An antenna question--43 ft vertical

[Wayne]

As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
metal roof. The antenna is fed with about 25 feet of RG-8, and there is a
tuner at the transmit end.

While I'm pretty happy with the antenna, I'd like to simplify the matching.

Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical?
( I assume the "4" side is on the antenna side.)

I'd expect a better coax to antenna match when the antenna feedpoint is a
high Z (example, at 30 meters), but I'd also expect a worse coax to antenna
match when the feedpoint is a low Z (example, at 10 meters).

Is that the way it works, or is there other magic involved?

[John S]

On 6/29/2015 10:48 AM, Wayne wrote:
> As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
> metal roof. The antenna is fed with about 25 feet of RG-8, and there is
> a tuner at the transmit end.

You use a 16ft vertical as a lead-in? For what and how is that done?

What are the dimensions of the metal roof?

> While I'm pretty happy with the antenna, I'd like to simplify the matching.

To what matching do you refer? You don't want to use the tuner, or is
there some other stuff you have not mentioned?

> Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
> vertical? ( I assume the "4" side is on the antenna side.)

You wrote that you were interested in a 16ft vertical. Now it is a 43ft
vertical?

> I'd expect a better coax to antenna match when the antenna feedpoint is
> a high Z (example, at 30 meters), but I'd also expect a worse coax to
> antenna match when the feedpoint is a low Z (example, at 10 meters).
>
> Is that the way it works, or is there other magic involved?

All this depends on your answers to the above questions.

[Wayne]

"John S" wrote in message news:mms4bb$u5h$1...@dont-email.me...

On 6/29/2015 10:48 AM, Wayne wrote:
>> As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
>> metal roof. The antenna is fed with about 25 feet of RG-8, and there is
>> a tuner at the transmit end.

>You use a 16ft vertical as a lead-in? For what and how is that done?

Grammatically, the description of the vertical is a lead in for the
question, not an actual antenna lead.

>What are the dimensions of the metal roof?

Somewhat irrelevant to my question. But it's about 20 by 35 feet.
I'm not looking for an analysis of the existing antenna.

>> While I'm pretty happy with the antenna, I'd like to simplify the
>> matching.

>To what matching do you refer? You don't want to use the tuner, or is there
>some other stuff you have not mentioned?

I want the tuner matching to be less awkward on some bands.
I'm willing to live with the existing high SWRs on the upper bands.

>> Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
>> vertical? ( I assume the "4" side is on the antenna side.)

>You wrote that you were interested in a 16ft vertical. Now it is a 43ft
>vertical?

Please disregard all about the 16 ft vertical. I'm asking about a 43 ft
vertical 1:4 unun.

> I'd expect a better coax to antenna match when the antenna feedpoint is
> a high Z (example, at 30 meters), but I'd also expect a worse coax to
> antenna match when the feedpoint is a low Z (example, at 10 meters).
>
> Is that the way it works, or is there other magic involved?

>All this depends on your answers to the above questions.

So, lets begin again, with no distractions.

What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical.

[Dave Platt]

In article <mmsaq8$r83$1...@dont-email.me>,

Wayne <mygarb...@verizon.net> wrote:
>So, lets begin again, with no distractions.
>
>What is the purpose (or benefit) of using a 1:4 unun on a 43 ft
>vertical.

http://www.eham.net/articles/21272 has a nice analysis.

It looks to me as if:

- Without a 4:1 unun, the antenna provides a very nice match at three
frequencies with in the HF band. At other frequencies, the SWR is
up over 10:1 much of the time - high enough that a coaxial feed
can be rather lossy.

- With a 4:1 unun, you do lose the excellent match at those three
frequencies... but the match gets better at most other
frequencies. The SWR across the HF band is much more uniform, and
lower on average... low enough to cut the coax losses somewhat and
(I think) within the matching range of many rigs' "line flattener"
built-in autotuners.

[Wayne]

"Dave Platt" wrote in message news:2sp86c-...@coop.radagast.org...

In article <mmsaq8$r83$1...@dont-email.me>,
Wayne <mygarb...@verizon.net> wrote:
>So, lets begin again, with no distractions.
>
>What is the purpose (or benefit) of using a 1:4 unun on a 43 ft
>vertical.

# http://www.eham.net/articles/21272 has a nice analysis.

# It looks to me as if:

# - Without a 4:1 unun, the antenna provides a very nice match at three
# frequencies with in the HF band. At other frequencies, the SWR is
# up over 10:1 much of the time - high enough that a coaxial feed
# can be rather lossy.

# - With a 4:1 unun, you do lose the excellent match at those three
# frequencies... but the match gets better at most other
# frequencies. The SWR across the HF band is much more uniform, and
# lower on average... low enough to cut the coax losses somewhat and
# (I think) within the matching range of many rigs' "line flattener"
# built-in autotuners.


Thanks Dave. I'll have to spend some more time studying it, but the article
is along the lines of what I was looking for.

I would assume that the 1:4 causes behavior just as you say....worse SWR at
nearly matched frequencies and better SWR elsewhere.

I'll have to pull out some textbooks and see how the math works out for a Z
seen through a 1:4 unun.

In practice, I've had good results with SWRs even in the 30:1 range with
short coax feeds.

More research...and thanks.

[Tom W3TDH]

I know that what I am about to say is provocative to some but I still think it is worth saying. If you look at the way that commercial and military radios are matched to antennas you will notice that most of the matching is done as close to the feed point as practical.

Since only the power that actually reaches the antenna can be radiated I have a hard time seeing the point of matching the transmitter to the feed line. Matching at the feed line connection point will prevent damage to the transmitter but if that were the main objective a dummy load would accomplish that.

When you couple the antenna to the load at the feed point you can have extremely low losses in the feed line. When you do the matching at the feed point you will transfer the most energy possible to the antenna and will get the highest available effective radiated power. Since the objective is the transfer of the highest practical amount of power to the antenna the place to do that is at the feed point were possible.

I do realize that it is often simpler and easier to match at the feed line connection but I felt obliged to point out that is is not the most effective place to do the job.

Tom

[rickman]

On 6/30/2015 12:40 PM, Tom W3TDH wrote:
>
> I know that what I am about to say is provocative to some but I still
> think it is worth saying. If you look at the way that commercial and
> military radios are matched to antennas you will notice that most of
> the matching is done as close to the feed point as practical.
>
> Since only the power that actually reaches the antenna can be
> radiated I have a hard time seeing the point of matching the
> transmitter to the feed line. Matching at the feed line connection
> point will prevent damage to the transmitter but if that were the
> main objective a dummy load would accomplish that.
>
> When you couple the antenna to the load at the feed point you can
> have extremely low losses in the feed line. When you do the matching
> at the feed point you will transfer the most energy possible to the
> antenna and will get the highest available effective radiated power.
> Since the objective is the transfer of the highest practical amount
> of power to the antenna the place to do that is at the feed point
> were possible.
>
> I do realize that it is often simpler and easier to match at the feed
> line connection but I felt obliged to point out that is is not the
> most effective place to do the job.

Has it occurred to you that it might be important to match impedance
both at the transmitter and at the antenna? When the feed line is not
impedance matched to the transmitter output the maximum power is not
transferred into the feed line. Then you have already lost power that
can't be recovered by the matching at the antenna even if it is perfect.

Your statements are not really provocative, they are just incomplete
and/or wrong.

--

Rick

[Jerry Stuckle]

On 6/30/2015 12:40 PM, Tom W3TDH wrote:

Tom, very close.

Yes, it's most effective to match the feedline to the antenna at the
antenna connection. But it's also important to match the transmitter to
the feedline.

This latter piece is often ignored because people will use a feedline
who's characteristic impedance matches the transmitter already (i.e. 50
ohm line for a 50 ohm transmitter).

However, there are exceptions. For instance, if you're feeding a 75 ohm
antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna
will provide a good match (ideally, 1:1). But there will be a 1.5:1
mismatch to a 50 ohm transmitter. In this case it would be better to
have the matching network at the transmitter.

You could also feed the antenna with 50 ohm feedline and place the
matching network at the antenna. The effect would still be a 1:1 SWR,
but the lower impedance of the coax would create higher i^2R losses; not
important if you're talking a short line, but a longer one would lower
output at the antenna.

--
==================
Remove the "x" from my email address
Jerry, AI0K
jstu...@attglobal.net
==================

[Wayne]

"Tom W3TDH" wrote in message
news:41c0c0c8-d9d7-4f97...@googlegroups.com...

# I know that what I am about to say is provocative to some but I still

think it is worth saying. If you look at the way that commercial and
military radios are matched to antennas you will notice that most of the

matching is done as close to the # feed point as practical.

# Since only the power that actually reaches the antenna can be radiated I

have a hard time seeing the point of matching the transmitter to the feed
line. Matching at the feed line connection point will prevent damage to the

transmitter but if # that were the main objective a dummy load would
accomplish that.

# When you couple the antenna to the load at the feed point you can have

extremely low losses in the feed line. When you do the matching at the feed
point you will transfer the most energy possible to the antenna and will get

the highest # available effective radiated power. Since the objective is

the transfer of the highest practical amount of power to the antenna the
place to do that is at the feed point were possible.

#I do realize that it is often simpler and easier to match at the feed line

connection but I felt obliged to point out that is is not the most effective
place to do the job.

I agree. However, the extra trouble of matching at the antenna feed point
may not necessarily provide a noticeable improvement.

On one of my antennas, I just provide a conjugate match for the
antenna-feedline at a tuner within the shack. Then I accept whatever
additional loss there is in the coax.

For my 25 foot run of RG8, there is about 0.5 dB of loss with a match at 30
MHz. From the charts, there will be about an additional 2.7 dB of loss if
the SWR is 20:1. For a 10:1 SWR the additional loss is around 0.9 dB.

Below 30 MHz, the numbers get smaller.

Of course, my automatic tuner (in the shack) fizzles out around SWR 5:1, but
my manual tuner can be used instead at higher SWRs.

In the case of the 43 foot vertical, it seems to me that a conjugate match
in the shack would still be optimum, but perhaps not significantly
beneficial.

On the 43 foot vertical it could be that the 4:1 unun provides a lower
amount of SWR induced additional feedline loss on high Z feedpoints. Of
course, the unun might increase the SWR induced additional feedline loss for
smaller Z.

That's why the original question, and what I'm trying to understand.

[Dave Platt]

In article <mmvbn2$gc6$1...@dont-email.me>,

Wayne <mygarb...@verizon.net> wrote:

>On the 43 foot vertical it could be that the 4:1 unun provides a lower
>amount of SWR induced additional feedline loss on high Z feedpoints. Of
>course, the unun might increase the SWR induced additional feedline loss for
>smaller Z.

From the charts, that kinda does appear to be the case.

Add into consideration the fact that losses go up with the square of
the current.

Another issue is the other aspect of SWR - voltage. If you're trying
to run "legal limit" or close to it, high SWR on the feedline coax
could exceed the voltage rating of the coax dielectric, and you'd get
arcing in the coax (or arcing at the connector between your feedline
and shack tuner).

Installing a hefty 4:1 unun right at the antenna would reduce the
worse-case voltage on the feedline, and inside the shack tuner
considerably. Depending on feedline length, that might be even more
of a consideration than the increase in peak feedline current and the
associated I^2*R losses.

[thomas.da...@gmail.com]

Rick

OK I'll buy incomplete and therefore wrong.

Now given a Fifty Ohm feed line connected to a transmitter that is designed for that impedance at the antenna connector does not the actual mismatch occur at the antenna feed point? Certainly that can be compensated for at the transmitter but isn't there a likelihood or at least a risk that you will loose significant effective radiated power in spite of adjusting the apparent feed line impedance to the transmitter? If I do the matching at the feed point will I not maximize the effective radiated power of the antenna by installing the tuner at the feed point.

I have already conceded that it is not as convenient to do the matching at the feed point. I do not allege that doing the matching at the transmitter end of the feed line is inherently ineffective only that there is a greater likelihood of loosing ERP needlessly and invisibly if the matching is done at transmitter end of the feed line. By this I mean to ask if I may well deceive the power meeter into showing more power out then I am actually getting. If any power lost is very likely to be insignificant at a practical level than help me to understand why that would be true and I will sell off my Icon AH-4, together with the control converter that allows my Yaesu FT-857D to control it, and my SGC SG-235 and go back to using the Yaesu FC-30 tuner with my FT-857D and the built in tuner on my Yaesu FT-1000.

This is especially important for me to get right with my FT-857D since it is the transceiver that I use for my personal go kit. If putting the Icon AH-4 on the mast and running the control line in addition to the coaxial cable is a waste of time I would really appreciate knowing that.

Thank you for helping with my education on this issue.

--
Tom Horne W3TDH

[Ian Jackson]

In message <0049f6c3-6b13-425b...@googlegroups.com>,
thomas.da...@gmail.com writes

It's indeed surprising how well things still work if you use the lowest
loss coax you can beg, steal or borrow, and do all the matching back in
the comfort of the shack.
--
Ian

[John S]

Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be
very high impedance at that frequency and a 1:4 unun will theoretically
bring that impedance down closer to the feed line impedance.

Does this help?

[Ian Jackson]

In message <mn0vln$hv1$1...@dont-email.me>, John S <Sop...@invalid.org>
writes

It was been pointed out to me that the figures for feeder loss with an
imperfect SWR are only correct when the length is fairly long (at least
an electrical wavelength?). How much loss does 25' of RG-8 really have
at 12MHz, when there's a halfwave hanging on the far end?
--
Ian

[Jeff Liebermann]

On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle
<jstu...@attglobal.net> wrote:

>Yes, it's most effective to match the feedline to the antenna at the
>antenna connection. But it's also important to match the transmitter to
>the feedline.
>
>This latter piece is often ignored because people will use a feedline
>who's characteristic impedance matches the transmitter already (i.e. 50
>ohm line for a 50 ohm transmitter).
>
>However, there are exceptions. For instance, if you're feeding a 75 ohm
>antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna
>will provide a good match (ideally, 1:1). But there will be a 1.5:1
>mismatch to a 50 ohm transmitter. In this case it would be better to
>have the matching network at the transmitter.

We may have had this discussion before. Matching a 75 ohm load to a
50 ohm source might be academically interesting, but the actual loss
is almost negligible. for a VSWR of 1.5, the return loss is 14dB and
the load mismatch attenuation is 0.177dB. That's about what I would
expect to lose in two coax connector pairs.

>You could also feed the antenna with 50 ohm feedline and place the
>matching network at the antenna. The effect would still be a 1:1 SWR,
>but the lower impedance of the coax would create higher i^2R losses; not
>important if you're talking a short line, but a longer one would lower
>output at the antenna.

True, but for roughly equivalent sizes of coax cables, the 75 ohm
cable has less loss and the equivalent 50 ohm cable. If you want to
handle high power, use 50 ohms. If you want low loss, use 75 ohms:
<http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm>
Note that these are for air dielectric cables.

Things are not so neat if we consider the dielectric. See the bottom
paragraph and graphs:
<http://www.microwaves101.com/encyclopedias/why-fifty-ohms>
Dielectric Dielectric const Minimum loss impedance
solid PTFE 2.2 50 ohms
foam PTFE 1.43 60
air 1.0 75
RG-6/u CATV 75 ohm foam coax still has slightly less loss than the
equivalent 50 ohm cable, but not as much as I've previously claimed.

This is cute:
<http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/>
"A good rule of thumb is that if the device being connected
via coaxial cable is a receiver of some kind, 75 Ohm Coax is ideal."


--
Jeff Liebermann je...@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

[rickman]

I have to plead ignorance. How can you deal with impedance mismatch of
the antenna and feed line by a network at the transmitter?

If you assume the transmitter is matched to the feed line, then yes, you
should only need matching at the antenna if that is mismatched. This
seems like a bit of a silly strawman, but maybe I am just not informed
and this is the typical scenario.

> I have already conceded that it is not as convenient to do the
> matching at the feed point. I do not allege that doing the matching
> at the transmitter end of the feed line is inherently ineffective
> only that there is a greater likelihood of loosing ERP needlessly and
> invisibly if the matching is done at transmitter end of the feed
> line. By this I mean to ask if I may well deceive the power meeter
> into showing more power out then I am actually getting. If any power
> lost is very likely to be insignificant at a practical level than
> help me to understand why that would be true and I will sell off my
> Icon AH-4, together with the control converter that allows my Yaesu
> FT-857D to control it, and my SGC SG-235 and go back to using the
> Yaesu FC-30 tuner with my FT-857D and the built in tuner on my Yaesu
> FT-1000.
>
> This is especially important for me to get right with my FT-857D
> since it is the transceiver that I use for my personal go kit. If
> putting the Icon AH-4 on the mast and running the control line in
> addition to the coaxial cable is a waste of time I would really
> appreciate knowing that.
>
> Thank you for helping with my education on this issue.
>


--

Rick

[Wayne]

"rickman" wrote in message news:mn14gi$bfp$1...@dont-email.me...

# I have to plead ignorance. How can you deal with impedance mismatch of
# the antenna and feed line by a network at the transmitter?

If there is a mismatch at the antenna (and there is no matching at the
antenna), then maximum power transfer will occur when the conjugate match is
applied at the transmitter end of the feedline.

Loss in the feedline will include the normal loss of a matched line, and
additionally the loss caused by a SWR other than 1:1.

With low loss feedline, and SWRs under perhaps 5:1, the additional loss is
small and often quite acceptable.

Why do it that way? You can have control in the shack, and you throw less
money at the problem.

In my own particular case, an automatic remotely tuned ATU would be a pain
to install/maintain.

[Wayne]

"Dave Platt" wrote in message news:ql7c6c-...@coop.radagast.org...

Good points. I'm running low powers at the moment, but there was a time
where I burned up a lot of stuff running a KW :)

[rickman]

When you say you have control in the shack, why do you need to do
anything other than install it? Would this matching network change with
use at different bands?

Also, why is this a lower cost solution?

--

Rick

[Wayne]

"Jeff" wrote in message news:mn195n$9i0$1...@speranza.aioe.org...

>> If there is a mismatch at the antenna (and there is no matching at the
>> antenna), then maximum power transfer will occur when the conjugate
>> match is applied at the transmitter end of the feedline.

>Surely a conjugate match will only match the load if the coax length is 1/2
>wavelength or multiple thereof, and the feeder is also lossless.

>Any other coax length will introduce a phase shift that will require a
>different match.

Yes, I'm assuming that the antenna tuner conjugate match is for the end of
the feedline, not for the antenna itself.

Wayne
W5GIE
exiled to W6 :)

[Roger Hayter]

Jeff <je...@ukra.com> wrote:

> > If there is a mismatch at the antenna (and there is no matching at the
> > antenna), then maximum power transfer will occur when the conjugate
> > match is applied at the transmitter end of the feedline.
>

> Surely a conjugate match will only match the load if the coax length is
> 1/2 wavelength or multiple thereof, and the feeder is also lossless.
>
> Any other coax length will introduce a phase shift that will require a
> different match.
>

> Jeff

You just have to match whatever impedance the aerial impedance has been
transformed to at the transmitter end. Then you will get maximum power
into the radiation resistance of the aerial (less the second order
losses in the feeder). A remaining reactive mismatch between the feeder
and the aerial will result in increased voltages and currents and
increased feeder loss (a second order effect at HF) but will not prevent
substantially full power transfer. We had this discussion about very
short aerials quite recently, You have to have a very extreme radiation
resistance for this not to work. Choosing a length of aerial with no
extreme values on the bands you are using is where we came in.

--
Roger Hayter

[Jerry Stuckle]

Jeff, do you always miss the forest for the trees? That was an EXAMPLE.
The same would be true if you were feeding a 300 ohm yagi with 300 ohm
twinlead and a transmitter with a 10 ohm output impedance.

And BTW - when calculating, you forgot about the transmitters which cut
back power to protect the finals. Many will do so even with a 1.5:1 SWR.

[John S]

Transmitter output impedance does not determine SWR.

[Jeff Liebermann]

On Wed, 01 Jul 2015 17:23:54 -0400, Jerry Stuckle
<jstu...@attglobal.net> wrote:

>Jeff, do you always miss the forest for the trees? That was an EXAMPLE.

Perhaps you missed my point. I don't care about VSWR as long as the
system is reasonably efficient, does not protect itself, and produces
adequate TX power. In my world, that means I'm concerned about
losses, not VSWR. The losses involved in using 75 ohm feedline and
coax in a 50 ohm system are negligible.

>The same would be true if you were feeding a 300 ohm yagi with 300 ohm
>twinlead and a transmitter with a 10 ohm output impedance.

One of my favorite methods of argumentation is to provide a ridiculous
and extreme example, and then use it as the basis for discussion. I
guess it's a form of "straw man" argument, where I'm now expected to
defend my point of view against your ridiculous and extreme example.
Please forgive me for not following your lead and continuing to be
more concerned with the 50/75 ohm problem:
<https://en.wikipedia.org/wiki/Straw_man>

>And BTW - when calculating, you forgot about the transmitters which cut
>back power to protect the finals. Many will do so even with a 1.5:1 SWR.

Could you provide me with one example of such a transmitter? I've
seen such radios on the bench, but they're usually mistuned or
misadjusted.

I did some Googling and found that the typical threshold for both AM
and FM broadcast xmitters is 1.5:1.
<https://www.google.com/#q=transmitter+high+vswr+threshold+1.5:1>
Same with most HF radios that I could find. The stuff I designed for
marine use (150 watts PEP) was set to operate up to 2.0:1 because at
the time, ATU's were just appearing and the typical vessel HF antenna
was problematic (23ft vertical with a dubious ground system). Today,
1.5:1 threshold would probably work.

Note that such a threshold does NOT mean that at 1.4999:1, the radio
would work normally, and at 1.5001:1 would shut down. For (marine)
radios that are expected to work with random antennas in emergencies,
shutting down at 1.5:1 is absurd. What is normally done is to slowly
reduce the drive starting at 1.5:1 until it gets to some point below
where either the final current is too high to maintain safe
dissipation, or the voltage across the final xsistor or FET is too
high to prevent breakdown. My guess(tm) is that's about 5:1 or more
with todays radios but I'll admit that I'm guessing and haven't
actually tried it. In the past, I used a test load, that someone else
built, that would provide a resistive 2:1 VSWR at 25/37.5/50/75/100
ohms, and an adjustable phase angle with a big variable capacitor and
roller inductor. I had to be VERY careful not to accidentally tune
the inductor and capacitor to resonance, or I would end up with a
short or open load. The dials had an accompanying chart that followed
varioius constant VSWR circles around the Smith chart.

The tricky part was not making the power amp work over a 4:1 impedance
range. The tricky part was making a VSWR sensor that would be fairly
flat over the entire 2 to 30 MHz range. Another headache was when a
mismatch caused the PA to draw more current. It wasn't final heating
that caused instabilities and odd behavior. It was my worthless bench
power supply that would detect the overcurrent and protect itself by
dropping the output voltage.

So, which radio shuts down at less than 1.5:1 VSWR? I know of a few
possible candidates, but I would like to see what you've observed
first.

[Jeff Liebermann]

On Tue, 30 Jun 2015 17:21:08 -0700, "Wayne" <mygarb...@verizon.net>
wrote:

># Since only the power that actually reaches the antenna can be radiated I
>have a hard time seeing the point of matching the transmitter to the feed
>line. Matching at the feed line connection point will prevent damage to the
>transmitter but if # that were the main objective a dummy load would
>accomplish that.

Yep. Note that the electric utility companies do not bother to match
the transmitter (generators) with the impedance of the transmission
lines and the load. That was one of the reasons that Edison and
Westinghouse has so much trouble with the experts when they proposed
electric power transmission. The experts assumed that the source had
to be matched to the load, which would cause the generators to
dissipate as much power as is dissipated in the load. Incidentally,
one reason Tesla/Westinghouse eventually went with 60 Hz instead of
133Hz, 400 Hz, or higher frequencies (which use less iron in the
xformers) was the danger of creating standing waves on the
transmission lines because of the mismatch. When the wavelength of 60
Hz (3100 miles or 5000 km) is longer than the width of the country,
it's a safe bet that there aren't going to be any standing waves.

So, why don't we run transmitters with lower than 50 ohm output
impedances? Well...
1. The gain of the PA stage would be reduced possibly requiring an
additional gain stage.
2. The current in the PA stage would increase, possibly causing the
power supply to complain.
3. The low pass harmonic filter will require physically larger parts.
4. The coax cable between the PA stage and the RF connector will need
to have a very thin dielectric to work at low impedances. Same with
the RF output connector.

Instead of dealing with these aforementioned hassles, it's probably
better to run the transmitter at some impedance that provides a
benefit and let everything else conform to that standard. That's
where the maximum power at 50 ohms for transmitters, and lowest loss
for 75 ohms (air dielectric) for CATV were derived. The rest of the
connected devices (PA, filter, antenna) simply conformed to these
standards.

[Jerry Stuckle]

Transmitter output impedance vs. feedline impedance does determine SWR
at one end of the system. If you have a mismatch, you will have a
non-1:1 SWR.

>> And BTW - when calculating, you forgot about the transmitters which cut
>> back power to protect the finals. Many will do so even with a 1.5:1 SWR.
>>
>

[Jerry Stuckle]

On 7/2/2015 8:43 AM, Jeff Liebermann wrote:
> On Wed, 01 Jul 2015 17:23:54 -0400, Jerry Stuckle
> <jstu...@attglobal.net> wrote:
>
>> Jeff, do you always miss the forest for the trees? That was an EXAMPLE.
>
> Perhaps you missed my point. I don't care about VSWR as long as the
> system is reasonably efficient, does not protect itself, and produces
> adequate TX power. In my world, that means I'm concerned about
> losses, not VSWR. The losses involved in using 75 ohm feedline and
> coax in a 50 ohm system are negligible.
>

I didn't miss your point. But you can't see the forest for the trees.

>> The same would be true if you were feeding a 300 ohm yagi with 300 ohm
>> twinlead and a transmitter with a 10 ohm output impedance.
>
> One of my favorite methods of argumentation is to provide a ridiculous
> and extreme example, and then use it as the basis for discussion. I
> guess it's a form of "straw man" argument, where I'm now expected to
> defend my point of view against your ridiculous and extreme example.
> Please forgive me for not following your lead and continuing to be
> more concerned with the 50/75 ohm problem:
> <https://en.wikipedia.org/wiki/Straw_man>
>

I prefer to use realistic examples to show a point. But you have to
nitpick with off-topic comments.

>> And BTW - when calculating, you forgot about the transmitters which cut
>> back power to protect the finals. Many will do so even with a 1.5:1 SWR.
>
> Could you provide me with one example of such a transmitter? I've
> seen such radios on the bench, but they're usually mistuned or
> misadjusted.
>

Many of the solid state finals amateur transmitters will start cutting
back well before 2:1 SWR. Even my early 80's era IC-720A would start
dropping power before then.

> I did some Googling and found that the typical threshold for both AM
> and FM broadcast xmitters is 1.5:1.
> <https://www.google.com/#q=transmitter+high+vswr+threshold+1.5:1>
> Same with most HF radios that I could find. The stuff I designed for
> marine use (150 watts PEP) was set to operate up to 2.0:1 because at
> the time, ATU's were just appearing and the typical vessel HF antenna
> was problematic (23ft vertical with a dubious ground system). Today,
> 1.5:1 threshold would probably work.
>

We're not talking AM and FM broadcast transmitters (which are immaterial
because the antenna system is tuned to get as close to a 1:1 match as
possible - much easier with one frequency). Pretty much the same with
marine use - a very limited band of frequencies. Additionally, the
limited power on marine radios allow you to use higher power finals so
they can dissipate the additional heat caused by a mismatch.

> Note that such a threshold does NOT mean that at 1.4999:1, the radio
> would work normally, and at 1.5001:1 would shut down. For (marine)
> radios that are expected to work with random antennas in emergencies,
> shutting down at 1.5:1 is absurd. What is normally done is to slowly
> reduce the drive starting at 1.5:1 until it gets to some point below
> where either the final current is too high to maintain safe
> dissipation, or the voltage across the final xsistor or FET is too
> high to prevent breakdown. My guess(tm) is that's about 5:1 or more
> with todays radios but I'll admit that I'm guessing and haven't
> actually tried it. In the past, I used a test load, that someone else
> built, that would provide a resistive 2:1 VSWR at 25/37.5/50/75/100
> ohms, and an adjustable phase angle with a big variable capacitor and
> roller inductor. I had to be VERY careful not to accidentally tune
> the inductor and capacitor to resonance, or I would end up with a
> short or open load. The dials had an accompanying chart that followed
> varioius constant VSWR circles around the Smith chart.
>

I never said shut down. I said cut back. But you can't read very well,
either, can you?

> The tricky part was not making the power amp work over a 4:1 impedance
> range. The tricky part was making a VSWR sensor that would be fairly
> flat over the entire 2 to 30 MHz range. Another headache was when a
> mismatch caused the PA to draw more current. It wasn't final heating
> that caused instabilities and odd behavior. It was my worthless bench
> power supply that would detect the overcurrent and protect itself by
> dropping the output voltage.
>
> So, which radio shuts down at less than 1.5:1 VSWR? I know of a few
> possible candidates, but I would like to see what you've observed
> first.
>

Once again, you can't read. But I know you're just foaming at the mouth
to contradict me, as you always do. But that's OK. I know what you
are, and I'm not going to bite.

Now go away, troll, and play with your CB radios. They're your speed.
Leave this discussion to adults - who can read and understand the points
being made.

And BTW - you can also go running to your mommy and tell her the mean
old man called you a troll.

[Jeff Liebermann]

On Thu, 02 Jul 2015 06:12:33 -0700, Jeff Liebermann <je...@cruzio.com>
wrote:
(...)

>where the maximum power at 50 ohms for transmitters, and lowest loss
>for 75 ohms (air dielectric) for CATV were derived. The rest of the
>connected devices (PA, filter, antenna) simply conformed to these
>standards.

Continuing from where I accidentally hit the "send" button...

It might be interesting to measure the output impedance of your HF
xmitter. All you need is a dummy load, and an RF voltmeter, RF probe
and voltmeter, or oscilloscope.
1. Turn down the xmitter RF output to some level where you won't blow
up your test equipment and so that it doesn't go into high VSWR
protect mode. My guess is about 10 watts is about right.
2. Measure the RF voltage across the output connector both with a
load (Vload) and without a load (Vno_load).
3. If measuring peak voltage, convert RMS by multiplying by 0.707. If
measuring peak-to-peak, divide by 2 and then multiply by 0.707.

Output_Impedance = 50 ohms (Vno_load - Vload) / Vload

It's been many years since I've done this, so I can't recall the range
of values that I obtained. I do recall that it was surprisingly large
and precipitated a few heated discussions in the lab. Also, the
output impedance will change with output power level but I don't
recall how much.

[John S]

A *resonant* half wave at 12MHz is about 36.7 feet long and it presents
an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The
current at the antenna end is 0.0245A while one watt is applied at the
source end. This means that the power applied to the antenna is about
0.687W. So, about 68% of the applied power reaches the antenna.

So, about 32% of the power is lost in the RG-8 for this example.

Does this help?

[John S]

Not true. Post some links to support your position, please.

[Jerry Stuckle]

Basic physics that anyone with even an inkling of AC theory should
understand. Any time you have an impedance mismatch in a system - in
this case, whether at the antenna or the transmitter end of a feedline -
you will not have perfect power transfer. What is not transferred will
be reflected. This causes an SWR greater than 1:1.

Try this - connect the output of an HF transmitter to an SWR bridge.
Now connect a piece of 75 ohm coax such as RG-59 to the output of the
SWR meter, and connect that to a 75 ohm resistive load. Do you think
the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1.

[Wayne]

"John S" wrote in message news:mn3jtf$kde$1...@dont-email.me...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
> In message <mn0vln$hv1$1...@dont-email.me>, John S <Sop...@invalid.org>
> writes
>> On 6/29/2015 3:47 PM, Wayne wrote:
>>>

<snipped to shorten>

>>
>> Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will
>> be very high impedance at that frequency and a 1:4 unun will
>> theoretically bring that impedance down closer to the feed line
>> impedance.
>>
>> Does this help?
>
> It was been pointed out to me that the figures for feeder loss with an
> imperfect SWR are only correct when the length is fairly long (at least
> an electrical wavelength?). How much loss does 25' of RG-8 really have
> at 12MHz, when there's a halfwave hanging on the far end?

# A *resonant* half wave at 12MHz is about 36.7 feet long and it presents
# an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The
# current at the antenna end is 0.0245A while one watt is applied at the
# source end. This means that the power applied to the antenna is about
# 0.687W. So, about 68% of the applied power reaches the antenna.

# So, about 32% of the power is lost in the RG-8 for this example.

I'm just trying to understand this, so let me ask a question about your
example.

Isn't the 32% lost a function of not having a conjugate match maximum power
transfer?
If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
wouldn't maximum power be transferred?
(Even with a SWR of about 21:1)

[rickman]

Transferred where? The match at the transmitter output only matches the
output to the line. There are still reflections from the mismatch at
the antenna. These reflections result in extra losses in the line as
well as power delivered back into the transmitter output stage
(especially with a perfect impedance match).

But I don't see anyone taking wavelength vs. feed line length into
account. If the wavelength is long compared to the feed line I believe
a lot of the "bad" stuff goes away. But then I am used to the digital
transmission line where we aren't really concerned with delivering
power, rather keeping a clean waveform of our (relatively) square waves.
So I guess a short feed line doesn't solve the SWR problems... or does
it?

--

Rick

[Ralph Mowery]

"Jerry Stuckle" <jstu...@attglobal.net> wrote in message
news:mn3lul$t1h$1...@dont-email.me...

>>
> Try this - connect the output of an HF transmitter to an SWR bridge.
> Now connect a piece of 75 ohm coax such as RG-59 to the output of the
> SWR meter, and connect that to a 75 ohm resistive load. Do you think
> the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1.
>
>

What you have described is a case of using the wrong swr bridge. You are
trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is
used it will show a 1:1 SWR.

The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no
reflected power.

[ji...@specsol.spam.sux.com]

Wayne <mygarb...@verizon.net> wrote:

<snip>

> In my own particular case, an automatic remotely tuned ATU would be a pain
> to install/maintain.

This part I do not understand at all.

At the antenna end is a box with a connector for the feed line and a
connector for the antenna. There is nothing to maintain there.

If you get an ATU that gets it's power through the coax, you put the
power injector in line with the feed line in the shack. There is
nothing to maintain there either and you do not need to run any extra
wires out to the antenna.




--
Jim Pennino

[rickman]

My knowledge of antenna systems is limited, but I do know that this is
correct, there will be no reflection from the antenna. If the
transmitter output is 50 ohms there will be a loss in this matching that
will result in less power being delivered to the feed line, but that
will not result in reflections in the feed line. I calculate the loss
to be -0.177 dB or 4%. How much loss would be expected in the feed line
itself if it is a moderate length?

--

Rick

[Jerry Stuckle]

On 7/2/2015 1:56 PM, Ralph Mowery wrote:

No, the SWR bridge is correct. The output of the transmitter is 50 ohms.

You are correct in that if a 75 ohm bridge is used, the indicated SWR
would be 1:1, because everything from that point on is 75 ohms.
However, the mismatch (and reflection) occurs on the transmitter side of
the bridge, not the antenna side. So the bridge will never see it. But
an accurate bridge will show lower power output due to the mismatch.

A mismatch is a mismatch, no matter where in the system it occurs. And
any mismatch will cause less than 100% power to be transferred. The
rest is reflected.

Just look at the specs of any amateur transceiver. They show an
impedance of 50 ohms. So a load of 50 ohms provides for maximum power
transfer; any other impedance causes a mismatch.

[Wayne]

wrote in message news:uhbg6c-...@mail.specsol.com...

The problem is with my own particular case. The antenna is a whip mounted
in the middle of a metal roof.

At my age, I shouldn't be wandering around on or climbing such a roof.

Once installed, any failure would require a trip to the roof. The ATU
would be exposed to extreme temperature and sunlight that might eventually
induce failures.

[Ralph Mowery]

"Jerry Stuckle" <jstu...@attglobal.net> wrote in message

news:mn41d5$bq1$1...@dont-email.me...

> On 7/2/2015 1:56 PM, Ralph Mowery wrote:
>> "Jerry Stuckle" <jstu...@attglobal.net> wrote in message
>> news:mn3lul$t1h$1...@dont-email.me...
>>>>

> You are correct in that if a 75 ohm bridge is used, the indicated SWR
> would be 1:1, because everything from that point on is 75 ohms.
> However, the mismatch (and reflection) occurs on the transmitter side of
> the bridge, not the antenna side. So the bridge will never see it. But
> an accurate bridge will show lower power output due to the mismatch.
>
> A mismatch is a mismatch, no matter where in the system it occurs. And
> any mismatch will cause less than 100% power to be transferred. The
> rest is reflected.
>
> Just look at the specs of any amateur transceiver. They show an
> impedance of 50 ohms. So a load of 50 ohms provides for maximum power
> transfer; any other impedance causes a mismatch.
>
> --

The real impedance of the transmitter is not 50 ohms. It is whatever the
device is used in the final stage and the poewr level. For a 100 watt
transmitter it is in the thousand ohm range and for solid state devices it
is very low. The matching circuit is often fixed to be 50 ohms,but could be
made for most any impedance. The older tube circuits were adjustable by the
user for a range of somewhat bleow 50 ohms to around 200 ohms. Could be
more or less depending on the design.

The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm
feedline and 75 ohm antenna is in the tuned circuits/matching circuit in the
transmitter. Whatever power comes out of the transmitter will make it to
the antenna minus the loss of the coax, but not additional loss due to swr.
The power comming out of a 50 ohm transmitter will be less due to mismatch,
but not because of swr of the antenna system which is 1:1.

[Wayne]

"rickman" wrote in message news:mn3pb4$ae6$1...@dont-email.me...

# Transferred where? The match at the transmitter output only matches the
# output to the line. There are still reflections from the mismatch at
# the antenna. These reflections result in extra losses in the line as
# well as power delivered back into the transmitter output stage
# (especially with a perfect impedance match).

Well, I put a few (unrealistic) qualifiers into my question: a transmitter
with a a 1063 ohm output (not 50), and a lossless RG-8.

Thus, the back and forth reflections would not have attenuation.
And the transmitter and load are conjugately matched for maximum power
transfer.

# But I don't see anyone taking wavelength vs. feed line length into
# account. If the wavelength is long compared to the feed line I believe
# a lot of the "bad" stuff goes away. But then I am used to the digital
# transmission line where we aren't really concerned with delivering
# power, rather keeping a clean waveform of our (relatively) square waves.
# So I guess a short feed line doesn't solve the SWR problems... or does
# it?

The attenuation at a given high SWR depends upon the the matched feedline
loss, as reflections encounter that loss with every forward or backward
trip.
Thus feedline length/attenuation should be considered.

As a young man I was given a problem of solving poor antenna performance on
an aircraft band fixed station antenna. The SWR at the transmitter was
close to 1:1, but the antenna didn't work well.
I climbed up on the tower and found that the coax had never been connected
to the antenna. That was with about 400 feet of coax at 120 MHz.

[Ian Jackson]

In message <mn44r7$pu8$1...@dont-email.me>, Wayne
<mygarb...@verizon.net> writes

>
>As a young man I was given a problem of solving poor antenna
>performance on an aircraft band fixed station antenna. The SWR at the
>transmitter was close to 1:1, but the antenna didn't work well.
>I climbed up on the tower and found that the coax had never been
>connected to the antenna. That was with about 400 feet of coax at 120
>MHz.

A length of coax, which has (say) at least 10dB loss at the frequency of
interest, can indeed make a superb SWR dummy load!
--
Ian

[Jerry Stuckle]

Incorrect. The real impedance of the transmitter is 50 ohms. The
impedance of the final stage may be above or below that, and is matched
to the 50 ohm standard. What happens before the match is unimportant.
The only important part of the discussion is the 50 ohm output.

> The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm
> feedline and 75 ohm antenna is in the tuned circuits/matching circuit in the
> transmitter. Whatever power comes out of the transmitter will make it to
> the antenna minus the loss of the coax, but not additional loss due to swr.
> The power comming out of a 50 ohm transmitter will be less due to mismatch,
> but not because of swr of the antenna system which is 1:1.
>
>

Incorrect. The connection between the 50 ohm transmitter and the 75 ohm
coax is also part of the antenna system. The system starts at the
transmitter output (actually the output of the final stage - but since
this is converted to the 50 ohm standard, you can effectively consider
the output of the matching network to be the start of the antenna
system), not the coax.

[ji...@specsol.spam.sux.com]

The age part I can understand; an inverted, cheap plastic trash can will
provide more than adequate protection against the elements.


--
Jim Pennino

[Ralph Mowery]

"Jerry Stuckle" <jstu...@attglobal.net> wrote in message

news:mn45vo$ult$1...@dont-email.me...

>> The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm
>> feedline and 75 ohm antenna is in the tuned circuits/matching circuit in
>> the
>> transmitter. Whatever power comes out of the transmitter will make it to
>> the antenna minus the loss of the coax, but not additional loss due to
>> swr.
>> The power comming out of a 50 ohm transmitter will be less due to
>> mismatch,
>> but not because of swr of the antenna system which is 1:1.
>>
>>
>
> Incorrect. The connection between the 50 ohm transmitter and the 75 ohm
> coax is also part of the antenna system. The system starts at the
> transmitter output (actually the output of the final stage - but since
> this is converted to the 50 ohm standard, you can effectively consider
> the output of the matching network to be the start of the antenna
> system), not the coax.
>
> --
>

So with my Icom 746 with a built in tuner, where does the system start ?
Within a small range 3 or 4 to 1, the internal SWR meter will show a 1:1
match, an external SWR meter will show a differant SWR if there is a
mismatch. If I hook up a good 75 ohm load and 75 ohm coax I will have no
SWR by definition . A 50 ohm bridge will show 1.5:1 but the internal
tuner/bridge will show 1:1.

Which SWR meter is correct ?

[rickman]

A friend of mine who is past 70 has had a TV antenna preamp on his roof
for some 50 years. He has been up there to check it at least once when
he couldn't get the digital TV signals as well any more. It was not the
problem. If a unit is constructed well, it should live a rich, full
life on the roof, protected from the abuse it might receive in your shack.

--

Rick

[rickman]

Your quoting style is very confusing. If you use > with a space at the
front of lines you are quoting it will show up the same as everyone
else's quotes.

Why will the reflections not have losses? Every load that is not an
infinite impedance will absorb some of the signal that would be
reflected. That applies to the transmitter output as well as the
antenna, no?

A matched impedance does not mean no losses. It means the maximum
transfer of power. These are not at all the same thing.

> # But I don't see anyone taking wavelength vs. feed line length into
> # account. If the wavelength is long compared to the feed line I believe
> # a lot of the "bad" stuff goes away. But then I am used to the digital
> # transmission line where we aren't really concerned with delivering
> # power, rather keeping a clean waveform of our (relatively) square waves.
> # So I guess a short feed line doesn't solve the SWR problems... or does
> # it?
>
> The attenuation at a given high SWR depends upon the the matched
> feedline loss, as reflections encounter that loss with every forward or
> backward trip.
> Thus feedline length/attenuation should be considered.
>
> As a young man I was given a problem of solving poor antenna performance
> on an aircraft band fixed station antenna. The SWR at the transmitter
> was close to 1:1, but the antenna didn't work well.
> I climbed up on the tower and found that the coax had never been
> connected to the antenna. That was with about 400 feet of coax at 120 MHz.

So how was the SWR 1:1?

--

Rick

[Ian Jackson]

In message <mn4e3j$1co$1...@dont-email.me>, rickman <gnu...@gmail.com>
writes

It's probably the go-and-return loss of (say) 20dB's worth of coax (at
120MHz), with the far end open (or short) circuit. That would give you
an RLR of 40dB (an SWR of 1.02), which probably far exceeds the
capability of an SWR meter to read.
http://bit.ly/1T8TwcF
<http://cgi.www.telestrian.co.uk/cgi-bin/www.telestrian.co.uk/vswr.pl>

--
Ian

[Jerry Stuckle]

A built in tuner is an entirely different story. The output from the
transmitter is not necessarily 50 homes at this point.

It's a shame they've dumbed down the exams so much that you don't need
to know anything to hold a ticket any more. Back when I got mine, even
the General Class was tougher than the Extra Class today - and it was
administered by the FCC, with no public question pool or cheat sheets.
You had to actually know something other than just memorizing a few answers.

And at the time, the Amateur Extra was harder than the First Class
Radiotelephone.

Nowadays, an 8-year-old can pass the Amateur Extra with virtually no
knowledge of math or electronics at all - just the ability to memorize a
few answers.

Learn something about electronics, AC power transfer and other theory.
Than maybe we can converse with a modicum of intelligence. As it is,
I'm tired of trying to teach an idiot who refuses to learn.

[Jeff Liebermann]

On Thu, 02 Jul 2015 14:55:53 -0400, Jerry Stuckle
<jstu...@attglobal.net> wrote:

It's easy enough to demonstrate that you're wrong.

1. Setup your favorite HF xmitter and attach a Tee connector to the
antenna connector, your favorite VSWR meter, a length of 50 ohm coax,
and a 50 ohm load.
2. Transmit and convince yourself that the VSWR is 1:1. Make sure
the transmitter is not into ALC.
3. Now, take another 50 ohm dummy load and connect it to the Tee
connector. The transmitter now sees 25 ohms, so the PA stage has half
the normal gain. You may need to increase the drive level to obtain
the same RF power as before.
4. Measure the VSWR again. It should also be 1:1.

Looking back towards the transmitter, the 50 ohm coax cable sees a 2:1
mismatch of 25 ohms (the alleged 50 ohms from radio in parallel with
another 50 ohms from the extra dummy load). I know this part works
because I've demonstrated it twice to the local non-believers.


Now, we go into uncharted territory and do it again with a 75 ohm coax
and a 75 ohm dummy load at the far end (antenna end) of the coax. Same
procedure.
1. Check the VSWR and it should be 1.5:1.
2. Connect a 50 ohm dummy load to the Tee connector, and measure the
VSWR again. It should still be 1.5:1.

Looking back towards the transmitter, the 75 ohm coax cable sees the
same 2:1 mismatch of 25 ohms. If you want to go further, I think it
can be demonstrated that almost any number of extra dummy loads at the
Tee connector will still produce the same 1.5:1 VSWR.

I'll try it on the bench, but I have other plans for the holiday
weekend. If I find time, and manage to get all the junk off my
workbench, I'll give try it.

[rickman]

I like the comment that the antenna "didn't work well". Lol. I wonder
how much better it worked when connected. The feed line would still
have a lot of loss one way. I wonder why the power amp wasn't closer to
the antenna.

Reminds me of a time I was working a job installing TV antennas and one
was up the side of the ridge near here. In the "old days" we would
unplug the TV to prevent shocks from a "hot" chassis set. This
installation also had a wall jack for the antenna connection. The guy
on the roof told me to plug the TV in and check the signal while the
turned the antenna. The image was a little snowy, but not bad. No
matter how they turned the antenna the image didn't get much better. I
looked behind the set and saw the three foot twin lead from the set
laying on the floor. I plugged it in and the picture was *perfect*! I
was amazed a short piece of lead in could receive as good a signal.

--

Rick

[John S]

No. As I said, one watt is applied at the source end. This is condition
defined by the example and has nothing to do with source matching. The
32% loss is due to transmission line loss. The mismatch at the load end
causes the high SWR which increases the line loss due to high current at
some point in the line as well as increased voltage at other point(s).

The impedance of the 1063+J0 load is transformed to 54+J192 ohms at the
source. However, at an electrical quarter wave away from the antenna,
the impedance is about 2.8+j0 ohms. So that point is a relative hot spot
in the line.

> If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
> wouldn't maximum power be transferred?
> (Even with a SWR of about 21:1)

With a lossless transmission line and one watt applied to one end the
other end will have one watt available. The only place the power can go
is into the antenna. To put one watt into 1063 ohms will require .0306
amps and 32.6V at the feed point.

Does this make sense?

[Ralph Mowery]

"Jerry Stuckle" <jstu...@attglobal.net> wrote in message

news:mn4hkl$br9$1...@dont-email.me...

> On 7/2/2015 6:08 PM, Ralph Mowery wrote:
>> "Jerry Stuckle" <jstu...@attglobal.net> wrote in message
>> news:mn45vo$ult$1...@dont-email.me...

> It's a shame they've dumbed down the exams so much that you don't need
> to know anything to hold a ticket any more. Back when I got mine, even
> the General Class was tougher than the Extra Class today - and it was
> administered by the FCC, with no public question pool or cheat sheets.
> You had to actually know something other than just memorizing a few
> answers.
>
> And at the time, the Amateur Extra was harder than the First Class
> Radiotelephone.
>

I had my First Phone at the age of 22 back in 1972. Passed the 2nd and
first the same day on the first try.

And back at you on trying to teach someone that will not learn. Seems that
several on here think you are wrong most of the time.

[Wayne]

"rickman" wrote in message news:mn4e3j$1co$1...@dont-email.me...

# Your quoting style is very confusing. If you use > with a space at the
# front of lines you are quoting it will show up the same as everyone
# else's quotes.

It's a problem with my newsreader not doing the proper job.



#Why will the reflections not have losses?

Because the assumption I posed was for a lossless line. In that case, with
a conjugate match on both ends, wouldn't there be maximum power transmission
regardless of the SWR?

.....just a question I'm posing to the group.

With no line losses, and a conjugate match, is the SWR of any consequence?

<A matched impedance does not mean no losses. It means the maximum
<transfer of power. These are not at all the same thing.


> # But I don't see anyone taking wavelength vs. feed line length into
> # account. If the wavelength is long compared to the feed line I believe
> # a lot of the "bad" stuff goes away. But then I am used to the digital
> # transmission line where we aren't really concerned with delivering
> # power, rather keeping a clean waveform of our (relatively) square waves.
> # So I guess a short feed line doesn't solve the SWR problems... or does
> # it?
>
> The attenuation at a given high SWR depends upon the the matched
> feedline loss, as reflections encounter that loss with every forward or
> backward trip.
> Thus feedline length/attenuation should be considered.
>
> As a young man I was given a problem of solving poor antenna performance
> on an aircraft band fixed station antenna. The SWR at the transmitter
> was close to 1:1, but the antenna didn't work well.
> I climbed up on the tower and found that the coax had never been
> connected to the antenna. That was with about 400 feet of coax at 120
> MHz.

# So how was the SWR 1:1?

It was a long run of coax at 120 MHz. The reflected wave was was attenuated
considerably by the time it returned to the source.

[rickman]

You aren't grasping the issue. Losses are *not* only in the
transmission line. When a reflected wave returns to the transmitter
output, it is not reflected 100%. If the output and transmission line
are matched exactly, 50% of the reflected wave reaching the output will
be reflected and 50% will be dissipated in the output stage.

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?

--

Rick

[John S]

Correction: 0.5+j0 at 1/4 wavelength away from the antenna.

[Jerry Stuckle]

On 7/2/2015 8:51 PM, Ralph Mowery wrote:
> "Jerry Stuckle" <jstu...@attglobal.net> wrote in message
> news:mn4hkl$br9$1...@dont-email.me...
>> On 7/2/2015 6:08 PM, Ralph Mowery wrote:
>>> "Jerry Stuckle" <jstu...@attglobal.net> wrote in message
>>> news:mn45vo$ult$1...@dont-email.me...
>> It's a shame they've dumbed down the exams so much that you don't need
>> to know anything to hold a ticket any more. Back when I got mine, even
>> the General Class was tougher than the Extra Class today - and it was
>> administered by the FCC, with no public question pool or cheat sheets.
>> You had to actually know something other than just memorizing a few
>> answers.
>>
>> And at the time, the Amateur Extra was harder than the First Class
>> Radiotelephone.
>>
>
> I had my First Phone at the age of 22 back in 1972. Passed the 2nd and
> first the same day on the first try.
>
> And back at you on trying to teach someone that will not learn. Seems that
> several on here think you are wrong most of the time.
>
>

I've got you there. First phone in 1970 at age 18 - before I started
college. Amateur I also passed Second and First the same day. My
Amateur Extra came 9 months later, but only because I had to hold a
General for two years before I could take the Extra exam.

Did you ever use your first phone? I was an engineer for one broadcast
station and chief engineer for another. I also repaired everything from
$40 cb sets to multi-million dollar mainframe computers. And even did
some digital design work back on the 70's.

But it's quite obvious from your updates that you have no idea what
you're talking about. You're an appliance operator with no
understanding of what's going on underneath the covers. And yes, I know
there are some idiotic trolls here who don't think I know what I'm
talking about. My EE professors would disagree with them.

--
==================
Remove the "x" from my email address

Jerry Stuckle
jstu...@attglobal.net
==================

[Jerry Stuckle]

On 7/2/2015 7:56 PM, Jeff Liebermann wrote:
> On Thu, 02 Jul 2015 14:55:53 -0400, Jerry Stuckle
> <jstu...@attglobal.net> wrote:
>
> It's easy enough to demonstrate that you're wrong.
>
> 1. Setup your favorite HF xmitter and attach a Tee connector to the
> antenna connector, your favorite VSWR meter, a length of 50 ohm coax,
> and a 50 ohm load.
> 2. Transmit and convince yourself that the VSWR is 1:1. Make sure
> the transmitter is not into ALC.
> 3. Now, take another 50 ohm dummy load and connect it to the Tee
> connector. The transmitter now sees 25 ohms, so the PA stage has half
> the normal gain. You may need to increase the drive level to obtain
> the same RF power as before.
> 4. Measure the VSWR again. It should also be 1:1.
>

Of course it will be. You have 50 ohms on one end. But you're not
measuring what the TRANSMITTER sees. The fact that you may have to
increase drive level indicates the circumstances have changed.

> Looking back towards the transmitter, the 50 ohm coax cable sees a 2:1
> mismatch of 25 ohms (the alleged 50 ohms from radio in parallel with
> another 50 ohms from the extra dummy load). I know this part works
> because I've demonstrated it twice to the local non-believers.
>

Sure. But you don't have power going from one resistor to the
transmitter and the other resister, so your measurement is meaningless -
and you are as full of crap as you normally are.

Stick your VSWR meter between the power source (the transmitter) and
BOTH loads (i.e. before the T). You will see a 2:1 SWR.

>
> Now, we go into uncharted territory and do it again with a 75 ohm coax
> and a 75 ohm dummy load at the far end (antenna end) of the coax. Same
> procedure.
> 1. Check the VSWR and it should be 1.5:1.
> 2. Connect a 50 ohm dummy load to the Tee connector, and measure the
> VSWR again. It should still be 1.5:1.
>
> Looking back towards the transmitter, the 75 ohm coax cable sees the
> same 2:1 mismatch of 25 ohms. If you want to go further, I think it
> can be demonstrated that almost any number of extra dummy loads at the
> Tee connector will still produce the same 1.5:1 VSWR.
>

Again, the dummy load is not producing any power, so adding something to
the other end of the T will have no effect. So once again your "test"
is meaningless and you are full of crap.

Now in this case if you connect the SWR bridge before the T, you will
show a 1.5:1 with one 75 ohm load, and a 1.3:1 SWR with two 75 ohm loads
(37.5 ohms).

SWR measurements are only valid when the VSWR meter is connected between
the power generator (transmitter) and the total load (one or both dummy
loads). Connecting between one leg of the T and the load only shows
VSWR for that leg - but not the entire system.

> I'll try it on the bench, but I have other plans for the holiday
> weekend. If I find time, and manage to get all the junk off my
> workbench, I'll give try it.
>

Go ahead - continue to mae a fool of yourself. You're real good at it.

[rickman]

On 7/3/2015 2:50 AM, Jeff wrote:
>
>> Are you suggesting that the conjugate match will reflect back to the
>> antenna 100% of the original reflected wave from the antenna?
>>
>

> Yes, it must.
>
> For example with an external ATU that provides a conjugate match it is
> clearly the case that if a 1:1 VSWR is achieved then no reflected power
> reaches the TX. (as shown on an SWR meter between the Tx and ATU.)

I am very certain that this assumption is not correct. I wish I had the
math to back me up. The only total reflection I am aware of is an open
circuit which of course absorbs no power at all.

Here is a point. The VSWR only shows no power being sent back to the
txmt output. That does not mean no power is absorbed from the reflected
wave by the matching circuit. I believe the example you gave was Z of
1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent
to a resistor? Resistors dissipate power don't they?

--

Rick

[Ian Jackson]

In message <mn5dj2$ovf$1...@dont-email.me>, rickman <gnu...@gmail.com>
writes

I have to admit that I am, to some extent, confused.

Maybe it helps to look at the situation from the point of view that the
matching circuit doesn't 'know' that there is a reflected wave. All it
sees is the impedance looking into the sending end of the coax - and
this is whatever is on the antenna end, transformed by the length of
coax. The load the matching unit sees could be replaced with the same
physical values of L, C and R, so there IS nowhere for a reflected wave
to exist.

Provided the TX sees a 50 ohm load when looking into the input of the
matcher, there will be no theoretical losses. However, a real-life
matcher WILL have loss, and so will the coax. Also, the coax will have a
loss greater than when it is matched, mainly because of the 'I -squared
R' (literal) hot-spots.
--
Ian

[Roger Hayter]

Surely it *is* the reflected wave that mediates the transformation of
the aerial impedance to what is seen at the transmitter end? The
transmitter sees the vector sum of all the waves traversing the
transmission line at that point. Or else how would it "know" what was
happening at the other end?

--
Roger Hayter

[Roger Hayter]

Jeff <je...@ukra.com> wrote:

> The point can be easily proved with a lossy feeder, the lossier the
> better. If your assumption is correct then the power delivered to the
> antenna would be the Tx power less the cable loss less the reflected
> power at the antenna mismatch, however it is the case that can be
> measured and seen on computer simulation that all of the power is
> delivered to the antenna except that which is dissipated in the cable
> loss (for the multiple reflections). Also it can been seen that with
> perfect components in the matching circuit no power is dissipated there.
>
> The 1063 ohms that you refer to is not resistive so with perfect Ls & Cs
> no power will be dissipated in it.
>
> In the real world the Cs and Ls in the matching unit will have some loss
> associated with them but that is a different story.
>
> Jeff


While conjugate matching is the way to transfer the maximum power from a
voltage (or current) generator to a load, it is not the way power
amplifiers are set up. The transmitter normally does not present a
match to signals from the aerial, hence the re-reflection.


--
Roger Hayter

[Ian Jackson]

In message <1m6ze5a.yzxqjf12a6hfkN%ro...@hayter.org>, Roger Hayter
<ro...@hayter.org> writes

I guess that until reflections are received back from the far end of the
coax, the transmitter will see the 50 ohms Zo (surge impedance) of the
coax. But once things have settled worn, the transmitter neither knows
nor cares what's at the far end. All it knows is that the load presented
to it isn't what it ought to be. But insert a matcher, and it will be as
happy as Larry. The system will work fine, but will suffer the penalty
of the additional SWR losses on the coax, and those of the matcher.
Provide these are not unacceptable, the benefit is that all the matching
can be done in the comfort of shack.
--
Ian

[John S]

On 6/29/2015 10:48 AM, Wayne wrote:
> As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
> metal roof. The antenna is fed with about 25 feet of RG-8, and there is
> a tuner at the transmit end.
>
> While I'm pretty happy with the antenna, I'd like to simplify the matching.
>
> Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
> vertical? ( I assume the "4" side is on the antenna side.)
>
> I'd expect a better coax to antenna match when the antenna feedpoint is
> a high Z (example, at 30 meters), but I'd also expect a worse coax to
> antenna match when the feedpoint is a low Z (example, at 10 meters).
>
> Is that the way it works, or is there other magic involved?

I think we strayed off the path to answering your original question.

The short answer is that you are correct and there is no magic involved.
A bit longer answer is:

A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at
30 meters. Using a 1:4 transformer at the feed point will reduce that to
253 + J 67 ohms. That is a bit closer to your 50 ohm line.

At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
1:4 transformer will reduce that to 37 + J 33 ohms.

There are several disclaimers I could include, but I think you
understand that the answers cannot be exact with the info presented.

I hope this helps.

[Ian Jackson]

In message <mn661m$r6d$1...@dont-email.me>, John S <Sop...@invalid.org>
writes

>On 6/29/2015 10:48 AM, Wayne wrote:
>> As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
>> metal roof. The antenna is fed with about 25 feet of RG-8, and there is
>> a tuner at the transmit end.
>>
>> While I'm pretty happy with the antenna, I'd like to simplify the matching.
>>
>> Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
>> vertical? ( I assume the "4" side is on the antenna side.)
>>
>> I'd expect a better coax to antenna match when the antenna feedpoint is
>> a high Z (example, at 30 meters), but I'd also expect a worse coax to
>> antenna match when the feedpoint is a low Z (example, at 10 meters).
>>
>> Is that the way it works, or is there other magic involved?
>
>I think we strayed off the path to answering your original question.
>
>The short answer is that you are correct and there is no magic
>involved. A bit longer answer is:
>
>A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at
>30 meters. Using a 1:4 transformer at the feed point will reduce that
>to 253 + J 67 ohms. That is a bit closer to your 50 ohm line.

A fixed-tuned TX will still need a matcher.

>
>At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
>1:4 transformer will reduce that to 37 + J 33 ohms.

A fixed-tuned TX will probably be reasonably happy with a direct
connection - although maybe even happier with a series capacitor of -J22

ohms.
>
>There are several disclaimers I could include, but I think you
>understand that the answers cannot be exact with the info presented.
>
>I hope this helps.

The question is really whether the losses with the 4:1 transformer, plus
those of any matcher at the TX end, exceed those when there is no
transformer (but with higher loss on the coax), plus a matcher. Put
another way, for short feeder lengths, is it better to use the
transformer?
--
Ian

[Ian Jackson]

In message <y1Xy5PAy...@g3ohx.demon.co.uk>, Ian Jackson
<ianREMOVET...@g3ohx.demon.co.uk> writes

>although maybe even happier with a series capacitor of -J22 ohms.

Sorry - somebody obviously swapped the '2' and '3' keys.


--
Ian

[Wayne]

"John S" wrote in message news:mn661m$r6d$1...@dont-email.me...

Thanks John.
Yes, we have strayed from the original question, but I have found the
discussion stimulating.
Perhaps a new thread should be started to address those subjects.

If I use EZNEC to model the 43 footer over perfect ground with a 3 inch
diameter radiator, I get impedances in the same ball park as you list.

If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna would
see as a feedline, if a 4:1 unun had 50 ohm coax on the other side), the SWR
plot becomes interesting.

The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR
getting below 2.5:1 around 29 MHz.

Is that a valid approach?

[John S]

On 7/3/2015 10:17 AM, Ian Jackson wrote:
> In message <mn661m$r6d$1...@dont-email.me>, John S <Sop...@invalid.org>
> writes
>> On 6/29/2015 10:48 AM, Wayne wrote:
>>> As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
>>> metal roof. The antenna is fed with about 25 feet of RG-8, and there is
>>> a tuner at the transmit end.
>>>
>>> While I'm pretty happy with the antenna, I'd like to simplify the
>>> matching.
>>>
>>> Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
>>> vertical? ( I assume the "4" side is on the antenna side.)
>>>
>>> I'd expect a better coax to antenna match when the antenna feedpoint is
>>> a high Z (example, at 30 meters), but I'd also expect a worse coax to
>>> antenna match when the feedpoint is a low Z (example, at 10 meters).
>>>
>>> Is that the way it works, or is there other magic involved?
>>
>> I think we strayed off the path to answering your original question.
>>
>> The short answer is that you are correct and there is no magic
>> involved. A bit longer answer is:
>>
>> A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms
>> at 30 meters. Using a 1:4 transformer at the feed point will reduce
>> that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line.
>
> A fixed-tuned TX will still need a matcher.

That was not part of the original question(s).

>>
>> At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
>> 1:4 transformer will reduce that to 37 + J 33 ohms.
>
> A fixed-tuned TX will probably be reasonably happy with a direct
> connection - although maybe even happier with a series capacitor of -J22
> ohms.

That was not part of the original question(s).

>> There are several disclaimers I could include, but I think you
>> understand that the answers cannot be exact with the info presented.
>>
>> I hope this helps.
>
> The question is really whether the losses with the 4:1 transformer, plus
> those of any matcher at the TX end, exceed those when there is no
> transformer (but with higher loss on the coax), plus a matcher. Put
> another way, for short feeder lengths, is it better to use the transformer?

That was not the question he asked. Please re-read the OP. I was trying
to address his original question(s) as best as I could. In addition I
also said that there were "several disclaimers I could include" which
may involve your personal concerns. I did not want to muddy the waters.

I think I answered Wayne's question(s), but I will wait to hear from him
to see if that is so.

[John S]

I have not done what you have done, but it sounds correct. I'll try to
verify what you have done when time permits.

I really think you know what you are doing. Don't forget that EZNEC can
use transmission lines, transformers, inductors, capacitors, resistors
and other stuff to help in your analysis. Although the true answers come
from the physical implementation, it is very helpful to use EZNEC to
gain insight into the situation. And, I think you know that as well.

[John S]

On 7/3/2015 10:37 AM, Wayne wrote:
>
>

Indeed! So have I.

> Perhaps a new thread should be started to address those subjects.

Please start one if you feel compelled.

> If I use EZNEC to model the 43 footer over perfect ground with a 3 inch
> diameter radiator, I get impedances in the same ball park as you list.

Ha! I used 1.5 inches. I will re-do.

> If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna
> would see as a feedline, if a 4:1 unun had 50 ohm coax on the other
> side), the SWR plot becomes interesting.

I've never done that. I will explore this set-up.

> The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR
> getting below 2.5:1 around 29 MHz.

Are we still considering a 10MHz to 30Mhz frequency sweep?

> Is that a valid approach?

You might be ahead of me on this.

[Wayne]

"John S" wrote in message news:mn6bla$l59$1...@dont-email.me...

Well, I have been running the SWR across 4 to 30 MHz, but mainly looking at
10 MHz and above.

As for EZNEC and transmission lines, I have never done that, but plan to
when I can. I don't follow how to do it. In the few cases I wanted the
info for a single frequency, I just used a Smith chart.

This thread has given me a lot to consider in improving my whip setup, but
details of the possibilities would run the thread off in the weeds :)

[Ian Jackson]

In message <mn6al2$g3r$1...@dont-email.me>, John S <Sop...@invalid.org>

You have certainly answered "Thus, the question: what is the purpose of
a 1:4 unun on a 43 foot vertical?" (ie to reduce the horrendous
mismatch). However, don't you think there's any virtue in wondering
whether, in the circumstances described (with the relatively short
feeder), it will be any better than a direct connection to the antenna,
and to do all the matching at the TX end? Also, would you use a
transformer if there was hardly any feeder at all, or (in an extreme
case) if the antenna was fed directly from the TX? I'm not advocating
anything - only wondering.
--
Ian

[Wayne]

"rickman" wrote in message news:mn4n79$q37$1...@dont-email.me...

I don't think I've ever heard that anywhere before. Could you elaborate?

>Are you suggesting that the conjugate match will reflect back to the
>antenna 100% of the original reflected wave from the antenna?

Well, yes. Minus losses in matching networks and transmission lines.

In examples with lossless lines and lossless matching networks, wouldn't it
be 100%.

[John S]

Yes, I agree that there is virtue in fully examining all the
possibilities. I may or may not use a transformer with a direct
connection to the antenna. It depends on my source's capabilities. Of
course, anything added to improve a match also causes a bit of loss.
It's all tradeoffs, as you well know.

[John S]

Okay. This data set is for a 43' carbon steel antenna on a perfect
ground fed with 25' of RG8A/U. No transformer.

Freq R X SWR

4.000 2.52 -12.74 21.140
5.000 20.16 20.82 2.977
6.000 28.78 -30.81 2.586
7.000 8.04 -5.76 6.307
8.000 5.68 12.25 9.337
9.000 6.08 30.66 11.344
10.000 8.95 56.02 12.703
11.000 20.21 104.84 13.685
12.000 137.06 277.80 14.297
13.000 126.21 -267.32 14.174
14.000 19.15 -95.40 12.417
15.000 11.09 -43.81 8.067
16.000 17.90 -11.07 2.949
17.000 56.99 -15.98 1.384
18.000 20.29 -31.49 3.568
19.000 9.29 -12.21 5.711
20.000 6.95 3.56 7.236
21.000 6.99 19.13 8.217
22.000 9.03 38.15 8.829
23.000 15.83 67.71 9.161
24.000 48.70 132.94 9.150
25.000 418.45 49.46 8.488
26.000 60.45 -119.25 6.589
27.000 28.69 -45.41 3.465
28.000 46.51 -9.99 1.245
29.000 62.03 -48.79 2.397
30.000 21.38 -41.26 4.116


This data set is the same except with a 1:4 transformer at the antenna.

Freq R X SWR
4.000 4.07 19.59 14.187
5.000 25.57 74.28 6.633
6.000 153.39 -83.56 4.058
7.000 25.51 -38.52 3.333
8.000 15.35 -6.11 3.312
9.000 16.11 18.43 3.567
10.000 26.59 49.05 3.970
11.000 87.00 102.27 4.497
12.000 180.75 -114.05 5.136
13.000 28.87 -72.84 5.814
14.000 11.18 -31.02 6.256
15.000 8.52 -4.23 5.914
16.000 13.30 23.05 4.607
17.000 50.10 61.48 3.196
18.000 107.24 -32.07 2.383
19.000 36.37 -30.20 2.135
20.000 22.85 -5.54 2.222
21.000 22.77 16.08 2.474
22.000 33.85 41.69 2.827
23.000 85.10 73.92 3.268
24.000 165.89 -60.03 3.790
25.000 39.38 -70.12 4.323
26.000 15.47 -31.46 4.605
27.000 12.03 -4.86 4.197
28.000 18.81 20.71 3.175
29.000 54.98 43.57 2.257
30.000 85.50 -14.78 1.786

Is this of any help?

[John S]

On 7/4/2015 1:34 AM, John S wrote:
> Okay. This data set is for a 43' carbon steel antenna on a perfect
> ground fed with 25' of RG8A/U. No transformer.

I made a mistake. I used 3" copper pipe. I will re-do with 3" carbon
steel pipe.

No transformer:
Freq R X SWR
4.000 2.61 -12.68 20.382
5.000 20.92 20.65 2.867
6.000 28.18 -30.18 2.600
7.000 8.03 -5.57 6.304
8.000 5.70 12.35 9.317
9.000 6.11 30.75 11.314
10.000 9.00 56.13 12.662
11.000 20.37 105.05 13.626
12.000 139.02 278.20 14.204
13.000 125.69 -265.07 14.021
14.000 19.40 -94.97 12.182
15.000 11.41 -43.56 7.809
16.000 18.65 -11.02 2.830
17.000 56.19 -17.88 1.426
18.000 19.92 -31.06 3.599
19.000 9.26 -12.02 5.720
20.000 6.96 3.68 7.228
21.000 7.02 19.23 8.195
22.000 9.09 38.27 8.792
23.000 15.99 67.89 9.102
24.000 49.53 133.22 9.057
25.000 412.93 41.45 8.343
26.000 60.97 -117.33 6.399
27.000 29.70 -44.72 3.324
28.000 48.31 -10.74 1.247
29.000 60.26 -49.68 2.445
30.000 21.05 -40.86 4.142

1:4 transformer:
Freq R X SWR
4.000 4.21 19.69 13.745
5.000 26.48 74.45 6.450
6.000 149.09 -82.20 3.972
7.000 25.55 -37.90 3.288
8.000 15.45 -5.81 3.285
9.000 16.25 18.68 3.549
10.000 26.91 49.36 3.954
11.000 88.45 102.31 4.478
12.000 177.47 -114.21 5.105
13.000 28.81 -72.20 5.756
14.000 11.31 -30.71 6.152
15.000 8.73 -3.99 5.767
16.000 13.78 23.29 4.469
17.000 51.75 60.81 3.109
18.000 104.23 -32.25 2.336
19.000 36.12 -29.39 2.111
20.000 22.93 -5.09 2.210
21.000 22.96 16.44 2.467
22.000 34.32 42.05 2.819
23.000 86.65 73.65 3.255
24.000 162.78 -61.65 3.764
25.000 39.08 -69.10 4.270
26.000 15.65 -30.95 4.510
27.000 12.38 -4.51 4.075
28.000 19.52 20.93 3.075
29.000 56.33 42.38 2.197
30.000 83.40 -15.44 1.755

[Wayne]

"John S" wrote in message news:mn8pdh$riv$1...@dont-email.me...

Yes, very interesting. Throw in 75 feet of cable, and things get "better".

The lowest SWR is about 2:1 at 19 MHz.
It is about 6:1 on 20 meters.

[John S]

On 7/4/2015 10:55 AM, Wayne wrote:

>
> Yes, very interesting. Throw in 75 feet of cable, and things get "better".
>
> The lowest SWR is about 2:1 at 19 MHz.
> It is about 6:1 on 20 meters.
>
>

Of course. More loss in the cable makes it "better" (but, you know that).

[John S]

On 7/4/2015 10:55 AM, Wayne wrote:
>

By the way, Wayne...

Are you aware of a companion Excel application for EZNEC called AutoEZ?
You can run many test cases in a few seconds using it. You can find it
on the EZNEC site.

It is how I generated the data I posted.

[Wayne]

"John S" wrote in message news:mn93ls$8fq$1...@dont-email.me...

Thanks, I'll look for that. I run the old wood burning version 3.0.

[rickman]

On 7/3/2015 3:27 PM, Wayne wrote:
>
>
> "rickman" wrote in message news:mn4n79$q37$1...@dont-email.me...
>
> On 7/2/2015 8:53 PM, Wayne wrote:
>>
>>
>> "rickman" wrote in message news:mn4e3j$1co$1...@dont-email.me...
>>
>> On 7/2/2015 3:52 PM, Wayne wrote:
>>>
>>>> Why will the reflections not have losses?
>>>>
>>> Because the assumption I posed was for a lossless line. In that case,
>>> with a conjugate match on both ends, wouldn't there be maximum power
>>> transmission regardless of the SWR?
>
>> You aren't grasping the issue. Losses are *not* only in the
>> transmission line. When a reflected wave returns to the transmitter
>> output, it is not reflected 100%. If the output and transmission line
>> are matched exactly, 50% of the reflected wave reaching the output
>> will be reflected and 50% will be dissipated in the output stage.
>
> I don't think I've ever heard that anywhere before. Could you elaborate?

I'm not so sure now. I think I mentioned before that I learned about
transmission lines in the digital context where source and loads are
largely resistive. Resistance dissipates power. So when matched the
source dissipates as much power as delivered to the load (or
transmission line). Likewise, matched impedance will not reflect power,
but rather it is all absorbed. That is what happens at the antenna for
sure. But I'm not clear about what this conjugate network is really.
If it is purely reactive, then it will not have losses other than the
parasitics.

I have to admit I am not fluent in the complex math of networks. So off
hand an impedance of 1063 -j0 says to me resistive. The imaginary part
implies phase shifting, no? With that term being 0 doesn't that say the
capacitive and inductive parts cancel out leaving only resistance? If
you can, please explain how I am wrong.

>> Are you suggesting that the conjugate match will reflect back to the
>> antenna 100% of the original reflected wave from the antenna?
>
> Well, yes. Minus losses in matching networks and transmission lines.
>
> In examples with lossless lines and lossless matching networks, wouldn't
> it be 100%.

I don't get how the matching network will reflect the wave from the
antenna 100%. Is that something you can explain?

--

Rick

[Jerry Stuckle]

Yes, he is correct.

Think of it this way, without the math. On the transmitter side of the
network, the match is 1:1, with nothing reflected back to the transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for the
signal to go is back to the antenna.

In a perfect system, all power is transferred to the antenna, even with
a large mismatch between the feedline and the antenna. However, that's
still not the same as having a match at the antenna, because reflected
signals most likely will arrive out of phase with the original signal.


-- ==================
Remove the "x" from my email address
Jerry, AI0K
jstu...@attglobal.net
==================

[Jeff Liebermann]

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
<jstu...@attglobal.net> wrote:
>Think of it this way, without the math. On the transmitter side of the
>network, the match is 1:1, with nothing reflected back to the transmitter.
>
>So you have a signal coming back from the antenna. You have a perfect
>matching network, which means nothing is lost in the network. The
>feedline is perfect, so there is no loss in it. The only place for the
>signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

<https://en.wikipedia.org/wiki/Impedance_matching>
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

--
Jeff Liebermann je...@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

[Jerry Stuckle]

On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
> On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
> <jstu...@attglobal.net> wrote:
>> Think of it this way, without the math. On the transmitter side of the
>> network, the match is 1:1, with nothing reflected back to the transmitter.
>>
>> So you have a signal coming back from the antenna. You have a perfect
>> matching network, which means nothing is lost in the network. The
>> feedline is perfect, so there is no loss in it. The only place for the
>> signal to go is back to the antenna.
>
> Wikipedia says that if the source is matched to the line, any
> reflections that come back are absorbed, not reflected back to the
> antenna:
>
> <https://en.wikipedia.org/wiki/Impedance_matching>
> "If the source impedance matches the line, reflections
> from the load end will be absorbed at the source end.
> If the transmission line is not matched at both ends
> reflections from the load will be re-reflected at the
> source and re-re-reflected at the load end ad infinitum,
> losing energy on each transit of the transmission line."
>

And you believe everything Wikipedia says? ROFLMAO.

But that also explains your ignorance.

--
==================
Remove the "x" from my email address

Jerry Stuckle
jstu...@attglobal.net
==================

[Wayne]

"Jeff Liebermann" wrote in message
news:1fqgpa9iktrru2cio...@4ax.com...

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
<jstu...@attglobal.net> wrote:
>>Think of it this way, without the math. On the transmitter side of the
>>network, the match is 1:1, with nothing reflected back to the transmitter.
>>
>>So you have a signal coming back from the antenna. You have a perfect
>>matching network, which means nothing is lost in the network. The
>>feedline is perfect, so there is no loss in it. The only place for the
>>signal to go is back to the antenna.

>Wikipedia says that if the source is matched to the line, any
>reflections that come back are absorbed, not reflected back to the
>antenna:

><https://en.wikipedia.org/wiki/Impedance_matching>
> "If the source impedance matches the line, reflections
> from the load end will be absorbed at the source end.
> If the transmission line is not matched at both ends
> reflections from the load will be re-reflected at the
> source and re-re-reflected at the load end ad infinitum,
> losing energy on each transit of the transmission line."

Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.

[Jeff Liebermann]

On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle

<jstu...@attglobal.net> wrote:

>On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
>> On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
>> <jstu...@attglobal.net> wrote:
>>> Think of it this way, without the math. On the transmitter side of the
>>> network, the match is 1:1, with nothing reflected back to the transmitter.
>>>
>>> So you have a signal coming back from the antenna. You have a perfect
>>> matching network, which means nothing is lost in the network. The
>>> feedline is perfect, so there is no loss in it. The only place for the
>>> signal to go is back to the antenna.
>>
>> Wikipedia says that if the source is matched to the line, any
>> reflections that come back are absorbed, not reflected back to the
>> antenna:
>>
>> <https://en.wikipedia.org/wiki/Impedance_matching>
>> "If the source impedance matches the line, reflections
>> from the load end will be absorbed at the source end.
>> If the transmission line is not matched at both ends
>> reflections from the load will be re-reflected at the
>> source and re-re-reflected at the load end ad infinitum,
>> losing energy on each transit of the transmission line."

>And you believe everything Wikipedia says? ROFLMAO.
>But that also explains your ignorance.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

[Jerry Stuckle]

With a perfect matching network and a perfect feedline (which is what we
are discussing), that is true. But I also know that is far beyond your
limited intelligence.

> Yet, when I have a perfectly matched load (antenna), all the power it
> is fed is radiated and nothing is reflected. You can't have it both
> ways because the reflected power from the load (antenna), becomes the
> incident power going towards the source (power amp). Matched and
> mismatched loads do NOT act differently depending on the direction of
> travel. If you claim were true, then transmitting into a matched
> antenna or dummy load would reflect all the power back towards the
> transmitter.
>

Nothing wrong with it at all - except your limited intelligence can't
understand simple physics. But then that's nothing new.

--
==================
Remove the "x" from my email address

Jerry, AI0K
jstu...@attglobal.net
==================

[Jerry Stuckle]

It's not clear because it depends on a lot of math on how things work.
Maybe it could use some editing, but I really don't think it can be
simplified much more - and certainly not enough for some idiots in this
newsgroup to understand.

But then they are just trolls who insist on showing their ignorance
based on a limited understanding of Ohm's law - and nothing higher.
Even the math behind Smith charts is beyond them.

[John S]

On 7/3/2015 1:06 PM, Wayne wrote:

>
> As for EZNEC and transmission lines, I have never done that, but plan to
> when I can. I don't follow how to do it.

Put a short piece of wire somewhere away from the antenna. Move your
source to this short piece of wire. Connect your transmission line
between the short wire and the antenna where you previously had the
source. Put the required line info into the transmission line box(es).
Start with a velocity factor of 1 and an attenuation of 0dB. You should
get the same results as before you moved your source. Then you can
adjust the Vf and loss based on the characteristics of the line that you
can find on line.

Cheers
John N1JLS

[ji...@specsol.spam.sux.com]

If the termination matches the line impedance, there is no reflection.

Both the antenna and the source are terminations.

This is a bit difficult to visualze with an RF transmitter, but is
more easily seen with pulses.

The wikipedia entry is correct as written.

In the real world, the output of an amateur transmitter will seldom
be exactly 50 Ohms unless there is an adjustable network of some
sort.



--
Jim Pennino

[Ian Jackson]

In message <2kum6c-...@mail.specsol.com>, ji...@specsol.spam.sux.com
writes

Being essentially a simple soul, that's how I sometimes try to work out
what happening.

>
>The wikipedia entry is correct as written.
>
>In the real world, the output of an amateur transmitter will seldom
>be exactly 50 Ohms unless there is an adjustable network of some
>sort.

I've always understood that the resistive part of a TX output impedance
was usually less than 50 ohms.

If a transmitter output impedance WAS 50 ohms, I would have thought that
the efficiency of the output stage could never exceed 50% (and aren't
class-C PAs supposed to be around 66.%?). Also, as much power would be
dissipated in the PA stage as in the load.
>
>
>

--
Ian

[John S]

BTW, Wayne, what are the dimensions of your metal roof? And how high?

[Ralph Mowery]

"Ian Jackson" <ianREMOVET...@g3ohx.demon.co.uk> wrote in message
news:YM5oXsEG$SmV...@g3ohx.demon.co.uk...

>> class-C PAs supposed to be around 66.%?). Also, as much power would be
> dissipated in the PA stage as in the load.

I think this is easy to disprove in practice. I have an amp that is
probably class B, but it does not mater about the class. If I adjust it to
an input of 2000 watts from the DC power supply, I get out 1200 watts to a
resistive dummy load. If the above is true, I should have to input 2400
watts to the final stage. Now can someone tell me where the extra 400 watts
are comming from ? This 400 extra watts is not even counting on any loss in
the circuits.

[Jerry Stuckle]

You'll be better off if you killfile the troll. You'll get a lot less
bad information and your life will become much easier.

>>
>> The wikipedia entry is correct as written.
>>
>> In the real world, the output of an amateur transmitter will seldom
>> be exactly 50 Ohms unless there is an adjustable network of some
>> sort.
>
> I've always understood that the resistive part of a TX output impedance
> was usually less than 50 ohms.
>
> If a transmitter output impedance WAS 50 ohms, I would have thought that
> the efficiency of the output stage could never exceed 50% (and aren't
> class-C PAs supposed to be around 66.%?). Also, as much power would be
> dissipated in the PA stage as in the load.

Fixed output amateur transmitters are a nominal 50 ohms. It can vary,
but that is due to normal variances in components, and the difference
can be ignored in real life.

But output impedance has little to do with efficiency. A Class C
amplifier can run 90%+ efficiency. It's output may be anything, i.e.
high with tubes but low with solid state. But the output impedance can
be converted to 50 ohms or any other reasonable impedance through a
matching network.

A perfect matching network will have no loss, so everything the
transmitter puts out will go through the matching network. Of course,
nothing is perfect, so there will be some loss. But the amount of loss
in a 1:1 match will not be significant.

Also, the amplifier generates the power; in a perfect world, 100% of
that power is transferred to the load. The transmitter doesn't
dissipate 1/2 of the power and the load the other 1/2. It's not like
having two resistors in a circuit where each will dissipate 1/2 of the
power.

BTW - the resistive part of the impedance is not the same as resistance.
For a simple case - take a series circuit of a capacitor, an inductor
and a 50 ohm resistor. At the resonant frequency, the impedance will
be 50 +j0 (50 ohms from the resistor, capacitive and inductive
reactances cancel). But the DC resistance is infinity. Again, a simple
example, but it shows a point.

[John S]

I encourage all of you to read Walter Maxwell's (W2DU) book "Reflections
III". It will explain everything about this.

Everything you are discussing has been put to bed.

Cheers.