Sal M. Onella wrote:
"James Barrett" wrote in message
news:WN6dnT9x6ZU6I7HanZ2dnUVZ_v-

snip

I thought that I wanted
an antenna with zero reflected energy or as close to that as possible.
Now it sounds like that is not always the case.

You were mostly right; this is the theoretical ideal, but reality forces
compromises on all of us.

To put it simply, yes, you want the most power to "jump off the antenna"
into space. Whatever doesn't jump off is dissipated (wasted) as heat
somewhere in the system. If too much is reflected back from the antenna and
dissipated within in your transmitter, the transmitter overheats ($$$) or it
reduces power to protect itself and nobody hears you. . .

As I said earlier, there's a lot of misinformation floating around.

A high SWR doesn't mean there's "reflected energy" which is going to be
dissipated anywhere, least of all in your transmitter. Except for
transmission line loss (which admittedly will be greater, although
usually insignificantly so, if the SWR is very high -- see the Antenna
Book), all the power leaving the transmitter will arrive at your
antenna. Thinking of waves of energy bouncing back and forth looking for
somewhere to be dissipated will lead you down paths that you won't be
able to reason your way out of. As has been clearly demonstrated on this
newsgroup over and over. Read the Antenna Book and other good texts, and
don't try to make up additional imaginary waves.

"Sal" is right about one thing, though. Most transmitters will reduce
output power if the SWR gets too high, which tells the transmitter that
the impedance it's seeing is beyond the range for which it's designed.
(The problem is that various places in the transmitter can encounter
voltages and/or currents too far above design values, or impedances
which might cause instability. It's not because there are waves of
"reflected energy" which dissipate themselves in the transmitter.) So
you do want to keep the SWR measured at the transmitter below that
value. There's no harm in having a very high SWR on the feedline,
however, as long as it has low matched loss.

Roy Lewallen, W7EL

Owen Duffy wrote:
"Sal M. Onella" wrote in
:

as heat somewhere in the system. If too much is reflected back from
the antenna and dissipated within in your transmitter, the transmitter
overheats ($$$) or it reduces power to protect itself and nobody hears
you.

Here we go again!

Yes, this misconception will never die. Is it really worth the trouble
continually trying to contradict it?

Roy Lewallen, W7EL

Owen Duffy wrote:
"Sal M. Onella" wrote in
:

as heat somewhere in the system. If too much is reflected back from
the antenna and dissipated within in your transmitter, the transmitter
overheats ($$$) or it reduces power to protect itself and nobody hears
you.

Here we go again!

If he said "sometimes overheats", he would be correct.
An SWR of 10:1 certainly *can* cause an over-current condition
in an unprotected transmitter assuming the reflected current
is in phase with the forward current at the transmitter.

However, just as likely is that the reflected voltage is in
phase with the forward voltage at the transmitter and an
over-voltage condition *can* result in punch-through of the
final transistor.

If over-current and over-voltage were not a problem caused
by reflected waves, protection of the finals would not be
necessary.

Note that the impedance seen by the transmitter above is a
*virtual* impedance, not an impedor. Virtual impedances are
only a *result* and not the cause of anything. Virtual impedances
are not the *cause* of over-current or over-voltage conditions.

Anyone who scoffs at virtual opens and virtual shorts being
the *cause* of the re-reflection of reflected energy cannot,
without contradicting himself, turn around and argue that
the virtual impedance seen by a transmitter is the *cause* of
the mismatch. One cannot have it both ways.
--
73, Cecil http://www.w5dxp.com

Roy Lewallen wrote:
A high SWR doesn't mean there's "reflected energy" which is going to be
dissipated anywhere, least of all in your transmitter.

Most transmitters will reduce
output power if the SWR gets too high, which tells the transmitter that
the impedance it's seeing is beyond the range for which it's designed.

Not readily apparent is the contradiction between these
two statements above which needs to be resolved. Reflected
energy cannot exist without energy, i.e. without ExB watts.
It is the energy in the reflected waves that is the *cause*
of the impedance "seen" by the transmitter. There is *zero*
dissipation in that virtual impedance so it is NOT a
real resistor - it is a dissipationless resistance.

The impedance seen by the transmitter is not a resistor
or inductor or capacitor, but instead is a *virtual*
impedance *caused by* the magnitude and phase of the
reflected wave with respect to the magnitude and phase
of the forward wave. The impedance seen by the
transmitter is:

Z = (Vfor+Vref)/(Ifor+Iref)

where the voltages and currents are phasors, each with
a magnitude and associated phase. There is NO resistor!
There is NO inductor! There is NO capacitor! There is
no power dissipation! Virtual impedances cannot cause
anything.

It is interesting to note that the very people who
support the virtual impedance seen by a transmitter as
being the cause of the conditions there are the same
people who rail loud and long against a virtual short
being able to cause 100% re-reflection. Why does a
virtual impedance cause things to happen only at a
transmitter but nowhere else?

If there's no "reflected energy", the transmitter will
see the characteristic impedance of the transmission
line, e.g. 50 ohms. So the transmitter CANNOT see any
impedance other than Z0 unless reflected energy is the
cause of the deviation away from Z0.

Depending upon the phase of the reflected energy, all
or some or none of the reflected energy may make its
way into the transmitter. The exact magnitude of
joules/sec making its way into the transmitter is:

P = P1 + P2 + 2*SQRT(P1*P2)cos(A)

where 'A' is the phase angle between the E-fields of
EMWave1 and EMwave2 and P1 = E1xB1 and P2 = E2xB2
The last term is known as the "interference term".
If it is negative, it represents destructive
interference. If it is positive, it represents constructive
interference. It should be obvious that 'P' above, can
assume any value between zero and a maximum value so
the amount of reflected energy flowing into the transmitter
can be anything from zero to that maximum value. If the
transmitter is looking into an ideal shorted 1/4WL stub,
the reflected energy flowing into the transmitter will
be zero just as the food-for-thought example demonstrates,
i.e. none of the available power is dissipated in the
transmitter. If the transmitter is looking into an ideal
open-circuit 1/4WL stub, all of the reflected energy will
flow into the transmitter, i.e. all of the available power
will be dissipated in the transmitter. How much depends
upon the relative magnitudes and phases of the forward
and reflected waves.
--
73, Cecil http://www.w5dxp.com

Roy Lewallen wrote:
Yes, this misconception will never die. Is it really worth the trouble
continually trying to contradict it?

Not if all you do is trade one old-wives tale
for another. One cannot understand these concepts
without understanding the conditions that cause
EM waves to interact.

The conditions that cause interaction between EM
waves are coherency and collinearity (in the same
direction in a transmission line).

The interaction of reflected EM waves can result
in zero or maximum reflected power being dissipated
in the transmitter - and anything in between.

The interaction of reflected EM waves at a thin-
film coating on glass can result in zero or
maximum reflected energy and anything in between
depending upon the thickness of the thin-film.
--
73, Cecil http://www.w5dxp.com

James Barrett wrote:



This thread has taken on a life of its own since I posted my first
question. I really don't understand any of this. I thought that I wanted
an antenna with zero reflected energy or as close to that as possible.
Now it sounds like that is not always the case. I need to learn all
about SWR and impedance in regards to Antennas, from start to finish. Is
there an easy-to-read tutorial out there for a beginner like me? Maybe
Ham Radio for Dummies has something about SWR and antennas?


I've been a ham since '58, and did 26 years in the navy in various areas
of communications and radar as an operator, technician, and manager. I have
done my best to learn about antennas and have quite a body of knowledge
accumulated, some of it good, some of it fallacious. Some of it remains
a complete mystery to me. There are a great many misconceptions when it
comes to antennas, feeders, SWR, and the like.

If I had paid attention to what some people say about SWR, conjugate
matches, and the like I would still be hesitant about putting up an
antenna. I'd still be looking for the perfect one.

Despite all the myths and misconceptions, many hams are surprised to find
out that regardless of how bad their antennas may seem in theory, in
practice they are getting out.

Put up an antenna, and 'tune' it for minimum SWR, then have fun. You
will find years of fun ahead and they will give you the time required
to learn more about antennas. Do not try to learn all there is -- you
won't get there. Instead, try to learn where the fallacies lie and
try to avoid them.

The hardest part is committing yourself to putting up an antenna which
you know from theory is not perfect. Once you get it up and find that you
can work the world with a few watts and a wet noodle, then you can
take the time to learn, experiment, and find an antenna that meets
most of your needs. Don't go looking for the perfect one -- it does
not exist.

Irv VE6BP (Heating the ionosphere and loving it!)

Irv Finkleman wrote:
I'd still be looking for the perfect one.

Antennas are like females - just try them out,
one by one, until you are satisfied. :-)
--
73, Cecil http://www.w5dxp.com

"Owen Duffy" wrote in message
...
"Sal M. Onella" wrote in
:

as heat somewhere in the system. If too much is reflected back from
the antenna and dissipated within in your transmitter, the transmitter
overheats ($$$) or it reduces power to protect itself and nobody hears
you.

Here we go again!

Owen

What did I say wrong?

"Roy Lewallen" wrote in message
...
Owen Duffy wrote:
"Sal M. Onella" wrote in
:

as heat somewhere in the system. If too much is reflected back from
the antenna and dissipated within in your transmitter, the transmitter
overheats ($$$) or it reduces power to protect itself and nobody hears
you.

Here we go again!

Yes, this misconception will never die. Is it really worth the trouble
continually trying to contradict it?

Roy Lewallen, W7EL

What did I say wrong?

"Cecil Moore" wrote in message
news
Roy Lewallen wrote:
Yes, this misconception will never die. Is it really worth the trouble
continually trying to contradict it?

Not if all you do is trade one old-wives tale
for another.

It's hardly an old wives' tale.

I mistakenly put a 2m antenna on my dual band HT and tried to use it for a
short QSO on a nearby 440 repeater. The other ham said I was barely making
the repeater, while my poor HT got so hot that I could barely hold it after
a minute's use.

The antenna was wrong and the heat was real -- whatever the theory behind
it.