On 7/2/2015 11:07 AM, John S wrote:
On 7/2/2015 8:14 AM, Jerry Stuckle wrote:
On 7/2/2015 6:31 AM, John S wrote:
On 7/1/2015 4:23 PM, Jerry Stuckle wrote:
On 7/1/2015 12:26 PM, Jeff Liebermann wrote:
On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle
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, 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.

Transmitter output impedance does not determine SWR.


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.




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

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.

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

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

"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)

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).

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

"Jerry Stuckle" wrote in message
...

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.

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.




--
Jim Pennino

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

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.

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

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

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.



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.

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

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

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.

"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.

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.

"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.

# 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.