In message , Wayne
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

On 7/2/2015 3:24 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...
On 7/2/2015 1:56 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...

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.


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.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

Wayne wrote:


wrote in message ...

Wayne 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.

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.

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


--
Jim Pennino

"Jerry Stuckle" wrote in message
...
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 ?

On 7/2/2015 3:17 PM, Wayne wrote:


wrote in message ...

Wayne 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.

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.

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

On 7/2/2015 3:52 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 12:18 PM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
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)

# 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.

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

In message , rickman
writes
On 7/2/2015 3:52 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 12:18 PM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
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)

# 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.

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?

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

On 7/2/2015 6:08 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...
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 ?




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.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

On Thu, 02 Jul 2015 14:55:53 -0400, Jerry Stuckle
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.

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

On 7/2/2015 6:53 PM, Ian Jackson wrote:
In message , rickman writes
On 7/2/2015 3:52 PM, Wayne wrote:

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'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

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