I am confused by an issue and hope some of you gurus can help.

I understand matching at the transmitter end. Using the "right" length of
line or an antenna tuner - oops, I mean conjugate matching device, I can get
the rig to put all or as much power as it can into the line. If I choose
ladder line, it has very low loss in the line itself. So far -- so good.

But what happens where the line hits the antenna? If the line is 50 or 450
ohm and the antenna is exhibiting 2 ohms, isn't there a big mismatch and a
lot of lost power? How much? Would the 450:2 mismatch lose more power than
a 50:2 mismatch and thereby give up the advantage gained by the low loss in
the 450 ohm ladder line. I don't see this quantified in the antenna
modeling programs.

I realize in multiband use, the mismatch will vary so there is not a whole
lot you can do except put the tuner at the junction of the antenna and line.
So, why isn't that the "normal" way to handle the problem?
--
Radio K4ia
Craig "Buck"
Fredericksburg, VA USA
FISTS 6702 cc 788 Diamond 64

Craig Buck wrote:
But what happens where the line hits the antenna? If the line is 50 or 450
ohm and the antenna is exhibiting 2 ohms, isn't there a big mismatch and a
lot of lost power?

"A lot" needs to be defined but the short answer is yes. That's why 2 ohm
antennas are not popular. A lot of antennas are designed with close to 50
ohm feedpoint impedances at resonance.

I realize in multiband use, the mismatch will vary so there is not a whole
lot you can do except put the tuner at the junction of the antenna and line.
So, why isn't that the "normal" way to handle the problem?

That is fairly normal now that remote autotuners are readily available.

Another way to handle the problem is to use very low-loss transmission
line, e.g. open-wire feedline which will tolerate a high SWR with relatively
low losses. The antenna *system* is then matched at the transmitter.

I personally use the open-wire feedline to transform the antenna impedance
to an impedance acceptable to my transmitter with no tuner required.
--
73, Cecil http://www.qsl.net/w5dxp



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On Wed, 7 Jan 2004 00:32:17 -0500, "Craig Buck" wrote:
But what happens where the line hits the antenna? If the line is 50 or 450
ohm and the antenna is exhibiting 2 ohms, isn't there a big mismatch and a
lot of lost power? How much? Would the 450:2 mismatch lose more power than
a 50:2 mismatch and thereby give up the advantage gained by the low loss in
the 450 ohm ladder line. I don't see this quantified in the antenna
modeling programs.

Hi Craig,

SWR is not inherently lossy, it contributes to the existing loss which
can only be found in ohmic resistance. That ohmic resistance is part
of the finals' bulk resistance, transmission line wire, and metal
construction of the antenna. If all such characteristics were 0 Ohms,
then any amount of SWR would exhibit no loss whatever.

However, the antenna also exhibits a "radiation resistance" in series
which is part of the mix. This resistance does not transform power to
heat, and thus does not qualify as loss in the conventional sense. In
other words, radiation resistance (Rr) does not diminish efficiency.
Even if we had 0 Ohms ohmic resistance, there would still be the
radiation resistance to eventually consume all the power.

Now backing up and then considering that there is some ohmic
resistance in these elements, in comparison to that radiation
resistance (Rr) we find an efficiency (which IS quantified in antenna
modeling programs). If you tried to drive a 2 Ohm (Rr) antenna with 2
Ohms of ohmic loss, you can guess where this leads. If you tried to
drive a 200 Ohm (Rr) antenna with 2 Ohms of ohmic loss, you can see
you've improved your DX. BOTH resistances consume power, but the
radiation resistance is a productive use.

These are extremes and ignore SWR which takes that ohmic loss and
multiplies it. Hence it is preferable to present a 50 Ohm source
(your transmitter is characterized as such at full power) to a 50 Ohm
transmission line, feeding a 50 Ohm Antenna. Within this scenario,
there still remains the ohmic loss that is strictly a matter of local
variables (poor connections, small diameter wire, iron wire instead of
copper, proximity of earth...).

One last conundrum: there are very few 50 Ohm antennas in nature or
off the shelf. What you see characterized as a 50 Ohm antenna is
derived through matching connections at the feed point that transform
the actual radiation resistance (Rr) to the anticipated transmission
line Z. The standard quarterwave vertical whip exhibits an Rr of 30
to 35 Ohms for the best of conditions, and the standard halfwave
dipole exhibits an Rr of 70 to 75 Ohms for its best of conditions.

Those native Rr's are often left untransformed because the SWR they
present are in fact meager (even 2:1 is fairly trivial). If you were
to examine such plumber's delights like the series available from GAP,
you would find that they have conspired to arrange elements to present
as nearly 50 Ohms at the different bands as they could; however, the
native Rr varies from 10 - 40 Ohms and is transformed to 50 Ohms by
the time it reaches the connector dangling below.

In this case, the SWR resides on the antenna itself, and it is the
ohmic loss of the antenna structure that determines how efficient the
system might ultimately be. GAP uses large metal tubes that in
comparison to #32 wire wrap wire is a slam dunk in comparison; and
thus the efficiency concerns are cast back into the losses of the
transmission line and source that are suitably matched (use a bigger
diameter coax, and a more expensive rig to do better).

73's
Richard Clark, KB7QHC

I realize in multiband use, the mismatch will vary so there is not a whole
lot you can do except put the tuner at the junction of the antenna and line.
So, why isn't that the "normal" way to handle the problem?
--
Radio K4ia
Craig "Buck"


Bacause it is "up there" where it is hard to reach, mostly matter of
convenience. In normal situations it is more convenient to tune things in the
shack.
Personally, I always try to have my antennas to match the coax, to avoid
losses, high SWR on the lines. You can design antennas to have 50 ohms or you
can find 50 ohms impedance on the antenna and use antenna to be the "matching
device". K7GCO has be advocating to feed the Yagis off center at 50 ohm point.
Contesters care about every watt of loss and try to minimize it. Every fraction
of dB counts.
Casual hams care mostly about 1:1 at the TX connector so the system
coax-antenna "looks good" to the transmitter. Few dBs don't matter.

Yuri, K3BU

I am focusing on the issue of power transfer at the junction of the line and
the antenna. Specifically, in a multiband dipole where the figures (all of
them) will vary wildly from band to band. Maximum power transfers at
resonance (oh no, let's not get into a war defining that). But I think it
is safe to say by anyone's definition, a multiband dipole usually not
operated at it's resonant frequencies. If the line is 450 ohm and the
antenna is 2 ohm or 20 ohm or 2000 ohm, there is not resonance.
Intuitively, I have got to think a 200:1 mismatch is significant.

So what is the loss at the antenna/line junction? I understand matching at
the transmitter end. I understand using low loss line. I don't understand
why the mismatch at the antenna junction is ignored.
--
Radio K4ia
Craig "Buck"
Fredericksburg, VA USA
FISTS 6702 cc 788 Diamond 64

On Wed, 7 Jan 2004 11:43:17 -0500, "Craig Buck" wrote:
[snip]
So what is the loss at the antenna/line junction? I understand matching at
the transmitter end. I understand using low loss line. I don't understand
why the mismatch at the antenna junction is ignored.

I suggest you download and run TLDetails and see for yourself. It is a
great freeware program which should answer most, if not all, of your
questions.

You can download it at: http://www.qsl.net/ac6la/tldetails.html

73
Danny, K6MHE

Hi Craig,

The 200:1 mismatch is "significant", but it does not directly cause any loss. In
handwaving fashion, this is how things work.

A mismatch allows some fraction of the power to pass through the connection
point, with the remainder reflected. (You can substitute voltage or current for
power. The numbers are different, but the principle is the same.)

Assume the transmitter supplies some level of power, say 100 W, to the
transmission line in a perfectly matched manner. A tuner will generally be
required. For purposes of this discussion, nothing passes from the transmission
line back to the transmitter.

The energy supplied by the transmitter has to go somewhere, and the only two
choices are to the antenna or to losses in the transmission line. If the line is
lossless then all of the energy goes into the antenna.

How does this happen when the junction between the line and the antenna reflects
most of the power?

The power level in the line increases so that even the small percentage
transferred to the antenna equals the same 100 W supplied by the transmitter.
There are typically long and loud arguments in this newsgroup on the exact
mechanism for this buildup in the transmission line, but it does happen within a
few cycles of RF. The resulting voltages and currents will be much higher than
those found in a fully matched system.

So far all is good. The antenna receives the full transmitter output, and there
are no added losses.

The problem comes from the higher losses that occur in even the "lossless"
transmission line when operating at high voltages and currents. In the case of
ladder line these losses may still remain quite small, but in the case of RG-58
they can become quite large. The transmission line may fail at lower power
levels than expected.

Soooo, the mismatch at the antenna junction cannot really be ignored, but its
impact is in the transmission line, not the junction itself. Unless the mismatch
is extreme the ladder line solution takes care of the loss problem.

73,
Gene, W4SZ





Craig Buck wrote:
I am focusing on the issue of power transfer at the junction of the line and
the antenna. Specifically, in a multiband dipole where the figures (all of
them) will vary wildly from band to band. Maximum power transfers at
resonance (oh no, let's not get into a war defining that). But I think it
is safe to say by anyone's definition, a multiband dipole usually not
operated at it's resonant frequencies. If the line is 450 ohm and the
antenna is 2 ohm or 20 ohm or 2000 ohm, there is not resonance.
Intuitively, I have got to think a 200:1 mismatch is significant.

So what is the loss at the antenna/line junction? I understand matching at
the transmitter end. I understand using low loss line. I don't understand
why the mismatch at the antenna junction is ignored.

Gene Fuller, W4SZ wrote:
"The problem comes from the higher losses that occur in even the
"lossless" transmission line when operating at high voltages and
currents."

That`s right. Gene put quotes around "lossless".

The power output of the transmitter equals thaat taken by the load
inspite of hier indicated forward power. The difference is a power that
continues to circulate, much as baggage on an airport carrousel
continues to circulate until it is taken away.

Best regards, Richard Harrison, KB5WZI

Craig Buck wrote:
If the line is 450 ohm and the
antenna is 2 ohm or 20 ohm or 2000 ohm, there is not resonance.
Intuitively, I have got to think a 200:1 mismatch is significant.

But consider a 9:1 mismatch using 450 ohm feedline. The impedance
at the current maximum point on the feedline is 50 ohms.

So what is the loss at the antenna/line junction? I understand matching at
the transmitter end. I understand using low loss line. I don't understand
why the mismatch at the antenna junction is ignored.

It's not ignored. It is taken into account by the losses in the feedline.
The power reflected by the antenna is not the antenna's problem. It is the
feedline's problem. A two ohm copper or aluminum antenna is probably very
efficient. It is just hard to feed directly.
--
73, Cecil http://www.qsl.net/w5dxp



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Buck, K4IA wrote:
"If the line is 450 ohm and the antenna is 2 ohm or 20 ohm or 2000 ohm,
there is not resonance."

A resonant dipole has a drivepoint impedance that varies from near zero
ohms at zero elevation above the earth to about 100 ohms at 0.3
wavelength above the earth. For harmonic resonances, the radiation
resistances and the drivepoint resistances are higher than at the first
(1/2-wave) resonance.

The antenna may be devoid of reactance (resonant) and have divergent
drivepoint resistances depending upon which resonance, 1st, 2nd, 3rd,
etc, and placement of the antenna with respect to its surroundings.
Resonance and impedance matching are two different things.

Best regards, Richard Harrison, KB5WZI