/2WL reflects
back the identical load impedance, no matter the transmission line
characteristic impedance (neglecting losses). If you have a 1/2WL 600
ohm line driving a 50 ohm load, then you have reflections (SWR), but at
the source end, you see 50 ohms (V and I in phase).
73 Gary N4AST

Hi Bob, I have never taught at the University level myself, this is an
"Amateur" newsgroup. If you look at the Smith Chart, 1/2WL reflects

OK __ I won't post here anymore. I once was an Amateur when I was 14.
I thought that I wanted to get back into the hobby now that I am retired.

wrote in message
ups.com...

Bob Agnew wrote:
I haven't taught this subject at the University level for a few years
now,
so my terminology is a little rusty. VSWR = (1 + abs(rho))/(1 -
abs(rho))

When the reflection coefficient rho is +/- 1 the VSWR is infinite. When
rho
= 0 the VSWR is 1:1. A VSWR of 1:1 corresponds to the unit circle at the
center of the Smith Chart.

Hi Bob, I have never taught at the University level myself, this is an
"Amateur" newsgroup. If you look at the Smith Chart, 1/2WL reflects
back the identical load impedance, no matter the transmission line
characteristic impedance (neglecting losses). If you have a 1/2WL 600
ohm line driving a 50 ohm load, then you have reflections (SWR), but at
the source end, you see 50 ohms (V and I in phase).
73 Gary N4AST

On Mon, 31 Jul 2006 16:30:21 -0700, "Bob Agnew"
wrote:

Hi Bob, I have never taught at the University level myself, this is an
"Amateur" newsgroup. If you look at the Smith Chart, 1/2WL reflects

OK __ I won't post here anymore. I once was an Amateur when I was 14.
I thought that I wanted to get back into the hobby now that I am retired.

Hi Bob,

You as a prof. are not alone here. We have several, some still in the
saddle, others retired, and all Hams. You and Gary are not wrong
either, simply posting at cross purposes when the topic was sidelined
into half wave lines, SWR, and a demand for an explanation for a
problem that was never offered in the first place. This was a troll
that bit you, not Gary.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Mon, 31 Jul 2006 16:30:21 -0700, "Bob Agnew"
wrote:

Hi Bob, I have never taught at the University level myself, this is an
"Amateur" newsgroup. If you look at the Smith Chart, 1/2WL reflects

OK __ I won't post here anymore. I once was an Amateur when I was 14.
I thought that I wanted to get back into the hobby now that I am retired.

Hi Bob,

You as a prof. are not alone here. We have several, some still in the
saddle, others retired, and all Hams. You and Gary are not wrong
either, simply posting at cross purposes when the topic was sidelined
into half wave lines, SWR, and a demand for an explanation for a
problem that was never offered in the first place. This was a troll
that bit you, not Gary.

73's
Richard Clark, KB7QHC

I have been an amateur continuously since I was 13: 1963.

Transmission line: 1/2 half wave repeats the load, 1/4 wave inverts the
load.
That I knew in Jr High.

Finally understood it when I studied E&M in college.

73
H.
NQ5H

www.hep.utexas.edu/mayamuon