RF transmitter output has impedance of 50 ohms and is connected to dipole
with a feedpoint impedance of 50 ohms via feeder with characteristic
impedance of 50 ohms. System is perfectly matched. I expect SWR meter to
show perfect match of 1:1.

Dipole has a standing wave on it. Ends of dipole are at high voltage. Dipole
is centre-fed with centre being high current point. Standing wave means that
a reflected wave exists. Wave is reflected from open ends of dipole. What
happens to the reflected wave? How does it vanish at centre of dipole? Why
does reflected wave not go along feeder into transmitter output? There
cannot be a reflected wave on feeder because SWR is 1:1.

David wrote:
RF transmitter output has impedance of 50 ohms and is connected to dipole
with a feedpoint impedance of 50 ohms via feeder with characteristic
impedance of 50 ohms. System is perfectly matched. I expect SWR meter to
show perfect match of 1:1.

Dipole has a standing wave on it. Ends of dipole are at high voltage. Dipole
is centre-fed with centre being high current point. Standing wave means that
a reflected wave exists. Wave is reflected from open ends of dipole. What
happens to the reflected wave? How does it vanish at centre of dipole? Why
does reflected wave not go along feeder into transmitter output? There
cannot be a reflected wave on feeder because SWR is 1:1.


If the reflection is exactly in phase with the next wave arriving
through the feed line, then it just raises the impedance the line
sees. In other words, the reflection acts as a large part of the feed
energy for the next cycle. It doesn't bounce into and out of the feed
line, it bounces back and forth from end to end of the dipole.
Actually there are two reflected waves going in opposite directions,
end to end, simultaneously.

On Mon, 4 Sep 2006 20:44:17 +0100, "David" nospam@nospam wrote:

RF transmitter output has impedance of 50 ohms and is connected to dipole
with a feedpoint impedance of 50 ohms via feeder with characteristic
impedance of 50 ohms. System is perfectly matched. I expect SWR meter to
show perfect match of 1:1.

Accepting the figures as an example and not necessarily a reality...

You have just described a point which is a junction between:
- a load where the ratio of voltage to current is 50+j0;
- a feedline whe
* the ratio of the voltage to current due to the forward travelling
wave must each be in the ratio 50+j0;
* the ratio of the voltage to current due to the reflected
travelling waves must each be in the ratio 50+j0.

Having regard for the sign of the traveling waves, the only solution
to those constraints / conditions is that the reflected travelling
wave must have zero amplitude.


Dipole has a standing wave on it. Ends of dipole are at high voltage. Dipole
is centre-fed with centre being high current point. Standing wave means that
a reflected wave exists. Wave is reflected from open ends of dipole. What
happens to the reflected wave? How does it vanish at centre of dipole? Why
does reflected wave not go along feeder into transmitter output? There
cannot be a reflected wave on feeder because SWR is 1:1.

Lets be clear that we are now talking about a single frequency.

At any point, the forward and reflected waves resolve to a single
voltage at that point, and a single current flowing at that point, and
the ratio of voltage to current is the impedance (and these are all
complex quantities, ie they have real and imaginery parts).

If the point you consider is the feedpoint, and the ratio of voltage
to current is 50+j0, then that is the impedance, it fully describes
the load at that frequency.

You talk of the "reflected wave" as if it has inertia, that it must
keep travelling when it reaches the junction of the feedline and the
antenna? You are not alone in speaking that way, but thinking that way
will get in the way of understanding what is happening. Next thing,
you will be thinking that the reflected wave must travel back to the
PA anode and will be absorbed there causing overheating.

Owen
--

"Owen Duffy" wrote:
You talk of the "reflected wave" as if it has inertia, that it must
keep travelling when it reaches the junction of the feedline and the
antenna? You are not alone in speaking that way, but thinking that way
will get in the way of understanding what is happening. Next thing,
you will be thinking that the reflected wave must travel back to the
PA anode and will be absorbed there causing overheating.
_______

If you write of reflected power existing on the transmission line between
the tx output connector and the antenna input connector, then yes -- with
sufficient tx output power and a poor enough match at the antenna feedpoint,
that reflected power can cause transmission line and/or tx PA component
failure.

I've analyzed and repaired many such failures of these installed systems in
my pre-retirement career, and know this first hand. Conventional theory
also supports this result.

RF

Visit http://rfry.org for FM transmission system papers.

On Mon, 4 Sep 2006 17:03:39 -0500, "Richard Fry"
wrote:

"Owen Duffy" wrote:
You talk of the "reflected wave" as if it has inertia, that it must
keep travelling when it reaches the junction of the feedline and the
antenna? You are not alone in speaking that way, but thinking that way
will get in the way of understanding what is happening. Next thing,
you will be thinking that the reflected wave must travel back to the
PA anode and will be absorbed there causing overheating.
_______

If you write of reflected power existing on the transmission line between
the tx output connector and the antenna input connector, then yes -- with
sufficient tx output power and a poor enough match at the antenna feedpoint,
that reflected power can cause transmission line and/or tx PA component
failure.

The mechanism is not "that the reflected wave must travel back to the
PA anode and will be absorbed there causing overheating".

Take an example of the 50 ohms load discussed, and an electrical half
wave of 70 ohm line connected to a transmitter designed for a 50 ohm
load. The transmitter behaves exactly as if that line were 50 ohms.
Though there is a reflected travelling wave on the line, it does not
travel back to the PA anode where it is absorbed and converted to
heat. At the tx end of the line, the forward and reflected wave
components resolve to a 50+j0 load, and the transmitter sees the same
50 ohm load as it would were 50 ohm line used. Increasing power or
increasing line Zo for a higher VSWR will not change the outcome of
this example.

Owen
--

Owen Duffy wrote:
Take an example of the 50 ohms load discussed, and an electrical half
wave of 70 ohm line connected to a transmitter designed for a 50 ohm
load. The transmitter behaves exactly as if that line were 50 ohms.
Though there is a reflected travelling wave on the line, it does not
travel back to the PA anode where it is absorbed and converted to
heat.

Of course not. Destructive interference redirects the energy
back toward the load at the impedance discontinuity. The same
thing happens with a 1/4WL thin-film on non-reflective glass.
--
73, Cecil http://www.w5dxp.com

"Owen Duffy" wrote
The mechanism is not "that the reflected wave must travel back to
the PA anode and will be absorbed there causing overheating".
Take an example of the 50 ohms load discussed, and an electrical half
wave of 70 ohm line connected to a transmitter designed for a 50 ohm
load. The transmitter behaves exactly as if that line were 50 ohms.
Though there is a reflected travelling wave on the line, it does not
travel back to the PA anode where it is absorbed and converted to
heat.
____________

Really, the mechanism is there -- only the unique circumstance you describe
protects the PA from seeing reflected power in such cases. Other line
lengths in this scenario could stress PA components beyond their ratings.

And even if the PA saw no reflected power because of a fortunate length of
transmission line connecting it to a mismatched antenna/load, that reflected
power still exists in the transmission line, and may cause its failure.

Manufacturers of the rigid coaxial line used in broadcast stations (e.g.,
Dielectric) require derating its maximum power rating inversely by the value
of the SWR existing in it . A power derating factor related to SWR also
applies to Andrew Heliax® and RG-type coax line. Deliberately setting up,
or tolerating reflected power on a transmission line is not done without
risk.

RF

Richard Fry wrote:
Really, the mechanism is there -- only the unique circumstance you
describe protects the PA from seeing reflected power in such cases.
Other line lengths in this scenario could stress PA components beyond
their ratings.

Yes, the exposure of the PA to reflected power depends
upon the phase of the reflected wave referenced to the
phase of the source wave. SWR doesn't tell the whole
story because phase is not reported by the SWR
measurement.

What is happening with 1/2WL of Z0=600 ohm feedline
connected to a 50 ohm load on one end and a 50 ohm
source on the other end is interference. Destructive
interference toward the source causes constructive
interference toward the load and the reflected energy
at the source is re-directed back toward the load.

What we want to avoid in our antenna systems is
constructive interference toward the source.
--
73, Cecil http://www.w5dxp.com

On Tue, 5 Sep 2006 06:48:57 -0500, "Richard Fry"
wrote:

"Owen Duffy" wrote
The mechanism is not "that the reflected wave must travel back to
the PA anode and will be absorbed there causing overheating".
Take an example of the 50 ohms load discussed, and an electrical half
wave of 70 ohm line connected to a transmitter designed for a 50 ohm
load. The transmitter behaves exactly as if that line were 50 ohms.
Though there is a reflected travelling wave on the line, it does not
travel back to the PA anode where it is absorbed and converted to
heat.
____________

Really, the mechanism is there -- only the unique circumstance you describe
protects the PA from seeing reflected power in such cases. Other line
lengths in this scenario could stress PA components beyond their ratings.


Richard,

There is not such a mechanism in the general case, the example I gave
shows that the reflected wave does not necessarily travel back to the
source where it is absorbed.

If you re-read my words "Next thing, you will be thinking that the
reflected wave must travel back to the PA anode and will be absorbed
there causing overheating." Remembering the context was a wave on the
dipole reflected from the o/c end, and the word "must" was used to
mean "necessarily".

Sure,transmission lines with VSWR may transform impedance, have higher
losses (if they are long enough), operate at higher voltages in places
(if they are long enough), operate with higher currents in places (if
they are long enough). Nothing I have said is in conflict with that or
suggests otherwise.

Transmitters operated at other than their rated load impedance may
operate at higher voltages, higher currents, different power etc and
may damage components. Nothing I have said is in conflict with that or
suggests otherwise.

But, the mechanism is not that the reflected wave *necessarily*
travels all the way back to the PA anode by virtue of some kind of
momentum (as sometimes expressed by some amateurs).

In the case raised by the OP, the reflected wave on the dipole and the
forward wave resolve (as in resolution of phasors) to an impedance of
50+j0 (OP's hypothetical example), and the constraints / conditions at
the feedline / feedpoint junction are fully satisfied with no
reflected wave on the feedline. (I used the term resolve, Cecil must
call it destructive interference.) The reflected wave on the dipole
does not have a momentum that *must* carry it to the PA anode to be
absorbed there.

Owen
--

Owen Duffy wrote:
Having regard for the sign of the traveling waves, the only solution
to those constraints / conditions is that the reflected travelling
wave must have zero amplitude.

Sorry Owen, that's not true. Assuming a 1/2WL dipole,
the reflected voltage has traveled 180 degrees. The
reflected current has traveled the same 180 degrees
plus experienced a 180 degree phase shift at the tip
of the dipole. Assuming 100 watts is being applied to
the antenna, the following conditions satisfy the
observed conditions on the dipole at the feedpoint.

Pfor = 500W, Vfor = 548V, Ifor = 0.91A

Pref = 400W, Vref = 490V, Iref = 0.81A

Pnet = 100W, Vnet = 58V, Inet = 1.72A

You talk of the "reflected wave" as if it has inertia, that it must
keep travelling when it reaches the junction of the feedline and the
antenna? You are not alone in speaking that way, but thinking that way
will get in the way of understanding what is happening.

It is well known that all EM waves contain energy and
momentum (inertia) so David is correct. The thing that
reverses the momentum of the reflected wave is destructive
interference at the Z0-match point. Anyone interested in
understanding how/why it happens is invited to peruse
my energy analysis article at: http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com